Bachelor Project - DTU Electronic Theses and...

Analysis of Adiabatic Liquid Piston Compressed Air Energy Storage, ALP-CAES

Bac

helo

r Pro

ject

Kristian Bo KahleMEK - TES - EP - 2011 - 09 July 2011

Preface

I would like to thank my supervisors Allan Schrøder Pedersen, Lars Reinholdtand Brian Elmegaard for giving me the opportunity to work with this new con-cept, in a field that is of great personal interest. I would also like to thank SteenSkaarup from the Department of Chemistry at DTU for his insight regardingcorrosion and the solubility of water in air. Finally a thank you to Fritz Cro-togino from KBB Underground Technologies GmbH, for providing insights onutilizing salt caverns.

Abstract

Energy storage is of interest today and in the future. This report presents aninitial analysis of Adiabatic Liquid Piston - Compressed Air Energy Storage(ALP-CAES). A Compressed Air Energy Storage (CAES) system using liquidpump and turbine technology, instead of compressors and gas turbines. Thesystem operates without the consumption of fossil fuels.

The results are strictly based on mathematical modelling. Results show,that it is a necessity to work at a low varying pressure range. By doing so, thetemperature gradient is minimized ensuring a low rate of heat transfer acrossthe gas-liquid interface. Utilizing a large underground cavern showed to be agreat way of limiting heat transfer to the surroundings. The thermodynamicefficiency of such a system thus showed to be close to 99%.

The most succesful setup showed a round-trip efficiency of 83.5%. Addition-ally, the analysis showed that the system is scalable. ALP-CAES thus has thepotential to become technologically competitive to other storage methods.

Resumé

Energilagring er relevant i dag og i fremtiden. Denne rapport indeholder enindledende analyse af Adiabatic Liquid Piston - Compressed Air Energy Storage(ALP-CAES). Et Compressed Air Energy Storage (CAES) system der benyttervæske- pumpe og turbine teknologi, i stedet for kompressorer og gasturbiner.Systemet drives uden forbrænding af fossile brændstoffer.

Resultaterne er baseret på matematisk modellering. Resultaterne viser, atdet er en nødvendighed at arbejde med et lavt varierende tryk. Ved at sørge fordette, minimeres temperaturgradienten der derved sikrer en lav varmeoverførselmellem gas-væske overfladen. At benytte en stor underjordisk kaverne, vistesig at være en god måde at begrænse varmeoverførsel til omgivelserne. Dentermodynamiske virkningsgrad af et sådant system viste sig at være tæt på99%.

Det mest succesfulde system viste en tur-retur effektivit på 83.5%. Deru-dover viste analysen, at systemet er skalerbart. ALP-CAES har således po-tentialer til at blive teknologisk konkurrencedygtig i forhold til andre energila-gringsmetoder.

Nomenclature

SymbolsE Energyg Acceleration of gravityh Convective heat transfer coefficientk Thermal conductivity

M Massm Mass flow rateP PressureQ Heat flowT Temperatureu Specific internal energyV VolumeV Volume flow rateVe Velocityw Humidity ratioW Workx coordinatez heightα Thermal diffusivityφ Relative humidityη Efficiencyρ Densityτ Time

Subscriptsa Air

com Compressionexp Expansion

g Gasi Initial

in In to systeml Liquid

out Out of systemrt Round-trips Siloxanetd Thermodynamicv Vaporw Waterx coordinate

Contents

1 Introduction 1

2 Problem statement 2

3 ALP-CAES 3

4 Adiabatic process 4

4.1 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44.2 Gas temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . 44.3 Stored energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

5 Cases 6

5.1 Case 1 - Large underground cavern as compression chamber . . . 65.2 Case 2 - Array of pressure vessels . . . . . . . . . . . . . . . . . . 75.3 Case 3 - Large underground cavern as pressure vessel . . . . . . . 8

6 Mathematical model 10

6.1 Application of mathematical model . . . . . . . . . . . . . . . . . 126.1.1 Case 1 - Large underground cavern as compression chamber 126.1.2 Case 2 - Array of pressure vessels . . . . . . . . . . . . . . 136.1.3 Case 3 - Large underground cavern as pressure vessel . . . 13

7 Heat transfer analysis 14

7.1 Case 1 - Large underground cavern as compression chamber . . . 147.1.1 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . 147.1.2 Heat transfer to surroundings . . . . . . . . . . . . . . . . 15

7.2 Case 2 - Array of pressure vessels . . . . . . . . . . . . . . . . . . 177.2.1 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . 187.2.2 Heat transfer to surroundings . . . . . . . . . . . . . . . . 18

7.3 Case 3 - Large underground cavern as pressure vessel . . . . . . . 207.3.1 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . 207.3.2 Heat transfer to surroundings . . . . . . . . . . . . . . . . 21

8 Psychrometrics 24

8.1 Case 1 - Large underground cavern as compression chamber, us-ing humid air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

9 Pressure Losses 27


5

10 Fluid machinery 29


11 Thermodynamic efficiency 37

11.1 Case 1 - Large underground cavern as compression chamber . . . 3711.2 Case 1 - Large underground cavern as compression chamber, us-

ing humid air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3811.3 Case 2 - Array of pressure vessels . . . . . . . . . . . . . . . . . . 3811.4 Case 3 - Large underground cavern as pressure vessel . . . . . . . 40

12 Round trip efficiency 41


13 Practical considerations 43


14 Discussion 45


15 Conclusion 48

15.1 Case 1 - Large underground cavern as compression chamber . . . 4815.2 Case 2 - Array of pressure vessels . . . . . . . . . . . . . . . . . . 4915.3 Case 3 - Large underground cavern as pressure vessel . . . . . . . 4915.4 Overall conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . 49

References 51

Appendix

1 Introduction

As the movement towards a more sustainable society progresses, with the useof renewable energy and less dependency on fossil fuels, a number of challengesmust be overcome. One of the challenges is energy storage. Energy productionfrom certain renewable sources experience large fluctuations. Fluctuations notgoverned by man, but nature. As the power production from renewables fluc-tuates, so does the energy consumption. With the large fluctuation in energyconsumption, it’s necessary to be able to save harvested energy for later use:also in a sustainable manner.

Energy storage today is a good way of regulating the energy supply, makingthe interaction between renewables and non-renewables better, and for levelingthe price of electricity. Energy storage is of increasing interest now and in thefuture.

Today, one efficient and reliable energy storage method is Pumped HydroElectricity Storage (PHES). In PHES water is pumped from a lower to a higherreservoir during overproduction of energy. This water is then led through awater turbine when the grid experiences a peak in demand. The round-tripefficiency of most PHES plants are around 65 per cent [20]. PHES plants requirea higher lying reservoir. Typically, this reservoir lies in mountanious regions.For this reason and due to the danish natural environment, PHES plants arenon-existent in Denmark.

Compressed Air Energy Storage (CAES) is a method that can be imple-mented in Denmark, utilizing large underground caverns. Implementing thisform of energy storage in Denmark has in recent years been thuroughly re-searched. The installed CAES plants around the world are, however, still relyingon fossil fuels.

Adiabatic Liquid Piston - Compressed Air Energy Storage (ALP-CAES) isan attempt to combine the high efficiency of PHES and CAES. In ALP-CAES,liquid is pumped into a closed pre-pressurized vessel containing a fixed amountof gas. Work is thereby carried out by compressing the gas with a liquid piston.When energy is needed, the liquid is forced through a water turbine, extractingthe work done by the expanding gas.

The hypothesis of the ALP-CAES principle is that due to the pre-pressurizedvessel the heat development will be low. It is also expected that the developedheat will be stored in the vessel wall and in the liquid piston, and later returnedto the air during expansion.

1

2 Problem statement

The ALP-CAES principle will be analyzed using three different cases. Theeffects of heat- development and transfer will be analyzed, as well as the ther-modynamic efficiency. Other factors, such as mass transport and pressure losseswill also be covered. At the end of the report an estimation of the round-tripefficiency will be made, taking pump and turbine types into consideration.

The focus of this project is an initial analysis of the ALP-CAES principle.This is to determine what factors are vital to the realization of this principleand assist further work on analyzing ALP-CAES. As part of the project, certainadvantages as well as disadvantages concerning fluid machinery and other prac-tical considerations are clarified. All modelling is done using the commercialsoftware Engineering Equation Solver (EES).

2

3 ALP-CAES

The principal behind the ALP-CAES system is a hybrid of Pumped Hydro Elec-tricity Storage (PHES) and Compressed Air Energy Storage (CAES) systems.In ALP-CAES, a closed container contains a fixed amount of gas. During over-production periods, a pump fills the container with liquid. The liquid then actsas a liquid piston, compressing the gas. The work done by the pump will bestored in the vessel as potential energy. When extra energy is needed duringlow production/high demand periods, the potential energy of the system is thenagain converted to mechanical energy by letting the pressure of the gas force theliquid through a turbine. The turbine is connected to a generator, supplying thegrid with electrical energy. The process comprises of three stages; Compression,Storage, and Expansion. See Figure 1.

Figure 1: ALP-CAES Principles three stages.

During compression of the gas, the gas will experience an increase in tem-perature. The heated gas will attempt to reach thermal equilibrium with itssurroundings, by transferring heat through the vessel walls and into the pumpedliquid. This causes a loss in the stored potential energy as the lowering of thegas temperature lowers the gas pressure. During storage, no work is carried outbut heat transfer will still be present. During expansion, the temperature of thegas will drop, and rate of heat transfer will be different than during compressionand storage. The effect of heat transfer will be separately analyzed in section 7.

3

4 Adiabatic process

The process will in each case neither be fully adiabatic nor will it be isother-mal. The mathematical expressions for an adiabatic process can, however, helppredict certain important factors, and are therefore briefly mentioned.

4.1 PressureThe pressure of an ideal gas after an adiabatic process is related to the volumeby [10]:

P1Vγ1 = P2V

γ2

P2 = P1

�V1

V2

�γ

The relative change in volume in an adiabatic process determines the finalpressure.

4.2 Gas temperatureThe change in temperature for an adiabatic process for an ideal gas, can beshown to be given by [10]:

T2 =

�P2

P1

�1− 1γ

T1

As will be demonstrated in the following sections, heat development duringcompression can be the major reason for energy losses. It is thus favorable tohave a low ratio between the final and initial pressure so that the tempera-ture increase is small. The lower the temperature difference of the air and thesurroundings, the lower the rate of heat transfer.

4.3 Stored energyWork done on a gas is expressed on differential form as [18]:

δW = −PdV

For an adiabatic process the total work done on an ideal gas, can be cal-culated by the initial and final state. The work done is given by the relation[10]:

W =1

γ − 1(P1 V1 − P2 V2)

It is the product of pressure and volume that is of interest when looking atstored energy.

4

It is evident that the optimal system has a high initial pressure to increaseenergy density. Moreover, a relatively small change in volume to minimize thefinal temperature. This will thereby minimize the rate of heat transfer.

In the model, the work done on the system is expressed as the work done bythe pump.

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5 Cases

The mathematical model explained in section 6 applies to different flow rates ofliquid, container size, etc.. As the results illustrate throughout this report, theimportance of heat transfer, compression ratio, etc. are very much dependenton the case to be analyzed. The cases that will be analyzed are:

• Case 1: Large underground cavern as compression chamber.

• Case 2: Array of pressure vessels.

• Case 3: Large underground cavern as pressure vessel

5.1 Case 1 - Large underground cavern as compressionchamber

Case 1 is a centralized energy storage system. The setting for Case 1 is assumedto be a cavern similar to those utilized at the natural gas storage facility atLl. Torup, Denmark. These caverns are washed out from a salt dome andare located about 1.5 km below the surface, with a volume of approximately 1 ·106m3. They are known to hold natural gas of high pressure up to approximately200 bar [1].

The cavern used in Case 1 is set to be cylindrical with a height of 300 metersand diameter of 65 meters. The bottom of the cave is set 1500 meters belowsurface. Please see Figure 2.

Figure 2: Case 1 setup utilizing a cavern as compression chamber and pressurevessel.

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The maximum pressure inside a salt dome cavern is estimated as a func-tion of depth. For a cavern of these dimensions, 1500m below the surface, themaximum pressure is Pmax � 216 bar [7]. The minimum pressure is approxi-mately one third of the maximum pressure, but this is not of concern since theminimum pressure will be much lower than that created by the 1.5 km liquidcolumn from the bottom of the cave to the surface. The gas/liquid combinationis air/water. More correctly, the pumped medium is a saturated saline solution,or brine. The brine is reused for every cycle to make sure that the cavern wallsdo not dissolve. The salinity of the returning brine during the washing of saltdomes is around 35% and saturated. The filling time of the cavern is set to eighthours at a volume flow rate of V = 2.0m3

/s. The cavern is pre-pressurized sothat the final pressure is the maximum pressure, Pmax.

5.2 Case 2 - Array of pressure vesselsCase 2 is a much smaller, localized, low energy system, storing energy for a singlehousehold. Compression could occur when the price of electricity is low, andstored until use during high demand periods where electricity is most expensive.In this small setting the structural stress to the vessel, due to the increase inpressure, must be taken into account. Therefore, an array of standard pressurevessels are used. The Pressure vessels chosen are the high pressure vessels typeBX. These have a capacity of 42.2 l, and are certified to hold pressures up to6000 psi � 414 bar. Please see appendix. An array of seven pressure vessels areset up, having one common pump/turbine. Please see Figure 3.

Figure 3: Case 2 setup with seven pressure vessels and one commonpump/turbine.

The gas/liquid combination is siloxane and nitrogen. The reason for usingthese are to prevent corrosion. Siloxane is a silicone compound used in hydraulic

7

applications. In EES, the name of the compound used is “Siloxane_2”.The thickness of each pressure vessel is about 2 cm. To keep as much heat

as possible within the system the cylinders are insulated. The filling time of thearray of pressure vessels is set to one hour at a flow rate of V = 2.5·10−5

m3/s =

25 cm3/s.

5.3 Case 3 - Large underground cavern as pressure vesselCase 3 is a modification of case 1, where the cavern is only used as a largepressure vessel. The liquid piston is instead in a smaller vessel, located close toground level. This smaller vessel is then connected with a pipe to the cavern.The pipes in the system have a diameter of 0.4m and a thickness of 2 cm. Pleasesee Figure 4:

8

Figure 4: Case 3 setup, having the compression chamber near ground level andutilizing a cavern as a pressure vessel.

The advantage of this system is that the 1.5 km water column is now elim-inated. This means that the initial pressure in the cavern will be almost equalto that at the pump, allowing more specific work to be done by the pump andsubsequently extracted from the turbine. The air/liquid combination is air andwater. To maximize the energy density and lower the ratio between the finaland initial pressure, the initial pressure is set at a value so that the final pres-sure after eight hours of compression is the maximum pressure allowable for thecavern. After compression, the compression chamber on ground is completelyfilled with water. This requires a volume flow rate of V = 0.136m3

/s.

9

6 Mathematical model

All models can be found in the appendix. The compression model is brieflyundergone in this section.

When modelling the ALP-CAES principle, the system is split into two dif-ferent control volumes and one control mass. Depending on the case, the initialvalues and states are different, but the basic model is the same. The controlvolumes and masses are shown in Figure 5:

Figure 5: Control mass and control volumes of the ALP-CAES system

The scenario as shown in Figure 5 is during compression. During expansionCV 2 is around the turbine.

The mathematical model is set up in EES. Initial conditions such as initialtemperature, pressure, vessel size, volumetric flow rate, etc. are set in thebeginning of the program. Since the container is closed, the mass of gas isconstant, but the total mass of the system increases as liquid is pumped intothe container at a given rate. The change in mass of the system is given by thedifferential equation:

dM

dτ= ρl V = m

This is rewritten by using separation of the variables:

dM = m dτ

M −Mi =

ˆ τ

0m dτ (1)

Where M is the total mass of the system and Mi the initial mass. In themodelling the initial mass is the sum of the mass of the gas and the initial liquid

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mass. Since the mass of gas does not change over time, the total mass can besplit into its two components

M = Ml +Mg

Knowing the mass of the fluids as a function of time, the assumption ofconstant liquid density is now made. This enables one to determine the vol-ume of each component. Of course liquid density is affected by the increase intemperature and pressure, but not only is it a reasonable assumption, it’s alsoa necessary assumption. The necessity of this assumption is because differen-tial equations, solved using EES, must be initial value problems. Meaning thatdifferential equations must be supplied with boundary conditions given at thestarting point of the integration. Temperature and pressure, which in realityaffect density, are determined by using a set of separate differential equations,dependent on the total solution of equation 1. The internal energy balancefor the Control Mass and the Control Volumes are set up. The first law ofthermodynamics for a control mass is defined by [18]:

d

dτ

�M

�u+

1

2Ve

2 + g z

��= Q+ W (2)

Work in compression of a gas is given by W = P · dVg, where the changein volume of the gas is given by dVg = −m/ρl. Potential energy for the air isneglected. U is the specific internal energy of the gas, and with no change ingas mass with respect to time, equation 2 rewrites to.

Mg (ug − ug,i) =

ˆ τ

0Q− P dVg dτ (3)

The specific internal energy ug is in equation 3 the only unknown. UsingEES’s thermodynamic properties library, the air temperature and pressure canbe found by adding the equations below:

ug,i = f(Ti;Pi)

ug = f(Tg;Pg)

Pg = f(Tg; ρg)

ρg =Mg

Vg

Note that the heat flow Q is actually a sum of heat flows, described in section7. For Control Volume 1, another internal energy balance is carried out. Thepressure of the gas is equal to the pressure of the compressing liquid in thevessel. The first law of thermodynamics for a control volume states[18]:

d

dτ

�M

�u+

1

2V

2e + g z

��+

��mout (hout +

1

2V

2e,out + g zout)

�

−��

min (hin +1

2V

2e,in + g zin)

�= Q+ W (4)

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During compression there is no liquid flow out of the system. The workdone by the liquid is equal but opposite to that done on the gas, so dVl = −dVg.Kinetic energy of the mass flow into the system is neglected. Equation 4 thussimplifies to:

d

dτ(M u) = Q+ min (hin + g zin)− P dVl

This rewrites to:

Ml ul −Ml,0·ul,i =

ˆ τ

0(Q+ min (hin + g zin)− P dVl)dτ (5)

The enthalpy in to the system is the enthalpy of the pumped liquid as itcrosses the boundary of CV 1. The enthalpy is calculated using the state priorto the pump, the efficiency of the pump and knowing the pressure Ppump on thedischarge side of the pump. The reason for introducing the variable Ppump isdescribed in section 6.1.

The specific internal energy of the compressing liquid is thus the only un-known of equation 5. Using the thermodynamic properties library, the followingis found:

Tl = f(ul;P )

ul,i = f(Ti;Pi)

The energy stored in the system is calculated by integrating the work doneby the pump over time. The work done by the pump is calculated by applyingequation 4 to CV 2.

The mathematical model explained above applies during compression. Dur-ing storage, the mass flow rate into the system is zero and thus only heat transferoccurs. During expansion, the same model is used, only with minor changes suchas new initial conditions, opposite flow direction, different flow rate, expansiontime. etc.

During expansion, the work extracted from the system is calculated by ap-plying equation 4 to a CV around the turbine.

6.1 Application of mathematical model6.1.1 Case 1 - Large underground cavern as compression chamber

When applying the mathematical model to Case 1, small variations to the modelare made. Mainly, the pressure right after the pump at ground surface is sig-nificantly lower than in the cavern, due to the depth (1.5 km). It is thereforeimportant to define the variable Ppump so that the enthalpy is of correct value:

Ppump = P + ρw g (zl − z0)

zl is the height of the liquid surface, and z0 is the distance from the pumpto the cavern bottom (the bottom of the cavern is set to height zero). With thedifference in elevation between the pump outlet and the liquid piston being so

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great, potential energy can not be neglected. The height zin is thus the heightfrom the pump, z0, to the center of mass of the water, 1

2 zl.The medium used for the gas is “Air_ha”. So the thermodynamic properties

for the gas is the high accuracy air and not the ideal gas “Air”. The EES libraryonly supports brine at atmospheric pressure. Therefore the pumping liquid usesthe thermodynamic properties of the medium water. The constant density ofthe brine is set to ρw = 1260 kg/m3.

Heat transfer between the entrapped air and brine is due to convection.Due to the thick semi-infinite walls of the cavern, heat transfer from the gas tothe surroundings is limited. The influence of heat transfer will be discussed insection 7.1.

6.1.2 Case 2 - Array of pressure vessels

In case 2, the changes in height throughout the system are negligible, thusthe variable Ppump � P and potential energy is not considered. Case 2 is anassembly of multiple vessels. When applying the mathematical model, all of theseven vessels are assumed to behave equally, hence only one vessel is actuallymodelled. The flow rate into each vessel is then 1

7 of the total flow rate. Theconstant density is set to ρs = 940 kg/m3.

Each cylinder is insulated to lower heat transferred to the surroundings.Heat transfer in Case 2 is analyzed in section 7.2.

6.1.3 Case 3 - Large underground cavern as pressure vessel

As in Case 1 it is necessary to define the variable Ppump. This pressure differenceis, however, not of as great magnitude since the height of the water column is 30times shorter. Potential energy is also accounted for. As previously mentioned,the pumped liquid is water. The constant density is set to ρw = 1005 kg/m3.The piping from the vessel to the compression chamber on land must also beincluded. The entrapped air in this 1300 meter long pipe is part of the air mass,as well as the entrapped air in the underground cavern and the compressionchamber.

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7 Heat transfer analysis

In this section, the influence of heat development and transfer will be analyzedfor each case. All calculations can be found in the appendix.


As heat develops due to the gas compression, heat will transfer into the com-pressing liquid and into the surrounding cave material. To determine the effectof each flow of heat, the two flows and their magnitude have been separatelyanalyzed. In this case, the compressed gas is air and the compressing liquid issaturated salt water.

7.1.1 Convection

During compression, the air will experience an increase in temperature. Theair will naturally try to reach thermal equilibrium with the surroundings, bytransferring heat to the cavern walls and through the air/brine interface. Thelatter is a case of convection. The following correlation for natural convectionis considered to fit the situation of Case 1 [17].

Nu = 0.27Ra14 , 105 ≤ Ra ≤ 1010 (6)

This correlation provides the convective heat transfer coefficient. For twofluids in contact, convection is thought to consists of heat transfer from air toa hypothetical infinitely thin membrane and then from the membrane to thewater. The membrane has no thermal resistance. Please see Figure 6.

Figure 6: Convection between two fluids. The hypothetical infinitely thin mem-brane, separating the two fluids, has no thermal resistance.

Correlation 6 applies to both the water- as well as the air side.

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The correlation for natural convection and the heat flow has been embed-ded into the EES program. Correlations for natural convection are based onempirical measurements, and the range of its validity is exceeded by the air athigh pressures. Therefore the overall heat transfer coefficient was manipulatedto determine if the validity and inaccuracy of empirical data affected the model.The final temperature and pressure of the air is shown in Table 1

Adiabatic Nat. Conv. Cor. Factor 500 Inc. Factor 1000 Inc.Ta [K] 310.2 310.2 309.9 309.7P [bar] 216 216 215.8 215.6

Table 1: Final temperature and pressure for an Adiabatic compression, a com-pression with the natural convection correlation, a compression with a factor500 and a factor 1000 increase of the natural convection correlation.

As seen above, the heat transfer from air to the compressing water is low.The final temperature of the compression is around 10 degrees higher than theinitial temperature of 300K. This gives a low temperature gradient throughoutthe compression, thus a slow rate of heat transfer. It is, however, evident thatthe overall heat transfer coefficient plays a noticeable role.

7.1.2 Heat transfer to surroundings

Due to the large mass surrounding the cavern, the walls are considered as asemi-infinite medium. Modelling heat transfer through a semi-infinite mediumis quite difficult. A semi-infinite medium can not be given a thermal resistance,as this is a case of transient conduction. Correlations for semi-infinite mediumsdo exist. The correlation, equation 7, describes the temperature in a semi-infinite medium at time τ as a function of distance x from the wall surface. Theboundary conditions is constant convection [17]:

Tx ,τ − Ti

Ta − Ti= erfc

�1

2

x√α τ

�− e

hxk +h2α τ

k2 erfc

�1

2

x√α τ

+h√α τ

k

�(7)

Tx,τ is the temperature at a given distance from the surface. The functionerfc is called the complimentary error function defined by [17]:

erfc(w) = 1− 2√π

ˆ w

0e−ν2

dν

The criteria for correlation 7 are constant convective heat transfer coefficient h,and constant air temperature Ta. Therefore this correlation can not be modelledinto the program. What has been done is to determine how much energy willat most be stored in the surrounding cavern material. Thermal properties ofthe mostly sodium chloride walls have been supplied, as well as a convectiveheat transfer coefficient of h = 1 W

m2K . The initial temperature of the cave is

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set to Ti = 300K and the air temperature set to the final temperature of thecompression of Ta = 310.2K. The time τ is the compression time of eight hoursplus the storage time of five hours. This gives a temperature distribution of thewall as shown in Figure 7.

Figure 7: Temperature distribution in semi-infinite solid.

As Figure 7 clearly shows, the cavern wall is already unaffected by the con-vection at a depth of 6 meters. To determine how much energy is stored inthe wall, the temperature distribution is integrated over the volume of the sur-rounding cavern material [17]:

Q = −ˆ

ρ c (Tx,τ − Ti) dV

The calculations can be found in the appendix. The total stored energy isfound to be:

Q = −2.47 · 1010 J (8)

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This corresponds to a drop in air temperature of

∆T = −0.12K

The inside of the caverns are, as mentioned earlier, modelled as cylindrical.In reality, however, the inside surface of the cavern is quite irregular. This, ofcourse, increases the surface area, increasing the rate of heat transfer. Figure 8shows the irregular shape of the two caverns used in the Huntorf CAES plant.

Figure 8: Sonar scan of the two caverns used in the CAES plant in Huntorf,Germany [9].

Due to the shape of caverns being somewhat irregular, it is assumed that theincreased surface area vastly increases the rate of heat transfer. This is done bymaking the cavern wall surface reach the same temperature as the air. Usingthe same correlation for the semi-infinite conduction, equation 7, shows that theenergy stored corresponds to a final temperature drop of ∆T = −0.44K.

7.2 Case 2 - Array of pressure vesselsThe advantage of the smaller system with less complexity, comes at the expenseof more heat loss to the surrounding environment. The compressing liquid piston

17

is siloxane and the entrapped gas is pure nitrogen.

7.2.1 Convection

Convection proves to be of much more significance in Case 2 compared to Case1. The correlation for Case 1 is again used, and as with Case 1 the convectiveheat transfer coefficient is manipulated. The results are seen in Figure 9:

Figure 9: Air temperature as a function of time, for different magnitudes of theconvective heat transfer coefficient.

The reason for the natural convection being of higher significance in Case 2 isthe driving temperature gradient. In Case 1, the difference in final temperaturewhen manipulating the overall heat transfer coefficient, was not as drastic asdepicted in Figure 9.The temperature at the end of the compression in Case 1was approximately 310K. At this temperature, it is also seen in Figure 9 thatthe magnitude of the convective heat transfer coefficient has little influence.


The major heat flows in Case 2 are the heat flow into siloxane and the heat flowto the surroundings through the pressure vessels’ steel walls and insulation. Theheat flow to the surroundings is modelled as cylindrical heat transfer throughthe cylinder surface. The area through which heat is transferred from the airdecreases as the cylinder is filled, while that from siloxane to the surroundings

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increases. Figure 10 shows the pressure vs. time for the system with the differentheat flows included in the model.

Figure 10: Pressure vs. time for system with different heat flows included inmodel.

Figure 10 clearly indicate that the heat flow out of the system is the greatestsource of potential energy loss in Case 2. The cylinders are not insulated inthe calculations shown in Figure 10. Figure 11 displays the influence of theinsulation thickness. The insulation material is chosen to be polyurethene witha thermal conductivity of k = 0.03 W

mK [17]:

19

Figure 11: Pressure vs. time at different insulation thickness.

Clearly the insulation is very important. Adding 6 cm of insulation yieldsa final pressure 3.3 bar higher compared to non-insulated cylinders. This willhave a significant influence on the efficiency.

7.3 Case 3 - Large underground cavern as pressure vesselThe compressing liquid piston is water and the entrapped gas is air.

7.3.1 Convection

The air in Case 3 experiences a very little change in volume during the processes.As the mathematics for an adiabatic process in section 4 showed, a small changein volume results in a small change in pressure. A small change in pressuremeans a small change in temperature. The temperature increase shows to bevery little. Setting the rate of heat transfer between the two media to zero, theinitial and final state of the air is:

20

Initial state Final stateTa [K] 300 300.7Tw[K] 300 300.4P [bar] 214.3 216.0

Table 2: Initial and final state of water and air in compression phase for Case 3if no heat transfer occurs.

As Table 2 shows, the rise in temperature for both water and air is verylittle. Manipulating the overall heat transfer coefficient as done in the formertwo cases, the final state of the air after compression is shown in Table 3.

Nat. Conv. Cor. Factor 500 Inc. Factor 1000 Inc.Ta [K] 300.7 300.7 300.7P [bar] 216 216 216

Table 3: Final temperature and pressure for a compression with the natural con-vection correlation, a compression with a factor 500 and a factor 1000 increaseof the natural convection correlation.

As could be expected, convection is negligible due to the low temperaturegradient.


Assuming that the whole system is in thermal equilibrium as the compressionprocess commences, the air will heat up the surrounding material slightly dur-ing compression. As done in section 7.1.2, the total amount of heat stored iscalculated. Calculations can be found in the appendix.

By integrating the temperature distribution of the surrounding cavern ma-terial, after being exposed to constant surface convection for 13 hours of com-pression and storage, the stored energy in the cavern walls are given. The heatlost to the cavern material is:

Qc = −1.69 · 109 J

The steel pipe leading down to the cavern is assumed to reach a constanttemperature throughout the material. The pipe thus experience a 0.9 degreetemperature increase. With the dimensions of the pipe given, this correspondsto a total heat loss of:

Qp = −8.93 · 106 J

The pipe is considered to be completely surrounded by soil, rock and sodiumchloride. The temperature on the outer surface of the pipe, and thus the surfaceof the surrounding material, is considered constant. The thermal propertiesgiven are those of sodium chloride, as the salt dome’s boundary starts as early

21

as 300m below ground [2]. The correlation for a semi infinite medium withconstant surface temperture is given by [17]:

T (x, t)− Ts

Ti − Ts= erf

�x

2�

(α τ

�

where the function erf is the error function given by [17]:

erf(w) =2√π

ˆ w

0e−ν2

dν

The temperature distribution after 13 hours of storage and compression isdepicted on Figure 12.

Figure 12: Temperature distribution of semi-infinite solid with constant surfacetemperature.

The temperature distribution is integrated over the surrounding volume.

22

The energy lost is found as:

Qe = −1.42 · 109 J

The total heat lost is therefore:

Q = Qc +Qp +Qe = −3.13 · 109

Corresponding to a temperature drop of the air of:

∆T = −0.013K

23

8 Psychrometrics

When a gas and a liquid are in contact, the two media will naturally try toreach vapor-liquid equilibrium. During compression, changes will occur in howmuch vapor a gas can contain. Dalton’s law states that the total pressure of agas is the sum of partial pressures of its components [18]:

P = Pv + Pa

When vapor-liquid equilibrium is reached, the partial pressure of the vaporis equal to the saturation pressure of the liquid at the given temperature. It canbe shown that the humidity ratio for water-vapor can be determined by [18]:

w = 0.622Pv

P − Pv(9)

For the partial pressure being the saturation pressure, equation 9 expressesthe humidity ratio at which the water-vapor mixture is saturated. So for anisothermal process, the humidity ratio for the saturated vapor must drop. Thismeans that during isothermal compression some of the vapor in the air willcondense. Figure 13 shows the saturated vapor line at pressures ranging from192 to 214 bar.

Figure 13: Mollier diagram showing saturation curves for vapor air mixture atfive different pressures

24

As Figure 13 clearly shows, temperature is the most influential variable onthe solubility of water vapor in air. In the following section, the influence ofhumid air will be analyzed for Case 1.

8.1 Case 1 - Large underground cavern as compressionchamber, using humid air

In this section the modelling of Case 1 is redone with humid air, using themedium “AirH2O” instead of the dry air “Air_ha”. If modeling using humid air,three independent variables must be supplied to determine the state variablesof the gas. Therefore it is not possible to model the compression, setting thehumidity ratio or the relative humidity as a free variable. Setting the relativehumidity as constant ranging from 100% to 20% gives results as in Figure 14.

Figure 14: Humidity ratio throughout compression for a constant relative hu-midity. Dotted colored lines are saturation curves at different pressures

The final pressure and temperature of the humid air at different relativehumidities are shown in Table 4.

25

φ 100% 80% 60% 40% 20% 0%

Ta [K] 307.3 307.3 307.3 307.3 307.3 307.3P [bar] 208.7 208.7 208.7 208.7 208.7 208.7

Table 4: final temperature and pressure of air after compression at differentrelative humidities.

As seen, the difference in final temperature and pressure lies beyond firstdecimal. It is noticed that for the relative humidity being zero, pressure andtemperature are significantly lower when using the medium “AirH2O” comparedto “Air_ha”. This means that the stored energy is less than that of Dry air,using the same amount of pumped fluid. The reason for this lies in propertiesof the fluid in the thermodynamic library. The thermodynamic efficiency of thetwo different media is determined in section 11.

The mathematical model used splits the system into its two media; waterand air. When using EES to model mass transport like this, it’s necessary tosplit water into its two phases, liquid and vapor, and treat them with a set ofseparate differential equations. This has not been done in this project. Insteadthe energy required to vaporize the transported mass is calculated, assuming theconstant relative humidity. This is done by multiplying the transported massby the latent heat of evaporation. Results are shown in Table 5:

φ 100% 80% 60% 40% 20%

Transportedmass [kg] 9926 7949 5962 3975 1987Heat required [MJ ] 24182 19366 14525 9683 4842

Temperature drop [K] 0.11 0.08 0.07 0.04 0.02

Table 5: Transported mass, heat required for evaporation and correspondingtemperature drop of air.

The heat required may seem quite substantial, but the temperature dropof the air is clearly insignificant. It is however important to notice that theenergy required to vaporize this water, does not necessarily leave the system.According to Steen Skaarup, the relative humidity will stay almost constant at100% throughout all of the process stages [25]. If this is the case, the vaporizedwater will simply condense during expansion, releasing the latent heat backto the air. A source off energy loss due to mass transport could be if thevaporized water condenses on the cavern walls, releasing the latent heat to thesurroundings. The effect of possible energy loss is treated in section 11.

Assuming that the absolute humidity of the air does not change throughoutthe compression, eg. no mass transport is present, the same results are foundas that shown in Table 4.

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9 Pressure Losses

In this section the pressure losses in each system are briefly examined. Calcu-lations can be found in the appendix.


For Case 1 the setup is thought to consist of a gate valve, a 90◦ elbow andthe 1.5 km pipe. Using data found in [12], the roughness of the steel pipingis 0.046mm and the equivalent length of the valve and elbow is 8 and 30,respectively.

The pressure loss in the system during compression is found to be Ploss =0.75 bar or in terms of head, 6.1m

During expansion the flow rate is 5.33m3/s. This pressure loss is higher due

to the higher velocity of the water. The pressure loss is found to be Ploss =4.69 bar or in terms of head, 37.9m.

9.2 Case 2 - Array of pressure vesselsFor Case 2 the system setup is divided into two parts, since each part experiencesdifferent flow rates. Part one before (and including) the gate valve and part twoafter the gate valve. Please see Figure 3.

Before the gate valve the system has two elbows, the gate valve and approx-imately four meters of piping. After the gate valve, there’s a splitter treated as6 45◦ elbows, another 6 45◦ elbows and approximately 11 meters of piping. Thepipes are 3/4” steel pipes. The diameter is chosen since these are what fit thegear pump to be used, as described in section 10.2.

The flow rate during compression is 0.000025m3/s. The equivalent length

of a 45◦ elbow is 16 [12]. The total pressure loss during compression is found tobe Ploss = 0.014 bar.

The expansion time is set to 30 minutes at a flow rate of 0.000045m3/s.

Using the same procedure as above, the pressure loss during expansion is foundto be Ploss = 0.025 bar.

9.3 Case 3 - Large underground cavern as pressure vesselAn advantage of Case 3 is, that the long pipe leading to the cavern only hasair flowing through them. The decrease in density means a decrease in pressureloss. The density of the air is at the final state 236 kg/m3. The flow rate of theair is approximated to be the same as the water flow rate.

The pipe from the pump entering the compression chamber is also set ata diameter of 0.4m. The 3-way valve is given the same equivalent Lengthas the gate valve. The total pressure loss during compression is found to bePloss = 0.06 bar, or in terms of head, 0.6m.

27

During expansion, the total pressure loss that the system experience is foundto be Ploss = 0.49 bar. In terms of head, 5m.

28

10 Fluid machinery

The main components for any ALP-CAES system are, of course, the pump andturbine. Many different types exist. The pumps selected for each case meet twocriteria; the desired discharge pressure and desired flow rate. Figure 15 showsthe approximate upper limit for different types of pump dependent on the twocriteria:

Figure 15: Approximate upper limit of pressure and flow rate for three pumpclasses [11].

It is seen that for a high pressure and flow rate a centrifugal pump is desired,whereas at a lower flow rate rotary and reciprocating pumps can be used. Bothrotary and reciprocating pumps are positive displacement pumps. Almost allpump types, with very few exceptions, are reversible. A reversible pump allowsthe flow direction to be reversed and act as a turbine extracting energy forelectricity production. In many existing PHES plants it is common to havea reversible pump [20]. The pumps chosen for each case will be reversible ifpossible.

When using pumps one must be aware of the Net Positive Suction Head(NPSH) to avoid cavitation. NPSH is defined as the difference in pressure bythe pump inlet and the saturation pressure for the liquid. For pumps, especiallycentrifugal pumps, the available NPSH value must be higher than the requiredNPSH value. The required NPSH is dependent on the pump design [12]. In the

29

model (Case 1 and 3), the liquid reservoir is located 10 meters above the pumpto increase the value of the NPSH available.

In the three following subsections fluid machinery for each case is brieflycovered. The actual setup may require multiple pumps being connected inseries or parallel. For Pumps in series the maximum head is the sum of theindividual pump discharge heads. For pumps in parallel, the sum of the pumpflow rates equal the total flow rate [11].


For Case 1, the flow rate during compression is set to 2.0m3/s. The flow rate

is chosen so that the final pressure after eight hours of compression reaches themaximum allowable pressure in the cavern of 216 bar. Due to the 1.5 km watercolumn, the pump only has to deliver the pressure right at the outlet of thepump. The pump must also overcome the pressure loss described in section 9.1.Including the pressure loss, the pump is required to deliver a head in the rangeof 61− 273m.

During expansion the turbine is set to operate at a flow rate of 5.33m3/s

and an expansion time of three hours. This flow rate and expansion time returnthe encaptured air to its initial state. Including the pressure losses, the rangeof head delivered at the turbine inlet is 230− 17m.

According to [20], a Francis type Turbine should be used under these condi-tions. The Francis turbine is shown in Figure 16:

30

Figure 16: Francis Turbine[14].

As mentioned, a reversible pump/turbine will be used if possible. Figure17 shows the performance characteristic curve for a reversible Francis typepump/turbine in both pump and turbine operation.

31

Figure 17: Performance characteristic curve for Francis pump/turbine withvarying head [24].

According to Figure 17, the heads percentage-wise variation in Case 1 by farexceeds that which is possible for the reversible pump/turbine.

A separate setup of pump and turbine is therefore needed.A Francis type turbine, as shown in Figure 16, has a performance charac-

teristic as shown in Figure 18. This is at constant specific speed with variableload. Load is the product of volume flow rate and pressure:

32

Figure 18: Constant specific speed efficiency curve for various turbine types atvarying percentage of full load [20].

The turbine efficiency curve in Figure 18 plays an important role in theround-trip efficiency determined in section 12.1.

The high flow rate during compression requires, according to Figure 15, acentrifugal pump. A drawing of a centrifugal pump is shown in Figure 19;

Figure 19: Centrifugal pump [14].

Centrifugal pumps do not increase the static pressure of the fluid at thedischarge side of the pump. The fluid is accelerated radially thereby increasingthe dynamic pressure. The increase in dynamic pressure results in a possible

33

rise of the fluid. As Figure 20 shows, the efficiency and discharge head is variedas the flow rate is varied;

Figure 20: Typical performace curve for a centrifugal pump at different specificspeeds [11].

As seen in Figure 20, the delivered head is constant at a certain flow rate fora centrifugal pump. For increasing the delivered head, the flow rate must drop.Clearly, the large variation in head in Case 1 is very unusual for centrifugalpumps.

Since the mathematical model has a constant flow rate, the pump is chosento deliver a constant head of 273m throughout the entire compression phase,working at the pumps Best Efficiency Point (BEP). This results in more energyconsumption than actually necessary and will, in the physical system, lead toa more intense entering of the cavern. As section 7.1.1 demonstrated, the finalstate of the compressed air is very much dependent on the overall heat transfercoefficient of the liquid-gas interface. Therefore, the more intense filling of thecave will most likely increase the rate of heat transfer and thereby furtherlydecrease the thermodynamic efficiency of the system.

In section 12.1, where the round trip efficiency of Case 1 is determined,the power comsumption by the pump is thus assumed constant throughout thecompression phase. Consuming the power needed to pump the brine at thegiven flow rate and with a discharge head of 273m with an overall efficiency of

34

90%.

10.2 Case 2 - Array of pressure vesselsSince Case 2 has a small flow rate in and out of the system, a rotary familypump is used: more precisely a gear pump. The working principle of a gearpump is quite simple and depicted in Figure 21.

Figure 21: Gear pump [14].

The gear pump is a positive displacement pump, meaning it simply movesthe fluid from a low pressure section to a high pressure section. This type ofpump is commonly used in hydraulic systems and thus the reason for choice ofpump. The data used for this pump is that obtained for the Kracht KP 0 gearpump. Please see [15] for data sheet. The total efficiency of the pump is givenas approximately 75%. Unfortunately, a characteristic could not be obtainedfrom Kracht. Neither from the pump working in pump or turbine operation.The efficiency is thus set constant in section 12.2.

10.3 Case 3 - Large underground cavern as pressure vesselThe high flow rate and discharge pressure of Case 3, just like Case 1, requires acentrifugal pump. The centrifugal pump in Case 3, however, will not experiencea large variation in discharge head. The maximum head that the pump mustdeliver throughout the entire compression phase is 2142.6m. This includespressure losses. The efficiency of the centrifugal pump at BEP is again assumedto be 90%.

When extracting work from a fluid at very high heads, the only reasonableturbine to use is the Pelton wheel turbine [20]. The Pelton turbine is an impulseturbine, meaning that the force applied to the impeller is discontinuous. In aPelton turbine the water is accelerated through one or more nozzles, creating ahigh velocity free stream jet. This stream strikes the impeller tangentially. Theimpeller consist of an arrangement of buckets. Please see Figure 22.

35

Figure 22: Principle of the Pelton type turbine with one jet stream [14].

The discharge side of the Pelton turbine is free. Meaning that the impelleris not submersed in the water. Therefore the Pelton turbine must be elevatedabove the water reservoir. This is accounted for when modelling. As Figure 18show, the Pelton wheel turbine is far more flexible than the Francis type turbine.The load variation does not influence efficiency drastically. The variation in headduring discharge is, including pressure losses, 2127− 2108m.

36

11 Thermodynamic efficiency

In this section the compression, storage and expansion models are used, wherethe pumps and turbines are considered ideal with an isentropic efficiency ofη = 1. Also, pressure losses in pipes and valves are neglected. This is done todetermine how much energy is retained within the system, where the only lossesare those which can not be avoided or minimized.


The compression time is set to eight hours. During these eight hours the totalenergy consumption by the pump is found to be:

Ein = 125293MJ = 34.8MWh

After compression, the energy is stored with no flow in or out of the system.Thus only heat transfer occurs. The storage time is set to five hours. The heatlost to the cavern walls during compression and storage is found as describedin section 7.1.2. After storage, the expansion model is used. The flow rate outthrough the water turbines is set at V = 5.33m3

/s. This flow rate is chosen sincethis corresponds to an expansion time of three hours, where the final pressureafter expansion is equal to the initial pressure of the compression. The energyextracted by the turbine is found to be:

Eout = −124059MJ = 34.5MWh

This gives a thermodynamic efficiency of:

ηtd =abs(Eout)

Ein

ηtd = 0.9901 (10)

So only 0.99% of the energy is lost as heat transferred to the cavern wallsand the brine. The thermodynamic efficiency assumes that the overall heattransfer coefficient for the gas/liquid interface is that found from equation 6.Fully blocking heat transfer between the two media gives an efficiency of:

ηtd = 0.9902

Thus 0.98% of the energy is transferred to the cavern walls. It is thus evidentthat heat transfer between the two media is of very little influence at a low valueof the overall heat transfer coefficient.

For the overall heat transfer coefficient being manipulated by a factor 1000.Neglecting heat loss in the cavern walls lowers the thermodynamic efficiency to:

ηtd = 0.9772

37

Assuming that the energy stored in the cavern walls is higher, as describedin the end of section 7.1.2, gives a thermodynamic efficiency of:

ηtd = 0.9612

Finally, assuming maximal heat loss to the compressing brine and cavern walls,the thermodynamic efficiency of the system lowers to:

ηtd = 0.9359

11.2 Case 1 - Large underground cavern as compressionchamber, using humid air

As seen in section 8.1 the final temperature and pressure when using “Air_H2O”instead of “Air_ha” was noticeably lower. The heat transferred to the cavernwall and compressing brine is also lower due to the smaller temperature gradient.Table 6 shows results at different constant relative humidities.

φ 100% 80% 60% 40% 20% 0%Ein [MJ ] 100408 100397 100388 100378 100368 100358Eout [MJ ] −100060 −99982 −99906 −99831 −99752 −99789

ηtd 0.9965 0.9959 0.9952 0.9946 0.9939 0.9943

Table 6: Results using humid air at different constant relative humiditiesthroughout compression and expansion.

The results shown in Table 6 assume, that the energy needed for the masstransport of water to air, is kept within the system and not lost to the surround-ings.

In Table 7 it is assumed that all the heat gone in to evaporation of water islost to the surroundings.

φ 100% 80% 60% 40% 20%

ηtd 0.9893 0.9912 0.9914 0.9910 0.9926

Table 7: Thermodynamic efficiency when all heat gone to mass transport is lostto the surroundings.

The thermodynamic efficiency is not significantly lowered when heat goneinto mass transport is lost to the surroundings. Also, comparing the thermody-namic efficiency of air and humid air, shows no significant difference. Becauseof these results, modelling using humid air will not be furtherly analyzed.

11.3 Case 2 - Array of pressure vesselsThe Heat transfer analysis in section 7.2 showed that Case 2 is far more sus-ceptible to lose energy in the form of the heat developed during compression,

38

compared to Case 1. The compression time for this case was set to one hour.The storage time, where only heat transfer occurs is also set to one hour. Theflow rate during expansion is set to 4.5 · 10−5

m3/s. This flow rate corresponds

to an expansion time of 30 minutes for the cylinders having 6 cm of insulation.Table 8 shows the results and thermodynamic efficiency for different thicknessof the insulation. Expansion time is slightly adjusted, so that the final pressureafter expansion is equal to the initial pressure of the compression process.

tiso [cm] 0 2 4 6 ∞Ein [kJ ] 1324 1327 1330 1331 1341Eout [kJ ] 1053 1129 1164 1188 1336

ηtd 0.795 0.8507 0.8752 0.8924 0.9962

Table 8: Case 2 results at different insulation thicknesses.

As Table 8 and Figure 10 clearly show, conduction through the vessel wallsis the greatest source of energy loss. The perfectly insulated system shows thatonly 0.38% of the energy goes into heating the siloxane piston. Figure 23 showshow the thermodynamic efficiency is drastically affected by the storage time.

Figure 23: Thermodynamic efficiency at different storage durations with 6 cmof insulation.

As seen the round trip efficiency drops fast during storage. The rate at whichthe round-trip efficiency drops is roughly 5% per hour.

39

11.4 Case 3 - Large underground cavern as pressure vesselDuring the eight hours of compression, the energy put into the system is:

Ein = 80471MJ = 22.4MWh

Energy lost during storage are lost to the surroundings as described in section7.3.2. The flow rate out through the water turbines during expansion is setto V = 0.36m3

/s, as this corresponds to an expansion time of three hours,where the final pressure after expansion is equal to the initial pressure of thecompression phase. The energy extracted by the turbine is found to be:

Eout = −79623MJ = 22.1MWh

This gives a thermodynamic efficiency of:

ηtd = 0.9895

The thermal energy lost is thus 1.05% of the energy put into the system.The thermal energy is lost to the cavern walls.

40

12 Round trip efficiency

This sections include calculations of the round trip efficiency, where pressurelosses and efficiency of pumps and turbines, as described in section 9 and section10, are included.


As described in section 10.1 the pump will under the given circumstances haveto deliver the final head of 273m throughout the compression stage. This isincluding the pressure loss in pipes, valves and elbows. The consumed energyin compression is found to be:

Epump = 254661MJ = 70.7MWh

Dividing the energy consumed by an ideal pump by the actual energy con-sumption, gives the overall efficiency of the compression stage:

ηcom = 0.492

During expansion, the head loss the system must overcome is 37.9m. Theactual variation in head is thus 229.9− 16.7m. For the expansion the efficiencycurve of Figure 18 is used. Curve fitting is done with Matlab and can be foundin the appendix. In the expansion model, the highest load in the expansion isset as 100% load. The energy extracted from the turbine is:

Eturbine = −77309MJ = −21.5MWh

The overall efficiency of the expansion stage is then:

ηexp = 0.623

The round trip efficiency is found as:

ηrt =abs(Eturbine)

Epump

ηrt = 0.3035

So for this case, including pressure losses and non-ideal equipment, the sys-tem returns 30.35% of the energy consumed in compression. This efficiency ap-plies for when using the natural convection correlation 6 and the energy storedin the wall is that found, equation8.

Assuming maximal heat loss, the round-trip efficiency of the system is foundto be:

ηrt = 0.2757

41

12.2 Case 2 - Array of pressure vesselsThe gear pump/turbine with a 75% overall efficiency is used in the compressionand expansion model. This efficiency is assumed to be constant throughout theprocess stages. The cylinders are insulated with 6 cm of polyurethene. Theconsumed energy during compression is then:

Epump = 1.775MJ = 1775 kJ = 0.49kWh

As Figure 23 clearly demonstrated, the energy lost to the surroundings isvery much dependent on the storage time. Table 9 below shows the round tripefficiency of Case 2 at different storage times:

Storage time 1hour 2hours 3hours

Eturbine [kJ ] 894.8 777.2 707.5ηrt 0.504 0.438 0.399

Table 9: Energy extracted and round-trip efficiency at different storage times.

As Table 9 demonstrates. The storage time is a critical factor on round tripefficiency for Case 2.

12.3 Case 3 - Large underground cavern as pressure vesselAs described in section 10.3, the pump must deliver the maximum head through-out all of the compression stage. The maximum head that the centrifugal pumpwill have to deliver is 2142.6m. This includes pressure losses. This results in aenergy consumption of the pump of:

Epump = 89820MJ = 25.0MWh

The overall efficiency of the compression stage is then:

ηcom = 0.8959

Close to that assumed for the pump. During expansion, the efficiency curveof a Pelton turbine from Figure 18 is used. The variation, however, proves tobe somewhat insignificant. After some optimization, the highest energy outputis found to be:

Eturbine = 75008MJ = 20.8MWh

The efficiency of the expansion stage shows to be:

ηexp = 0.9420

The overall round-trip efficiency of case 3 is thus:

ηrt = 0.8351

The setup in Case 3 thus returns 83.51% of the energy consumed in thepumping stage.

42

13 Practical considerations

In this section some practical considerations are taken into account. This sec-tions discusses some of the problems that will or may occur, that should betaken into consideration in future work.


Case 1 has its advantage of a massive volume, ability to contain a high pressureand resilient to heat loss. Some of the practical problems that will most likelyoccur in Case 1 all boil down to the fact that the cave material is soluble in water.If fresh water is being pumped into the system, the water will during expansionreturn as a saline solution with a high concentration of sodium chloride. Heavymetals and other elements are also present in small amounts, which may causeenvironmental concerns [16]. Pressure and temperature of the water have littleinfluence on the rate of the dissolving of sodium chloride. The turbulence of thesolvent flow is more important. For instance, many people have tried to drop acube of sugar into a cup of coffee. If the coffee is not mixed with a spoon, therate at which all of the sugar is completely dissolved is very low. This wouldrequire the coffee to sit for hours. If the cube of sugar is instead dropped intothe coffee and stirred with a spoon, the sugar is completely dissolved in a matterof seconds. The same applies for Case 1. The more turbulent the flow is, thefaster the sodium chloride walls dissolve [25].

If fresh water were to be used for every cycle, the returning water would befully saturated brine, thus returning approximately 0.35 kg of NaCl for everykilogram of pumped water. With NaCl having a density of about 2200 kg/m3,the approximate increase in volume of the cave is 15%, relative to the pumpedwater . To ensure no increase in cavern volume, the brine should be reused forevery cycle [6]. Case 1 has been modelled with a brine as the working liquid.

The salinity of the water will require corrosion-resistant materials in thepumps and turbines. This is, and has been, of great interest in many appli-cations. Research for the development of specially designed pump/turbines forseawater has been conducted in 1998 by Toshiba [13]. The worlds first pumpedhydro storage plant using the sea as a lower reservoir was built in Japan in 1999[4]. Ireland is also setting up the GLINSK project. A PHES plant using seawa-ter to store up to one third of the wind power generated electricity. Commissionof the facility is planned for 2013[5].

Clearly corrosion resistant seawater pump/turbines have been produced andutilized. However, a saturated saline solution has a salinity of about 35% bymass. Most seawater around the world has a salinity of 3-4% by mass.

Howard A. Porte states [22]:

“A literature survey was made to determine the effect of environ-mental variables on the corrosion rates of metals submerged in seawater. The most important variables were found to be dissolved oxy-

43

gen concentration, velocity, and temperature. Other factors whichinfluence the corrosion rates are pH, salinity, and micro-organisms.”

The above mentioned survey was conducted towards metals submerged in water.A survey regarding corrosion rate in pumps/turbines has not been found, butas seen above, many factors influence corrosion. The velocity of the water ina pump/turbine will be very high and thereby of high influence. The increasein salinity of up to a factor 10, compared to regular seawater, will not equal afactor 10 increase in the corrosion rate. The increase in rate will be far less [25].

Structural considerations should also be made regarding the somewhat freehanging pipe needed to reach the bottom of the cave. This includes it ownweight and especially some modal analysis should be made. A more sturdysetup would be having the pipe enter the cave from below.

13.2 Case 2 - Array of pressure vesselsThe outer dimensions of the seven pressure vessel setup would be a nearly cylin-drical set up of a height of about 130 cm and a diameter of approximately 140 cm.Additionally there’s the need of a liquid basin along with piping, pump/turbineand valves. Each of the pressure vessels has an empty weight of 136 kg. Thefinal setup will therefore reach a weight well over 1 ton and require a lot ofspace.

The setup will, however, have a high life cycle, since corrosion is not ofconcern.

13.3 Case 3 - Large underground cavern as pressure vesselCase 3 requires a high pressure compression chamber of large volume. The largerthe volume, the higher the energy capacity of the system. In the calculations,the vessel was set to a diameter of 10m and a height of 50m. The hoop stressfor a thin walled cylinder is given by [21]:

σ =P r

t

Calculations show, that when using a high tensile strength steel for the vessel,it requires an absolute minimum vessel thickness of around 25 cm. This wouldbe expensive in labor and materials. Another possibility is to build a Prestressedconcrete pressure vessel. Prestressed concrete pressure vessels have been used aspressurized water reactors in nuclear power plants and lots of research has goneinto optimising the designs1. Concrete is an anisotropic material that showsincredible strength when compressed. The tensile strength, however, is low.The concrete is thus armored with pre-stressed steel, lowering the hoop stressof the concrete when loaded with an internal pressure.

1Interesting design on this field include patent numbers: 4313902 and 3390211

44

14 Discussion

In this section each case will be discussed. The discussion is based on the resultsof the model analysis and practical considerations.


Unfortunately, the heat transfer to the surrounding semi-infinite material hadto be analyzed separately and was not embedded into the model. This wasnecessary as heat transfer through a semi-infinite medium is a case of transientconduction. Analyzing this heat transfer separately induces inaccuracies.

The hypothesis of the ALP-CAES principle was that the compression wouldbe adiabatic near isothermal. This proved to be somewhat true. The heattransfer to the compressing liquid showed to be almost negligible. However,the heat transfer to the compressing liquid is very dependent on the gradientand more importantly; the overall heat transfer coefficient. The overall heattransfer coefficient was determined using a correlation for natural convection.This proved to yield a low convective heat transfer coefficient. It is by the authorbelieved, that the convective heat transfer coefficient will be very low for a slowand calm filling. The intensity of the filling will, however, increase the magnitudeof the overall heat transfer coefficient. Manipulation of the overall heat transfercoefficient showed a noticeable influence on the thermodynamic efficiency. Moreextensive analysis of the fluid flow will be necessary to determine the influenceof convection.

The heat development was expected to be absorbed in the vessel walls andlater released during expansion. Due to the heat transfer between the air andcavern walls being analyzed separately, this was not modelled during the expan-sion process but was instead considered as a loss.

The psycrometric analysis turned out to be an estimate and not actual mod-elling. The estimate, however, showed that the losses due to humidification areindeed negligible.

It is important to note, on further work, that the energy density of thiscase can not be significantly increased. The minimum pressure must be high,to be able to lift the 1.5 km liquid column and structural considerations of thecavern sets a maximum allowable pressure of Pmax = 216 bar. Water turbinesalso require a certain minimum pressure. The operating pressure range is thussmall. Hence the maximal amount of energy that can be stored in this case isin the order of that found. That is in the order of 34.8MWh.

14.2 Case 2 - Array of pressure vesselsAs with Case 1, there’s room for improvement if performance characteristics areobtained for the fluid machinery. This includes pump and turbine operation.However, taking the stored energy, thermodynamic efficiency and the size of thissetup into account makes this highly unlikely to ever be realized. The extracted

45

energy (ideally and after one hour of storage) is roughly 8.5% of the energy anaverage danish household consumes per day. When taking non-ideal machineryinto account the output energy is approximately 2% [3]. Also, some energywould have to go into automatic controlling of the setup. As demonstrated, thethermodynamic efficiency drops at a rate of approximately 5% per storage hour.Hence Case 2 also has very poor storage potential.

An alike setup, which proved to be feasible, has been researched and ex-perimented [19]. This, however, is done diabatically, almost isothermally andhas interaction with super capacitors . The research also showed that a 30◦

temperature increase of the compressed gas induces a 5% efficiency drop. Thisrealization emphasizes what was found in this project; It is an absolute necessityto not loose heat to the surroundings.

To further minimize the energy consumption during compression, one couldequip the vessels with passive interior heat exchangers (metal plates or rods),improving the heat transfer from the nitrogen to the siloxane. This would lowerthe gas temperature and thus compression would require less energy, improvingthe efficiency. Minimizing the energy consumption of compression, by the meansof creating near isothermal conditions, is a method used in various compressordesigns2.

14.3 Case 3 - Large underground cavern as pressure vesselOn further work, calculations on the compression chamber should be made. Ves-sels in comparable scale and strength, to act as compression chambers, were notfound. It is thus important to analyze the possibility of a prestressed concretepressure vessel, steel vessel, or an alternative solution.

Figure 24 below shows the output energy and corresponding efficiencies asthe compression chamber volume is increased.

2Some isothermal compressor designs include the following patent numbers: 462776, 586100and 4242878

46

Figure 24: Output energy from turbine, and corresponding thermodynamic andround-trip efficiency, vs. increasing compression chamber volume.

The cavern size in the calculations, depicted in Figure 24, is kept constant.Also, initial pressure is decreased as compression chamber volume is increased,so that the final pressure is always the maximum allowable for the cavern. So asthe volume of the compression chamber is increased, so is the pressure range. Itis thus important to note that the round-trip efficiency can be improved. Thiscan be done by implementing a compression model with variable flow rate, sincethe greatest loss in energy is during the pumping stage.

By comparison, the Huntorf CAES plant has a power output of approxi-mately 290MW for two hours. A total of 580MWh [8]. It is, however, im-portant to note that the Huntorf plant is still dependent on fossil fuels. Theefficiency of CAES plants are thus only comparable to ALP-CAES by exergyanalysis. An exergy analysis of the Huntorf plant, shows to have an exergetic ef-ficiency of 43% [8]. Since there’s no addition of heat in the ALP-CAES principle,the exergetic efficiency and round-trip efficiency are interchangeable terms.

The efficiency of Case 3 is thus much higher than that seen in the existingHuntorf CAES plant.

As Figure 24 suggest, Case 3 has the potential to be scaled and thus increasecapacity, with an efficiency still superior to the Huntorf CAES plant.

47

15 ConclusionIn this section, the findings of each are concluded. Finally, an overall conclusionis made.

15.1 Case 1 - Large underground cavern as compression

chamber

Case 1 showed to be resilient to heat losses to its surroundings. The analysisshowed that the heat loss to the cavern walls and the compressing brine, wouldat most induce a loss of 6.41% of the total energy put into the system. The heattransfer- and thermodynamic efficiency analysis also showed, that the heat lostis very dependent on what assumptions are made. The highest thermodynamicefficiency of the system was found to be ηtd = 0.9901.

The round-trip efficiency of the system was ηrt = 0.3035 assuming minimalheat loss. This can, however, be greatly improved. The greatest loss in theprocess clearly showed to be in the pumping stage, where the pump worked ata constant flow to deliver the maximum head. The pump thus consumed morethan double the amount of energy required to bring the system to its final state.Centrifugal pumps, which showed to be the most suitable pumps under givenconditions, vary in delivered head dependent on the flow rate. In future work,using this case, a model with varying flow rate should be implemented. To varythe flow rate, using the model built in this project, requires the flow rate to varyover time. It can not vary as a function of discharge head. This is due to theprogram structure and since EES solves initial value problems. The consumedenergy in the compression phase was found to be 254661MJ = 70.7MWh.

The expansion process showed better results. A model with varying flowrate may, however, increase the efficiency of the expansion stage. The expansionstage had an overall efficiency of 62.3%. The energy output of the system was77309MJ = 21.5MWh.

The necessity of using brine as the pumped medium brings material require-ments. To ensure low maintenance and high life cycle, the materials must bevery resistant to corrosion. The high salinity of the water has an influence onthe corrosion rate. However, salinity is of less importance compared to fluidvelocity, dissolved oxygen and temperature. Fluid velocity will be high.

Case 1 was also modelled using the properties of humid air. Assuming aconstant relative humidity throughout the process, enabled one to determinethe influence of heat gone into evaporation of water. This clearly showed thatthe temperature drop of the air, if all heat gone into evaporation of water waslost to the surroundings, was indeed insignificant.

The high liquid column from the bottom of the cave to the surface provedto be a blocking factor for capacity. The maximum energy capacity of Case 1showed to be in the order of 34.8MWh

48

15.2 Case 2 - Array of pressure vessels

Case 2 showed to be very susceptible to lose energy in the form of heat. Theheat development due to compression showed to escape through the cylindersquite rapidly. The thermodynamic efficiency was thus low even though the rateat which heat dissipated could be lowered by the means of insulation. The roundtrip efficiency fell as storage time was increased. This was at a rate of roughly5% per hour.

For one hour of storage the round-trip efficiency was 50.4%. This can, how-ever, be optimized if more specific data for the fluid machinery is used. Theextracted energy was, ideally, in the order of 8.5% of the daily average danishhousehold consumption. Including losses in pumps, this figure dropped to 2%.

15.3 Case 3 - Large underground cavern as pressure vessel

Case 3 showed the best results troughout the analysis. Thermodynamically itwas due to the low temperature gradient, as a result of low variation in pressureduring compression and expansion. A low variation in pressure also enablesthe fluid machinery to work at or close to its best efficiency point. This iscertainly of high significance with centrifugal pumps. The overall efficiency ofthe pumping stage showed to be near 90%. Regarding the subsequent energyextraction, the high discharge pressure enables the use of the super effective,and flexible, Pelton wheel turbines. The round-trip efficiency of Case 3 showedto be η = 0.8351 with a total energy output of 75008MJ = 20.8MWh.

Further work on Case 3 should, according to the author, mainly be the sizingand design of the compression chamber.

15.4 Overall conclusion

Heat transfer through the gas/liquid interface showed to be negligible at lowtemperature gradients or with a calm liquid surface during filling. The formercan be achieved by making sure that the gas experiences little variation inpressure. The greatest source of decline in thermodynamic efficiency, was clearlyenergy lost to the surroundings. It is thus vital to the ALP-CAES principlethat heat transfer to the surroundings is very much limited. This is the case ifutilizing a large underground cavern.

Another advantage of small variation in pressure, is in regard to the fluidmachinery. Pumps, centrifugal pumps especially, show best performance if thedischarge head does not vary. This allows the pump to work close to or at thebest efficiency point, ensuring a high overall pumping efficiency.

Pre-pressurizing the system, increases the specific work done by and on themachinery. This thereby increases the capacity and hence the energy density ofthe system. This was clearly demonstrated with Case 3. Another advantage ofthe high pressure is the possibility of using a Pelton wheel turbine for extraction.This is a highly efficient and flexible turbine design.

49

An increase of the volume of the compression chamber in Case 3, showedthat the ALP-CAES principle has the potential to be scaled, still maintaininga high efficiency and thus become technologically competitive to other storagemethods.

50

References[1] http://gaslager.energinet.dk/da/om-gaslageret/sider/default.aspx.

[2] http://gaslager.energinet.dk/da/om-gaslageret/sider/geologisk-udvikling.aspx.

[3] http://www.dongenergy.dk/privat/energiforum/tjekditforbrug/typiskelforbrug/pages/typiskelforbrug.aspx.

[4] http://www.electricitystorage.org/esa/technologies/pumped_hydro/.

[5] http://www.organicpower.ie/content/projects/glinsk.htm.

[6] Personal correspondance with Fritz Crotogino.

[7] Fritz Crotogino. Large scale energy storage in geological formations. Pre-

sentation at Pan European Energy Storage Forum, September 2010.

[8] Brian Elmegaard. Caes - compressed air electricity storage. PresentationJanuary 14th 2009.

[9] Fritz Crotogino et al. Huntorf caes: More than 20 years of succesful oper-ation. March 2001.

[10] Hugh D. Young et al. University Physics, volume 1. Pearson, 12th editionedition, 2006.

[11] Igor J. Karassik et al. Pump Handbook. McGraw-Hill, Two Penn Plaza,New York, NY, fourth edition, 2008.

[12] Robert W. Fox et al. Introduction to fluid mechanics. John Wiley andSons, Inc, seventh edition, 2009.

[13] Tetsuo Fujihara et al. Development of pump turbine for seawater pumped-storage power plant. Hitachi Review, 47(5):199–202, 1998.

[14] Zoeb Husain et al. Basic Fluid Mechanics and Hydraulic Machines. BSPublications, 2008.

[15] http://kracht.eu/uploads/tx_ttproducts/datasheet/KP0_GB_05 10.pdf.

[16] HWR. Prognose for sammensætning af oppumpet saltbrine fra ll. torupgaslager. Technical report, Rambøll Danmark A/S, Teknikerbyen 31, DK-2830 Virum, August 2009.

[17] Frank P. Incropera. Inroduction to Heat Transfer. John Wiley and Sons,Inc, USA, fourth edition, 2002.

[18] Poul Scheel Larsen. Teknisk Termodynamik. DTU-Tryk, Kgs. Lyngby, thirdedition, 2008.

51

[19] S. Lemofouet and A. Rufer. Hybrid energy storage system based on com-pressed air and supercapacitors with maximum efficiency point tracking.2005.

[20] P.K. Nag. Power Plant Engineering. Tata McGraw-Hill, New Delhi, thirdedition, 2009.

[21] Arne Gudmann Nielsen. Eksempelsamling til styrkelære 2 noter, December1999.

[22] Howard A. Porte. The effect of environment on the corrosion of metals insea water - - a literature survey. Technical report, Naval civil engineeringLaboratory, Port Hueneme, California, July 1967.

[23] Y.R. Rashid. Ultimate strength analysis of prestressed concrete pressurevessels. Nuclear Engineering and Design, 7(4):334 – 344, 1968.

[24] R. S. Steltzer and R. N. Walters. Estimating reversible pump-turbine char-acteristics. Technical report, Engineering and Research Center, Bureau ofReclamation, Denver, CO, December 1977.

[25] Meeting with Steen Skaarup. Associate Professor Department of Chem-istry DTU.

52

Appendix

Overview

A EES Code

B Energy stored in cavern wall

C Pressure losses

D Turbine efficiency

E Pressure vessel information

File:Compression_case1.EES 12-07-2011 11:46:08 Page 1EES Ver. 8.643: #0780: Department of Energy Engineering, Tech. Univ. of Denmark

$integraltable time:1000;V_a;T_a;P;T_w;M_w; {CompressionCase 1 - Large underground Cavern as compression chamberAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Vessel conditions"z_surface=1500 [m] "bottom of cave is z=0"P_overtryk=5,5 [bar] "5 bar minimumpressure at surface"P_0=P_atm+z_surface*g*rho_w+P_overtryk*1e5 [Pa] "Pressure in cave"T_0=300 [K] "Start temperature of air"T_w_0=300 [K] "Initial Temperature of water"D=65 [m] "Diameter"H=300 [m] "Height"A=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A*H "Volume of cave"V_a_0=V_c_0-V_w_0 "Initial volume of air"V_w_0=2*A "Initial volume water" g=9,81 [m/s^2] "Gravity" P_max=0,18*(z_surface-H) "Maximum pressure in cavern"P_min=1/3*P_max "Minimal pressure in cavern" P_atm=101300 [Pa] "Atmospheric pressure" "Pump conditions"Q_pump=2 "Flowrate of water"Q_pump=m_dot/rho_w "Mass flow rate"tau_full=8*3600 [s] "filling time"h_reservoir=10 "height of water resevoir to pump"rho_w_0=1260 "Density of water"P_pump2=P_atm+rho_w_0*g*h_reservoir "Pressure at pump inlet" "Air mass"rho_a_0=Density(Air_ha;T=T_0;P=P_0) "Initial density of air"M_a=rho_a_0*V_a_0 "Iniitial mass" "Initial water mass"rho_w=1260 "Density of water"M_w0=rho_w*V_w_0 "Initial total mass"M_t0=M_w0+M_a "dM/dt=m_dot"M-M_t0=integral(m_dot;time;0;tau_full) "Mass of water"M_w=M-M_a


"Volume of water"V_w=M_w/rho_w "Volume of air"V_a=V_c_0-V_w "Height of water surface in cavern"z_w=4*V_w/(D^2*Pi) "dV/dt"dV_a=-m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air" M_a*(u_a-u_a0)=integral(Q_dot_1-P*dV_a;time;0;tau_full) u_a0=IntEnergy(Air_ha;T=T_0;P=P_0) "initial int. energy" u_a=IntEnergy(Air_ha;T=T_a;P=P) "Int. energy, determines temperature" rho_a=M_a/V_a "Air density" P=Pressure(Air_ha;T=T_a;v=1/rho_a) "Air pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w0*u_w0=integral(-Q_dot_1+m_dot*(h_w+g*(z_surface-1/2*z_w))-P*dV_w;time;0;tau_full) T_w=temperature(Water;u=u_w;P=P) "Temperature of water"u_w0=IntEnergy(Water;T=T_w_0;P=P_0) "initial internal energy" P_pump=P+rho_w*g*(z_w-z_surface) "Pressure at pump outlet"Head_pump=P_pump/(rho_w*g) "Head at pump" "_________________________" "Work done on system by pump" "Pump work"eta_pump=(h_w0-h_sw)/(h_w0-h_w)h_w0=Enthalpy(Water;T=T_0;P=P_pump2)s_w0=Entropy(Water;T=T_0;P=P_pump2) " FOR Round-Trip Efficiency"eta_pump=0,90 "Efficiency of centrifugal pump"Head_loss=6,1 "Head loss"h_sw=Enthalpy(Water;s=s_w0;P=P_pump_max) "Enthalpy after pump, if no entropy increase"P_pump_max=(266,9+Head_loss)*g*rho_w "266,9 Is the maximum head of ideal"T_pump=Temperature(Water;h=h_w;P=P_pump_max) "Temperature after pump""___________________" {"FOR TDE"


eta_pump=1 "eta_pump=1 for thermodynamic eff"h_sw=Enthalpy(Water;s=s_w0;P=P_pump) "Enthalpy after pump if no entropy increase "T_pump=Temperature(Water;h=h_w;P=P_pump) "Temperature after pump""___________________"} W_pump=m_dot*(h_w-h_w0)*1e-6 "Pump Power in MW"E_s=integral(W_pump;time;0;tau_full) "Energy used"E_s_t=max(E_s) "% In MJ" "__________________________" "Heat transfer" "air -> water"Q_dot_1=UA*(T_w-T_a) "HEat transfer air->water" UA=UA_1 "UA value"UA_1=1/(R_1a+R_1b) "Definition of UA value"R_1a=1/(h_1a*A_water) "Definition of resitances"R_1b=1/(h_1b*A_water) "Deinition of resistance"A=A_water "Heat transfer area" "Natural convection, hot air to cold 'membran' "Nusselt_1a=0,27*Rayleigh_1a^(1/4) "nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_D^3/nu_1a "Grashoff number"L_D=A_water/(D*Pi) "Equivalent length"mu_1a=Viscosity(Air_ha;T=T_a;P=P) "Dynamic visosity"Prandtl_1a=Prandtl(Air_ha;T=T_a;P=P) "Prandtl number"nu_1a=mu_1a/rho_a "Kinemtaic visc."beta_1a=1/T_a "Thermal exp. coeff."k_a=Conductivity(Air_ha;T=T_a;P=P) "Thermal conductivity"Nusselt_1a=h_1a*L_D/k_a "Nusselt number"T=abs(T_a-T_w) "Temp. difference" "Natural convection, hot 'membran' to cold water"Nusselt_1b=0,27*(Rayleigh_1b)^(1/4) "Nusselt number"Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayleigh number"Grashof_1b=g*beta_1b*T*L_D^3/nu_1b "Grashof number"mu_1b=Viscosity(Water;T=T_w;P=P) "Dyanmic viscosity"Prandtl_1b=Prandtl(Water;T=T_w;P=P) "prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(Water;T=T_w;P=P) "Volumetric exp. coeff."k_b=Conductivity(Water;T=T_w;P=P) "Thermal conductivity"Nusselt_1b=h_1b*L_D/k_b "Convective heat transfer coeff"

SOLUTIONUnit Settings: SI K Pa J mass degA = 3318 Awater = 3318 1a = 0,003224

1b = 0,0003036 D = 65 [m] dVa = -2 dVw = 2 pump = 0,9 Es = 254661 Es,t = 254661 g = 9,81 [m/s2] Grashof1a = 1,179E+10


Grashof1b = 1,788E+08 H = 300 [m] Headloss = 6,1 Headpump = 266,9 h1a = 0,1991 h1b = 1,803 hreservoir = 10 hsw = 115854 hw = 116205 hw0 = 112696 ka = 0,03849 kb = 0,608 LD = 16,25 M = 2,924E+08 1a = 0,00002425

1b = 0,0008357 Ma = 2,114E+08 m = 2520 [kg/s]]Mt0 = 2,198E+08 Mw = 8,094E+07 Mw0 = 8,362E+06 Nusselt1a = 84,05 Nusselt1b = 48,19 1a = 1,068E-07

1b = 6,632E-07 P = 2,160E+07 Prandtl1a = 0,7961 Prandtl1b = 5,673 P0 = 1,919E+07 [Pa] Patm = 101300 [Pa]Pmax = 216 Pmin = 72 Povertryk = 5,5 [bar]Ppump = 3,299E+06 Ppump2 = 224906 Ppump,max = 3,374E+06 Q1 = -5521 Qpump = 2 [m3/s] Rayleigh1a = 9,390E+09 Rayleigh1b = 1,014E+09 a = 227 a,0 = 213,8

w = 1260 w,0 = 1260 R1a = 0,001514 R1b = 0,0001671 sw0 = 392,8 T = 9,281

full = 28800 [s] time = 28800 T0 = 300 [K]Ta = 310,2 Tpump = 300,1 Tw = 300,9 Tw,0 = 300 [K] UA = 594,9 UA1 = 594,9 ua = 184332 ua0 = 178787 uw = 114322 uw0 = 110926 Va = 931256 Va,0 = 988856 [m3]Vc,0 = 995492 Vw = 64237 Vw,0 = 6637 [m3]Wpump = 8,842 zsurface = 1500 [m] zw = 19,36

0 10 20 30 40 50 60 70 800

100000

200000

300000

400000

500000

600000

time

time,

P

P[1]=Pressure(Air;T=T[1];u=u[1])Integral Table

time Va Ta P Tw Mw

Row 1 0 988856 300 1,919E+07 300 8,362E+06 Row 2 1000 986856 300,3 1,927E+07 300,3 1,088E+07 Row 3 2000 984856 300,7 1,934E+07 300,4 1,340E+07


Integral Tabletime Va Ta P Tw Mw

Row 4 3000 982856 301 1,942E+07 300,5 1,592E+07 Row 5 4000 980856 301,3 1,950E+07 300,6 1,844E+07 Row 6 5000 978856 301,7 1,958E+07 300,7 2,096E+07 Row 7 6000 976856 302 1,965E+07 300,7 2,348E+07 Row 8 7000 974856 302,4 1,973E+07 300,8 2,600E+07 Row 9 8000 972856 302,7 1,981E+07 300,8 2,852E+07 Row 10 9000 970856 303 1,989E+07 300,8 3,104E+07 Row 11 10000 968856 303,4 1,997E+07 300,8 3,356E+07 Row 12 11000 966856 303,7 2,005E+07 300,8 3,608E+07 Row 13 12000 964856 304,1 2,014E+07 300,8 3,860E+07 Row 14 13000 962856 304,4 2,022E+07 300,9 4,112E+07 Row 15 14000 960856 304,8 2,030E+07 300,9 4,364E+07 Row 16 15000 958856 305,1 2,038E+07 300,9 4,616E+07 Row 17 16000 956856 305,5 2,047E+07 300,9 4,868E+07 Row 18 17000 954856 305,9 2,055E+07 300,9 5,120E+07 Row 19 18000 952856 306,2 2,064E+07 300,9 5,372E+07 Row 20 19000 950856 306,6 2,072E+07 300,9 5,624E+07 Row 21 20000 948856 306,9 2,081E+07 300,9 5,876E+07 Row 22 21000 946856 307,3 2,090E+07 300,9 6,128E+07 Row 23 22000 944856 307,6 2,099E+07 300,9 6,380E+07 Row 24 23000 942856 308 2,107E+07 300,9 6,632E+07 Row 25 24000 940856 308,4 2,116E+07 300,9 6,884E+07 Row 26 25000 938856 308,7 2,125E+07 300,9 7,136E+07 Row 27 26000 936856 309,1 2,134E+07 300,9 7,388E+07 Row 28 27000 934856 309,5 2,143E+07 300,9 7,640E+07 Row 29 28000 932856 309,9 2,153E+07 300,9 7,892E+07 Row 30 28800 931256 310,2 2,160E+07 300,9 8,094E+07

File:Storage_case1.EES 12-07-2011 11:57:00 Page 1EES Ver. 8.643: #0780: Department of Energy Engineering, Tech. Univ. of Denmark

$integraltable time:1000;T_a;P;T_w;Q_dot_1 {StorageCase 1 - Large underground Cavern as compression chamberAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Vessel conditions"z_surface=1500 [m] "bottom of cave is z=0"u_a_s=184332 "initial internal energy of air from compression"u_w_0=114322D=65 [m] "Diameter"H=300 [m] "Height"A=(D^2/4*Pi) "Crossectional area of cave" g=9,81 [m/s^2] "Gravity" P_atm=101300 [Pa] "Atmospheric pressure" tau_full=5*3600 [s] "storage time"rho_w=1260 "Air mass"V_a_0=931256 "Initial air volume"M_a=2,114e8 "Air mass"M_a=rho_a*V_a_0 "Air density"T_0=temperature(Air_ha;v=1/rho_a;u=u_a_s) "Initial Temperature"P_0=Pressure(Air_ha;T=T_0;v=1/rho_a) "Initial Pressure" "Initial water mass"M_w=7,731e7 "water mass"T_w_0=Temperature(WateR;u=u_w_0;P=P) "Initial Temperature, water" "-----------------------------------------" "Internal energy balance air"M_a*(u_a-u_a0)=integral(Q_dot_1;time;0;tau_full) u_a0=IntEnergy(Air_ha;T=T_0;P=P_0) "Initial internal energy air" u_a=IntEnergy(Air_ha;T=T_a;P=P) "Determines temperature" P=Pressure(Air_ha;T=T_a;v=1/rho_a) "Determines pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*(u_w-u_w0)=integral(-Q_dot_1;time;0;tau_full) "Water" T_w=temperature(Water;u=u_w;P=P) "Temperature of water"u_w0=IntEnergy(Water;T=T_w_0;P=P_0) "Initial internal energy water" "Heat transfer""air -> water"


Q_dot_1=UA_1*(T_w-T_a) "Heat transfer air to water interface" Q_1=Integral(Q_dot_1;time;0;tau_full) "Total amount of heat transferred" UA_1=1/(R_1a+R_1b) "Definition of UA value"R_1a=1/(h_1a*A_water) "Resistance air"R_1b=1/(h_1b*A_water) "Resistance water"A=A_water "Heat transferable area" "Natural convection, hot air to cold membran"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_D^3/nu_1a "Grashiof number"L_D=A_water/(D*Pi) "Equivalent lentght"mu_1a=Viscosity(Air_ha;T=T_a;P=P) "Dynamic viscosity"Prandtl_1a=Prandtl(Air_ha;T=T_a;P=P) "Prandtl number"nu_1a=mu_1a/rho_a "Nusselt number"beta_1a=1/T_a "Thermal expansion coefficient"k_a=Conductivity(Air_ha;T=T_a;P=P) "Thermal conductivity"Nusselt_1a=h_1a*L_D/k_a "convective heat transfer coefficient"T=abs(T_a-T_w) "Temp. difference" "Natural convection, hot membran to cold water"Nusselt_1b=0,27*(Rayleigh_1b)^(1/4) "Nusselt number"Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayleigh number"Grashof_1b=g*beta_1b*T*L_D^3/nu_1b "Grashof number"mu_1b=Viscosity(Water;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(Water;T=T_w;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic visc."beta_1b=VolExpCoef(Water;T=T_w;P=P) "Vol. exp. coeff."k_b=Conductivity(Water;T=T_w;P=P) "Conductivity"Nusselt_1b=h_1b*L_D/k_b "Conv. heat transfer coeff."


1b = 0,0003036 D = 65 [m] g = 9,81 [m/s2]Grashof1a = 1,178E+10 Grashof1b = 1,787E+08 H = 300 [m]h1a = 0,199 h1b = 1,803 ka = 0,03849 kb = 0,608 LD = 16,25 1a = 0,00002425

1b = 0,0008357 Ma = 2,114E+08 Mw = 7,731E+07 Nusselt1a = 84,04 Nusselt1b = 48,18 1a = 1,068E-07

1b = 6,632E-07 P = 2,160E+07 Prandtl1a = 0,7962 Prandtl1b = 5,673 P0 = 2,160E+07 [Pa] Patm = 101300 [Pa]Q1 = -9,932E+07 Q1 = -5517 Rayleigh1a = 9,384E+09 Rayleigh1b = 1,014E+09 a = 227 w = 1260 R1a = 0,001514 R1b = 0,0001672 T = 9,276

full = 18000 [s] time = 18000 T0 = 310,1 [K]Ta = 310,1 Tw = 300,9 Tw,0 = 300,9 [K]UA1 = 594,8 ua = 184332 ua0 = 184332 ua,s = 184332 uw = 114323 uw0 = 114322 uw,0 = 114322 Va,0 = 931256 [m3] zsurface = 1500 [m]


0 10 20 30 40 50 60 70 800

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time Ta P Tw Q1

Row 1 0 310,1 2,160E+07 300,9 -5518 Row 2 1000 310,1 2,160E+07 300,9 -5518 Row 3 2000 310,1 2,160E+07 300,9 -5518 Row 4 3000 310,1 2,160E+07 300,9 -5518 Row 5 4000 310,1 2,160E+07 300,9 -5518 Row 6 5000 310,1 2,160E+07 300,9 -5518 Row 7 6000 310,1 2,160E+07 300,9 -5518 Row 8 7000 310,1 2,160E+07 300,9 -5518 Row 9 8000 310,1 2,160E+07 300,9 -5518 Row 10 9000 310,1 2,160E+07 300,9 -5518 Row 11 10000 310,1 2,160E+07 300,9 -5518 Row 12 11000 310,1 2,160E+07 300,9 -5518 Row 13 12000 310,1 2,160E+07 300,9 -5518 Row 14 13000 310,1 2,160E+07 300,9 -5518 Row 15 14000 310,1 2,160E+07 300,9 -5517 Row 16 15000 310,1 2,160E+07 300,9 -5517 Row 17 16000 310,1 2,160E+07 300,9 -5517 Row 18 17000 310,1 2,160E+07 300,9 -5517 Row 19 18000 310,1 2,160E+07 300,9 -5517

File:Expansion_case1.EES 12-07-2011 11:51:59 Page 1EES Ver. 8.643: #0780: Department of Energy Engineering, Tech. Univ. of Denmark

$integraltable time:1000;V_a;T_a;P;T_w;M_w;P_turbine;Head_turbine; {ExpansionCase 1 - Large underground Cavern as compression chamberAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Initial internal energies from storage"u_e=184048-u_cavern "initial Internal energy air"u_w_e=115065 "initial Internal energy of water"u_cavern=116,3 "Internal energy lost to surroundings" "Vessel conditions"z_surface=1500 [m] "bottom of cave is z=0"D=65 [m] "Diameter"H=300 [m] "Height"A=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A*H "Volume of cave"V_w_0=2*A "Volume of water before compression"g=9,81 [m/s^2] "Gravity" P_atm=101300 [Pa] "Atmospheric pressure"h_reservoir=10 "Height of water resevoir" P_turbine2=P_atm+rho_w*g*h_reservoir "Pressure at turbine outlet" "Turbine conditions"Q_turbine=5,33 "Flowrate of water"Q_turbine=m_dot/rho_w "Mass flow rate" tau_empty=3*3600[s] "Expansion time" "Air mass"M_a=2,114E+08 "Air mass, from storage model"V_a_e=931256 "air volume, from storage model"rho_a_e=M_a/V_a_e "Air density" "Initial temperatures"T_e=Temperature(Air_ha;u=u_e;v=1/rho_a_e)T_w_e=Temperature(Water;P=P_e;u=u_w_e) "Initial water mass"rho_w=1260M_w_e=8,094e7 "From storage model"M_w_e=rho_w*V_w_e "Initial pressure"P_e=Pressure(Air_ha;T=T_e;v=1/rho_a_e)"Initial total mass"M_t_e=M_w_e+M_a "dM/dt=m_dot"M-M_t_e=integral(-m_dot;time;0;tau_empty)


"Mass of water"M_w=M-M_a "Volume of water"V_w=M_w/rho_w "Volume of air"V_a=V_c_0-V_w "Height of water surface"z_w=4*V_w/(D^2*Pi) "change in volume. dV/dt"dV_a=m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air"M_a*(u_a-u_a0)=integral(Q_dot_1-P*dV_a;time;0;tau_empty) u_a0=IntEnergy(Air_ha;T=T_e;P=P_e) "initial int. energy air" u_a=IntEnergy(Air_ha;T=T_a;P=P) "finds temperature" rho_a=M_a/V_a "Density of air" P=Pressure(Air_ha;T=T_a;v=1/rho_a) "Pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w_e*u_w0=integral(-Q_dot_1-m_dot*(h_w_t+g*(z_surface-1/2*z_w))-P*dV_w;time;0;tau_empty) T_w=Temperature(Water;u=u_w;P=P) "Water temperature"u_w0=IntEnergy(Water;T=T_w_e;P=P_e) "Initial internal energy" P_turbine=P+rho_w*g*(z_w-z_surface) "pressure at turbine inlet""_________________________" "!Work done by system on turbine""FOR RTE""turbine work"Head_loss=37,9 "Head loss"P_turbine_loss=P_turbine-Head_loss*rho_w*g "Pressure at turbine, after loss"s_w_t=Entropy(Water;h=h_w_t;P=P_turbine_loss) "Entropy"h_w_t=Enthalpy(Water;T=T_w;P=P_turbine_loss) "Enthalpy in to system"P_turbine_max=(267,9-Head_loss)*rho_w*g "Maximum turbine pressure, after losses"Load_per=P_turbine_loss/P_turbine_max*100 "Maximum pressure set to 100% turbine load" "Correlation for the efficiency of a francis turbine. Found using MATLAB"eta_francis = p1*z^7 + p2*z^6 +p3*z^5 + p4*z^4 +p5*z^3 + p6*z^2 +p7*z + p8 z = (load_per-mu_f)/sigma_fmu_f= 43,846sigma_f= 33,86p1 = 0,00069809


p2 = 0,0019064p3 = 0,013802p4 = -0,068269p5 = 0,060016p6 = -0,073648p7 = 0,20217p8 = 0,76286 W_turbine_2=abs(W_turbine) "Absolute value of power"Head_eff=P_turbine_loss/(rho_w*g) "Effictive head at turbine inlet"N_s=N*sqrt(W_turbine_2*1e3)/Head_eff^(5/4) "definition of specific speed. Power is in kW"N_s=350 "Specific speed for turbine, to fit with figure for efficiency" "_________________________" {"FOR TDE"s_w_t=Entropy(Water;h=h_w_t;P=P_turbine) "Entropy"h_w_t=Enthalpy(Water;T=T_w;P=P_turbine) "Enthalpy out of system"eta_francis=1"__________________________"} h_sw0=Enthalpy(Water;s=s_w_t;P=P_turbine2) "Enthalpy if no increase in entropy"eta_francis=(h_w_t-h_w0)/(h_w_t-h_sw0) "To find enthalpy after turbine"W_turbine=m_dot*(h_w0-h_w_t)*1e-6 "Power of turbine, in MW"E_s=integral(W_turbine;time;0;tau_empty) "Energy extracted from turbine, MJ"abs(E_s)=E_s_turbineE_s_pump=125292 "Energy consumed in pump"eta_rt=E_s_turbine/E_s_pump "Round trip efficiency" Head_turbine=P_turbine/(rho_w*g) "Heah at turbine inlet" "__________________________"Q_dot_1=UA*(T_w-T_a) "Heat transfer accross air/brine interface" UA=UA_1UA_1=1/(R_1a+R_1b) "Definition on UA value"R_1a=1/(h_1a*A_water) "Resistance air"R_1b=1/(h_1b*A_water) "Resistance water"A=A_water "Heat transferrable surface area""Natural convection air"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_D^3/nu_1a "Grashof number"L_D=A_water/(D*Pi) "Equivalent length"mu_1a=Viscosity(Air_ha;T=T_a;P=P) "dynamic Viscosity"Prandtl_1a=Prandtl(Air_ha;T=T_a;P=P) "Prandtl number"nu_1a=mu_1a/rho_a "Kinematic visc."beta_1a=1/T_a "Thermal expansion coeff"k_a=Conductivity(Air_ha;T=T_a;P=P) "Thermal conductivity"Nusselt_1a=h_1a*L_D/k_a "convective heat transfer coefficient"T=abs(T_a-T_w) "Temp. difference" "Natural convection water"Nusselt_1b=0,27*(Rayleigh_1b)^(1/4) "Nusselt number"


Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayleigh number"Grashof_1b=g*beta_1b*T*L_D^3/nu_1b "Grashof number"mu_1b=Viscosity(Water;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(Water;T=T_w;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(Water;T=T_w;P=P) "Thermal exp. coeff."k_b=Conductivity(Water;T=T_w;P=P) "Thermal conductivity"Nusselt_1b=h_1b*L_D/k_b "convective heat transfer coefficient"


1b = 0,0002962 D = 65 [m] dVa = 5,33 dVw = -5,33 francis = 0,2208 rt = 0,5843 Es = -73206 Es,pump = 125292 Es,turbine = 73206 g = 9,81 [m/s2] Grashof1a = 9,329E+08 Grashof1b = 1,340E+07 H = 300 [m] Headeff = 12,95 Headloss = 37,9 Headturbine = 50,85 h1a = 0,1012 h1b = 0,9448 hreservoir = 10 hsw0 = 113594 hw0 = 113543 hw,t = 113528 ka = 0,03678 kb = 0,606 Loadper = 5,628 LD = 16,25 M = 2,198E+08

1a = 0,00002331 1b = 0,0008477 f = 43,85 Ma = 2,114E+08 m = 6716 Mt,e = 2,923E+08 Mw = 8,409E+06 Mw,e = 8,094E+07 N = 874,6 Nusselt1a = 44,7 Nusselt1b = 25,34 1a = 1,090E-07

1b = 6,728E-07 Ns = 350 P = 1,914E+07 p1 = 0,0006981 p2 = 0,001906 p3 = 0,0138 p4 = -0,06827 p5 = 0,06002 p6 = -0,07365 p7 = 0,2022 p8 = 0,7629 Prandtl1a = 0,805 Prandtl1b = 5,783 Patm = 101300 [Pa] Pe = 2,155E+07 [Pa]Pturbine = 628480 Pturbine2 = 224906 Pturbine,loss = 160013 Pturbine,max = 2,843E+06 Q1 = 219,4 Qturbine = 5,33 [m3/s]Rayleigh1a = 7,510E+08 Rayleigh1b = 7,754E+07 a = 213,8

a,e = 227 w = 1260 R1a = 0,002979 R1b = 0,000319 f = 33,86 sw,t = 395,8 T = 0,7236 empty = 10800 [s] time = 10800 Ta = 299,5 [K] Te = 309,6 Tw = 300,2 Tw,e = 301 UA = 303,2 UA1 = 303,2 ua = 178403 ua0 = 183932 ucavern = 116,3 ue = 183932 uw = 111810 uw0 = 115065 uw,e = 115065 Va = 988818 Va,e = 931256 Vc,0 = 995492 Vw = 6674 Vw,0 = 6637 [m3]Vw,e = 64238 Wturbine = 0,09655 Wturbine,2 = 0,09655 z = -1,129 zsurface = 1500 [m] zw = 2,011 Integral Table

time Va Ta P Tw Mw Pturbine Headturbine

[K]

Row 1 0 931254 309,6 2,155E+07 301 8,094E+07 3,244E+06 262,4 Row 2 1000 936584 308,6 2,130E+07 301 7,422E+07 2,980E+06 241,1 Row 3 2000 941914 307,7 2,106E+07 301 6,751E+07 2,721E+06 220,1 Row 4 3000 947244 306,7 2,083E+07 300,9 6,079E+07 2,467E+06 199,6 Row 5 4000 952574 305,7 2,060E+07 300,9 5,408E+07 2,217E+06 179,4 Row 6 5000 957904 304,8 2,037E+07 300,8 4,736E+07 1,972E+06 159,5 Row 7 6000 963234 303,8 2,015E+07 300,7 4,065E+07 1,731E+06 140 Row 8 7000 968564 302,9 1,993E+07 300,7 3,393E+07 1,493E+06 120,8


Integral Tabletime Va Ta P Tw Mw Pturbine Headturbine

[K]

Row 9 8000 973894 302 1,972E+07 300,6 2,721E+07 1,260E+06 102 Row 10 9000 979224 301,1 1,951E+07 300,5 2,050E+07 1,031E+06 83,43 Row 11 10000 984554 300,2 1,931E+07 300,4 1,378E+07 806038 65,21 Row 12 10800 988818 299,5 1,914E+07 300,2 8,409E+06 628480 50,85

File:Compression_psycro2.EES 12-07-2011 11:49:39 Page 1EES Ver. 8.643: #0780: Department of Energy Engineering, Tech. Univ. of Denmark

$integraltable time:1000;P;T_a;T_w;T_pump;M_w;P_pump;cp {CompressionCase 1 - Large underground Cavern as compression chamber, using humid airAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Vessel conditions"z_surface=1500 [m] "bottom of cave is z=0"P_overtryk=5,5 [bar] "5 bar minimumpressure at surface"P_0=P_atm+z_surface*g*rho_w+P_overtryk*1e5 [Pa] "Pressure in cave"T_0=300 [K] "Start temperature, water and air"D=65 [m] "Diameter"H=300 [m] "Height"A=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A*H "Volume of cave"V_a_0=V_c_0-V_w_0 "Initial volume of air"V_w_0=2*A "Initial volume water" g=9,81 [m/s^2] "Gravity" P_atm=101300 [Pa] "Atmospheric pressure" "Psycrometric variables"x=HumRat(AirH2O;T=T_a;r=rh;P=P) "Humidity ratio"rh=1 "Relative humidity, 0...1" delta_x=x-x_0 "Change in hum. ratio"delta_M=M_a*delta_x "Mass transported" h_l0=Enthalpy(Water;x=0;T=T_w)h_l1=Enthalpy(Water;x=1;T=T_w)h_l=h_l1-h_l0 "Latent heat of evaporation"h_lm=integral(h_l;time;0;tau_full)/tau_full "Average value of above" E_v=h_lm*delta_M*1e-6 "Heat required to vaporize mass transport, in MJ"Drop_T=-E_v*1e6/(M_a*cp) "Cooresponding temperature drop if heat lost to surrounding"cp=Cp(AirH2O;T=T_a;r=rh;P=P)u_loss=u_a-E_v*1e6/M_a "loss in specifc internal energy" "Air mass"x_0=HumRat(AirH2O;T=T_0;r=rh;P=P_0) "Initial Humidity ratio"rho_a_0=Density(AirH2O;T=T_0;P=P_0;w=x_0) "Initial density"M_a=rho_a_0*V_a_0 "Initial mass of humid air" "Pump conditions"Q_pump=2,0 [m^3/s] "Flow rate of water"Q_pump=m_dot/rho_w "mass flow rate"tau_full=8*3600 "filling time"h_reservoir=10 "height of water resevoir to pump"P_pump2=P_atm+rho_w*g*h_reservoir "Pressure at pump inlet" "Initial water mass"rho_w=1260 "Constant brine density"


M_w0=rho_w*V_w_0 "Initial mass of water" "Initial total mass"M_t0=M_w0+M_a "dM/dt=m_dot"M-M_t0=integral(m_dot;time;0;tau_full) "Mass of water"M_w=M-M_a "Volume of water"V_w=M_w/rho_w "Volume of air"V_a=V_a_0-V_w+6636,6 "6636,6 is the initial water volume V_w_0. Adding the variable V_w_0 gives a too small change in variable for EES to calculate" "Height of water surface"z_w=4*V_w/(D^2*Pi) "change in volume dV/dt"dV_a=-m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air"M_a*(u_a-u_a0)=integral(Q_dot_1-P*dV_a;time;0;tau_full) u_a0=IntEnergy(AirH2O;T=T_0;w=x;P=P_0) "Initial int. energy" u_a=IntEnergy(AirH2O;T=T_a;w=x;P=P) "Internal energy air_" rho_a=M_a/V_a "Air density" rho_a=Density(AirH2O;T=T_a;r=rh;P=P) "Determines pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w0*u_w0=integral(-Q_dot_1+m_dot*(h_w+g*(z_surface-1/2*z_w))-P*dV_w;time;0;tau_full) {!h_w=Enthalpy(Water;T=T_0;P=P_pump)}T_w=temperature(Water;u=u_w;P=P) "Temperature of water"u_w0=IntEnergy(Water;T=T_0;P=P_0) "initial Internal energy water" P_pump=P+rho_w*g*(z_w-z_surface) "Pressure at pump outlet" "_________________________" "Work done on system by pump" "Pump work"eta_pump=1 "Efficiency of pump"eta_pump=(h_w0-h_sw)/(h_w0-h_w) "determining enthalpy after pump"h_w0=Enthalpy(Water;T=T_0;P=P_pump2) "enthalpy prior to pump"


s_w0=Entropy(Water;T=T_0;P=P_pump2) "Entropy prior to pump"h_sw=Enthalpy(Water;s=s_w0;P=P_pump) "Enthalpy if no entropy increase"T_pump=Temperature(Water;h=h_w;P=P_pump) "Temp. of water after pump "W_pump=m_dot*(h_w-h_w0)*1e-6 "in MW"E_s=integral(W_pump;time;0;tau_full) "Stored energy"E_s_t=max(E_s) "In MJ" "__________________________""Heat transfer" "air ->brine"Q_dot_1=UA*(T_w-T_a) "Heat transfer from air->brine" UA=UA_1UA_1=1/(R_1a+R_1b) "Definition of overall heat transfer coefficient"R_1a=1/(h_1a*A_water) "Resitance, air"R_1b=1/(h_1b*A_water) "Resistance, water"A=A_water "Heat transfer Area" "Natural convection, air"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_D^3/nu_1a "Grashof number"L_D=A_water/(D*Pi) "Equivalent length"mu_1a=Viscosity(AirH2O;T=T_a;w=x;P=P) "dynamic Viscosity"Prandtl_1a=Prandtl(Air;T=T_a) "Prandtl number"nu_1a=mu_1a/rho_a "Kinematic viscosity"beta_1a=1/T_a "Volumetric expansion coeff."k_a=Conductivity(AirH2O;T=T_a;w=x;P=P) "Conductivity"Nusselt_1a=h_1a*L_D/k_a "Convective heat transfer coefficient"T=abs(T_a-T_w) "Temperature difference between media" "Natural convection water"Nusselt_1b=0,27*(Rayleigh_1b)^(1/4) "Nusselt Number"Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayeigh number"Grashof_1b=g*beta_1b*T*L_D^3/nu_1b "Grashof number"mu_1b=Viscosity(Water;T=T_w;P=P) "dynamic Viscosity water"Prandtl_1b=Prandtl(Water;T=T_w;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic visc."beta_1b=VolExpCoef(Water;T=T_w;P=P) "Volumetric exp. coeff."k_b=Conductivity(Water;T=T_w;P=P) "Conductivity"Nusselt_1b=h_1b*L_D/k_b "Convective heat transfer coeff."


1b = 0,0002997 cp = 1007 D = 65 [m]M = 9926 x = 0,00004504 DropT = -0,109

dVa = -2 dVw = 2 pump = 1 Es = 100408 Es,t = 100408 Ev = 24182 g = 9,81 [m/s2] Grashof1a = 1,168E+10 Grashof1b = 1,284E+08 H = 300 [m] h1a = 0,132 h1b = 1,662 hl = 2,436E+06 hl0 = 114388 hl1 = 2,550E+06 hlm = 2,436E+06 hreservoir = 10 hsw = 115051


hw = 115051 hw0 = 112696 ka = 0,02619 kb = 0,6071 LD = 16,25 M = 3,013E+08

1a = 0,00001891 1b = 0,0008432 Ma = 2,204E+08 m = 2520 [kg/s]] Mt0 = 2,287E+08 Mw = 8,094E+07 Mw0 = 8,362E+06 Nusselt1a = 81,93 Nusselt1b = 44,48

1a = 7,992E-08 1b = 6,692E-07 P = 2,087E+07 Prandtl1a = 0,7258 Prandtl1b = 5,736 P0 = 1,919E+07 [Pa]Patm = 101300 [Pa] Povertryk = 5,5 [bar] Ppump = 2,573E+06 Ppump2 = 224906 Q1 = -2765 Qpump = 2 [m3/s]Rayleigh1a = 8,478E+09 Rayleigh1b = 7,366E+08 rh = 1

a = 236,6 a,0 = 222,8 w = 1260 R1a = 0,002283 R1b = 0,0001814 sw0 = 392,8 T = 6,813 full = 28800 [s] time = 28800 T0 = 300 [K] Ta = 307,266350554 Tpump = 300 Tw = 300,5 UA = 405,9 UA1 = 405,9 ua = -53468 ua0 = -58699 uloss = -53578 uw = 112659 uw0 = 110926 Va = 931256 Va,0 = 988856 [m3] Vc,0 = 995492 Vw = 64237 Vw,0 = 6637 [m3] Wpump = 5,934 x = 0,0001595 x0 = 0,0001145 zsurface = 1500 [m] zw = 19,36

00,00001 0,00004 0,00007 0,0001 0,00013 0,00016 0,00019 0,00022300

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00,000005 0,00004 0,00007 0,00010,000125 0,00016 0,00019 0,00022300

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00,000005 0,00004 0,00007 0,00010,000125 0,00016 0,00019 0,00022300

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00,000008 0,000044 0,00008 0,000112 0,000148 0,000184 0,00022300

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T a


time P Ta Tw Tpump Mw Ppump cp

Row 1 0 1,919E+07 300,000000000 300 300 8,362E+06 676025 1007 Row 2 1000 1,925E+07 300,242367735 300,1 300 1,088E+07 737959 1007 Row 3 2000 1,930E+07 300,485479525 300,2 300 1,340E+07 800180 1007 Row 4 3000 1,936E+07 300,729245896 300,2 300 1,592E+07 862657 1007 Row 5 4000 1,941E+07 300,973566954 300,2 300 1,844E+07 925348 1007 Row 6 5000 1,947E+07 301,218663340 300,3 300 2,096E+07 988342 1007 Row 7 6000 1,952E+07 301,464552243 300,3 300 2,348E+07 1,052E+06 1007 Row 8 7000 1,958E+07 301,711008338 300,3 300 2,600E+07 1,115E+06 1007 Row 9 8000 1,964E+07 301,958128748 300,3 300 2,852E+07 1,179E+06 1007 Row 10 9000 1,969E+07 302,206175544 300,3 300 3,104E+07 1,243E+06 1007 Row 11 10000 1,975E+07 302,454798395 300,3 300 3,356E+07 1,307E+06 1007 Row 12 11000 1,981E+07 302,703998169 300,3 300 3,608E+07 1,372E+06 1007 Row 13 12000 1,986E+07 302,954208302 300,4 300 3,860E+07 1,437E+06 1007 Row 14 13000 1,992E+07 303,205030376 300,4 300 4,112E+07 1,502E+06 1007 Row 15 14000 1,998E+07 303,456438474 300,4 300 4,364E+07 1,568E+06 1007 Row 16 15000 2,004E+07 303,708745118 300,4 300 4,616E+07 1,634E+06 1007 Row 17 16000 2,010E+07 303,961799604 300,4 300 4,868E+07 1,700E+06 1007 Row 18 17000 2,016E+07 304,215449426 300,4 300 5,120E+07 1,766E+06 1007 Row 19 18000 2,021E+07 304,469882650 300,4 300 5,372E+07 1,833E+06 1007 Row 20 19000 2,027E+07 304,725203465 300,4 300 5,624E+07 1,900E+06 1007 Row 21 20000 2,033E+07 304,981129137 300,4 300 5,876E+07 1,967E+06 1007 Row 22 21000 2,039E+07 305,237719634 300,4 300 6,128E+07 2,035E+06 1007 Row 23 22000 2,046E+07 305,495341431 300,4 300 6,380E+07 2,103E+06 1007 Row 24 23000 2,052E+07 305,753577822 300,4 300 6,632E+07 2,171E+06 1007 Row 25 24000 2,058E+07 306,012429761 300,4 300 6,884E+07 2,239E+06 1007


Integral Table

time P Ta Tw Tpump Mw Ppump cp

Row 26 25000 2,064E+07 306,272315099 300,4 300 7,136E+07 2,308E+06 1007 Row 27 26000 2,070E+07 306,532897835 300,4 300 7,388E+07 2,378E+06 1007 Row 28 27000 2,076E+07 306,794106105 300,4 300 7,640E+07 2,447E+06 1007 Row 29 28000 2,082E+07 307,056228240 300,4 300 7,892E+07 2,517E+06 1007 Row 30 28800 2,087E+07 307,266350554 300,5 300 8,094E+07 2,573E+06 1007

File:Expansion_psycro.EES 12-07-2011 11:56:12 Page 1EES Ver. 8.643: #0780: Department of Energy Engineering, Tech. Univ. of Denmark

$integraltable time:1000;V_a;T_a;P;T_w;M_w;P_turbine;W_turbine;Q_dot_1 {ExpansionCase 1 - Large underground Cavern as compression chamber, using humid airAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Vessel conditions" E_s_pump=100408 "Stored energy"rh=1 "relative humidity"T_e=307,26635-T_cavern "initial temp"T_cavern=0,12 "Temperature drop due to losses"cor=0 "Correction time for tau_empty"T_w_e=300,5 "Intial water temp" z_surface=1500 [m] "bottom of cave is z=0"D=65 [m] "Diameter"H=300 [m] "Height"A=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A*H "Volume of cave"V_w_0=2*A "Volume of water before compression"g=9,81 [m/s^2] "Gravity"M_w0=rho_w*V_w_0 "Initial water mass" P_atm=101300 [Pa] "Atmospheric pressure" "Turbine conditions"Q_turbine=5,33 "Flowrate of water"Q_turbine=m_dot/rho_w "Mass flow rate" tau_full=8*3600 "filling time"tau_empty=3*3600-cor "Expansion time" h_reservoir=10 "Height of water reservoir"P_turbine2=P_atm+rho_w*g*h_reservoir "Pressure at turbine outlet" "Psycrometric variables" {rh=RelHum(AirH2O;T=T_a;w=x;P=P)}x=HumRat(AirH2O;T=T_a;r=rh;P=P) "Humidity ratio" "Air mass"M_a=2,204E+08 "Air mass"V_a_e=931256 "initial air volume"rho_a_e=M_a/V_a_e "Intial air density" "Initial water mass"rho_w=1260 "Fluid density"M_w_e=8,094e7 "initial Mass of water"M_w_e=rho_w*V_w_e "Initial mass of water" "Initial pressure"rho_a_e=density(AirH2O;T=T_e;P=P_0;w=x)"Initial total mass"


M_t_e=M_w_e+M_a "dM/dt=m_dot"M-M_t_e=integral(-m_dot;time;0;tau_empty) "Mass of water"M_w=M-M_a "Volume of water"V_w=M_w/rho_w "Volume of air"V_a=V_c_0-V_w "Height of water surface"z_w=4*V_w/(D^2*Pi) "change in volume dV/dt"dV_a=m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air"M_a*(u_a-u_a0)=integral(Q_dot_1-P*dV_a;time;0;tau_empty) u_a0=IntEnergy(AirH2O;T=T_e;w=x;P=P_0) "initial int. energy" u_a=IntEnergy(AirH2O;T=T_a;P=P;w=x) "Internal energy" rho_a=M_a/V_a "density of water" rho_a=Density(AirH2O;T=T_a;r=rh;P=P) "Determines pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w_e*u_w0=integral(-Q_dot_1-m_dot*(h_w_t+g*(z_surface-1/2*z_w))-P*dV_w;time;0;tau_empty) h_w_t=Enthalpy(Water;T=T_w;P=P_turbine) "enthalpy of brine as it leaves system"T_w=Temperature(Water;u=u_w;P=P) "Water temperature"u_w0=IntEnergy(Water;T=T_w_e;P=P_0) "Intitial int. energy" P_turbine=P+rho_w*g*(z_w-z_surface) "pressure at turbine inlet""_________________________" "Work done on system by pump" "turbine work" s_w_t=Entropy(Water;h=h_w_t;P=P_turbine) "Entropy of water at turbine"h_sw0=Enthalpy(Water;s=s_w_t;P=P_turbine2) "Enthalpy of water after expansion if no entropy increase"eta_tur=1 "Efficiency of turbine"eta_tur=(h_w_t-h_w0)/(h_w_t-h_sw0) "Enthalpy of water after turbine" W_turbine=m_dot*(h_w0-h_w_t)*1e-6 "power extracted"


E_s=integral(W_turbine;time;0;tau_empty) "Extracted energy" abs(E_s)=E_s_turbine eta=E_s_turbine/E_s_pump "Thermodynamic Efficiency" "__________________________" Q_dot_1=UA_1*(T_w-T_a) "Heat transfer between media" UA_1=1/(R_1a+R_1b) "Definition of UA value"R_1a=1/(h_1a*A_water) "Resistance air"R_1b=1/(h_1b*A_water) "resistance water"A=A_water "Heat transfer surface""Natural convection air"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_D^3/nu_1a "Grashof number"L_D=A_water/(D*Pi) "equivalent length"mu_1a=Viscosity(Air_ha;T=T_a;P=P) "Dynamic viscosity"Prandtl_1a=Prandtl(Air_ha;T=T_a;P=P) "Prandtl number"nu_1a=mu_1a/rho_a "Kinematic viscosity"beta_1a=1/T_a "Volumetric exp. coefficient"k_a=Conductivity(Air_ha;T=T_a;P=P) "Thermal conductivity"Nusselt_1a=h_1a*L_D/k_a "Nusselt number"T=abs(T_a-T_w) "Temp. difference" "Natural convection water"Nusselt_1b=0,27*(Rayleigh_1b)^(1/4) "Nuselt number"Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayleigh number"Grashof_1b=g*beta_1b*T*L_D^3/nu_1b "Grashof number"mu_1b=Viscosity(Water;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(Water;T=T_w;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(Water;T=T_w;P=P) "Volumetric expansion"k_b=Conductivity(Water;T=T_w;P=P) "Thermal conductivity" Nusselt_1b=h_1b*L_D/k_b


1b = 0,0002903 cor = 0 D = 65 [m]dVa = 5,33 dVw = -5,33 = 0,997

tur = 1 Es = -100106 Es,pump = 100408 Es,turbine = 100106 g = 9,81 [m/s2] Grashof1a = 5,434E+08 Grashof1b = 7,243E+06 H = 300 [m] h1a = 0,08844 h1b = 0,812 hreservoir = 10 hsw0 = 110489 hw0 = 110489 hw,t = 110939 ka = 0,03681 kb = 0,6048 LD = 16,25 M = 2,288E+08

1a = 0,00002333 1b = 0,0008613 Ma = 2,204E+08 m = 6716 Mt,e = 3,013E+08 Mw = 8,409E+06 Mw0 = 8,362E+06 Mw,e = 8,094E+07 Nusselt1a = 39,04 Nusselt1b = 21,82 1a = 1,046E-07 1b = 6,836E-07


P = 1,919E+07 Prandtl1a = 0,8046 Prandtl1b = 5,886 P0 = 2,087E+07 Patm = 101300 [Pa] Pturbine = 673327 Pturbine2 = 224906 Q1 = -107,2 Qturbine = 5,33 [m3/s]Rayleigh1a = 4,372E+08 Rayleigh1b = 4,263E+07 rh = 1

a = 222,9 a,e = 236,7 w = 1260 R1a = 0,003407 R1b = 0,0003712 sw,t = 385,4 T = 0,4051 empty = 10800 [s] full = 28800 [s]time = 10800 Ta = 299,9 [K] Tcavern = 0,12 Te = 307,1 Tw = 299,5 Tw,e = 300,5 UA1 = 264,7 ua = -58892 ua0 = -53665 uw = 108784 uw0 = 112854 Va = 988818 Va,e = 931256 Vc,0 = 995492 Vw = 6674 Vw,0 = 6637 [m3] Vw,e = 64238 Wturbine = -3,021 x = 0,0001138 zsurface = 1500 [m] zw = 2,011 Integral Table

time Va Ta P Tw Mw Pturbine Wturbine Q1

[K]

Row 1 0 931254 307,1 2,087E+07 300,5 8,094E+07 2,569E+06 -15,79 -3647 Row 2 1000 936584 306,4 2,070E+07 300,5 7,422E+07 2,383E+06 -14,54 -3198 Row 3 2000 941914 305,8 2,054E+07 300,4 6,751E+07 2,199E+06 -13,3 -2768 Row 4 3000 947244 305,1 2,038E+07 300,3 6,079E+07 2,018E+06 -12,08 -2359 Row 5 4000 952574 304,4 2,022E+07 300,3 5,408E+07 1,838E+06 -10,87 -1971 Row 6 5000 957904 303,7 2,006E+07 300,2 4,736E+07 1,661E+06 -9,675 -1607 Row 7 6000 963234 303 1,991E+07 300,1 4,065E+07 1,486E+06 -8,496 -1267 Row 8 7000 968564 302,4 1,975E+07 300,1 3,393E+07 1,313E+06 -7,33 -953,4 Row 9 8000 973894 301,7 1,960E+07 300 2,721E+07 1,142E+06 -6,178 -670,4 Row 10 9000 979224 301,1 1,945E+07 299,8 2,050E+07 972787 -5,039 -423,2 Row 11 10000 984554 300,4 1,931E+07 299,7 1,378E+07 805678 -3,913 -222,8 Row 12 10800 988818 299,9 1,919E+07 299,5 8,409E+06 673327 -3,021 -107,2


$integraltable time:100;P;T_a;T_w;M_w {CompressionCase 2 - Array of pressure vesselsAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Vessel conditions"P_over=110 [bar]P_0=P_over*1e5 [Pa] "Pressure in vessel"T_0=300 [K] "Start temperature, water and air"D=0,2068 [m] "Diameter"H=1,257 [m] "Height"t_vessel=0,02162 [m] "Vessel thickness"t_iso=0,060[m] "Isolation thickness"sigma=P*d/t_vessel*1e-6 "radial stress"A=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A*H "Volume of cave"V_a_0=V_c_0-V_w_0 "Initial volume of air"V_w_0=0,01*A "Initial volume water" g=9,81 [m/s^2] "Gravity" P_atm=101300 [Pa] "Atmospheric pressure" "Pump conditions"Q_pump_array=0,000025 [m^3/s] "Flowrate of water"Q_pump_array=m_dot_array/rho_w "mass flow rate"m_dot_array=m_dot*7 "mass flow rate into one cylinder" tau_full=3600 [s] "filling time" "Air mass"rho_a_0=Density(Nitrogen;T=T_0;P=P_0) "Initial density nitrogen"M_a=rho_a_0*V_a_0 "Mass of nitrogen" "Initial water mass"rho_w=940 "Denisty of siloxane"M_w0=rho_w*V_w_0 "Initial mass of siloxane" "Initial total mass"M_t0=M_w0+M_a "Mass balance, dM/dt=m_dot"M-M_t0=integral(m_dot;time;0;tau_full) "Mass of water"M_w=M-M_a "Volume of liquid"V_w=M_w/rho_w "Volume of air"V_a=V_a_0-V_w


"dV/dt"dV_a=-m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air"M_a*(u_a-u_a0)=integral(Q_dot_1+Q_dot_2+Q_dot_3-P*dV_a;time;0;tau_full) u_a0=IntEnergy(Nitrogen;T=T_0;P=P_0) "Initial internal energy"u_a=IntEnergy(Nitrogen;T=T_a;P=P) "Internal energy, finds temperature of Nitrogen"rho_a=M_a/V_a "Density" P=Pressure(Nitrogen;T=T_a;v=1/rho_a) "Pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w0*u_w0=integral(-Q_dot_1+m_dot*h_w-P*dV_w;time;0;tau_full) T_w=temperature(siloxane_2;u=u_w;P=P) "Temperature siloxane"u_w0=IntEnergy(siloxane_2;T=T_0;P=P_0) "Initial internal energy" "_________________________" "Work done on system by pump" eta_pump=0,75 "pump efficiency, eta=1 for thermodynamic efficiency."eta_pump=(h_w0-h_sw)/(h_w0-h_w) "Finding enthalpy after pump"h_w0=Enthalpy(siloxane_2;T=T_0;P=P_atm) "Enthalpy prior to pump"s_w0=Entropy(siloxane_2;T=T_0;P=P_atm) "Entropy after pump"h_sw=Enthalpy(siloxane_2;s=s_w0;P=P) "Enthalpy after pump if no entropy increase"T_pump=Temperature(siloxane_2;h=h_w;P=P) "Temperature after pump"W_pump=m_dot_array*(h_w-h_w0)*1e-6 "Power in MW"E_s=integral(W_pump;time;0;tau_full) "Energy stored"E_s_t=max(E_s) "In MJ" "__________________________" A=A_water "Heat transfer area"D_t=D+t_vessel+t_iso "Outer diameter"

"Heat transfer"T_inf=299,9 "Temperature of surroundings" Q_dot_2=UA_2*(T_inf-T_a) "Heat flow through vessel sides" 1/UA_2=R_2a+R_2b+R_2c+R_2d "Ua value"R_2a=1/(h_1a*A_walls) "Resistance, convection inside"R_2b=ln((1/2*D+t_vessel)/(1/2*D))/(2*Pi*H_air*k_steel) "Cylindrical conduction steel"R_2c=ln((1/2*D+t_vessel+t_iso)/(1/2*D+t_vessel))/(2*Pi*H_air*k_iso) "Conduction through insulation"R_2d=1/(h_conv_air*A_outer_walls) "Resistance, convection outsie"H_air=V_a*4/(D^2*Pi) "Height of air"A_walls=H_air*Pi*D "Transferable area"A_outer_walls=H_air*P*(D+t_vessel+t_iso) "Outer area"k_steel=15 "Thermal conductivity steel"k_iso=0,03 "Thermal conductivity Polyurethene"


h_conv_air=0,5 "natural convection on outside" Q_dot_3=UA_3*(T_inf-T_a) "Heat transfer through top"1/UA_3=R_3a +R_3b+R_3c "UA value"R_3a=t_vessel/(k_steel*A_water) "Resistance, through steel"R_3b=t_iso/(k_iso*A_water) "Resistance through insulation"R_3c=1/(h_conv_air*A_water) "Convection outside"R_3d=1/(h_1a*A_water) "Convection inside" "air -> water"Q_dot_1=UA_1*(T_w-T_a) "Heat transfer to water"UA_1=1/(R_1a+R_1b) "UA value"R_1a=1/(h_1a*A_water) "REsistance air"R_1b=1/(h_1b*A_water) "Resistance water""Natural convection, hot air to cold membran"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_1^3/nu_1a "Grashof number"L_1=A_water/(D*Pi) "Equivalent lenght"mu_1a=Viscosity(N2;T=T_a) "Dynamic viscosity"Prandtl_1a=Prandtl(N2;T=T_a) "Prandtl number"nu_1a=mu_1a/rho_a "Kinematic viscosity"beta_1a=1/T_a "Thermal exp. coeff"k_a=Conductivity(Air;T=T_a) "Thermal conductivity"Nusselt_1a=h_1a*L_1/k_a "convective heat transfer coeff"T=abs(T_a-T_w) "Temperature difference" "Natural convection, hot membran to cold water"Nusselt_1b=0,27*abs(Rayleigh_1b)^(1/4) "Nusselt number water"

Rayleigh_1b=Grashof_1b*Prandtl_1b "raylegih number"Grashof_1b=g*beta_1b*T*L_1^3/nu_1b "Grashoff number"mu_1b=Viscosity(siloxane_2;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(siloxane_2;T=T_a;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(siloxane_2;T=T_w;P=P) "Volumetric exp. coeff"k_b=Conductivity(siloxane_2;T=T_w;P=P) "thermal conductivity"Nusselt_1b=h_1b*L_1/k_b "convective heat transfer coeff" "Energy lost to heat transfer"Q_1=Integral(Q_dot_1;time;0;tau_full) "Heat lost to siloxane"Q_2=Integral(Q_dot_2;time;0;tau_full) "Heat lost to surroundings"Q_3=Integral(Q_dot_3;time;0;tau_full) "Heat lost to surroundings"Q_t=(Q_1+Q_2+Q_3)*1e-6*(m_dot_array/m_dot) "Total heat lost"

SOLUTIONUnit Settings: SI K Pa J mass degA = 0,03359 Aouter,walls = 5,047E+06 Awalls = 0,555 Awater = 0,03359

1a = 0,002815 1b = 0,0009643 D = 0,2068 [m] dVa = -0,000003571 dVw = 0,000003571 Dt = 0,2884


pump = 0,75 Es = 1,774 Es,t = 1,774 g = 9,81 [m/s2]Grashof1a = 1607 Grashof1b = 32,34 H = 1,257 [m] h1a = 0,8996 h1b = 2,678 Hair = 0,8542 hconv,air = 0,5 hsw = 24271 hw = 31418 hw0 = 2829 ka = 0,02968 kb = 0,1091 kiso = 0,03 ksteel = 15 L1 = 0,0517 M = 17,53

1a = 0,00002017 1b = 0,001804 Ma = 5,131 m = 0,003357 [kg/s]]marray = 0,0235 Mt0 = 5,447 Mw = 12,4 Mw0 = 0,3157 Nusselt1a = 1,567 Nusselt1b = 1,269

1a = 1,127E-07 1b = 0,00000192 P = 2,049E+07 Prandtl1a = 0,7063 Prandtl1b = 15,07 P0 = 1,100E+07 [Pa]Patm = 101300 [Pa] Pover = 110 [bar]Q1 = -1394 Q2 = -21318 Q3 = -780,3 Q1 = -1,074 Q2 = -12,47 Q3 = -0,4649 Qpump,array = 0,000025 [m3/s] Qt = -0,1644 Rayleigh1a = 1135 Rayleigh1b = 487,4

a = 178,8 a,0 = 122,5 w = 940 R1a = 33,09

R1b = 11,12 R2a = 2,003 R2b = 0,002358 R2c = 2,434 R2d = 3,962E-07 R3a = 0,04291 R3b = 59,54 R3c = 59,54 R3d = 33,09 = 196 sw0 = -341,3 T = 47,49 [K]

full = 3600 [s] time = 3600 T0 = 300 [K] Ta = 355,3 Tinf = 299,9 tiso = 0,06 [m]Tpump = 309,3 tvessel = 0,02162 [m]Tw = 307,8 UA1 = 0,02262 UA2 = 0,2252 UA3 = 0,008394 ua = 233567 ua0 = 200543 uw = 7718 uw0 = -961,5 Va = 0,02869 Va,0 = 0,04188 [m3]Vc,0 = 0,04222 Vw = 0,01319 Vw,0 = 0,0003359 [m3] Wpump = 0,0006718 P[1]=Pressure(Air;T=T[1];u=u[1])Integral Table

time P Ta Tw Mw

Row 1 0 1,110E+07 300,2 300 0,3157 Row 2 100 1,126E+07 301,5 302,6 0,6514 Row 3 200 1,142E+07 302,8 303,4 0,9872 Row 4 300 1,158E+07 304 304 1,323 Row 5 400 1,176E+07 305,3 304,3 1,659 Row 6 500 1,193E+07 306,6 304,5 1,994 Row 7 600 1,211E+07 308 304,7 2,33 Row 8 700 1,229E+07 309,3 304,9 2,666 Row 9 800 1,248E+07 310,6 305 3,001


Integral Tabletime P Ta Tw Mw

Row 10 900 1,267E+07 312 305,1 3,337 Row 11 1000 1,286E+07 313,3 305,2 3,673 Row 12 1100 1,306E+07 314,7 305,3 4,009 Row 13 1200 1,327E+07 316,1 305,4 4,344 Row 14 1300 1,348E+07 317,5 305,5 4,68 Row 15 1400 1,370E+07 318,9 305,6 5,016 Row 16 1500 1,392E+07 320,4 305,7 5,351 Row 17 1600 1,415E+07 321,8 305,8 5,687 Row 18 1700 1,438E+07 323,3 305,9 6,023 Row 19 1800 1,462E+07 324,7 305,9 6,359 Row 20 1900 1,487E+07 326,2 306 6,694 Row 21 2000 1,513E+07 327,8 306,1 7,03 Row 22 2100 1,539E+07 329,3 306,2 7,366 Row 23 2200 1,566E+07 330,9 306,3 7,701 Row 24 2300 1,594E+07 332,4 306,4 8,037 Row 25 2400 1,623E+07 334 306,5 8,373 Row 26 2500 1,652E+07 335,7 306,6 8,709 Row 27 2600 1,683E+07 337,3 306,7 9,044 Row 28 2700 1,714E+07 339 306,8 9,38 Row 29 2800 1,746E+07 340,7 306,9 9,716 Row 30 2900 1,780E+07 342,4 307 10,05 Row 31 3000 1,814E+07 344,1 307,1 10,39 Row 32 3100 1,851E+07 345,9 307,2 10,72 Row 33 3200 1,887E+07 347,7 307,3 11,06 Row 34 3300 1,926E+07 349,6 307,4 11,39 Row 35 3400 1,965E+07 351,4 307,5 11,73 Row 36 3500 2,006E+07 353,3 307,7 12,07 Row 37 3600 2,049E+07 355,3 307,8 12,4


{StorageCase 2 - Array of pressure vesselsAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} T_0=355,3 "End of compression temp"T_w0=307,8 "end of compression water temp"t_iso=0,06 "Isolation thickness" tau_storage=1*3600 "Storage time" D=0,2068 [m] "Diameter"H=1,257 [m] "Height"A=(D^2/4*Pi) "Crossectional area"g=9,81 "Gravity"V_a=0,02869 "Volume of air"P_0=Pressure(Nitrogen;T=T_0;v=1/rho_a) "Initial pressure"rho_a=M_a/V_a "Density of air"M_a=5,131 "Mass of air" t_vessel=0,02162 [m] "Vessel thickness" M_w=12,4 "mass of water"

rho_w=940 "Density" "Storage"$integraltable time:1000;T_a;T_w;PM_a*(u_a-u_a0)=integral(Q_dot_1+Q_dot_2+Q_dot_3;time;0;tau_storage) u_a0=IntEnergy(Nitrogen;T=T_0;P=P_0) "initial internal energy" u_a=IntEnergy(Nitrogen;T=T_a;P=P) "Finds temperature" P=Pressure(Nitrogen;T=T_a;v=1/rho_a) "Pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*(u_w-u_w0)=integral(-Q_dot_1;time;0;tau_storage) T_w=temperature(Siloxane_2;U=u_w;P=P) "Temperature of water" u_w0=IntEnergy(Siloxane_2;T=T_w0;P=P_0) "Initial internal energy" "__________________________"A=A_water "Heat transferable area between media"T_inf=299,9 "Surroundings temperature" "Energy lost to heat transfer"Q_1=Integral(Q_dot_1;time;0;tau_storage)Q_2=Integral(Q_dot_2;time;0;tau_storage)Q_3=Integral(Q_dot_3;time;0;tau_storage)Q_t=(Q_1+Q_2+Q_3)*1e-6*7 "total heat lost"


"____________________________________" Q_dot_2=UA_2*(T_inf-T_a) 1/UA_2=R_2a+R_2b+R_2c+R_2d "Ua value"R_2a=1/(h_1a*A_walls) "Resistance, convection inside"R_2b=ln((1/2*D+t_vessel)/(1/2*D))/(2*Pi*H_air*k_steel) "Cylindrical conduction steel"R_2c=ln((1/2*D+t_vessel+t_iso)/(1/2*D+t_vessel))/(2*Pi*H_air*k_iso) "Conduction through insulation"R_2d=1/(h_conv_air*A_outer_walls) "Resistance, convection outsie"H_air=V_a*4/(D^2*Pi) "Height of air"A_walls=H_air*Pi*D "Transferable area"A_outer_walls=H_air*P*(D+t_vessel+t_iso) "Outer area"k_steel=15 "Thermal conductivity steel"k_iso=0,03 "Polyurethene"h_conv_air=0,5 "natural convection on outside" Q_dot_3=UA_3*(T_inf-T_a) "Heat transfer through top"1/UA_3=R_3a +R_3b+R_3c "UA value"R_3a=t_vessel/(k_steel*A_water) "Conduction through top"R_3b=t_iso/(k_iso*A_water) "Resistance through insulation"R_3c=1/(h_conv_air*A_water) "Convection outside"R_3d=1/(h_1a*A_water) "Convection inside" "air -> water"Q_dot_1=UA_1*(T_w-T_a) "Heat transfer to water"UA_1=1/(R_1a+R_1b) "UA value"R_1a=1/(h_1a*A_water) "REsistance air"R_1b=1/(h_1b*A_water) "Resistance water""Natural convection, hot air to cold membran"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_1^3/nu_1a "Grashof number"L_1=A_water/(D*Pi) "Equivalent lenght"mu_1a=Viscosity(N2;T=T_a) "Dynamic viscosity"Prandtl_1a=Prandtl(N2;T=T_a) "Prandtl number"nu_1a=mu_1a/rho_a "Kinematic viscosity"beta_1a=1/T_a "Thermal exp. coeff"k_a=Conductivity(Air;T=T_a) "Thermal conductivity"Nusselt_1a=h_1a*L_1/k_a "convective heat transfer coeff"T=abs(T_a-T_w) "Temperature difference" "Natural convection, hot membran to cold water"Nusselt_1b=0,27*abs(Rayleigh_1b)^(1/4) "Nusselt number water"Rayleigh_1b=Grashof_1b*Prandtl_1b "raylegih number"Grashof_1b=g*beta_1b*T*L_1^3/nu_1b "Grashoff number"mu_1b=Viscosity(siloxane_2;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(siloxane_2;T=T_a;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(siloxane_2;T=T_w;P=P) "Volumetric exp. coeff"k_b=Conductivity(siloxane_2;T=T_w;P=P) "thermal conductivity"Nusselt_1b=h_1b*L_1/k_b "convective heat transfer coeff"

SOLUTIONUnit Settings: SI K Pa J mass deg


A = 0,03359 Aouter,walls = 4,848E+06 Awalls = 0,5549 Awater = 0,03359 1a = 0,002906 1b = 0,0009703 D = 0,2068 [m] g = 9,81 Grashof1a = 1300 Grashof1b = 24,9 H = 1,257 [m] h1a = 0,8303 h1b = 2,584 Hair = 0,8542 hconv,air = 0,5 ka = 0,02888 kb = 0,1091 kiso = 0,03 ksteel = 15 L1 = 0,0517 1a = 0,00001968

1b = 0,001803 Ma = 5,131 Mw = 12,4 Nusselt1a = 1,487 Nusselt1b = 1,224 1a = 1,100E-07

1b = 0,000001919 P = 1,968E+07 Prandtl1a = 0,7069 Prandtl1b = 16,97 P0 = 2,049E+07 Q1 = -3282 Q2 = -39486 Q3 = -1498 Q1 = -0,7664 Q2 = -9,601 Q3 = -0,3713 Qt = -0,3099 Rayleigh1a = 918,8 Rayleigh1b = 422,4 a = 178,8

w = 940 R1a = 35,86 R1b = 11,52 R2a = 2,17 R2b = 0,002359 R2c = 2,435 R2d = 4,125E-07 R3a = 0,04291 R3b = 59,54 R3c = 59,54 R3d = 35,86 T = 36,31

storage = 3600 time = 3600 T0 = 355,3 Ta = 344,1 Tinf = 299,9 tiso = 0,06 [m]tvessel = 0,02162 [m] Tw = 307,8 Tw0 = 307,8 UA1 = 0,02111 UA2 = 0,217 UA3 = 0,008394 ua = 224954 ua0 = 233582 uw = 8000 uw0 = 7735 Va = 0,02869 Integral Table

time Ta Tw P

Row 1 0 355,3 307,8 2,049E+07 Row 2 1000 351,9 307,8 2,024E+07 Row 3 2000 348,7 307,8 2,001E+07 Row 4 3000 345,8 307,8 1,980E+07 Row 5 3600 344,1 307,8 1,968E+07


$integraltable time:100;V_a;V_frac;T_a;P;T_w;M_w {ExpansionCase 2 - Array of pressure vesselsAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Vessel conditions"T_e=344,9 "initial nitrogen temperature"T_w_e=318,2 "Intitial siloxane temperature"E_s_pump=1,775 "Energy consumed by pump during compression"tau_empty=1820 [s] "filling time" V_frac=V_a/V_c_0 "Fraction of water/total volume"D=0,2068 [m] "Diameter"H=1,257 [m] "Height"t_vessel=0,02162 [m] "Vessel thickness"t_iso=0,06[m] "Isolation thickness"sigma=P*d/t_vessel*1e-6 "radial stress"A=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A*H "Volume of cave"V_w_0=0,01*A "Volume of water before compression"g=9,81 [m/s^2] "Gravity" P_atm=101300 [Pa] "Atmospheric pressure" "Turbine conditions"Q_turbine_array=0,000045 [m^3/s] "Flowrate of siloxane"Q_turbine_array=m_dot_array/rho_w "Mass flow rate"M_dot_array=m_dot*7 "Mass flow rate in one cylinder" "Nitrogen mass"M_a=5,131V_a_e=0,02869 "Nitrogen volume"rho_a_e=M_a/V_a_e "Nitrogen density" "Initial pressure"P_e=Pressure(Air_ha;T=T_e;v=1/rho_a_e) "Initial pressure" "Initial siloxane mass"rho_w=940 "Denisty of siloxane"M_w_e=12,4 "Siloxane mass"M_w_e=rho_w*V_w_e "Volume" "Initial total mass"M_t_e=M_w_e+M_a "mass balance dM/dt=m_dot"M-M_t_e=integral(-m_dot;time;0;tau_empty) "Mass of water"


M_w=M-M_a "Volume of water"V_w=M_w/rho_w "Volume of air"V_a=V_c_0-V_w-V_w_0 "dV/dt"dV_a=m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air" M_a*(u_a-u_a0)=integral(Q_dot_1+Q_dot_2-P*dV_a;time;0;tau_empty) u_a0=IntEnergy(Nitrogen;T=T_e;P=P_e) "Initial internal energy" u_a=IntEnergy(Nitrogen;T=T_a;P=P) "Internal energy to find temperature" rho_a=M_a/V_a "Nitrogen density" P=Pressure(Nitrogen;T=T_a;v=1/rho_a) "Pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w_e*u_w0=integral(-Q_dot_1+Q_dot_3-m_dot*h_w_t-P*dV_w;time;0;tau_empty) h_w_t=Enthalpy(Siloxane_2;T=T_w;P=P) "Enthalpy siloxane at turbine outlet"T_w=Temperature(Siloxane_2;u=u_w;P=P) "Siloxane temperature"u_w0=IntEnergy(Siloxane_2;T=T_w_e;P=P_e) "internal energy siloxane" "_________________________" "Work done on system by pump" "turbine work" s_w_t=Entropy(Siloxane_2;h=h_w_t;P=P) "Entropy prior to turbine"h_sw0=Enthalpy(Siloxane_2;s=s_w_t;P=P_atm) "Enthalpy priop to turbine"eta_tur=1 "Efficiency of turbine"eta_tur=(h_w_t-h_w0)/(h_w_t-h_sw0) "finds enthalpy after turbine" W_turbine=m_dot_array*(h_w0-h_w_t)*1e-6 "Turbine power" E_s=integral(W_turbine;time;0;tau_empty) "Energy extracted" abs(E_s)=E_s_turbine eta=E_s_turbine/E_s_pump "Efficiency" "__________________________"A=A_water "Heat transfer area across nitrogen-siloxane interface"


T_inf=299,9 "Temperature of surroundings" Q_dot_2=UA_2*(T_inf-T_a) "Heat transfer through sides" 1/UA_2=R_2a+R_2b+R_2c+R_2d "UA value"R_2a=1/(h_1a*A_walls) "Inside conv. resistance"R_2b=ln((1/2*D+t_vessel)/(1/2*D))/(2*Pi*H_air*k_steel) "Cylindrical conduction steel"R_2c=ln((1/2*D+t_vessel+t_iso)/(1/2*D+t_vessel))/(2*Pi*H_air*k_iso) "Conduction through insulation"R_2d=1/(h_conv_air*A_outer_walls) "Resistance outside convection"H_air=V_a*4/(D^2*Pi) "Height of air"A_walls=H_air*Pi*D "INside area of cylinder"A_outer_walls=H_air*P*(D+t_vessel+t_iso) "Outside are cylinder"k_steel=15 "Thermal conductivity steel"k_iso=0,03 "Polyurethene"h_conv_air=0,5 "Convective heat transfer coefficient" Q_dot_3=UA_3*(T_inf-T_a) "Heat transfer through top"1/UA_3=R_3a +R_3b+R_3c "Ua value"R_3a=t_vessel/(k_steel*A_water) "Conduction through top"R_3b=t_iso/(k_iso*A_water) "Resistance, insulation"R_3c=1/(h_conv_air*A_water) "Resistance convection"R_3d=1/(h_1a*A_water) "Resistance convection, inside" "air -> water"Q_dot_1=UA_1*(T_w-T_a) "Heat transfer across fluids"UA_1=1/(R_1a+R_1b) "UA value"R_1a=1/(h_1a*A_water) "Resistance Nitrogen"R_1b=1/(h_1b*A_water) "Resitance siloxane""Natural convection, hot air to cold membran"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_1^3/nu_1a "Grashoff number"L_1=A_water/(D*Pi) "Equivalent lenght"mu_1a=Viscosity(N2;T=T_a) "Dynamic viscosity"Prandtl_1a=Prandtl(N2;T=T_a) "Prandtl number nitorgen"nu_1a=mu_1a/rho_a "Kinematic viscosity"beta_1a=1/T_a "Vol. exp coef."k_a=Conductivity(N2;T=T_a) "Thermal conductivity"Nusselt_1a=h_1a*L_1/k_a "Convective heat trnasfer coeff. N2"T=abs(T_a-T_w) "Temperature difference" "Natural convection, hot membran to cold water"Nusselt_1b=0,27*abs(Rayleigh_1b)^(1/4) "Nusselt number"Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayleigh number"Grashof_1b=g*beta_1b*T*L_1^3/nu_1b "Grashof number"mu_1b=Viscosity(siloxane_2;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(siloxane_2;T=T_a;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(siloxane_2;T=T_w;P=P) "Vol. exp. coefficient"k_b=Conductivity(siloxane_2;T=T_w;P=P) "Thermal conductivity siloxane"Nusselt_1b=h_1b*L_1/k_b "Convective heat transfer coefficient"

SOLUTIONUnit Settings: SI K Pa J mass deg


A = 0,03359 Aouter,walls = 3,801E+06 Awalls = 0,7813 Awater = 0,03359

1a = 0,003445 1b = 0,00105 D = 0,2068 [m] dVa = 0,000006429 dVw = -0,000006429 = 0,6746

tur = 1 Es = -1,197 Es,pump = 1,775 Es,turbine = 1,197 g = 9,81 [m/s2] Grashof1a = 903,2 Grashof1b = 21,88 H = 1,257 [m]h1a = 0,6601 h1b = 2,824 Hair = 1,203 hconv,air = 0,5 hsw0 = 23653 hw0 = 23653 hw,t = 35315 ka = 0,0251 kb = 0,1072 kiso = 0,03 ksteel = 15 L1 = 0,0517 M = 6,533 1a = 0,00001722

1b = 0,001603 Ma = 5,131 m = 0,006043 marray = 0,0423 Mt,e = 17,53 Mw = 1,402 Mw,e = 12,4 Nusselt1a = 1,36 Nusselt1b = 1,361 1a = 1,356E-07

1b = 0,000001705 P = 1,096E+07 Prandtl1a = 0,7119 Prandtl1b = 29,52 Patm = 101300 [Pa] Pe = 1,891E+07 [Pa]Q1 = 0,4712 Q2 = 2,625 Q3 = 0,08087 Qturbine,array = 0,000045 [m3/s]Rayleigh1a = 643 Rayleigh1b = 646

a = 127 a,e = 178,8 w = 940 R1a = 45,1

R1b = 10,54 R2a = 1,939 R2b = 0,001675 R2c = 1,729 R2d = 5,261E-07 R3a = 0,04291 R3b = 59,54 R3c = 59,54 R3d = 45,1 = 104,8 sw,t = -273,4 T = 26,22

empty = 1820 [s] time = 1820 Ta = 290,3 [K] Te = 344,9 Tinf = 299,9 tiso = 0,06 [m]tvessel = 0,02162 [m] Tw = 316,5 Tw,e = 318,2 UA1 = 0,01797 UA2 = 0,2725 UA3 = 0,008394 ua = 192280 ua0 = 226633 uw = 23631 uw0 = 23679 Va = 0,04039 Va,e = 0,02869 Vc,0 = 0,04222 Vfrac = 0,9567 Vw = 0,001491 Vw,0 = 0,0003359 [m3]Vw,e = 0,01319 Wturbine = -0,0004933 Integral Table

time Va Vfrac Ta P Tw Mw

[K]

Row 1 0 0,02869 0,6796 346,3 1,983E+07 318,4 12,4 Row 2 100 0,02934 0,6948 342,1 1,905E+07 318,2 11,8 Row 3 200 0,02998 0,7101 338 1,831E+07 318,1 11,19 Row 4 300 0,03062 0,7253 334,1 1,762E+07 318 10,59 Row 5 400 0,03126 0,7405 330,4 1,697E+07 317,8 9,983


Integral Tabletime Va Vfrac Ta P Tw Mw

[K]

Row 6 500 0,03191 0,7557 326,8 1,637E+07 317,7 9,379 Row 7 600 0,03255 0,771 323,4 1,580E+07 317,6 8,774 Row 8 700 0,03319 0,7862 320,1 1,526E+07 317,5 8,17 Row 9 800 0,03384 0,8014 316,9 1,476E+07 317,4 7,566 Row 10 900 0,03448 0,8166 313,9 1,429E+07 317,3 6,961 Row 11 1000 0,03512 0,8319 310,9 1,385E+07 317,2 6,357 Row 12 1100 0,03576 0,8471 308,1 1,342E+07 317,1 5,753 Row 13 1200 0,03641 0,8623 305,3 1,302E+07 317 5,149 Row 14 1300 0,03705 0,8775 302,7 1,264E+07 316,9 4,544 Row 15 1400 0,03769 0,8928 300,1 1,228E+07 316,8 3,94 Row 16 1500 0,03834 0,908 297,6 1,194E+07 316,7 3,336 Row 17 1600 0,03898 0,9232 295,2 1,161E+07 316,7 2,731 Row 18 1700 0,03962 0,9384 292,9 1,131E+07 316,6 2,127 Row 19 1800 0,04026 0,9537 290,7 1,101E+07 316,5 1,523 Row 20 1820 0,04039 0,9567 290,3 1,096E+07 316,5 1,402

File:Compression_Case_3.EES 12-07-2011 11:44:10 Page 1EES Ver. 8.643: #0780: Department of Energy Engineering, Tech. Univ. of Denmark

$integraltable time:1000;V_a;V_frac;T_a;P;T_w;T_pump; {CompressionCase 3 - Large underground Cavern as pressure vesselAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Vessel conditions"P_over=213,3 [bar] "Initial pressure over atmospheric."P_0=P_atm+P_over*1e5 "Initial pressure of air"T_0=300 [K] "Start temperature air"T_w_0=300 "Initial temperature of water"D=65 [m] "Diameter"H=300 [m] "Height"A_c=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A_c*H "Volume of cave" D_p=0,4 "Diameter of pipe"H_p=1200 "Height of pipe"V_p_0=D_p^2/4*Pi*H_p "Volume of pipe" D_t=10 "Tank diameter"H_t=50 "Tank height"A=D_t^2*Pi/4 "Crossectional area of tank"V_t_0=A*H_t "Tank volume" V_w_0=1*A "Initial water volume" V_total=V_c_0+V_p_0+V_t_0 "Total volume"V_a_0=V_c_0+V_p_0+(V_t_0-V_w_0) "initial air volume" g=9,81 [m/s^2] "Gravity" P_atm=101300 [Pa] "Atmospheric pressure" "Pump conditions"Q_pump=(V_t_0-V_w_0)/tau_full "Flowrate of water"Q_pump=m_dot/rho_w "mass flow rate"tau_full=8*3600 [s] "filling time"h_reservoir=10 "height of water resevoir to pump"rho_w_0=995 "Initial water density"P_pump2=P_atm+rho_w_0*g*h_reservoir "Pressure at pump inlet" "Air mass"rho_a_0=Density(Air_ha;T=T_0;P=P_0) "Initial air density"M_a=rho_a_0*V_a_0 "Mass of air" "Initial water mass"rho_w=1005 "Density of water in compression chamber"M_w0=rho_w*V_w_0 "Initial water mass" "Initial total mass"M_t0=M_w0+M_a


"dM/dt=m_dot"M-M_t0=integral(m_dot;time;0;tau_full) "Mass balance" "Mass of water"M_w=M-M_a "Volume of water"V_w=M_w/rho_w "Volume of air"V_a=V_total-V_w "Volume of air in tank"V_a_t=V_t_0-V_w "Volume fraction of air in tank"V_frac=V_a_t/V_t_0 "Height of water surface in tank"z_w=4*V_w/(D^2*Pi) "change in volume dV/dt"dV_a=-m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air"M_a*(u_a-u_a0)=integral(Q_dot_1-P*dV_a;time;0;tau_full) u_a0=IntEnergy(Air_ha;T=T_0;P=P_0) "Initial internal energy" u_a=IntEnergy(Air_ha;T=T_a;P=P) "internal energy to find temperature" rho_a=M_a/V_a "Density of air" P=Pressure(Air_ha;T=T_a;v=1/rho_a) "Pressure" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w0*u_w0=integral(-Q_dot_1+m_dot*(h_w+g*(H_t-1/2*z_w))-P*dV_w;time;0;tau_full) T_w=temperature(Water;u=u_w;P=P) "Temperature of water, total mass"u_w0=IntEnergy(Water;T=T_w_0;P=P_0) "initial internal energy water" P_pump=P+rho_w*g*(z_w-H_t) "Pressure at pump outlet"Head_pump=P_pump/(rho_w*g) "Head at pump outlet" "_________________________" "Work done on system by pump"{"for Thermodynamic efficiency"h_sw=Enthalpy(Water;s=s_w0;P=P_pump) "Enthalpy of water"T_pump=Temperature(Water;h=h_w;P=P_pump) "temperature of water at pump outlet"eta_pump=1 "Isentropic efficiency of turbine"


"____________________"} "RTE"eta_pump=0,90 "Efficiency of centrifugal pump"Head_loss=0,6 "Head loss during compression"h_sw=Enthalpy(Water;s=s_w0;P=P_pump_max) "Enthalpy after pump, if no entropy increase"P_pump_max=(2142+Head_loss)*rho_w*g "2142 is the maximum head"Head_max=P_pump_max/(rho_w*g) "maximum head"T_pump=Temperature(Water;h=h_w;P=P_pump_max) "Temperature of water after pump""______________________" "Pump work"eta_pump=(h_w0-h_sw)/(h_w0-h_w) "Definition of pump efficiency, finds enthalpy after pump"h_w0=Enthalpy(Water;T=T_w_0;P=P_pump2) "Enthalpy prior to pump"s_w0=Entropy(Water;T=T_w_0;P=P_pump2) "Entropy prior to pump"W_pump=m_dot*(h_w-h_w0)*1e-6 "in MW"E_s=integral(W_pump;time;0;tau_full) "Energy stored in MJ"E_s_t=max(E_s) "In MJ"E_s_t_MWh=E_s_t*1/3600 "In MWh" "__________________________" "Heat transfer" "air -> water"Q_dot_1=1*UA_1*(T_w-T_a) "Heat transfer between the air/water interface" UA_1=1/(R_1a+R_1b) "Definition of UA value"R_1a=1/(h_1a*A_water) "Definition of resistance, air"R_1b=1/(h_1b*A_water) "Definition of resistance, water"A=A_water "Heat transferable area" "Natural convection, hot air to cold membran"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_D^3/nu_1a "Grashof number"L_D=A_water/(D_t*Pi) "Equivalent lenght"mu_1a=Viscosity(Air_ha;T=T_a;P=P) "dynamic viscosity of air"Prandtl_1a=Prandtl(Air_ha;T=T_a;P=P) "Prandtl number"nu_1a=mu_1a/rho_a "Kinematic viscosity"beta_1a=1/T_a "volumetric expansion coefficient approximation"k_a=Conductivity(Air_ha;T=T_a;P=P) "Thermal conductivity air"Nusselt_1a=h_1a*L_D/k_a "Convective heat transfer coefficient"T=abs(T_a-T_w) "Temperature difference" "Natural convection, hot membran to cold water"Nusselt_1b=0,27*(Rayleigh_1b)^(1/4) "Nusselt number water"Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayleigh number"Grashof_1b=g*beta_1b*T*L_D^3/nu_1b "Grashof number"mu_1b=Viscosity(Water;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(Water;T=T_w;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(Water;T=T_w;P=P) "Volumetric expansion"k_b=Conductivity(Water;T=T_w;P=P) "Thermal conductivity"


Nusselt_1b=h_1b*L_D/k_b "nusselt number"

SOLUTIONUnit Settings: SI K Pa J mass degA = 78,54 Ac = 3318 Awater = 78,54

1a = 0,003326 1b = 0,0003047 D = 65 [m]dVa = -0,1336 dVw = 0,1336 Dp = 0,4 Dt = 10 pump = 0,9 Es = 89820 Es,t = 89820 Es,t,MWh = 24,95 g = 9,81 [m/s2]Grashof1a = 1,746E+06 Grashof1b = 19727 H = 300 [m]Headloss = 0,6 Headmax = 2143 Headpump = 2142 h1a = 0,1428 h1b = 1,201 Hp = 1200 hreservoir = 10 hsw = 133573 Ht = 50 hw = 135895 hw0 = 112672 ka = 0,03837 kb = 0,6083 LD = 2,5 M = 2,386E+08

1a = 0,0000241 1b = 0,0008332 Ma = 2,347E+08 m = 134,3 [kg/s]] Mt0 = 2,348E+08 Mw = 3,947E+06 Mw0 = 78933 Nusselt1a = 9,301 Nusselt1b = 4,934

1a = 1,022E-07 1b = 8,291E-07 P = 21600226 Prandtl1a = 0,8064 Prandtl1b = 5,655 P0 = 2,143E+07 [Pa]Patm = 101300 [Pa] Pover = 213,3 [bar] Ppump = 2,112E+07 Ppump2 = 198910 Ppump,max = 2,112E+07 Q1 = 3,51 Qpump = 0,1336 [m3/s] Rayleigh1a = 1,408E+06 Rayleigh1b = 111555

a = 235,7 a,0 = 234,8 w = 1005 w,0 = 995 R1a = 0,08919 R1b = 0,01061

sw0 = 392,8 T = 0,3502 full = 28800 [s]time = 28800 T0 = 300 [K] Ta = 300,7 Tpump = 301,0 Tw = 301,0 Tw,0 = 300 UA1 = 10,02 ua = 175746 ua0 = 175393 uw = 114890 uw0 = 110747 Va = 995643 Va,0 = 999491 [m3] Va,t = -5,995E-15 Vc,0 = 995492 Vfrac = -1,526E-18 Vp,0 = 150,8 Vtotal = 999570 Vt,0 = 3927 Vw = 3927 Vw,0 = 78,54 [m3]Wpump = 3,119 zw = 1,183 P[1]=Pressure(Air;T=T[1];u=u[1])Integral Table

time Va Vfrac Ta P Tw Tpump

Row 1 0 999491 0,98 300 21431300 300,0 301,0 Row 2 1000 999358 0,946 300 21437127 300,7 301,0 Row 3 2000 999224 0,9119 300 21442957 300,8 301,0 Row 4 3000 999091 0,8779 300,1 21448790 300,9 301,0 Row 5 4000 998957 0,8439 300,1 21454625 300,9 301,0 Row 6 5000 998823 0,8099 300,1 21460463 300,9 301,0 Row 7 6000 998690 0,7758 300,1 21466305 300,9 301,0 Row 8 7000 998556 0,7418 300,2 21472148 301,0 301,0 Row 9 8000 998422 0,7078 300,2 21477994 301,0 301,0 Row 10 9000 998289 0,6737 300,2 21483844 301,0 301,0 Row 11 10000 998155 0,6397 300,2 21489695 301,0 301,0 Row 12 11000 998022 0,6057 300,3 21495550 301,0 301,0 Row 13 12000 997888 0,5717 300,3 21501407 301,0 301,0 Row 14 13000 997754 0,5376 300,3 21507268 301,0 301,0


Integral Tabletime Va Vfrac Ta P Tw Tpump

Row 15 14000 997621 0,5036 300,3 21513130 301,0 301,0 Row 16 15000 997487 0,4696 300,3 21518996 301,0 301,0 Row 17 16000 997353 0,4356 300,4 21524865 301,0 301,0 Row 18 17000 997220 0,4015 300,4 21530736 301,0 301,0 Row 19 18000 997086 0,3675 300,4 21536610 301,0 301,0 Row 20 19000 996953 0,3335 300,4 21542487 301,0 301,0 Row 21 20000 996819 0,2994 300,5 21548366 301,0 301,0 Row 22 21000 996685 0,2654 300,5 21554248 301,0 301,0 Row 23 22000 996552 0,2314 300,5 21560134 301,0 301,0 Row 24 23000 996418 0,1974 300,5 21566022 301,0 301,0 Row 25 24000 996284 0,1633 300,5 21571912 301,0 301,0 Row 26 25000 996151 0,1293 300,6 21577806 301,0 301,0 Row 27 26000 996017 0,09528 300,6 21583702 301,0 301,0 Row 28 27000 995883 0,06125 300,6 21589601 301,0 301,0 Row 29 28000 995750 0,02722 300,6 21595503 301,0 301,0 Row 30 28800 995643 -1,526E-18 300,7 21600226 301,0 301,0

File:Expansion_Case_3.EES 12-07-2011 11:51:12 Page 1EES Ver. 8.643: #0780: Department of Energy Engineering, Tech. Univ. of Denmark

$integraltable time:1000;V_w;P;V_a;T_a;T_w;M_w;Head_turbine;P_turbine_loss;load_per;eta_pelton {ExpansionCase 3 - Large underground Cavern as pressure vesselAuthor: Kristian Kahlestudent number: s073003Technical University of [email protected]} "Initial state of fluids from compression model"u_e=175746-u_cavern "Internal energy of air"u_w_e=114890 "initial int. energy of water"u_cavern=13,35 "Lost to surroundings" "Initial pressure"P_e=Pressure(Air_ha;u=u_e;v=1/rho_a_e) "Initial pressure""Initial temperatures"T_e=Temperature(Air_ha;u=u_e;v=1/rho_a_e) "Initial pressure"T_w_e=Temperature(Water;u=u_w_e;P=P) "Initial temperature" "Air mass"M_a=2,34676488583e8 "Mass of air"V_a_e=995643 "Air volume"rho_a_e=M_a/V_a_e "Air density" "Vessel conditions"D=65 [m] "Diameter"H=300 [m] "Height"A_c=(D^2/4*Pi) "Crossectional area of cave"V_c_0=A_c*H "Volume of cave" D_p=0,4 "Diameter of pipe"H_p=1200 "Height of pipe"V_p_0=D_p^2/4*Pi*H_p "Volume of pipe" D_t=10 "Tank diameter"H_t=50 "Tank height"A=D_t^2*Pi/4 "Tank crosssectional area"V_t_0=A*H_t "Tank volume" V_w_0=1*A "Volume of water before compression"V_total=V_c_0+V_p_0+V_t_0 "Total volume"V_a_0=V_c_0+V_p_0+(V_t_0-V_w_0) "Total initial air volume" g=9,81 [m/s^2] "Gravity" P_atm=101300 "Atmospheric pressure" h_reservoir=10 "Height of reservoir" "Turbine conditions"Q_turbine=0,36 "Flowrate of water"Q_turbine=m_dot/rho_w "Mass flow rate" tau_empty=3*3600 [s] "Expansion time. hours - minor correction to return to air to initial pressure"


"Initial water mass"rho_w=1005 "Water density in compression chamber"M_w_e=3,947e6 "Initial water mass"M_w_e=rho_w*V_w_e "Initial water volume" "Initial total mass"M_t_e=M_w_e+M_a "dM/dt=m_dot"M-M_t_e=integral(-m_dot;time;0;tau_empty) "Mass balance" "Mass of water"M_w=M-M_a "Volume of water"V_w=M_w/rho_w "Volume of air"V_a=V_total-V_w "Volume fraction of water in tank"V_frac=V_w/V_t_0 "Height of water surface"z_w=4*V_w/(D^2*Pi) "Change in volume dV/dt"dV_a=m_dot/rho_wdV_w=-dV_a "-----------------------------------------" "Internal energy balance air"M_a*(u_a-u_a0)=integral(Q_dot_1-P*dV_a;time;0;tau_empty) u_a0=IntEnergy(Air_ha;T=T_e;P=P_e) "Initial internal energy" u_a=IntEnergy(Air_ha;T=T_a;P=P) "Internal energy to determine temperature of air" rho_a=M_a/V_a "Density of air" P=Pressure(Air_ha;T=T_a;v=1/rho_a) "Pressure of air" "Internal energy balance water. Q_dot_1 opposite sign"M_w*u_w-M_w_e*u_w0=integral(-Q_dot_1-m_dot*(h_w_t+g*((H_t+h_reservoir)-1/2*z_w))-P*dV_w;time;0;tau_empty) T_w=Temperature(Water;u=u_w;P=P) "Temperature of water"u_w0=IntEnergy(Water;T=T_w_e;P=P_e) "initial internal energy of water" P_turbine=P+rho_w*g*(z_w-H_t-h_reservoir) "Pressure at turbine inlet"Head_turbine=P_turbine/(rho_w*g) "Head at turbine inlet""_________________________""Work done by system on turbine"


"For round trip efficiency""turbine work"Head_loss=5 "Head loss"P_turbine_loss=P_turbine-Head_loss*rho_w*g "Pressure at turbine, after loss"s_w_t=Entropy(Water;h=h_w_t;P=P_turbine_loss) "Entropy"h_w_t=Enthalpy(Water;T=T_w;P=P_turbine_loss) "Enthalpy in to system"P_turbine_max=3,625e7 "Maximum turbine pressure, is found by the optimization function. The operating load is around 60%"Load_per=P_turbine_loss/P_turbine_max*100 "Load percentage" "Variable efficiency found using matlab. Only valid in the load percentage range of 30-80%"eta_pelton = p1*Load_per^2 + p2*Load_per +p3 "Coefficients:"p1 = -3,8393e-005p2 = 0,0044232p3 = 0,79739"__________________________" {"For Thermodynamic efficiency"s_w_t=Entropy(Water;h=h_w_t;P=P_turbine) "Entropy at turbine inlet"h_w_t=Enthalpy(Water;T=T_w;P=P_turbine) "Enthalpy at turbine inlet"eta_pelton=1 "Efficiency of turbine""__________________________"} h_sw0=Enthalpy(Water;s=s_w_t;P=P_atm) "Enthalpy if no increase in entropy"eta_pelton=(h_w_t-h_w0)/(h_w_t-h_sw0) "definition of efficiency to determine enthalpy after turbine"W_turbine=m_dot*(h_w0-h_w_t)*1e-6 "Power of turbine, in MW"E_s=integral(W_turbine;time;0;tau_empty) "Extracted energy"abs(E_s)=E_s_turbineE_s_pump=89820 "Energy used in compression from compression program"eta_rt=E_s_turbine/E_s_pump "Round trip efficiency" "__________________________" Q_dot_1=UA_1*(T_w-T_a) "Heat transfer between the two media" UA_1=1/(R_1a+R_1b) "Definition of UA value"R_1a=1/(h_1a*A_water) "Definition of resistance, air"R_1b=1/(h_1b*A_water) "Definition of resistance, water"A=A_water "Crossectional area""Natural convection, hot air to cold membran"Nusselt_1a=0,27*Rayleigh_1a^(1/4) "Nusselt number air"Rayleigh_1a=Grashof_1a*Prandtl_1a "Rayleigh number"Grashof_1a=g*beta_1a*T*L_D^3/nu_1a "Grashof number"L_D=A_water/(D*Pi) "Equivalent length"mu_1a=Viscosity(Air_ha;T=T_a;P=P) "Dynamic viscosity, air"Prandtl_1a=Prandtl(Air_ha;T=T_a;P=P) "Prandtl number air"nu_1a=mu_1a/rho_a "Kinematic viscosity"beta_1a=1/T_a "Volumetric expansion coefficient"k_a=Conductivity(Air_ha;T=T_a;P=P) "Thermal conductivity"Nusselt_1a=h_1a*L_D/k_a "Nusselt number"T=abs(T_a-T_w) "Temperature difference between media" "Natural convection, hot membran to cold water"Nusselt_1b=0,27*(Rayleigh_1b)^(1/4) "Nusselt number water"Rayleigh_1b=Grashof_1b*Prandtl_1b "Rayleigh number water"


Grashof_1b=g*beta_1b*T*L_D^3/nu_1b "Grashof number"mu_1b=Viscosity(Water;T=T_w;P=P) "Dynamic viscosity"Prandtl_1b=Prandtl(Water;T=T_w;P=P) "Prandtl number"nu_1b=mu_1b/rho_w "Kinematic viscosity"beta_1b=VolExpCoef(Water;T=T_w;P=P) "Volumetric expansion coefficient"k_b=Conductivity(Water;T=T_w;P=P) "Thermal conductivity"Nusselt_1b=h_1b*L_D/k_b "Nusselt number"

SOLUTIONUnit Settings: SI K Pa J mass degA = 78,54 Ac = 3318 Awater = 78,54

1a = 0,003334 1b = 0,0003044 D = 65 [m]dVa = 0,36 dVw = -0,36 Dp = 0,4 Dt = 10 pelton = 0,9248 rt = 0,8351 Es = -75008 Es,pump = 89820 Es,turbine = 75008 g = 9,81 [m/s2] Grashof1a = 18472 Grashof1b = 208,2 H = 300 [m] Headloss = 5 Headturbine = 2113 h1a = 0,2967 h1b = 2,501 Hp = 1200 hreservoir = 10 hsw0 = 114902 Ht = 50 hw0 = 116456 hw,t = 135569 ka = 0,03825 kb = 0,6082 Loadper = 57,34 LD = 0,3846 M = 2,347E+08 1a = 0,00002404 1b = 0,0008335 Ma = 2,347E+08 m = 361,8 Mt,e = 2,386E+08 Mw = 39560 Mw,e = 3,947E+06 Nusselt1a = 2,983 Nusselt1b = 1,582 1a = 1,023E-07 1b = 8,294E-07 P = 21427815 p1 = -0,00003839 p2 = 0,004423 p3 = 0,7974 Prandtl1a = 0,807 Prandtl1b = 5,658 Patm = 101300 [Pa] Pe = 21598438,7377 [Pa] Pturbine = 2,084E+07 Pturbine,loss = 20787094 Pturbine,max = 36250000 Q1 = 21,17 Qturbine = 0,36 [m3/s] Rayleigh1a = 14908 Rayleigh1b = 1178

a = 234,8 a,e = 235,7 w = 1005 R1a = 0,04291 R1b = 0,005091 sw,t = 400,5 T = 1,016 empty = 10800 [s] time = 10800 Ta = 300 [K] Te = 300,6 Tw = 301 Tw,e = 301 UA1 = 20,83 ua = 175376 ua0 = 175733 ucavern = 13,35 ue = 175733 uw = 114839 uw0 = 114876 uw,e = 114890 Va = 999531 Va,0 = 999491 Va,e = 995643 Vc,0 = 995492 Vfrac = 0,01002 Vp,0 = 150,8 Vtotal = 999570 Vt,0 = 3927 Vw = 39,36 Vw,0 = 78,54 [m3] Vw,e = 3927 Wturbine = -6,915 zw = 0,01186 Integral Table

time Vw P Va Ta Tw Mw Headturbine Pturbine;loss

[K]

Row 1 0 3927 21598451 995643 300,6 301 3,947E+06 2132 20969282 Row 2 1000 3567 21582554 996003 300,6 301 3,585E+06 2130 20952314 Row 3 2000 3207 21566675 996363 300,5 301 3,223E+06 2128 20935366 Row 4 3000 2847 21550817 996723 300,5 301 2,862E+06 2127 20918439 Row 5 4000 2487 21534980 997083 300,4 301 2,500E+06 2125 20901532 Row 6 5000 2127 21519163 997443 300,3 301 2,138E+06 2123 20884645 Row 7 6000 1767 21503365 997803 300,3 301 1,776E+06 2122 20867778 Row 8 7000 1407 21487588 998163 300,2 301 1,414E+06 2120 20850931 Row 9 8000 1047 21471830 998523 300,1 301 1,053E+06 2118 20834104 Row 10 9000 687,4 21456093 998883 300,1 301 690800 2116 20817297


Integral Tabletime Vw P Va Ta Tw Mw Headturbine Pturbine;loss

[K]

Row 11 10000 327,4 21440375 999243 300 301 329000 2115 20800510 Row 12 10800 39,36 21427815 999531 300 301 39560 2113 20787094

Integral TableLoadper pelton

Row 1 57,85 0,9248 Row 2 57,8 0,9248 Row 3 57,75 0,9248 Row 4 57,71 0,9248 Row 5 57,66 0,9248 Row 6 57,61 0,9248 Row 7 57,57 0,9248 Row 8 57,52 0,9248 Row 9 57,47 0,9248 Row 10 57,43 0,9248 Row 11 57,38 0,9248 Row 12 57,34 0,9248

B Energy stored in cavern wall

> >

(8)(8)

(4)(4)

> >

> >

> >

(2)(2)

> >

> >

> >

> >

> >

(7)(7)

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(3)(3)

(1)(1)> >

> >

(5)(5)

(6)(6)

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> >

restart:with(plots):alpha:=k/(rho*c);k:=5;rho:=2165;c:=50;h:=1;T_a:=310.2;T_i:=300;t:=13*3600;

! :=k" c

k := 5

" := 2165c := 50h := 1

T_a := 310.2T_i := 300t := 46800

M_a:=2.114*10^8;M_a := 2.114000000 108

Cp_a:=1006;Cp_a := 1006

r1:=65/2;

r1 :=652

z:=300;z := 300

'(T_x_t-T_i)/(T_a-T_i)=erfc(x/(2*sqrt(alpha*t)))-(exp(h*x/k+h^2*alpha*t/k^2)*(erfc(x/(2*sqrt(alpha*t))+h*sqrt(alpha*t)/k)))';

T_x_tKT_iT_aKT_i

= erfc12

x

! tK e

h xk

Ch2 ! t

k2 erfc

12

x

! tCh ! tk

(T_x_t-T_i)/(T_a-T_i)=erfc(x/(2*sqrt(alpha*t)))-(exp(h*x/k+h^2*alpha*t/k^2)*(erfc(x/(2*sqrt(alpha*t))+h*sqrt(alpha*t)/k)));

0.09803921569 T_x_tK 29.41176471 = erfc1

312 x 11258

K e15

xC936

10825 erfc1

312 x 11258 C

62165

11258

T_x_t:=evalf(solve(%,T_x_t));

(8)(8)

> >

> >

T_x_t := 10.20000000 erfc 0.3400760347 x

K 10.20000000 e0.2000000000 xC 0.08646651270 erfc 0.3400760347 xC 0.2940518878C 300.0000000

smartplot( (8) );

xK10 K5 0 5 10

302

304

306

308

310

312

314

316

plot((8),x=0..7,y=300..303,title="Temperature distribution in semi-infinite solid",titlefont = ["ROMAN", 14],labels=["Distance in meters","Temperature in Kelvin"],labeldirections = ["horizontal", "vertical"],labelfont = ["ROMAN", 12]);

(8)(8)

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> > (10)(10)

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(9)(9)

Distance in meters0 1 2 3 4 5 6 7

Tem

pera

ture

in K

elvi

n

300

301

302

303Temperature distribution in semi-infinite solid

#T_x_t2 x forskudt'(T_x_t2-T_i)/(T_a-T_i)=erfc((x-r1)/(2*sqrt(alpha*t)))-(exp(h*(x-r1)/k+h^2*alpha*t/k^2)*(erfc((x-r1)/(2*sqrt(alpha*t))+h*sqrt(alpha*t)/k)))';

T_x_t2KT_iT_aKT_i

= erfc12

xK r1

! tK e

h xK r1k

Ch2 ! t

k2 erfc

12

xK r1

! tCh ! tk

latex( '(9)' );{\frac {{\it T\_x\_t2}-{\it T\_i}}{{\it T\_a}-{\it T\_i}}}={\it erfc} \left( 1/2\,{\frac {x-{\it r1}}{ \sqrt{\alpha\,t}}} \right) -{{\rm e}^{{\frac {h \left( x-{\it r1} \right) }{k}}+{\frac {{h}^{2}\alpha\,t}{{k}^{2}}}}}{\it erfc} \left( 1/2\,{\frac {x-{\it r1}}{ \sqrt{\alpha\,t}}}+{\frac {h \sqrt{\alpha\,t}}{k}} \right) T_x_t2:=solve(%,T_x_t2);

T_x_t2 := 10.20000000 erfc 0.3400760347 xK 11.05247113

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K 10.20000000 e0.2000000000 xK 6.413533487 erfc 0.3400760347 xK 10.75841924C 300.0000000

r2:=r1+6;

r2 :=772

Q:=-int(int(int(rho*c*(T_x_t2-T_i)*x,x=r1..r2),theta=0..2*Pi),L=0..z);

Q := K2.465755344 1010

#Temperature drop of air

Q=M_a*Cp_a*Delta_T;K2.465755344 1010 = 2.126684000 1011 Delta_T

solve(%);K0.1159436637

deltaU:=Q/M_a;deltaU := K116.6393256

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restart:with(plots):#Cavernalpha:=k/(rho*c);k:=5;rho:=2165;c:=50;h:=1;T_a:=300.7;T_i:=300;t:=13*3600;

! :=k" c

k := 5

" := 2165c := 50h := 1

T_a := 300.7T_i := 300t := 46800

M_a:=2.347*10^8;M_a := 2.347000000 108

Cp_a:=1006;Cp_a := 1006

r1:=65/2;

r1 :=652

r2:=r1+6;

r2 :=772

z:=300;z := 300

#T_x_t x forskudt'(T_x_t-T_i)/(T_a-T_i)=erfc((x-r1)/(2*sqrt(alpha*t)))-(exp(h*(x-r1)/k+h^2*alpha*t/k^2)*(erfc((x-r1)/(2*sqrt(alpha*t))+h*sqrt(alpha*t)/k)))';

T_x_tKT_iT_aKT_i

= erfc12

xK r1

! tK e

h xK r1k

Ch2 ! t

k2 erfc

12

xK r1

! tCh ! tk

T_x_t:=solve((7),T_x_t);T_x_t := 0.6999999998 erfc 0.3400760347 xK 11.05247113

K 0.6999999998 e0.2000000000 xK 6.413533487 erfc 0.3400760347 xK 10.75841924C 300.

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Q_c:=-int(int(int(rho*c*(T_x_t-T_i)*x,x=r1..r2),theta=0..2*Pi),L=0..z);

Q_c := K1.692185039 109

#Pipe "Properties of stainless steel AISI 302 [Incropera]k_s:=15.1;rho_s:=8055;c_s:=480;

k_s := 15.1rho_s := 8055c_s := 480

r_p:=0.4/2;r_p := 0.2000000000

z_p:=1200;z_p := 1200

t_p:=0.02;t_p := 0.02

V_pipe:=evalf(z_p*Pi*(r_p+t_p)^2-z_p*Pi*r_p^2);V_pipe := 31.66725395

m_p:=rho_s*V_pipe;m_p := 2.550797306 105

Q_p:=(T_i-T_a)*m_p*c;Q_p := K8.927790570 106

#Semi infinite solid constant temperaturerp2:=r_p+t_p;

rp2 := 0.2200000000'(T_x_t2-T_a)/(T_i-T_a)=erf((x-rp2)/(2*sqrt(alpha*t)))';

T_x_t2KT_aT_iKT_a

= erf12

xK rp2

! tT_x_t2:=evalf(solve(%,T_x_t2));T_x_t2 := K0.6999999998 erf 0.3400760347 xK 0.07481672764 C 300.7000000

limit(-.6999999998*erf(0.4511618428e-1*x-0.9925560545e-2),x=infinity);

K0.6999999998

plot((19),x=0..10,y=300..302,title="Temperature distribution in semi-infinite solid",titlefont = ["ROMAN", 14],labels=["Distance inmeters","Temperature in Kelvin"],labeldirections = ["horizontal", "vertical"],labelfont = ["ROMAN", 12]);

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Distance in meters0 2 4 6 8 10

Tem

pera

ture

in K

elvi

n

300

300.5

301

301.5

302Temperature distribution in semi-infinite solid

Q_c2:=-int(int(int(rho*c*(T_x_t2-T_i)*x,x=rp2..6),theta=0..2*Pi),L=0..z_p);

Q_c2 := K1.430814806 109

Q:=Q_p+Q_c+Q_c2;Q := K3.131927636 109

Q=M_a*Cp_a*Delta_T;K3.131927636 109 = 2.361082000 1011 Delta_T

solve(%);K0.01326479824

Delta_u:=Q/M_a; #Change in specific internal energyDelta_u := K13.34438703

C Pressure losses

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Restart:#Pressure losses Case 1Q_c:=2.0;

Q_c := 2.0dia:=1;

dia := 1Vel_c:=evalf(Q_c/(dia^2*Pi/4));

Vel_c := 2.546479089rho:=1260;

! := 1260mu:=0.00084;

µ := 0.00084RE_c:=Vel*rho*dia/mu;

RE_c := 3.819718633 106

e:=0.046; #Roughness(millimeters);e := 0.046

Rel:=e/(dia*1000);Rel := 0.00004600000000

#Friction factor obtained from Moody diagram;f_c:=0.012;

f_c := 0.012L:=1500;

L := 1500CV:=8;

CV := 8El:=30;

El := 30P_loss_com:=1/2*rho*Vel_c^2*f_c*(L/dia+CV+El);

P_loss_com := 75397.74540#ExpansionQ_e:=5.33;

Q_e := 5.33Vel_e:=evalf(Q_e/(dia^2*Pi/4));

Vel_e := 6.786366772RE_e:=Vel_e*rho*dia/mu;

RE_e := 1.017955016 107

f_e:=0.0105;f_e := 0.0105

P_loss_exp:=1/2*rho*Vel_e^2*f_e*(L/dia+CV+El);P_loss_exp := 4.685552832 105

#In terms of headH_loss_com:=evalf(P_loss_com/(rho*9.81));

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H_loss_com := 6.099845106H_loss_exp:=evalf(P_loss_exp/(rho*9.81));

H_loss_exp := 37.90716335

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Restart:#Pressure losses Case 2 compression, part 1Q_1:=0.000025;

Q_1 := 0.000025dia:=0.009525;

dia := 0.009525Vel_1:=evalf(Q_1/(dia^2*Pi/4));

Vel_1 := 0.3508489318rho:=940;

! := 940mu:=0.00175;

µ := 0.00175RE_1:=Vel_1*rho*dia/mu;

RE_1 := 1795.043378#From moody diagram, this Reynoldsnumber -> laminar flor

#Friction factor obtained from Moody diagram;f_1:=64/RE;

f_1 := 0.03565373449L_1:=4;

L_1 := 4CV:=8;

CV := 8El:=30;

El := 30P_loss_com_1:=1/2*rho*Vel_1^2*f_1*(L_1/dia+CV+2*El);

P_loss_com_1 := 1006.505852#Pressure losses Case 2 compression, part 2Q_2:=1/7*Q_1;

Q_2 := 0.000003571428571Vel_2:=evalf(Q_2/(dia^2*Pi/4));

Vel_2 := 0.05012127596RE_2:=Vel_2*rho*dia/mu;

RE_2 := 256.4347682 #From moody diagram, this Reynoldsnumber -> laminar flowf_2:=64/RE_2;

f_2 := 0.2495761415L_2:=11;

L_2 := 11El_45:=16;

El_45 := 16P_loss_com_2:=1/2*rho*Vel_2^2*f_2*(L_2/dia+12*El_45);

P_loss_com_2 := 396.8863946

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P_loss_com:=P_loss_com_1+P_loss_com_2;P_loss_com := 1403.392247

Restart:#Pressure losses Case 2 expansion, part 1Q_1:=0.000045;

Q_1 := 0.000045dia:=0.009525;

dia := 0.009525Vel_1:=evalf(Q_1/(dia^2*Pi/4));

Vel_1 := 0.6315280771rho:=940;

! := 940mu:=0.00175;

µ := 0.00175RE_1:=Vel_1*rho*dia/mu;

RE_1 := 3231.078079 #From moody diagram, this Reynoldsnumber means the flow is a transition zone. f=64/RE#Friction factor obtained from Moody diagram;f_1:=64/RE_1;

f_1 := 0.01980763028L_1:=4;

L_1 := 4CV:=8;

CV := 8El:=30;

El := 30P_loss_exp_1:=1/2*rho*Vel_1^2*f_1*(L_1/dia+CV+2*El);

P_loss_exp_1 := 1811.710534#Pressure losses Case 2, part 2Q_2:=1/7*Q_1;

Q_2 := 0.000006428571429Vel_2:=evalf(Q_2/(dia^2*Pi/4));

Vel_2 := 0.09021829673RE_2:=Vel_2*rho*dia/mu;

RE_2 := 461.5825827#From moody diagram, this Reynoldsnumber is in the transition zone between laminar and turbulentf_2:=64/RE_2;

f_2 := 0.1386534120L_2:=11;

L_2 := 11El_45:=16;

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P_loss_exp_2:=1/2*rho*Vel_2^2*f_2*(L_2/dia+12*El_45);P_loss_exp_2 := 714.3955100

P_loss_exp:=P_loss_exp_1+P_loss_exp_2;P_loss_exp := 2526.106044

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Restart:#Pressure losses Case 3 compressionQ:=0.1336;

0.1336dia:=0.4;

0.4Vel:=evalf(Q/(dia^2*Pi/4));

1.063155020rho_a:=236;

236mu_a:=0.000024;

0.000024RE_a:=Vel*rho_a*dia/mu_a;

4.181743079 106

e:=0.046; #Roughness(millimeters);0.046

Rel:=e/(dia*1000);0.0001150000000

#Friction factor obtained from Moody diagram;f_a:=0.012;

0.012L_a:=1200;

1200CV:=8;

8El:=30;

30P_loss_com_a:=1/2*rho_a*Vel^2*f_a*(L_a/dia+2*El);

4897.538610rho_w:=1005;

1005mu_w:=0.0008;

0.0008RE_w:=Vel*rho_w*dia/mu_w;

5.342353975 105

f_w:=0.015;0.015

L_w:=50;50

P_loss_com_w:=1/2*rho_w*Vel^2*f_w*(L_w/dia+El+CV);1388.698985

P_loss_com:=P_loss_com_w+P_loss_com_a;6286.237595

Head_loss:=P_loss_com/(rho_w*9.81);0.6376108850

Restart:#Pressure losses Case 3 expansionQ:=0.356;

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0.356dia:=0.4;

0.4Vel:=evalf(Q/(dia^2*Pi/4));

2.832957986rho_a:=236;

236mu_a:=0.000024;

0.000024RE_a:=Vel*rho_a*dia/mu_a;

1.114296808 107

e:=0.046; #Roughness(millimeters);0.046

Rel:=e/(dia*1000);0.0001150000000

#Friction factor obtained from Moody diagram;f_a:=0.014;

0.014L_a:=1200;

1200CV:=8;

8El:=30;

30P_loss_exp_a:=1/2*rho_a*Vel^2*f_a*(L_a/dia+2*El);

40570.62863rho_w:=1005;

1005mu_w:=0.0008;

0.0008RE_w:=Vel*rho_w*dia/mu_w;

1.423561388 106

f_w:=0.0135;0.0135

L_w:=50;50

P_loss_exp_w:=1/2*rho_w*Vel^2*f_w*(L_w/dia+El+CV);8874.373575

P_loss_exp:=P_loss_exp_w+P_loss_exp_a;49445.00220

head_loss:=P_loss_exp/(rho_w*9.81);5.015189314

D Turbine efficiency

!"#$"%& '#(%$)* !"+&,%-.. ./01 ./011. ./1 ./10. ./1.2 ./1.23. ./00 ./145. ./0625 ./1422. ./0 ./146. ./370 ./1-27. ./55 ./0124. ./262 ./022-2 ./65 ./0-2-. ./762 ./322 ./4 ./54. ./.- ./.-

8"9:;)#$+":)*:&<":%,&"9:"==)$)"%$>:=,#:=#(%$)*:&?#@)%"A+?":)*:&<":%,&"9:"==)$)"%$>:=,#:B"+&,%:&?#@)%"

.:

./4:

./6:

./5:

./0:

-:

.: 4.: 6.: 5.: 0.: -..:

!"#$%&

#'()*+,))

-%.#%&/01%)23)3455)5206()*7,)

!"#$%&#')23)3.0&#$8)/4.9$&%)

y = p1*z^7 + p2*z^6 + p3*z^5 + p4*z^4 + p5*z^3 + p6*z^2 + p7*z + p8

where z is centeredand scaled:

z = (x-mu)/sigmamu = 43.846sigma = 33.86

Coefficients: p1 = 0.00069809 p2 = 0.0019064 p3 = 0.013802 p4 = -0.068269 p5 = 0.060016 p6 = -0.073648 p7 = 0.20217 p8 = 0.76286

Norm of residuals = 0.0057593

y = p1*x^2 + p2*x + p3

Coefficients: p1 = -3.8393e-005 p2 = 0.0044232 p3 = 0.79739

Norm of residuals = 0.0036105

E Pressure vessel information

High-Pressure

BYBXMED.E

L.B.L.B.I.

DD-1CBA 4X

A B AA A-2 A-3A-1C A-5

Low-Pressure

LP2.5 LP5

Specialty Gas Cylinder Information

Y

TA

C2H2

C(Al)B(Al) 4X(Al)D-1(Al)

HCI, Bulk Electronic Gases

CI2, H2S, SO2, CH3Cl, C2H5Cl

APC2386/1-10 10/16/02 1:58 PM Page 7

Specialty Gas Cylinder DimensionsNominal* Average Average

Product Number Dimensions Tare InternalDescription DOT (Excluding Valve and Cap) Weight Volume

Size Digits Specification in (cm) lb (kg) ft3 (L)

High PressureA 01 3AA2400 9 x 55 (23 x 140) 137 (62) 1.76 (49.8)B 02 3AA2265 9 x 51 (23 x 130) 119 (54) 1.55 (43.9)C 03 3A2015 7 x 33 (18 x 84) 57 (26) 0.56 (15.9)D-1 04 3A2015 7 x 19 (18 x 48) 26 (12) 0.26 (7.4)D 05 3AA2015 4 x 17 (10 x 43) 9 (4) 0.10 (2.8)4X 46 3AA2015 4 x 13 (10 x 33) 6.6 (3) 0.075 (2.12)L.B.I. X6 3E1800 2 x 12 (5 x 30) 2 (0.7) 0.015 (0.43)L.B. 06 3E1800 2 x 12 (5 x 30) 2 (0.9) 0.015 (0.43)Medical E 07 3AA2015 4 x 26 (10 x 66) 14 (6) 0.16 (4.5)BX 88 3AA6000 10 x 51 (25 x 130) 300 (136) 1.49 (42.2)BY 89 3AA3500 9 x 51 (23 x 130) 187 (85) 1.53 (43.3)Low PressureA 09 3A480 10 x 49 (25 x 124) 85 (39) 1.93 (54.7)B 10 3A480 10 x 36 (25 x 91) 90 (41) 1.28 (36.2)C 11 3A480 8 x 22 (20 x 56) 33 (15) 0.53 (15.0)AA 08 4AA480 15 x 52 (38 x 132) 160 (73) 4.46 (126.3)A-1 91 4BW240 16 x 50 (41 x 127) 75 (34) 3.83 (108.5)A-2 90 4BW240 22 x 48 (56 x 122) 167 (76) 7.64 (216.4)A-3 92 4BA240 12 x 45 (30 x 114) 48 (22) 2.31 (65.4)A-5 81 4BW240 30 x 57 (76 x 145) 315 (143) 16.00 (453.0)LP2.5 92 4B240 9 x 17 (23 x 43) 14 (6) 0.4 (11.3)LP5 93 4B240 12 x 18 (30 x 46) 18 (8) 0.77 (21.8)C2H2

A 18 8/8AL 12 x 41 (30 x 104) 185 (84) 2.36 (66.8)HCI, Bulk Electronic GasesY 37 3A1800 24 x 90 (61 x 229) 1,108 (503) 15.83 (448)H2ST 38 106A800X 30 x 82 (76 x 208) 2,254 (1,022) 25.82 (731)SO2, C2H5Cl, Cl2, CH3CIT 45 106A500X 30 x 82 (76 x 208) 1,400 (635) 25.64 (726)AluminumA(AI) 31 3AL2216 10 x 52 (25 x 132) 90 (41) 1.64 (46.4)B(AI) 28 3AL2015 8 x 48 (20 x 122) 48 (22) 1.04 (29.5)C(AI) 29 3AL2216 7 x 33 (18 x 84) 32 (15) 0.56 (15.8)D-1(AI) 30 3AL2216 7 x 16 (18 x 41) 15 (7) 0.21 (5.9)4X(AI) 34 3AL1800 4 x 10 (10 x 26) 3.3 (1.6) 0.057 (1.61)NickelB 61 3BN400 7 x 45 (18 x 14) 88 (40) 0.65 (18.4)D-1 56 3BN400 7 x 22 (18 x 56) 48 (22) 0.28 (8.0)D-2 58 3BN400 5 x 15 (12 x 38) 10 (4) 0.10 (2.9)Stainless Steel55 gallon 52 UN1A1 24 x 45 (61 x 114) 175 (79) 7.35 (208.2)10 gallon 50 UN1A1 14 x 29 (35 x 74) 50 (23) 1.34 (37.8)5 gallon 51 UN1A1 9 x 24 (23 x 61) 25 (11) 0.67 (18.9)

*These dimensions are not exact. They should not be used for engineering drawings or equipment specifications.

APC2386/1-10 10/16/02 1:58 PM Page 8

Approximate ScottDimensions Air BOC Alphagaz Specialty

(inches) Products AGA Airgas (Airco) (Liquid Air) Praxair Matheson MG Solkatronics Gases

High Pressure Steel24 x 90 Y — — — — TO — — — —9 x 55 A 049 300 300 49 T 1L 300 49 K9 x 51 B 044 200 200 44 K 1A 200 44 A7 x 33 C 016 80 80 16 Q 2 80 16 B7 x 19 D-1 007 35 30 7 G 3 35 7 C4 x 17 D 003 7 12 3 F 4 10 3 D2 x 12 L.B. LBR L.B. L.B. L.B. L.B. L.B. L.B. — L.B.4 x 26 E 005 E E MEDE ANE 3L E — ER10 x 51 BX 485 3HP 500 44H 6K 1U 3HP — —9 x 51 BY — — — 44H 3K 1H 2HP — —Aluminum10 x 52 A(AI) — — — AT — — — — —8 x 48 B(AI) A31 150A 150A 30AL AS 1R 150AL 29A AL7 x 33 C(AI) A16 80A 80A 22AL AQ 2R 80AL — BL7 x 16 D-1(AI) A07 33A 30A 7AL AG 3R 33AL — CL

Specialty Gas Cylinder Size Comparison Chart

Additional Supply Modes — Bulk Specialty Gases and Chemicals

M size cylinder 7 x 43 inches

Many Air Products specialty gases and chemicals can be supplied in bulk quantity.Products available in bulk quantity are identified throughout the catalog by the symbols shown below:

Tank trucks are used for over-the-road transportation of cryogenic liquids. Liquidproduct is then transfilled to cryogenic storagetanks at customer locations.

Tube trailers (T.T.) provide over-the-road shipment of high-pressure gases, gaseouschemicals, and gas mixtures. The trailers serve as on-site storage systems at customerlocations.

Cryogenic liquids such as nitrogen and helium are supplied in dewars (low-pressurecryogenic tanks) for larger requirements nearcustomers’ point of use.

If you are considering bulk supply, a representative from Air Products can discussyour requirements and the economics of alternate supply systems.

APC2386/1-10 10/16/02 1:58 PM Page 9

Cylinder IdentificationPackaging and ColorAir Products uses a unicolor paint scheme toidentify specialty gas cylinders. Here are thehighlights of our cylinder packaging and colorcodes.

• Virtually all steel cylinder bodies are painteduniformly dark blue and covered with a protective plastic diamond mesh.

• A cylinder neck ring is permanently fixedbelow the base of the valve. Each cylinderneck ring is color-coded to help identifycylinder contents and gas category (e.g.,yellow for corrosive, red for flammables).

• A color-coded shoulder label indicates theproduct’s shipping name and identificationnumber. On pure products, a grade label isalso applied to the cylinder shoulder. Thecolor-coded label border correlates withneck ring color for product identification.The shoulder label also specifies gas gradeinformation.

• Some cylinders are painted with a verticalstencil identifying cylinder contents.

MarkingsAir Products specialty gas cylinders arestamped with markings designed to indicateownership, specifications, pressure ratings,and other important data. Air Products also utilizes a bar code label for product identification and tracking.

1. Cylinder Specification:• DOT—Department of Transportation

(previously ICC – Interstate Commerce Commission), which is the regulatory body that governs the use of cylinders.

• Specification of the cylinder type of material of construction (e.g., 3AA).

• Service or working pressure in pounds per square inch (e.g., 2,265 psi).

2. Cylinder Serial Number:• The letters SG precede the serial

numbers for Specialty Gas cylinders.

8. Cylinder Manufacturer’s Inspection Marking

9. Cylinder Tare (Empty) Weight:• This value is preceded by the letters TW.

3. Registered Owner Symbol:• Symbol used to indicate the original

owner of the cylinders.

• APROINC is a Registered Owner Symbol for Air Products.

4. Date of Manufacture:• This date (month-year) also indicates

the original hydrostatic test.

5. Neck Ring Identification:• The cylinder neck ring displays the

name of the current owner of the cylinder.

6. Retest Markings:• The format for a retest marking is:

Month – Facility – Year – Plus Rating – Star Stamp.

• The + symbol (Plus Rating) indicates that the cylinder qualifies for 10% overfill.

• The ! symbol (Star Stamp) indicates that the cylinder meets the requirements for 10-year retest.

7. CylinderTrak™ Bar Code Label:• The CylinderTrak bar code label provides

a unique cylinder identifier and is used by computer systems to track cylinders throughout the fill process. As an optional service, we have the capability of tracking cylinders to and from customers.

5

68

7

6

12349

NON-FLAMMABLE GAS

2

OXIDIZER

5.1

POISON

6

POISON GAS

2

8

CORROSIVE

FLAMMABLE GAS

2

APC2386/1-10 10/16/02 1:58 PM Page 10

DTU Mechanical EngineeringSection of Thermal Energy SystemsTechnical University of Denmark

Nils Koppels Allé, Bld. 403DK- 2800 Kgs. Lyngby DenmarkPhone (+45) 45 25 41 31Fax (+45) 45 88 43 25

www.mek.dtu.dk

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