#### Transcript Entropy and Closed Systems

# 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

### Reading Assignment:

Read Chapter 6, Sections 9-10

### 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*

### Area under T-S

An

### energy transfer by heat to a closed system

during an

### internally reversible

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**

### 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 1 st 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 Σ*Q*/*T *for each process. Usually *σ *will be determined from *σ *= ∆*S *- Σ*Q/T*

•

### 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 **

*adiabatically*

**.**

► ► ► 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 )

### 1

*kg*

### 1.9427

### 7.4665

2595.2

*kJ*

) 2391.84

674.86

1.9427

### 5.5238

*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*

*Q*

*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

*W*

23

*Q*

23

*U m m*

Net Work over both processes:

*m*

23

*Q*

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

## End of Slides for Lecture 25

31