The Second Law of ThermodynamicsEntropy and Work

Chapter 7c

Work Done during a ProcessIn Chapter 4 we found the work done

in a closed system due to moving boundaries and expressed it in terms of the fluid properties

In the processes describing the behavior of many engineering devices there are no moving boundaries

2

1PdvWb

For example, a steam turbine does not have any moving boundaries

Steam Turbine Work

Steam turbines do produce work

Work Done During a ProcessIt would be useful to be able to

express the work done during a steady flow process, in terms of system properties

Recall that steady flow systems work best when they have no irreversibilities

dpedkedhwq outrevinrev ,, Energy Balance for a steady flow device

Tdsqrev

vdPdhTds

dpedkevdPw outrev ,

,

e

rev out iw vdP ke pe

Often the change in kinetic energy and potential energy is 0

“All” we have to do now is integrate!!

,

e

rev out iw vdP

In order to integrate, we need to know the relationship between v and P

For solids and liquids

v is constant

,rev out e iw v P P ke pe

,

e

rev out iw vdP ke pe

Steady flow of a liquid through a pipe or a nozzle

,rev out e iw v P P ke pe There is no work!!

2 2

02

e ie i e i

V Vv P P g z z

Bernoulli’s equation

Steady Flow of a Liquid through a pump or a turbine

,rev out e iw v P P

Note that the work term is smallest when v is small, so for a pump (which uses work) you want v to be small. For a turbine (which produces work) you want v to be big.

Or..

,rev in e iw v P P

Compressor Work

,

e

rev in iw vdP

We integrated this equation for v = constant, which is good for liquids – but what about gases?Consider an ideal gas, at constant TP

RTv

, ln erev in

i

Pw RTP

Remember, this is only true for the isothermal case, for an ideal gas

Compressor Work

,

e

rev in iw vdP

Another special case is isentropic

We derived the isentropic relationships earlier in this chapterCPvk

kk PCv 11 Rearrange to find v, plug in and integrate

1 11

1 1

, 11

k kk e i

rev ink

P Pw C

Now its “just” algebra, to rearrange into a more useful form

1 1 1 1

, 11

k k k ke e i i

rev ink

C P P C P Pw

, 1 11 1 1

e i e ie e i irev in

k k

R T T kR T Tv P v Pwk

vPC

CPvkk

k

11

1 11

1 1

, 11

k kk e i

rev ink

P Pw C

, 1i e i

rev ini i

kRT T Tw T Tk

k

k

PP

TT

1

1

2

1

2

1

, 11

kk

i erev in

i

kRT Pw

k P

Remember, this equation only applies to the isentropic case, for an ideal gas, assuming constant specific heats

Compressor Work

,

e

rev in iw vdP

Another special case is polytropic

Back in Chapter 4 we said that in a polytropic process Pvn is a constantCPvn

This is exactly the same as the isentropic case, but with n instead of k!!

, 1 11 1 1

e i e ie e i irev in

n n

R T T nR T Tv P v Pwn

1

, 11

nn

i erev in

i

nRT Pw

n P

The area to the left of each line represents the work, vdP

Note, it takes less work for an isothermal process

You should compress isothermallycompress isothermally, and you should use an isentropicisentropic process in a turbineturbine!!

Pv Diagram for Isentropic, Polytropic and Isothermal compression, for the same final and initial pressures

Inlet

Exit

How do you keep a compression process isothermal?

The gas will heat up as it is compressed, so it needs to be cooled

Intercooling is difficultInstead, multistage compression is

more common, with cooling between steps

Two stage Compressor

How do you decide how to break up the compression load?You save the most work by

intercooling, when each compressor carries the same load

1

, 11

kk

i erev in

i

kRT Pwk P

Since you cool back to T1 between stages, the only things that change in this equation are the P’s

1

, 11

kk

i erev in

i

kRT Pw

k P

For the work done by each stage to be equal, the pressure ratio must be equal

1

11

kk

i xstage

i

kRT Pw

k P

1

11

kk

i e

x

kRT Pk P

e x

x i

P PP P

Or…x e iP P P

Isentropic Efficiency of Steady Flow Devices

Efficiency We’d like a measure of efficiency to

compare real devices to the best we can do

There are always irreversibilities that downgrade performance

Most steady flow devices are intended to operate under adiabatic conditions

If a device is reversible and adiabatic, it is isentropic

Real devices are never really isentropic

Isentropic EfficiencyLets compare how well real devices

work to how well comparable isentropic devices workSame inlet conditionsSame outlet conditionsTurbine, Compressor and Nozzle

actual result isentropic resultII

Turbines

work turbineisentropic work turbineactual

Turbine

s

aTurbine w

w

esi

eaiTurbine hh

hh

ei hhw

pekehmWQ

Remember, the work done in a turbine can be found from the energy balance

Isentropic Efficiencies of Compressors and PumpsRatio of the work required to raise the

pressure of a gas to a specified value, in a isentropic manner, to the actual work

a

scompressor w

w

Note that this equation is arranged so that it is always less than one!!

eai

esiCompressor hh

hh

ei hhw

pekehmWQ

Remember, the work done by a compressor can be found from the energy balance

eai

esiCompressor hh

hh

Applies to both gases and liquids

)( ies PPvw Isentropic work for a liquid

eai

ieCompressor hh

PPv

)( Only applies

to a liquid

Sometimes compressors are cooled intentionally – Why?

Cooling reduces the specific volume, resulting in less work required for compression

For compressors that are intentionally cooled, the isothermal model is more realistic

a

tCompressor w

w

Isentropic Efficiency of NozzlesThe objective of a nozzle is to increase

the kinetic energy of the gasUsually, the inlet velocity is low

enough that we can consider it to have zero kinetic energy

22eaV 22

esV

keh

2

2

22

22

es

ea

ies

iea

s

aNozzle

V

V

VV

VVkeke

esi

eaiNozzle hh

hh

Entropy Balance

Entropy Balance

There is no such thing as the conservation of entropy

Entropy of the universe always increases

Real processes always generate entropy

system theofentropy total

in the Change

GeneratedEntropyTotal

OutEntropyTotal

InEntropyTotal

systemgenoutin SSSS

Often called the entropy balance

systemgenoutin SSSS

12 SS

Remember, entropy is a state function. It doesn’t change unless the state of the system changes!!!

How Does Entropy Enter and Leave a System?Heat Transfer

S=Q/T (T=constant) If T is not constant S=Qk/Tk

There is no entropy transfer with work!!

Mass FlowSmass=ms

Entropy Generation

Closed Systems

12 SSSTQ

genk

kThere is no mass transfer in a closed system

In an isolated system there is no heat transfer, so the equation becomes…

12 SSSSgen

The Universe is an Isolated System

gssurroundinsystemgen SSSS

Control Volumes

cvgeneeiik

k SSsmsmTQ

cvgeneeiik

k SSsmsmTQ

Simplify this equation, depending on the process

Surface area for heat transfer is 30 m^2Thermal Conductivity is 0.69 W/(m C)

Inside

T= 27 C

Outside

T=0 C

Inside Surface T = 20 C

Outside Surface T = 5 C

Consider Example 7-17Entropy Generation in a Wall

Determine the rate of heat transfer through the wall

W1035

xTkAQ

Determine the rate of entropy generation in the wall

systemgenoutin SSSS 0

The state of the system does not change with time

0

genout

out

in

in STQ

TQ

0K 278 W1035

K 293 W1035

genS

W/K191.0genS

Determine the rate of entropy generation for the process

systemgenoutin SSSS 0

Consider an extended system – wall plus the surrounding air

0

genout

out

in

in STQ

TQ

0K 273 W1035

K 300 W1035

genS

W/K341.0genS

Where is entropy generated?

Reducing the Cost of Compressed AirSkimRepair Air LeaksInstall High Efficiency MotorsUse a small motor at high capacity,

instead of a large motor at low capacity

Use outside air for compressor intakeReduce the air pressure setting

SummarySteady Flow work for a reversible process

,

e

rev out iw vdP ke pe

SummaryFor incompressible substances

The work term is smallest when v is small, so for a pump (which uses work) you want v to be small. For a turbine (which produces work) you want v to be big.

,rev out e iw v P P ke pe

Summary

•We looked at three special cases of the work equation• Isothermal• Isentropic• Polytropic

SummaryIsothermal Compression

, ln erev in

i

Pw RTP

SummaryIsentropic Compression

1

, 11

kk

i erev in

i

kRT Pwk P

, 11 1

e ie e i irev in

k

kR T Tv P v Pwk

SummaryPolytropic Compression

1

, 11

nn

i erev in

i

nRT Pwn P

, 11 1

e ie e i irev in

n

nR T Tv P v Pwn

Summary

•The work input to a compressor can be reduced by using multistage compression with intercooling. For maximum savings from the work input, the pressure ratio across each stage of the compressor must be the same.

Summary

•Most steady-flow devices operate under adiabatic conditions, and the ideal process for these devices is the isentropic process.

In the relations above, h2a and h2s are the enthalpy values at the exit state for actual and isentropic processes, respectively.

SummaryIsentropic or Adiabatic Efficiency

Actual turbine work wa h1 - h2aIsentropic turbine work ws h1 - h2s

= = =~

Isentropic compressor work ws h2s - h1Actual compressor work wa h2a - h1

= = =~

Actual KE at nozzle exit V2a h1 - h2aIsentropic KE at nozzle exit h1 - h2s2V2s

= = =~2

SummaryEntropy Balance

SummaryEntropy Balance – Rate form

SummaryEntropy Balance – Steady-flow