ideal gas entropy

8/9/2019 ideal gas entropy

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Unit 9 Ideal gas entropy November 2, 20

ME 370 - Thermodynamics

Unit NineUnit NineEntropyEntropy

Calculations for Ideal GasesCalculations for Ideal GasesMechanical Engineering 370

Thermodynamics

Larry Caretto

November 2, 2010

2

Outline

Quiz eight solution

Review Goals for unit nine

Calculating entropy with ideal gases

Constant and variable heat capacity

Isentropic calculations

Use of equations and ideal gas tables

Air tables versus tables with less data

3

Review

Entropy is a property

The maximum work is done in areversible process

With sign convention for work on an inputdevice this is the minimum work input

In an adiabatic reversible process themaximum work is done when s = 0 Find inlet/initial entropy at initial state

Final/outlet state determined by entropy

and another final/outlet property4

First Unit Nine Goal

As a result of studying this unit you

should be able to compute entropy

changes in ideal gases for any change

of state

using constant heat capacities

using equations that give heat capacities

as a function of temperature

using ideal gas tables

5

Second Unit Nine Goal

As a result of studying this unit you

should be able to compute the end

states of isentropic processes in ideal

gases using constant heat capacities

using equations that give heat capacities

as a function of temperature

using ideal gas tables

6

Derive Ideal Gas Entropy

Entropy defined as ds = (du + Pdv)/T

Multiply by T: Tds = du + Pdv

Tds Pdv = du = d(h Pv) = dh Pdv

vdP

Ideal gas: Pv = RT, du = cvdT, dh = cpdT

For ideal gases ds = cvdT/T + Rdv/v

For ideal gases ds = cpdT/T RdP/P

Integrate last two equations to get s

so Tds = dh vdP


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7

Ideal Gas Entropy Change

dS = cvdT/T + Rdv/v = cpdT/T RdP/P

Integrate between two states (1 and 2)

1

212 ln

2

1v

vR

T

dTcss

T

T

v

1

212 ln

2

1P

PR

T

dTcss

T

T

p

8

Handling cp(T) = cv(T) + R

Easiest case is constant heat capacity

If we have an equation for cp(T) or cv(T)

we can integrate cpdT/T or cvdT/T

We only need one equation since cp(T) and

cv(T) are related by cp(T) = cv(T) + R

Ideal gas tables give integral of cpdT/T

called so as described next

1

2

1

2

1

2

1

212 lnlnlnln

PPR

TTc

vvR

TTcss pv

9

Ideal Gas Entropy Tables

Define T

T

p

o

T

dTTcTs

0'

')'()(

1

0

2

0

2

1

'

')'(

'

')'(

'

')'()()( 12

T

T

p

T

T

p

T

T

poo

T

dTTc

T

dTTc

T

dTTcTsTs So that

1

2

1212

ln)()(P

PRTsTsss

oo

10

Example

Air is heated from 300 K to 500 K at

constant pressure. What is s?

Compute result using air tables and

repeat with constant heat capacity

1

2

1

2

1

2

1

212 lnlnlnln

P

PR

T

Tc

v

vR

T

Tcss

pv

21

1212

1

212 ln)()(ln)()(

Tv

TvRTsTs

P

PRTsTss oooo

11

Example Answers

Air is heated from 300 K to 500 K at

constant pressure. What is s?

From table A-17, page 936, so(300 K) =

1.70203 kJ/kgK and so(500 K) =2.21952 kJ/kgK; s = 0.51749 kJ/kgK

Assuming constant cp = 1.005 kJ/kgKgives s = cpln(T2/T1) =1.005 ln(500/300) = 0.51338 kJ/kgK

12

Second Example

Air is heated from 300 K, 100 KPa to500 K, 200 KPa. What is s using theair tables?

kPakPa

KkgkJKsKss oo

100200ln287.0)300()500(

2ln287.070203.121952.2

Kkg

kJ

KkgKkg

kJs

Kkg

kJs

31856.0


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13

Third Example

Air is heated from 300 K, 100 KPa to500 K, 200 KPa. What is s assuming

constant heat capacity?

Kkg

kJ

Kkg

kJ

Kkg

kJs

31444.019893.51338.0

kPa

kPa

Kkg

kJ

K

K

Kkg

kJ

P

PR

T

Tcs p

100

200ln

287.0

300

500ln

005.1lnln

1

2

1

2

Kkg

kJs

31856.0 Air table result

14

Derive Isentropic Ideal Gas

Start with ds = cpdT/T RdP/P

For ds = 0, cpdT/T = RdP/P Integrate for ds = 0 and constant cp to

get cpln(T2/T1) = R ln(P2/P1) giving

ln(T2/T1) = ln(P2/P1)R/Cp or T2/T1 =

(P2/P1)R/Cp

R/cp = (cp cv)/cp = (cp/cv 1)/ (cp/cv)

R/cp = (k 1)/k, where k = cp/cv

15

Ideal Gas Isentropic Processes

Final result: T2/T1 = (P2/P1)(k-1)/k

T/P(k-1)/k = constant

Can derive similar equations for other

variables

T2/T1 = (v1/v2)(k-1) => Tvk-1 = constant

P2/P1 = (v1/v2)k => Pvk = constant

Apply only to ideal gases with constant

heat capacities in isentropic processes

16

Variable Heat Capacity

Set s2 s1 = 0 in equations below forideal gas isentropic process

0ln1

212

2

1

v

vR

T

dTcss

T

T

v

0ln1

212

2

1

P

PR

T

dTcss

T

T

p

17

Variable Heat Capacity II

What if we have an equation for cP(T) orcv(T) = cP(T) R?

12ln

2

1v

vR

T

dTc

T

Tv

1

2ln2

1 P

PR

T

dTc

T

T

p

Can solve for volume or pressure ratios if

T1 and T2 are given

A trial-and-error solution is required if T1or T2 are unknown

18

so for Isentropic Processes

s2 s1 = 0 for isentropic process

121212

ln)()( PPRTsTsss oo

1212 ln)()( PPRTsTs oo

Given T1 and P2/P1 find so(T1) in tables Compute so(T2) = s

o(T1) + R ln(P2/P1)

Interpolate in tables to find T2 that gives so

Need to solve by iteration if given v2/v1instead of P2/P1 Use so(T2) = s

o(T1) + R ln(v1T2/v2T1)


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19

Pr(T) for Isentropic Process

For s2 = s1 1212 ln)()( PPRTsTs oo

Solve for P2/P1and define Pr(T)= Aexp[so(T)/R] A is an arbitrary

constant thatcancels

1

2

)(

)()()(

1

2

1

212

1

2

12

ln)()(

TP

TP

e

ee

P

P

PP

RTsTs

r

r

R

Ts

R

Ts

R

TsTs

oo

o

o

oo

We can tabulate the function Pr(T) suchthat Pr(T2) = (P2/P1)Pr(T1) relates theend states for s2 = s1

20

vr(T) for Isentropic Process

2

1

2

2

1

1

1

2

2

1

2

1

2

1

1

2

1

2

1

2

Tv

Tv

TP

T

TP

T

TP

TP

T

T

v

v

v

v

T

T

P

P

TP

TP

r

r

r

r

r

r

r

r

We can tabulate the function Pr(T) suchthat Pr(T2) = (P2/P1)Pr(T1) relates the

end states for s2 = s1 If we know v2/v1 = (T2/T1)(P1/P2) can use

Tabulate function vr(T) such that vr(T2) =(v2/v1)vr(T1) gives end states for s2 = s1

21

Variable Heat Capacity Tables

For ideal gas, isentropic processes, with

variable heat capacities we can define

Pr(T), and vr(T) such that

v2/v1 = vr(T2)/ vr(T1)

P2/P1 = Pr(T2)/ Pr(T1)

Values of Pr(T), and vr(T) given in ideal

gas tables

We still use Pv = RT at individual data

points

Relates end

states such that

s(T2,P2) = s(T1,P1)

22

Example Problem

Adiabatic, steady-flow air compressor

used to compress 10 kg/s of air from

300 K, 100 kPa to 1 MPa. What is the

minimum work input?

Minimum work input for adiabatic

process is when process is isentropic

Steady first law for one inlet, one outlet,

negligible KE and PE:

)( inoutu hhmWQ

23

Example Continued

What is outlet state for maximum work?

Use ideal gas tables for air on page 934

Pr(300 K) = 1.3860

P2/P1 = Pr(T2)/ Pr(T1) so that Pr(T2) = Pr(T1) P2/P1 = 1.3860(1000/100)

What is T with Pr= 13.860

Pr(570 K) = 13.50; Pr(580 K) = 14.38

Interpolate to get Pr(T2) = 13.86 at T2 =

574.1 K

24

Example Concluded

Use enthalpy data from ideal gas tables

hin = h(300 K) = 300.19 kJ/kg

hout = h(574.1 K) = 579.86 kJ/kg (interpolated)

MWkJ

sMW

kg

kJ

kg

kJ

s

kg

hhmW outinu

797.21000

187.57919.30010

)(

Negative value shows work input

Work input of 2.797 MW is minimum


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25

Repeat with Constant cp

Here we use data for air at 300 K: k =

1.4, cp

= 1.005 kJ/kgK (page 911)

T2 = T1(P2/P1)(k 1)/k = (300 K)[(1000

kPa)/(100 kPa)](1.4 1)/1.4 = 579.2 K

MWkJ

sMWKK

Kkg

kJ

s

kg

TTcmhhmWoutinpoutinu

806.21000

12.579300

005.110

)()(

26

Class Exercise

Air in an adiabatic piston-cylinder device

with V1

= 0.1 m3, P1

= 50 kPa, and T1

=

300 K is compressed to V2 = 0.01 m3.

Find the minimum work input

Do the calculation for variable heat

capacity and repeat with constant cp Variable: v2/v1 = vr(T2)/ vr(T1) and w = u

Constant: T2/T1 = (v1/v2)(k-1) and w =cvT

27

Constant Heat Capacity Solution

State data: V1 = 0.1 m3, P1 = 50 kPa,

T1 = 300 K, V2 = 0.01 m3

First law: Q = U + W = m(u2 u1) + Wso W = m(u1 u2) for Q = 0

Second law: Maximum work inadiabatic process is when s = 0

Properties:Air as ideal gas withconstant heat capacity (cp = 1.005kJ/kgK, cv = 0.718 kJ/kgK, R = 0.287

kJ/kgK,k = 1.4 at 300K, p 911)28

Constant Heat Capacity Solution II

Ideal gas equations: PV = mRT, du =

cvdT, dh= cpdT; for s = 0 with constantheat capacity, T2 = T1(v1/v2)

k-1

Solution:

Km

mK

v

vTT

k

6.75301.0

1.0)300(

14.1

3

31

1

212

1

2

2121

T

T

vv TTmcdTcmuumW

29

Constant Heat Capacity Solution III

Solution concluded:

kJKKKkg

kJkg

TTmcdTcmuumW

T

T

vv

9.186.753300718.0

05807.0

1

2

2121

kg

K

Kkg

mkPa

mkPa

RT

VPm 05807.0

300287.0

1.0503

3

1

11

Minimum work input is 18.9 kJ

30

Variable Heat Capacity Solution

State data: V1 = 0.1 m3, P1 = 50 kPa,

T1 = 300 K, V2 = 0.01 m3

First law: Q = U + W = m(u2 u1) + W

so W = m(u1 u2) for Q = 0 Second law: Maximum work in

adiabatic process is when s = 0

Properties: Use ideal gas tables for air

on page 936


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31

Variable Heat Capacity Solution II

Ideal gas equations:

PV = mRT,

u = u(T2) u(T1)

h = h(T2) h(T1)

For s = 0, vr(T2) = vr(T1)(v2/v1) = vr(T1)(V2/V1)

Solution:At T1 = 300 K vr= 621.2

vr(T2) = vr(T1)(V1/V2) = 621.2(0.01 m3) / (0.1

m3) = 62.12

In tables vr(730 K) = 62.13 so T2 = 730 K

2121

TuTum

uumW

32

Variable Heat Capacity Solution III

Solution concluded:

kJkg

kJ

kg

kJkgTuTumW

7.18

07.53607.21405807.021

kgK

Kkg

mkPa

mkPa

RT

VPm 05807.0

300287.0

1.0503

3

1

11

Minimum work input is 18.7 kJ

Air tables: u(T1 = 300 K) = 214.07 kJ/kgK

and u(T2 = 730 K) = 536.07 kJ/kgK and

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