EGR 334 ThermodynamicsChapter 6: Sections 6-8

Lecture 25: Entropy and closed system analysis Quiz Today?

Today’s main concepts:• Heat transfer of an internally reversible process

can be represented as an area on a T-s diagram.• Learn how to evaluate the entropy balance for a

closed system

Reading Assignment:

Homework Assignment:

Read Chapter 6, Sections 9-10

Problems from Chap 6: 36, 38, 59, 66

3

Energy Balance:

Recall from last time:

syssys sys

dEQ W

dt

sysdE

dt

Q W

Entropy Balance:

sys gen

QS

T

Entropy Rate Balance:

sysgen

dS Q

dt T

sys sys sysE Q W

Energy Rate Balance:

sysdS

dt

Q

T

gen

Entropy Balance for Closed Systems

• Unlike mass and energy balances, the entropy balance doesn’t represent a conserved quantity.

= 0 (no irreversibilities present within the system)> 0 (irreversibilities present within the system)< 0 (impossible)

s:

where the subscript b indicates the integral is evaluated at the system boundary.

b

QS

T

change of entropy

entropy transfer

entropy production

Change of Entropy of the System

• If two or more properties of the end states of a process are known, then the Change of Entropy per unit mass, ∆s, is completely defined and discernible.

b

QS

T

As seen in the previous lecture, ∆S = S2-S1, represents a difference of state properties and may be evaluated by -- looking up s values on substance property tables.-- applying Tds relationships for ideal gas

Entropy Transfer

• Consider the Entropy Transfer term of the Entropy Balance. On a differential basis it can expressed

• This expression indicates that when a closed system undergoing an internally reversible process receives energy by heat transfer, the system experiences an increase in entropy. Conversely, when energy is removed by heat transfer, the entropy of the system decreases. From these considerations, we say that entropy transfer accompanies heat transfer. The direction of the entropy transfer is the same as the heat transfer.

Entropy and Heat Transfer

which can be integrated from state 1 to state 2,

►Rearranging the differential expression gives

►In an internally reversible, adiabatic process (no heat transfer), entropy remains constant. Such a constant-entropy process is called an isentropic process.

2

int

1rev

Q TdS

Entropy and Heat Transfer

An energy transfer by heat to a closed system during an internally reversible process is represented by an area on a temperature-entropy diagram:

2

int

1

Area under T-Srev

Q TdS

Consider how this integral would represented on a T-S diagram:

96.6&7 : Entropy and Closed Systems

For the Carnot cycle:

Carnot Work:

PdVW

Carnot Heat Transfer:

Q TdS

In the Carnot Cycle, only the constant temperature processes contribute to Heat (and Entropy) transfer.

What does the area represent?What does the area represent?

106.6&7 : Entropy and Closed Systems

For the Carnot cycle:

int rev.

QdS

T

TdSQ

3

23

2

Q TdS

1

41

4

Q TdS

12 23 34 41Q Q Q Q Q (note that Q41 is negative)

00

Entropy Balance for Closed Systems

That s has a value of zero when there are no internal irreversibilities and is positive when irreversibilities are present within the system leads to the interpretation that s accounts for entropy produced (or generated) within the system by action of irreversibilities.

Expressed in words, the entropy balance ischange in the amount of entropy contained

within the system during sometime interval

net amount ofentropy transferred in

across the system boundaryaccompanying heat transferduring some time interval

amount ofentropy produced

within the system during sometime interval

+

rev. intT

QS

A general approach to analyze a Closed System with entropy balance.

•Step 1: Identify properties at each state including T, p, v, u, x, and s.

•Step 2: Apply 1st law and attempt to evaluate ∆U, Q, and W for each process.

• Step 3: Write the 2nd Law (Entropy balance) and attempt to evaluate ∆S and ΣQ/T for each process. Usually σ will be determined from σ = ∆S - ΣQ/T

•Step 4: Explain the significance of σ.

Since the expansion occurs adiabatically,

Q and entropy balance reduces to give

Example: One kg of water vapor contained within a piston-cylinder assembly, initially at 5 bar, 400oC, undergoes an adiabatic expansion to a state where pressure is 1 bar and the temperature is (a) 200oC, (b) 100oC. Using the entropy balance, determine the nature of the process in each case.

b12

T

QSS

1

2

0

→ m(s2 – s1) = s

Find property values: using m = 1 kg and Table A-4 get s1 = 7.7938 kJ/kg∙K.

Boundary

Example 1 continued:

(a) At p = 1 bar and T = 200 deg C.

Table A-4 gives, s2 = 7.8343 kJ/kg∙K.

(Since s is positive, irreversibilities are present

within the system during expansion )

(b) At p = 1 bar and T = 100 deg C.

Table A-4 gives, s2 = 7.3614 kJ/kg∙K.

(Since s is negative, expansion (b) is impossible.

It cannot occur adiabatically.)

then s = (1 kg)(7.8343 – 7.7938) kJ/kg∙K = 0.0405 kJ/K

= s m(s2 – s1)

then s = (1 kg)(7.3614 – 7.7938) kJ/kg∙K = –0.4324 kJ/K

= s m(s2 – s1)

Example 1 continued:

► Since s cannot be negative and

► For expansion in (part b) DS is negative, then

► By inspection the integral must be negative and so heat transfer from the system must occur in expansion (b).

Just a little more analysis of part b) The result of part b was that s is negative. The process cannot occur adiabatically.

= +< 0 ≥ 0< 0

Entropy Rate Balance for Closed Systems

Some problems are presented in the form of a closed system entropy rate balance given by

dt

dS

where

the time rate of change of the entropy of the system

j

j

T

Qthe time rate of entropy transfer through the portion of the boundary whose temperature is Tjtime rate of entropy production due to irreversibilities within the system

Example 2: An inventor claims that the device shown generates electricity at a rate of 100 kJ/s while receiving a heat transfer of energy at a rate of 250 kJ/s at a temperature of 500 K, receiving a second heat transfer at a rate of 350 kJ/s at 700 K, and discharging energy by heat transfer at a rate of 500 kJ/s at a temperature of 1000 K. Each heat transfer is positive in the direction of the accompanying arrow. For operation at steady state, evaluate this claim.

kJ/s 3502 Q

kJ/s 2501 Q

+

–

T1 = 500 K

T2 = 700 K

T3 = 1000 K kJ/s 5003 Q

kJ/s 3502 Q

kJ/s 2501 Q

+

–

T1 = 500 K

T2 = 700 K

T3 = 1000 K kJ/s 5003 Q

Applying an entropy rate balance at steady state

Example 2 continued:

kJ/s 100 kJ/s 500kJ/s 350kJ/s 250eW

eWQQQdt

dE 3210

3

3

2

2

1

10T

Q

T

Q

T

Q

dt

dS

Applying an energy rate balance

at steady stateSolving

The claim is in accord with the first law of thermodynamics.

Solving

K

kJ/s0.5

K

kJ/s5.05.05.0

K 1000

kJ/s 500

K 700

kJ/s 350

K 500

kJ/s 250

Since σ is negative, the claim is not in accord with the 2nd Law of Thermodynamics and is therefore denied.

∙

0

0

19

T

s

12

Example 3 (6.14): One kilogram of water contained in a piston-cylinder assembly, initially at 160°C, 150 kPa undergoes an isothermal compression process to saturated liquid. For the process, W= –471.5 kJ. Determine for the process.(a) Sketch the process on a T-s diagram.(b) The heat transfer, in kJ(c) The change in entropy in kJ/K(d) The entropy generated in the process.

State 1: T1= 160 oC p1 = 150 kPa = 1.50 bar super heated vapor

State 2: T2= T1 =160 oC sat. liquid.

T

s

20

Example 3 (6.14): One kilogram of water contained in a piston-cylinder assembly, initially at 160°C,150 kPa undergoes an isothermal expansion from compression process to saturated liquid. For the process, W= –471.5 kJ. b) find the heat transfer

Look up the rest of the state properties.

U Q W

WuumQ 12

State 1 2

T (°C) 160 160

p (kPa) 150 617.8

x SH 0

u (kJ/kg 2595.2

674.86

s (kJ/kg K)

7.4665

1.9427

1 674.86 2595.2 / ( 471.5 ) 2391.84 /Q kg kJ kg kJ kJ K (c) To find the change in entropy

1 1.9427 7.4665 / 5.5238kJ

S kg kJ kg KkgK

2 1( )S m s s

State 1 2

T (°C) 160 160

p (kPa) 1.50

x SH 0

u (kJ/kg

s (kJ/kg K)

Then apply the 1st Law.

21

Example 3 (6.14): One kilogram of water contained in a piston-cylinder assembly, initially at 160°C,150 kPa undergoes an isothermal compression process to saturated liquid. For the process, W= –471.5 kJ. Determine for the process.

(d) To find the entropy generation

QS

T

2 1 5.5238 /S m s s kJ kg K

State 1 2

T (°C) 160 160

p (kPa) 150 617.8

x SH 0

u (kJ/kg 2595.2

674.86

s (kJ/kg K)

7.4665

1.9427

2391.845.5239 /

(160 273)

Q kJkJ K

T K

What does this mean? Process had no irreversibilities

where

5.5238 ( 5.5239) 0.0001 0 /Q

S kJ KT

226.11 : Isentropic Processes

States may be given as having the same entropy (two-phase, saturated vapor, superheated vapor)

Any process where the entropy does not change is called isentropic.

23

Consider isentropic process for an ideal gas

6.11 : Isentropic Processes

1

2lnv

vR

T

dTTcs V

2

1

lnPc T dT ps R

T p

1

k

kRcP

1

k

RcV

1

2ln1 v

vR

T

dT

k

Rs

1

2

1

2 lnln1

1

v

v

T

T

k 11

1

2

2

1

k

v

v

T

T

2

1

ln1

pkR dTs R

k T p

2 2

1 1

ln ln1

T pk

k T p

( 1)

2 2

1 1

k kT p

T p

and

2 1

1 2

kp v

p v

thus

1 1 2 2k kp v p v

An isentropic process is a type of polytropic process

thus

Starting with

and

Next with

and

24

Example (6.27): Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from State 1 where T1 = 290 K, p1 = 1 bar.

Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant

Process 2 – 3 : Isentropic expansion to p3 = 1 bar.

(a) Sketch the two processes on T-s coordinates(b) Determine the temperature at State 2 in K(c) Determine the net work in kJ

Air

T

S

1 bar

2

1

3

5 bar

290 K

T

S

25

Example (6.27): Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from state 1 where T1 = 290 K, p1 = 1 bar.

Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constantProcess 2 – 3 : Isentropic expansion to p3 = 1 bar.

(a) Sketch the two processes on T-s coordinates(b) Determine the temperature at state 2 in K(c) Determine the net work in kJ

1 1.19 1 1.19

22 1

1

5290 375

1

n np

T T K Kp

Process 1-2:

1 1 2 2N Np V p V

1 1 2 2

1 2

p V p V

T T

polytropic: ideal gas:1/

2 1

1 2

NV p

V p

2 1 2

1 2 2

V p T

V p T

1/

1 1 2

2 2 2

Np p T

p p T

26

Example (6.27):Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant

Process 2 – 3 : Isentropic expansion to p3 = 1 bar.

Find the state properties State 1 2 3

T (K) 290 375

p (bar) 1 5 1

u (kJ/kg) 206.91 268.075

s °(kJ/kg K) 1.66802 1.92657

State 1: p1 = 1 bar T1 = 290 K

State 2: p2 = 5 bar T2 = 375 K

State 3: p3 = 1 bar s3 = s2

From Table A-22: at T = 290 K: u = 206.91 so = 1.66802From Table A-22: at T = 375 K: u = 268.075 so = 1.92657

State 1 2 3

T (K) 290 375

p (bar) 1 5 1

u (kJ/kg)

s °(kJ/kg K)

27

Example (6.27):Process 2 – 3 : Isentropic expansion to p3 = 1 bar.

State 1 2 3

T (K) 290 375 236.83

p (bar) 1 5 1

u (kJ/kg) 206.91 268.075 168.86

s °(kJ/kg K) 1.66802 1.92657 1.46466

At State 3:

33 2 3 2

2

lnp

s s s T s T Rp

33 2

2

lnp

s T s T Rp

3

1(1.92657 / ) (0.2870 / ) ln 1.46466 /

5

bars T kJ kg K kJ kg K kJ kg K

bar

33 2

2

0 lnp

s T s T Rp

From Table A-22: for so=1.46466 find T and u T3 = 236.82 K and u3 = 168.86 kJ/kg

State 1 2 3

T (K) 290 375

p (bar) 1 5 1

u (kJ/kg) 206.91 268.075

s °(kJ/kg K) 1.66802 1.92657

28

Example (6.27): (c) Determine the net work in kJ

State 1 2 3

T (K) 290 375 636.82

p (bar) 1 5 1

u (kJ/kg) 206.91 268.075 168.86

s °(kJ/kg K) 1.66802 1.92657 1.46466

2 12 2 1 112 1 1

mR T Tp V p VW pdV

n n

12

0.287 / (375 290)

1 1.19

kJ kg K KW

m

Process ΔU Q W

1 – 2 +61.2

-67.2

-128.4

2 – 3 Process 1 – 2:

12 12 12U Q W

12 2 1( )U m u u

12 268.08 206.91U

m

61.17 /kJ kg 128.39 /kJ kg

12 12 12 61.17 ( 128.39) 67.22 /Q U W

kJ kgm m m

where

Process ΔU Q W

1 – 2

2 – 3

29

Example (6.27):State 1 2 3

T (K) 290 375 236.82

P (bar) 1 5 1

u (kJ/kg) 206.91 268.075

168.86

s °(kJ/kg K) 1.66802

1.92657

1.46468

3

23

2

0Q Tds

23 23 23 0 ( 99.22) 99.22 /W Q U

kg kgm m m

kg

kJ

kg

kJ

kg

kJ

m

W

m

W

m

Wnet 17.2922.9939.1282312

Process ΔU Q W

1 – 2 +61.2

-67.2

-128.4

2 – 3 -99.2 0 +99.2

Process 2-3:

23 23 23U Q W

23 3 2( )U m u u where

23 168.86 268.08 99.22 /U

kJ kgm

Net Work over both processes:

Process ΔU Q W

1 – 2 +61.2

-67.2

-128.4

2 – 3

306.8 : Directionality of Processes

Second Law statement: “It is impossible for a system to operate such that entropy is destroyed.”

This can be seen in the entropy balance, but first look at the energy balance. 0

system isolatedE

system environment0E E

Energy is neither created nor destroyed.

rev. intSystem Isolated T

QS

system environmentS S

31

End of Slides for Lecture 25