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©2006 Halim Gürgenci, The University of Queensland (Australia) and Bogazici University (Istanbul, Turkey)

Energy and Environment

Professor Halim Gürgenci(visitor from the University of Queensland)

Rm 4245; Phone : (212) 359 [email protected]


REFERENCESThe first textbook was always on my desk when I was preparing these lecture notes. I

also list wikipedia, which was the source for many of the images and animations plus many useful tidbits of information.

Specific references are supplied on every slide except when the slide content is part of common mechanical engineering lore and can be found in any relevant textbook.

• Fay, James A, and Golomb, Dan S, Energy and the Environment, Oxford University Press (2002).

• Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Main_Page)


Module 3

Thermodynamic Principles of Energy Conversion


Mechanical Energy


Internal and Chemical Energy


Other Energy Forms


First Law of thermodynamics


Second LawWe can never achieve 100% conversion of all of our heat to useful work. Work is generated through an exchange between a hot and cold reservoir.

This equation is 1st Law.

The 2nd Law states that heat cannot be converted 100% to work. When heat is converted to work, there will ALWAYS be losses. So, there will always be OH & QL.


Thermodynamic PropertiesEnthalpy, h e pv= +

pConstant-pressure specific heat, cp

hT∂

=∂

vConstant-volume specific heat, cv

eT∂

=∂

Gibbs Free Energy, f h Ts e pv Ts≡ − = + −

For a reversible process,Tds de pdv

dh vdpdh sdT df

= += −= + −

These equations are often applied to irreversible processes between two known states. But the integration should be performed along a reversible path between these two given states.


Combustion of Solid Fuels

Air-Fuel Ratios and Heating Values


Burning fossil fuels2 2 2n m x yC H O CO H Oz+ → +

Fossil fuels are usually compounds of C and H atoms (e.g. CH4 for methane; C8H18 for octane; C3H8 for propane, etc). These are also referred to as hydrocarbon fuels.

Coal is a solid hydrocarbon fuel made up of mostly carbon but sometimes including other hydrocarbons. Most liquid hydrocarbon fuels (gasoline, kerosene, diesel fuel) are produced from crude oil. It is convenient to describe a liquid fuel in terms of a single hydrocarbon even though most liquid fuels are mixtures of many hydrocarbons. Gasoline is usually considered to be octane (C8H18) and diesel fuel is considered to be dodecane (C12H26). The composition of a hydrocarbon fuel may also be given in terms of percentages of C and H.

In the above equation for the chemical reaction describing the combustion of a fossil fuel CnHm, 1 mole of fuel is combined with x moles of O2 to produce y moles CO2 and z moles of water vapour. The values of x, y, and z are to be found by balancing the number of atoms on both sides of the equation. The value of x when both sides of the equation are perfectly balanced is also called the stoichiometric ratio for this fuel. According to wikipedia, The word is derived from the Greek word stoikheion ("element") and metriā ("measure," from metron).


Burning fossil fuels2 2 2n m x yC H O CO H Oz+ → +

x = ?

y= ?

z = ?


Stoichiometric oxygen/fuel ratio2 2 24 2n m

m mC H n O nCO H O⎛ ⎞ ⎛ ⎞+ + → +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

4O molecules mn

Fuel molecules= +Stoichiometric ratio (in terms of moles or volume)

Stoichiometric ratio (in terms of mass)2 32 8

12 1.008O mass n m

Fuel mass n m+

=+

Usually, the fossil fuels are combusted in air rather than pure oxygen. Therefore, it has more practical sense to consider the air/fuel ratio rather than the oxygen/fuel ratio. When we do this, we will treat air as a mixture of oxygen (21%) and nitrogen (79%). The trace gases in the air will be included by an adjustment of the molecular weight of nitrogen (atmospheric nitrogen is assigned a molecular weight of 28.16 as compared with 28.013 for pure nitrogen).


Stoichiometric air/fuel ratioWhen burning in air (21% O2 + 79% N2), there is 3.7619 (=79/21) N2 for every O2

2222 23.76 3.764 4 2 4n m

m mn N n Nm mC H n O nCO H O⎛ ⎞ ⎛ ⎞+ + + →⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Therefore, the molar stoichiometric air-fuel ratio is 4.76 (n + m/4). Similar to pure oxy-fuel combustion, the mass stoichiometric ratio can be determined by incorporating the molecular masses:

( )( ) m3.7619 28.16 32 n+

32 84/ 4.31912 1.008 12 1.008st

n mA Fn m n m

⎛ ⎞× + ⎜ ⎟ +⎝ ⎠= =+ +

If the actual A/F ratio is higher than the stoichiometric ratio, it is called a lean mixture and rich mixture otherwise. Combustion is not complete in “rich” mixtures and unburned hydrocarbons remain in the product or exhaust gas.


The Heating ValueConsider a hypothetical example where the combustion takes place at constant pressure and all of the heat produced during combustion is used to heat water through a perfect heat exchanger. Also assume that the heat transfer is complete and cools the combustion products down to the temperature of the charge (reactants) before the combustion.

The Heating Value is the heat transferred to the water in this scenario, usually expressed per mass of fuel. If water is a product, two heating values may be defined: the Lower Heating Value (LHV) with the water in vapour phase; and the Higher Heating Value (HHV) with the water condensed. Since the heat of condensation of water is usually not available to process, the LHV has more meaning. However, fuel manufacturers like to quote the HHV to make their fuel look better.


Table of Heating Values

Created by using the data in Table 3.1 of Fay & Golomb, Energy and the Environment, 2002.

SymbolLHV

(MJ/kg-fuel)Df

(MJ/kg-fuel) FuelLHV

(MJ/kg-fuel)H2 119.96 117.63 Comm. Nat. Gas 36-42C 32.76 32.83 Gasoline 47.4

CH4 50.04 51.02 No 2 Oil 45.5CO 10.10 9.18 Anthracite coal 32-34

C3H8 46.33 47.80 Bituminous coal 28-36C2H6-O 27.73 28.90 Lignite 14-18C8H18 44.79 Wood 21C12H26 44.48 Manure 13


Adiabatic Flame TemperatureThe Adiabatic Flame Temperature is the maximum temperature that can be achieved by combusting a given fuel starting at a given temperature and pressure for the reactants. It refers to a combustion process where no work is done and no heat is lost to the outside environment and the enthalpy of combustion is entirely used to heat the products in a reversible process.

The adiabatic flame temperature is obtained at the stoichometric air-fuel ratio. The temperature of the products is reduced at leaner mixtures because of the need to heat the extra air.

If the specific heat of the product gas is known, the maximum adiabatic flame temperature can be estimated by using the following equation:

( )r p p adft rper unit product massh h c T T⎡ ⎤− = × −⎣ ⎦

Where the enthalpy difference term on the left-hand-side is the HHV expressed in terms of the product mass.

Solve Exercises 3.3 and 3.4.


Gaseous Mixtures-volume fractions

Volume fraction

Mole fraction

Molecular Weight

Mass per kmole of mixture

Mass fraction

CO2 12% 0.12 x 44 5.28 18%O2 4% 0.04 x 32 1.28 4%N2 82% 0.82 x 28 22.96 76%CO 2% 0.02 x 28 0.56 2%

100% 30.08 100%

A furnace gas has the following composition by volume.

CO2:12%; O2:4%; N2:82%; CO:2%

Calculate the molecular weight of the mixture and the mass fractions.

Volume fraction

CO2 12%O2 4%N2 82%CO 2%

100%


Gas mixtures mass fractionsA furnace gas has the following composition by mass.

CO2:18%; O2:4%; N2:76%; CO:2%

Calculate the molecular weight of the mixture and the mass fractions.

Mass fraction

Molecular Weight

kmol/kg-mixture

Mole fraction

Volume fraction

CO2 18% / 44 0.004091 12% 12%O2 4% / 32 0.00125 4% 4%N2 76% / 28 0.027143 82% 82%CO 2% / 28 0.000714 2% 2%

100% 0.033198 100% 100%

Mass fraction

CO2 18%O2 4%N2 76%CO 2%

100%


Ideal Heat Engine Cycles


The Carnot CycleSadi Carnot is a French engineer who expressed the foundations of the 2nd Law of Therodynamics in 1824.

The 2nd Law of thermodynamics states that heat cannot be fully converted to work. To produce work, one must transfer heat between a hot and a cold reservoir.

What is then the maximum amount of work that can be achieved for a given hot and cold reservoir temperatures, TH and TC? Carnot answered this question:

1 cth

h h

w Tq T

η = = −


Rankine Cycle

• 1 2 Water is pumped to the boiler pressure• 2 3 Water is heeated to its boiling point at constant (boiler) pressure• 3 4 Water boils at constant pressure and is converted to steam saturated at the boiler pressure• 4 5 Steam is superheated at constant pressure• 5 6 Superheated steam runs through a turbine and its energy is converted to turbine kinetic energy; its

pressure drops to condenser pressure; the steam is saturated at this point at that pressure• 6 1 The saturated steam is condenses to water and is fed into the water pump and the cycle restarts.


Rankine Cycle Efficiency

• Ideal Rankine cycle efficiencies are in the range 30-45% depending on the cycle improvements.

• These improvements could be one of the following:– Supercritical boilers (heating steam above the critical point of water). For high-pressure poinlers, the

temperature of the superheated steam may reach up to 550oC– Multi-stage cascading turbines where the steam is converted to mechanical work at different pressure

at each stage. The steam is reheated after each individual turbine stage. Three turbine stages are common.

– Turbine exhaust steam is used to heat the feedwater pump (called regenerative feed water heating)• In spite of all these improvements, the real power plant efficiencies rarely exceed 40% and are

typically around 30-35%, based on the net mechanical power output divided by the fuel heating value.

( ) ( )5 6 2 1 6 1

5 2 5 2

1netth

H

h h h hw h hq h h h h

η− − − −

= = = −− −


Otto Cycle Processes

Source : http://www.grc.nasa.gov/WWW/K-12/airplane/ottoa.html

Image Source : Wikipedia

Strokes1 : 1 2 Intake (Induction)2 : 2 3 Compression3 : 3 5 Combustion (Power)4 : 6 3 Exhaust


Otto Cycle

Drawing Source : http://www.grc.nasa.gov/WWW/K-12/airplane/otto.html


Otto Cycle Efficiency

• The efficiency increases by increasing the compression ratio, r=v2/v3. What is the limit against using higher compression ratios?

• The best efficiencies for real gasoline (Otto) and diesel engines with losses are 28% and 39%, respectively (Fay & Golomb, p 53).

• What is the main difference between Otto and Diesel cycles?• How to we adjust to the load in an Otto-cycle engine? How is it done in a Diesel engine?• Why do the diesel engines usually have efficiencies higher than comparable Otto-cycle

engines?• What is a knock? What causes it?• Do diesels knock? Explain the reasons.

1

3

2

1thvv

γ

η−

⎛ ⎞= − ⎜ ⎟

⎝ ⎠


Brayton Cycle


1

1

2

1 ;

kk

pth

v

cp kp c

η

−

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

Thermal efficiency, ηth < 33% for simple Brayton cycles for practical reasons like temperature and material limitations. Regenerative heating of the compressor inlet gas will increase this. Other important factors include the aerodynamic efficiencies of the turbine and compressor blades, the blade reliability and the temperature resistance of the turbine blade material.


Combined Cycle Plant


The thermal efficiency of a combined cycle power plant is normally in terms of the net power output of the plant as a percentage of the lower heating value (LHV) of the fuel.

Generating only electricity, power plant efficiencies of up to 59% can be achieved.

cc g s g sη η η η η= + −In the case of combined heat and power generation, the efficiency can increase to about 85%.


Refrigeration / Heat Pump

The above is a heat engine in reverse. Instead of producing work from the difference between the heat exchanged from a hot to a cold reservoir, this transfers heat from a cold to a hot reservoir. We have to add work to do this. The less work is needed, the better it is. A Coefficient Of Performance (C.O.P.) is used to measure the performance of a heat pump/refrigerator. The COP is defined as the useful heat quantity (the heat removed for a refrigerator or the heat added for a heat pump) divided by the work added to the cycle.

1 4

2 1

Crefrig

Q h hCOPW h h

−= =

−

2 4

2 1

Hhp

Q h hCOPW h h

−= =

−

Note that the same machine can be used as an air conditioning system in summer and as a heater (heat pump) in winter by swapping the evaporator and the condenser (called reverse-cycle A/C units)


Fuel Cells• What if we try to convert the internal energy of a fuel directly to electrical energy without

burning it? Since we are not trying to transform heat into another form of energy, the Carnot efficiency is no longer the limit.

• Fuel Cells are theoretically able to convert all of the “free energy” in the fuel to electricity• Fuel Cell principle was invented by a Welsh judge Sir William Grove in 1839. He tried to

reverse the electrolysis of water and combined hydrogen and oxygen to produce electricity and water. This is the basis of a simple fuel cell. The term “fuel cell” was coined later in 1889 by Ludwig Mond and Charles Langer, who attempted to build the first practical device using air and industrial coal gas(1).

• The theoretical limit on the fuel cell power is the free energy of the fuel. This is dictated by the following form of the second law of thermodynamics:

(1) http://www.corrosion-doctors.org/Biographies/GroveBio.htm

f h T sΔ = Δ − Δ

Where f is the Gibbs free energy for the fuel. If the process is reversible, Δs=0, and Δf = Δh. The first law says that w = Δh. Therefore,

w f≤ Δ

The inequality applies when there are irreversibilities.


Fuel Cell Operation - 1Consider the hydrogen-oxygen fuel cell. The hydrogen fuel is fed into the porous anode and is ionised when in contact with the electrolyte

{ } { }2 2 2el aH H eφ φ+ −→ +

Where φel and φa are the electric potentials for the electrolyte and the anode, respectively. The H+ ions that move to the cathode are oxidised:

{ } { } 2 212 22el cH e O H Oφ φ+ −+ + →

Where φc is the cathode electric potential. The net effect is found by adding these two equations:

{ }2 2 21 22 a cH O H O e φ φ−+ → + −

The last term on the right is electrical potential difference that can be achieved by this fuel cell.


We scale both the numerator and the denominator with the Avogadro’s number to convert from molecules to moles. The electron charge multiplied by the Avogadro number is the Faraday constant (=96485 Coulomb/mole). MH2 is the molecular mass (g) of one mole of H2.

The work, w, should be less than Δf (the 2nd law). This leads to an upper limit for the electric potential difference that can be achieved by a H2-O2 fuel cell.

Fuel Cell Operation - 2For each H2 molecule, two electrons go through the cell electric potential difference. The electrical work done by these electrons is

( )2 e c aw q φ φ= −

Where qe is the electron charge in Coulombs. The work per unit mass of fuel can be found by dividing both sides by the mass of one hydrogen molecule:

( ) ( )2 2

2 2e avoc a c a

H avo H

q N Fwm N M

φ φ φ φ×= − = −

×

( ) 2 2

2H H

c a

f MF

φ φΔ ×

− ≤


Fuel Cell Efficiency in Practice• We have seen that, theoretically, a fuel cell can produce 100% of the free energy of the fuel to

electricity. Since the free energies of the fuels are close to the lower heating values, one can say that, based on the fuel heating value, a fuel cell are potentially capable of delivering 100% efficiency.

• Unfortunately, unless the fuel cell is operating at very low power levels, the process is higly irreversible and the practically achievable fuel cell efficiencies are around 45%.

• In addition to pure hydrogen, there are Hydrocarbon fuels for fuel cells, including diesel, methanol (Direct-methanol fuel cells) and chemical hydrides. The waste product with these types of fuel is carbon dioxide.

• For more information on fuel cells one can try the following:– http://en.wikipedia.org/wiki/Fuel_cells– Hirschenhofer, J.; “How the fuel cell produces power”, Aerospace and Electronic Systems Magazine,

IEEEVolume 7, Issue 11, Nov. 1992 Page(s):24 - 25 – Hirschenhofer, J H, et al, Fuel Cell Handbook, 4th edition, Report DE-AC21-94MC31166, US Dept of

Energy, Washington (1998)


Fuel EfficiencyThe energy conversion efficiencies

(a.k.a. the thermal efficiency or the fuel efficiency) are limited by

(a) The 1st and 2nd laws of thermodynamics

(b) Process irreversibilities

(c) Engineering and design issues

If we know the thermal efficiency of a process and the heating value of the fuel, the fuel consumpotion can be estimated by the following equation:

( ) ( )( )

/ sec/f

th

W kWm kg

LHV kJ kgη=

×

Power Plant Type Eff. %Coal-fired power plants

Steam power plant (6.2 MPa and 480 deg C) 30%Steam power plant (31 MPa and 560 deg C) 42%

Nuclear power plantSteam at 7 MPa, 286 deg C 33%

Automotive gasoline engine 25%Automotive diesel engine 35%Gas turbine electric power 30%Combined cycle electric power plant 45%Fuel cell electric power 45%

Table 3.2 of Fay & Golomb, Energy and the Environment, 2002.

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