Thermofluids 3a - Chapter 10 - 2015

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)

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




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.



Steps in the Carnot Engine Cycle



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:



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.



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.



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.

IrreversibilityReal Steady State Process


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



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.



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



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



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.



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



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



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

IrreversibilityControl Mass


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



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



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



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

Availability


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

AvailabilityControl Mass


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

AvailabilityControl Mass: Steady State Process


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.



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



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.



Rate of Irreversibility = Rate of destruction of availability

~ Rate of entropy increase



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



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.



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



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)



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.

Availability & 2ND Law Efficiency2ND Law Efficiency


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.



In the case of a heat exchanger we would have:

In general we would have:

433

121law2nd

m

m

source

CVsource

source

wantedlaw2nd

I

Availability & 2ND Law EfficiencyExergy Balance


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



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



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



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



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



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



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

Thermofluids 3a - Chapter 10 - 2015

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Transcript of Thermofluids 3a - Chapter 10 - 2015