1 Tutorial 6

ECE 309 Introduction to Thermodynamics and Heat Transfer

Tutorial # 6

Entropy

Problem 1 Air is compressed steadily by a 5-kW compressor from 100 kPa and 17oC to 600 kPa and 167oC at a rate of 1.6 kg/min. During this process, some heat transfer takes place between the compressor and the surrounding medium at 17oC. Determine

(a) the rate of entropy change of air and (b) the rate of entropy generation during this process.

AIR

COMPRESSOR

100 kPa

17oC

600 kPa

167oC

5 kW

17oC

Solution: Step 1: Write the given data from the problem statement State 1: P1 = 100 kPa, T1 = 17oC State 2: P2 = 600 kPa, T2 = 167oC Surrounding temperature: Tsurr = 17oC Work input to the compressor: W& =- 5 kW (negative sign indicates work is done on the system) Mass flow rate: skgkgmmm /02666.0min/6.1

21==== &&&

Step 2: Write what we are asked to solve for:

(a) rate of entropy change of air: ?=!airS&

(b) rate of entropy generation during the process: ?=genS&

Step 3: State the assumption(s): (1) At the specified conditions air can be assumed as an ideal gas (2) Assuming steady state steady flow process (SSSF) (3) Changes in kinetic and potential energies to be negligible

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Step 4: Write the Energy balance (First Law of Thermodynamics) and Entropy Balance (Second Law of Thermodynamics) equations for the system shown in Figure P1 Energy Balance (First Law of Thermodynamics):

( )12hhmWQ !=! &&& (1.1)

for ideal gas, we know that

dTCdh avp,= (1.2)

( )12, TTCmWQ avp !+= &&& (1.3) Entropy Balance (Second Law of Thermodynamics):

( )! "=+12ssmS

T

Qgen

&&&

(1.4)

Again for ideal gas, we know that

1

2

1

2,12 lnln

P

PR

T

TCss avp !=! (1.5)

where R is gas constant

! ""#

$%%&

'(=+

1

2

1

2, lnln

P

PR

T

TCmS

T

Qavpgen

&&&

(1.6)

Step 5: Solve for the unknown quantities From Eq. (1.3), calculate the heat transfer during the process

(1.7)

Note: In the above calculation of heat transfer, Cp,av is found from Table A-2b at Tav= 365 K (a) Rate of entropy change (

airS&! )

( ) !!"

#$$%

&'='=(

1

2

1

2

,12 lnlnP

PR

T

TCmssmS avpair

&&& (1.8)

where

12ss ! is replaced by Eq. (1.5)

( ) ( )( )( ) kWKKkgkJskgkWQ 96101.0290440/010.1/02666.05 !=!"+!=&

3 Tutorial 6

( ) ( ) ( ) !!"

#$$%

&'('=)

kPa

kPaKkgkJ

K

KKkgkJskgSair

100

600ln/287.0

290

440ln/010.1/02666.0& (1.9)

KkWS

air/00248.0!=" &

(b) Rate of entropy generation during the process ( genS& ) Using Eq. (1.6) and result (a), we can find the rate of entropy generation during the process

K

kWKkW

T

QSS

surr

airgen290

96101.0/00248.0

!!!=!"=

&&& (1.10)

KkWSgen /00083.0=&

4 Tutorial 6

Problem 2 Steam enters a turbine at 30 bars and 400°C with a velocity of 160 m/s. Saturated vapor exits at 100oC with a velocity of 100 m/s. At steady state, the turbine develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat transfer between the turbine and its surroundings occurs at an average outer surface temperature of 500K Determine the rate at which entropy is produced within the turbine per kilogram of steam flowing, in KJ/kgK. Neglect the change in potential energy between inlet and exit. Solution: Step 1: Write the given data from the problem statement State 1: P1 = 30 bars, T1 = 400oC, V1 = 160 m/s State 2: Saturated vapor, T2 = 100oC,V2 = 100 m/s Figure P2 Boundary temperature: Tb = 500 K Work output from the turbine: W = 540 kJ/kg Step 2: Write what we are asked to solve for:

Rate of entropy produced per kg of steam flowing within the turbine: ?=genS Step 3: State the assumption(s):

(1) Assuming steady state steady flow process (SSSF) (2) Change in potential energy between the inlet and outlet is negligible (3) Heat transfer between the turbine and the surroundings occurs at a boundary

temperature Tb Step 4: Write the Energy balance (First Law of Thermodynamics) and Entropy Balance (Second Law of Thermodynamics) equations for the system shown in Figure P2 Energy Balance (First Law of Thermodynamics):

!!"

#$$%

& '+'='

2

2

1

2

2

12

VVhhmWQ &&& (2.1)

2

2

1

2

2

12

VVhh

m

W

m

Q !+!+=

&

&

&

&

(2.2)

2

2

1

2

2

12

VVhhW

m

QQ

!+!+==

&

&

(2.3)

5 Tutorial 6

( ) !!"

#$$%

&

!!!!

"

#

$$$$

%

&

!!"

#$$%

&!!"

#$$%

& '+!!"

#$$%

&'+!!

"

#$$%

&=

mN

kJ

s

mkg

N

s

m

kg

kJ

kg

kJQ

3

2

2

222

10

1

1

1

2

1601009.32301.2676540 (2.4)

kgkJQ 6.22!= {Note: From Table A-4, @ T1 = 400oC and P1 = 30 bars, h1 = 3230.9 kJ/kg and s1 = 6.9212 kJ/kg K From Table A-2, @ T2 = 100oC and x2 = 1, h2 = 2676.1 kJ/kg and s2 = 7.3549 kJ/kg K } Entropy Balance (Second Law of Thermodynamics):

( )12ssmS

T

Qgen

b

!=+ &&&

(2.5)

( )b

gen

T

m

Q

ssm

S !!"

#$$%

&

''=&

&

&

&

12 (2.6)

( )( )

K

kg

kJ

Kkg

kJSgen

500

6.22

9212.63549.7

!!"

#$$%

&'

'!!"

#$$%

&'= (2.7)

KkgkJSgen /4789.0=