Thermofluids 3a - Chapter 10 - 2015
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Transcript of Thermofluids 3a - Chapter 10 - 2015
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ENABLING OBJECTIVESWhat do we aim to achieve in this section?
Define and understand the concept of available energy Identify the difference between energy and availability
(exergy)
Understand the difference between the first and secondlaw efficiency
Understand that destruction of exergy is due to entropy Understand the basis of the exergy equation i.e. energy
and entropy
THERMOFLUIDS 3A11 (TMS 3A11)
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Available EnergyKey questions
What is the potential of producing work from some heatsource or energy supply?
How best can we optimise production of work in a heatengine?
Common sources of thermal energy include: Coal
Crude and Refined oil
Radioactive materials
Solar
Biomass
THERMOFLUIDS 3A11 (TMS 3A11)
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Available EnergyKey questions
THERMOFLUIDS 3A11 (TMS 3A11)
To answer the first question, What is the potential of producing work fromsome heat source or energy supply?, recall that maximum work can beobtained in a heat engine that is completely reversible i.e. a (Sadi)
CARNOT cycle heat engine.
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Available EnergyKey questions
THERMOFLUIDS 3A11 (TMS 3A11)
Steps in the Carnot Engine Cycle
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Available EnergyKey questions
THERMOFLUIDS 3A11 (TMS 3A11)
Represented on a T-s diagram we have the configuration shown below.
0.. QQW EHrev
Work done, from first law:
From 2ND Law we already
know that:
T
QTQ
T
Q
T
Q 00
0
0
T
T1QW 0H.E.rev
Thus:
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Available EnergyKey questions
THERMOFLUIDS 3A11 (TMS 3A11)
Thus the available portion of Q i.e. the energy that can be
converted into work is:
T
T1 0
To maximise available energy:
1. To must be as low as possible (usually ambient temperature)
2. T as high as possible (subject to practical limitations)
The potion of energy that cannot be converted into useful work and
is rejected to the environment is referred to as the unavailable
energy.
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Available EnergyKey questions
THERMOFLUIDS 3A11 (TMS 3A11)
In cases where heat transfer is through constant pressure systems
typical of most heat exchangers (e.g. boilers and condensers), T-s
diagram changes to that shown below.
In this case computation of the energy transfer involves an integral.
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Available EnergyKey questions
THERMOFLUIDS 3A11 (TMS 3A11)
Thus from the second law the change in entropy is given by:
0
0rev
T
Q
T
Q
T
QS
Hence low temperature heat transfer for a heat engine is given by:
STQ oo
Meaning that the reversible work for such a process is given by:
STQW 0H.E.rev
where the unavailable portion of work is ToS.
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Consider a steady state process as shown below:
Assume the following:
1. Fluid enters c.v. at state i
2. Fluid exits at state e
3. Amount of heat received by
the fluid in c.v. is q per unit
mass from heat source at TH
4. Fluid does work w per unit
mass
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
From the first law and assuming that the potential and kinetic energies
are negligible we have energy balance as:
whhq ie
From the second law, the entropy balance becomes:
0sT
qss
dt
dS
m
1gen
Hie
net
To answer the second question, How best can we optimise production of work in a heat engine? one requires to know quantitatively, the level of irreversibility of a process.
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
First consider a reversible process with the same inlet and exit conditions
as the irreversible process. In this process we note that:
a. The entropy change is zero.
b. In terms of the entropy balance equation, it follows that there is an
extra negative term to bring the positive generated entropy to zero.
c. This negative term can only be a result of heat exchange with the
environment
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Thus the entropy balance for the reversible process is:
0T
q
T
qss
dt
dS
m
1
o
revo
Hie
net
It follows that:
H
oieo
revo
T
TqssTq
But from energy balance (1St Law) we have:
ierevo
rev hh)q(qw
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Substituting for the heat transfer terms gives:
H
0ieie0
rev
T
T1qhhssTw
This work represents the theoretical upper limit for the work per
unit mass flow that could be produced by a control volume
undergoing a steady-state process from state i to state e in which
heat q is transferred at TH, with all the processes occurring in an
environment at temperature To.
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
The work done by a real process operating between the same states
is:
)h(h-qw ie
Because of irreversibilities (internal and external), the real work will be
less than the reversible work.
Thus we can define a quantity of irreversibility per unit mass
measured as the difference between the ideal and real work i.e.
irreversibility
wwi rev
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Significance of irreversibility:
1. Measures inefficiencies in an actual process
2. To improve actual work output, focus on inefficiencies or
irreversibilities
3. For a completely reversible process, irreversibility is zero and non-
zero for real processes
4. The above derivation has assumed no kinetic and potential energies
which can be included into the energy balance equation if necessary
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IrreversibilityReal Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Substituting the appropriate terms we have:
gen0
net/real0
Hie0
H
0ie0
H
0ie0
H
0ie0
sT
dt
dS
m
1T
T
qssT
qT
TssT
wT
T1qwqssT
wT
T1qwqssTi
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IrreversibilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
Consider the control mass that changes from state 1 to state 2 and
receives an amount of heat 1Q2 from a reservoir at TH while producing
work 1W2 as shown in the figure below:
From the 1ST law, the energy balance neglecting kinetic and potential
energy changes, we have:
211221 WUUQ
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IrreversibilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
And the 2ND law, entropy balance gives:
As was the case for the steady state process, it is necessary to compare
the real and ideal reversible processes with the same change of state and
same heat transfer while doing work.
0S
T
QSS
SSS
2gen1
H
2112
surrCMnet/real
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IrreversibilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
For a reversible process, the 2ND law (entropy balance) is:
leading to:
0T
Q
T
QssS
o
revo
H
21ienet/rev
H
o21ieo
revo
T
TQssTQ
and the 1ST law (energy balance) for the reversible process is:
12revo21
rev21
QQW UU
which can be expressed as:
H
o211212o
rev21 T
T1QUUSSTW
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IrreversibilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
The difference between this quantity and the actual work done is a
measure of the level of irreversibility in the process i.e.:
which can be reduced to:
Again this work represents the theoretical upper limit for the work
that could be produced by a control mass undergoing a change of
state from 1 to 2 in which heat is transferred at TH, when the
process is taking place in an environment at temperature To.
21rev2121
WWI
2gen1o21 STI
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Availability
THERMOFLUIDS 3A11 (TMS 3A11)
Some of the key questions that we need answered when designing
power generation systems are:
1. What is the maximum work that can be done by a mass of a fluid at
a given state?
2. What final state will give the maximum reversible work?
To answer these questions, it is important to note the following:
a. When the mass comes into equilibrium with the environment, it is
no longer capable of doing any work.
b. If a mass at a given state undergoes a process that produces work
in a reversible way until it is in equilibrium with the environment
then it would have produced maximum work.
Control Mass
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AvailabilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
Availability is then defined as the ability of a mass at a given state to
produce maximum work.
The implications of equilibrium with the environment are:
Same temperature
Same pressure
Same chemical composition
No external forces acting on the system
Zero velocity
Zero potential energy
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AvailabilityControl Mass: Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Consider a mass going through a steady state process, maximum
reversible work is obtained when the mass entering a control volume at
a certain state exits in equilibrium with the environment.
Reversible work for a control volume is
H
0e0TOTei0TOTi
rev
T
T1qsThsThw
Availability due to the flow Contribution from
heat transfers
Reversible work will be maximum when he = ho.
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AvailabilityControl Mass: Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Flow availability is the maximum reversible work per unit mass without
the addition of heat. This is also termed exergy i.e.
o0002
0 gZsThgZV2
1sTh
State of mass at
entry
State of mass at exit
(equilibrium with the
environment)
Therefore, Reversible work (wrev) = decrease in flow availability
+ reversible work from heat
transfer at TH
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AvailabilityControl Mass: Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
As before irreversibility for a control volume is the difference between
the reversible work and the actual work i.e.
In summary,
Irreversibility = decrease in availability of mass flows
+ decrease in availability of each heat transfer
- increase of availability of the surroundings
receiving the actual work.
CVjCVj
0eeiiCV WQ
T
T1mmI
The summations are included to take into account multiple flow streams
into and out of the control volume and multiple heat transfers.
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AvailabilityControl Mass: Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Rate of Irreversibility = Rate of destruction of availability
~ Rate of entropy increase
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AvailabilityControl Mass: Steady State Process
THERMOFLUIDS 3A11 (TMS 3A11)
Now consider a control mass in which the volume can change and some
work can be exchanged with the environment but such work would not be
available as useful work.
H
021202101
rev21
T
T1qsTesTew
Reversible work including kinetic and potential energy contributions is
This work is the maximum amount of reversible work that can be
obtained between the two given states.
The highest value is obtained when the final state is in equilibrium with
the environment, i.e. e2 = eo = uo + gZo, KE = 0 and s2 = so
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AvailabilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
Work against the surroundings is
And the maximum available work is
01100surr vvPvvPw o
H
021000000
surrrevmax
maxavail
T
T1qvvPsTesTe
www
The above equation gives the maximum available work at a
given state with 1Q2 available from source at TH.
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AvailabilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
The nonflow availability is defined as the maximum available
work from a state without the heat transfers included i.e.
0000000
000000
sTvPesTvPe
vvPsTesTe
Initial State of
Mass
And the irreversibility of a control mass is given by
120
act21j
j
02121 VVPWQ
T
T1mI
Final State of Mass
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AvailabilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
Thus the irreversibility is equal to the decrease inavailability of the CM plus the decrease in availability of the
heat transfers at reservoirs minus the increase in availability
of the surroundings that received the actual work
The less irreversibility associated with a given change ofstate, the greater the amount of work that will be done (or the
smaller amount of work that will be required)
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AvailabilityControl Mass
THERMOFLUIDS 3A11 (TMS 3A11)
Key Conclusions
Availability is one of our natural resources
This availability is available in oil, coal, natural gas,uranium and other natural reserves
The greater the irreversibilities in our processes, thegreater the decrease in availability
To minimise cost of producing work requires minimisingirreversibilities in our processes.
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
Previous definitions of process efficiency are all first-law efficiencies
(i.e. a comparison of two energy quantities). They were comparisons of
ideal performance and actual performance based on same initial state
but different exit states. Second Law efficiency would be the actual
work output of the process divided by the decrease in availability from
the same inlet state to the same exit state. The example of a turbine is
ei
alaw2nd
w
For devices not involving the production/input of work, the definition of
a second-law efficiency refers to the accomplishment of the goal of the
process relative to the process input in terms of availability changes or
transfers.
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
In the case of a heat exchanger we would have:
In general we would have:
433
121law2nd
m
m
source
CVsource
source
wantedlaw2nd
I
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Availability & 2ND Law EfficiencyExergy Balance
THERMOFLUIDS 3A11 (TMS 3A11)
Exergy is the concept unifying flow and non-flow availability. Consider
exergy as the maximum work available at a given state of mass i.e.
From the derivation of the control mass availability we have
)35.10.(ssmTvvmPeemm
ooooo Eq
m
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
Rate of change of exergy noting that the final state is constant, we
have:
(Eq10.38)n destructioExergyST
flowbyTransfermm
workdaryshaft/bounbyTransferdt
dVPW
TatheatbyTranferQT
T-1
dt
d
geno
eeii
oc.v.
c.v.o
Therefore,
Rate of Exergy storage = Transfer by heat + Transfer by
shaft/boundary work + Transfer by
flow Exergy destruction
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
For this case there is no build up of mass, energy, entropy or exergy
and hence the balance equation becomes:
Expressed in terms of intensive properties yields:
gen0eeiiCV0
CV STmmQT
T1W
gen0eiCV
0 sTqT
T1w
Noting the definition of availability terms leads to:
gen0rev
gen0ei0eTOTiTOTCV0
sTw
sTssThhqT
T1w
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
For a control mass going through a process starting at state 1 and
ending at state 2:
With a single heat transfer taking place at TH, the available work is
i.e. the work is equal to the exergy from heat transfer, lowered by the
increase in the stored exergy and the amount of exergy destroyed in
the process.
2
1gen21012021CV
012 STVVPWdtQ
T
T1
gen2101221H
012021 STQ
T
T1VVPW
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
Example
Steam enters a well insulated turbine at 800oC and 10 MPa at a flow
rate of 2.5 kg/s. The steam exits at 50 kPa. The isentropic efficiency of
the turbine is 0.9332. Assuming a reference environment at 25oC,
determine the rate at which availability enters the turbine, including that
associated with flow work and the availability destruction rate from an
availability balance.
Solution
i
e
Tw
T
s
i
e
10 MPa
50 kPa
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
Assumptions:
i. Steady-state and steady flow
ii. Adiabatic process (Qcv = 0)
iii. Negligible kinetic and potential energies
Analysis
Known parameters: Ti, Pi, Pe, m, isent,tTo determine: Availability inflow and availability destroyed
Availability inflow is given by:
)22.10(gZsThgZV2
1sTh o000
2
0
If kinetic and potential energies are negligible, we have
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Availability & 2ND Law Efficiency2ND Law Efficiency
THERMOFLUIDS 3A11 (TMS 3A11)
0i0oi s-sThhm
If kinetic and potential energies are negligible, we have
Getting properties from the steam tables we have:
MW4.78
Watts0.3672-7.4077298104.924114.82.5
To obtain the rate of availability destruction we use:
kW214
kW7.40777.6952298(2.5)
ssThhmWST eioeicvgenodestroyed