# EGR 334 Thermodynamics Chapter 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

### Homework Assignment:

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

Recall from last time: Energy Balance: 

E sys

Q sys

W sys

Energy Rate Balance:

dE sys dt

Q sys

W sys

Entropy Balance : 

S sys

 

Q T

 

gen

Entropy Rate Balance:

dS sys dt

 

Q T

 

gen Q T Q dE sys dt W dS sys dt

gen

3

### Entropy Balance for Closed Systems

change of entropy  

Q T b

entropy transfer   entropy production

where

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

b

:

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

= 0 > 0 < 0

(no irreversibilities present within the system) (irreversibilities present within the system) (impossible)

### Change of Entropy of the System

 

Q T

b

  As seen in the previous lecture, ∆S = S 2 -S 1 , 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 • 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 .

## 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

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

isentropic

process.

► Rearranging the differential expression gives which can be integrated from state 1 to state 2 ,

Q

int

rev

 2 1 

TdS

## Entropy and Heat Transfer

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

Q

int

rev

 2 1 

TdS

 

 An

during an

process is

### represented by an area on a temperature-entropy diagram

:

6.6&7 : Entropy and Closed Systems For the Carnot cycle: What does the area represent?

Carnot Work:

W

 

PdV

What does the area represent?

Carnot Heat Transfer:

Q

 

TdS

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

6.6&7 : Entropy and Closed Systems For the Carnot cycle:

Q

23  3 2 

TdS Q

Q dS

Q

 

T TdS Q

12 0 

Q

23 int rev.

Q

34 0 

Q

41

Q

41  (note that Q 41 is negative) 1 4 

TdS

10

Entropy Balance for Closed Systems

That  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  accounts for entropy produced (or generated) within the system by action of irreversibilities.

S

  

Q

 

T

int rev.

Expressed in words, the entropy balance is

change

in the amount of entropy contained within the system during some time interval net amount of entropy

transferred in

across the system boundary accompanying heat transfer during some time interval

+

amount of

entropy produced within

the system during some time interval

### Step 1

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

### Step 2

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

### Step 4

: Explain the significance of σ.

Example:

One kg of water vapor contained within a piston cylinder assembly, initially at 5 bar, 400 o C , undergoes an adiabatic expansion to a state where pressure is 1 bar and the temperature is

(a)

200 o C ,

(b)

100 o C . Using the entropy balance, determine the nature of the process in each case.

Since the expansion occurs adiabatically, Q  and entropy balance reduces to give

0

S

2 

S

1  

1 2

  

Q T

  b  

m(s 2 – s 1 ) =

 Boundary

Find property values: using m = 1 kg

and

Table A-4

get

s

1 = 7.7938 kJ/kg∙K

.

Example 1 continued:

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

Table A-4

then gives,

s

2 = 7.8343 kJ/kg∙K

.  

m(s 2

s 1 )

= (1 kg)(7.8343 – 7.7938) kJ/kg∙K =

0.0405 kJ/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

then gives,

s

2 = 7.3614 kJ/kg∙K

 

m(s 2

s 1 )

= (1 kg)(7.3614 – 7.7938) kJ/kg∙K .

= –0.4324 kJ/K

(

Since s is negative, expansion (b) is impossible. It cannot occur adiabatically.)

Example 1 continued: Just a little more analysis of part b) The result of part b was that

is negative . The process cannot occur

.

► ► ► Since 

< 0 = < 0 + ≥ 0

cannot be negative and For expansion in (part b ) 

S

is negative, then By inspection the integral must be negative and so heat transfer

from

the system

must

occur in expansion (b) .

Entropy Rate Balance for Closed Systems

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

where

dS dt

the time rate of change of the entropy of the system

T j j

the time rate of entropy transfer through the portion of the boundary whose temperature is

T j

  

time 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 rate of

350 kJ/s

at

700 K

, and

discharging

a second heat transfer at 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.

250 kJ/s

T

1

350 kJ/s

T

2

500 kJ/s

T

3

Example 2 continued:

Applying an energy rate balance at steady state

dE

0

0 

dt

1  2 

Q

3  

e

Solving 

e

 250 kJ/s  350 kJ/s  500 kJ/s 

100 kJ/s

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

Applying an entropy rate balance at steady state

dS

0

dt

0 

Q

1

T

1 

T

2 2 

T

3 3    Solving       250 kJ/s 500 K  350 kJ/s 700 K  500 kJ/s 1000 K        0 .

5  0 .

5  0 .

5  kJ/s K  

0.5

kJ/s K

Since σ is negative, the claim is

not

in accord with the 2 nd Thermodynamics and is therefore denied.

Law of

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) (d) The change in entropy in kJ/K The entropy generated in the process.

19 State 1: T 1 = 160 o C p 1 = 150 kPa = 1.50 bar super heated vapor 2 1 State 2: T 2 = T 1 =160 o C sat. liquid.

s 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.

Then apply the 1 st Law. 617.8

Q

### Q W

m

u

2 

u

1  

W Q

 (c) To find the change in entropy ( 2 

s

1 )  



 2595.2

 

kJ

)   2391.84

674.86

1.9427

 

### kJ kgK

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.

State 1 2

(d) To find the entropy generation  where  2 

### Q T

s

1    5.5238

 T (°C) p (kPa) x u (kJ/kg 160 150 SH 2595.2

s (kJ/kg K) 7.4665

160 617.8

0 674.86

1.9427

 

Q T

  2391.84

kJ

  5.5239

K

      

### T

What does this mean? Process had no irreversibilities 21

6.11 : 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.

22

6.11 : Isentropic Processes Consider isentropic process for an ideal gas Starting with and 

s

 

c V c V

 

dT T

R

R

ln  

v

2

v

1

k

 1 

s

k R

 1 

dT T

R

ln    

v

2

v

1   thus

k

1  1 ln  

T

2

T

1  

T T

2 1        

v

2

v

1    1 ln 

k

 

v

 1 

v

1 2   An isentropic process is a type of polytropic process  

p

2

p

1 and Next with and 

c P T c P

 

R

ln  

kR k

 1

p

2

p

1  

k kR

 1 

dT T

R

ln  

p

2

p

1  

k k

 1 ln  

T

2

T

1    ln  

p

2

p

1  

v v

2 1  

k

thus  

T

2

T

1

p

2

p

1   (

k

 1)

k p v

1 1

k

p v

2 2

k

23

Example (6.27): State 1 where T 1 Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from = 290 K , p 1 = 1 bar . Process 1 – 2 : Compression to p 2 = 5 bar during which pV 1.19

= constant Process 2 – 3 : Isentropic expansion to p 3

(a)

= 1 bar .

Sketch the two processes on T-s coordinates

(b) (c) Determine the temperature at State 2 in K Determine the net work in kJ Air 24 2 5 bar 3 1 290 K 1 bar S S

25 Example (6.27): state 1 where T 1 Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from = 290 K, p 1 = 1 bar. Process 1 – 2 : Compression to p 2 = 5 bar during which pV 1.19

= constant Process 2 – 3 : Isentropic expansion to p 3 = 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 Process 1-2: polytropic:

p V

1 1

N V

2 

p

1

p V

2 2

N

 1/

N

ideal gas:

p V

1 1

V

2

T

1  

p T

1 2

p V

2 2

T

2

V

1

p

2  

p

1

p

2  1/ 

N

p T

2 2

V

1

p T

2 2

T

2 

T

1  

p

2

p

1   

n

 1 

n

  290

K

 5    375

K

Example (6.27): Process 1 – 2 : Compression to p 2 = 5 bar during which pV Process 2 – 3 : Isentropic expansion to p 3 = 1 bar.

1.19

= constant Find the state properties State 1: p 1 = 1 bar T 1 = 290 K State 2: p 2 = 5 bar T 2 = 375 K 206.91

1.66802

268.075

1.92657

State 3: p 3 = 1 bar s 3 = s 2 From Table A-22: at T = 290 K: u = 206.91

s o = 1.66802

From Table A-22: at T = 375 K: u = 268.075

s o = 1.92657

26

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

1 1 5 5 236.83

1 1 168.86

At State 3: 1.46466

s

3 0 

s

2

s

   3   3    3   2    (1.92657

R

   ln

R

  2 ln

p

3

p

2    

R

ln  

p p

2 3  

p

3  

p

2 )    (0.2870

/  ) ln 1

bar

 1.46466

/  5

bar

From Table A-22: for s o =1.46466 find T and u T 3 = 236.82 K and u 3 = 168.86 kJ/kg

28 Example (6.27): (c) Determine the net work in kJ +61.2 -67.2

-128.4

State

T (K) p (bar) u (kJ/kg)

1

290 1 206.91

Process 1 – 2: 

U

12 

Q

12 

W

12 s °(kJ/kg K) 1.66802

where 

U

12  ( 2 

u

1 ) 

U

12

m

  61.17

W

12

W

12

m

 

pdV

  0.287

2

375 5 268.075

1.92657

3

636.82

1 168.86

1.46466

p V

2 2 1  

n p V

1 1     2 1 

n

T

1 

K

   128.39

/ 

Q

12

m

 

U

12

m

W

12

m

   67.22

/

29 Example (6.27): -99.2

0 +99.2

State

T (K) P (bar) u (kJ/kg)

1

290 1 206.91

2

375 5 268.075

Process 2-3: 

U

23 

Q

23 

W

23 s °(kJ/kg K) 1.66802 1.92657

where 

U

23 

U

23 

m

 ( 3 

u

2 )   99.22

23 

23  

### U m m

Net Work over both processes:

23 

23 99.22

 3 2 

### Tds

 0 /

W net m

W

12

m

W

23

m

  128 .

39

kJ kg

 99 .

22

kJ kg

  29 .

17

kJ kg

3

236.82

1 168.86

1.46468

6.8 : 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.

E

### 0

isolated system 

E

system  

E

environment  0 Energy is neither created nor destroyed.

S

Isolated System   

Q T

int rev.

  30 

S

system  

S

environment   

31