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ME 200 L27: Control Volume Entropy Balance ME 200 L27: Control Volume Entropy Balance

Kim See’s Office ME Gatewood Wing Room 2172 Please check your HW and Examination Grades on Blackboard

Please pick up all graded Home Work and Examinations from Room 2172 before Friday, 4/4/14

https://engineering.purdue.edu/ME200/ThermoMentor© Program

Spring 2014 MWF 1030-1120 AMJ. P. Gore

gore@purdue.eduGatewood Wing 3166, 765 494 0061

Office Hours: MWF 1130-1230TAs: Robert Kapaku rkapaku@purdue.edu

Dong Han han193@purdue.edu

Entropy Balance Equation

► Like Internal Energy, Enthalpy, and Specific Volume, Entropy Like Internal Energy, Enthalpy, and Specific Volume, Entropy is an extensive property of a working substance.is an extensive property of a working substance.

► Analogous to and must apply simultaneously with the Analogous to and must apply simultaneously with the Conservation of Energy andConservation of Energy and

► Conservation of MassConservation of Mass

CVi i e e j CV irr

i e j irr

dSm s m s Q T

dt

CVi i e e j g

i e j g

dEm h m h Q W

dt

CVi e j

i e j

dmm m M

dt

No Internal Source of mass!

Substitute Entropy Definition into First Law

► Like Internal Energy, Enthalpy, and Specific Volume, Entropy Like Internal Energy, Enthalpy, and Specific Volume, Entropy is an extensive property of a working substance.is an extensive property of a working substance.

► Analogous to and must apply simultaneously with the Analogous to and must apply simultaneously with the Conservation of Energy andConservation of Energy and

► Conservation of MassConservation of Mass

0 i i e e CV i e CVj jm s m s Q T m( s s ) Q T

0 i e CVi e

m m m; m m Constant

0 i i e e j CV e i CV j CV CVm h m h Q W m( h h ) m q m w

Entropy Generation using Tables: Example 1Entropy Generation using Tables: Example 1

4

Given: Steam at 120oC, 0.7 bar is pressurized through a diffuser to 1 bar, 160oC and negligible velocity. Find: Find the change in entropy of steam in kJ/kg-K and comment on whether the diffuser can be adiabatic and the resulting impact.Assumptions: Change in PE neglected, No heat transfer, No work doneother than flow work, Steady state, Steady flow, Mass is conserved.

2 2

2

2 2

2000 2796 2 2719 6 390 892

4 120 0 7 7 6395

160 1

CVi e CVj

CVCV CV i i e e

H ,L B,S I E

CV ii e e i i

I E

oI

oE

dSm( s s ) Q T

dt

dE V VQ W m ( h gZ ) m ( h gZ )

dt

dm Vm m ;h h ;V ( . . ) . m / s

dt

Table A : s s( C, . bar ) . kJ / kg K

s s( C, b

7 6597

7 6597 7 6395 0 0202CV e i

ar ) . kJ / kg K

/ m ( s s ) . . . kJ / kg K

5

State 1: 0.7bar, 100 C

State 2: 1 bar, 160 C

T-s Diagram and Diffuser Action

State 2: 1 bar, h2>h1, s2>S1

State 1: 0., 7bar, 100 C

On the T-s diagram drawn to scale State 1 and State 2

2

1

Entropy Generation Calculation: Example 2Entropy Generation Calculation: Example 2

7

Given: 0.5 kg/s of steam at 280oC, 20 bar is expanded in a turbine to 1 bar in a constant entropy process. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K. Find: Find the work produced by the steam in kW and show the processes on a T-s diagram. If the process was not a constant entropy process and resulted in saturated steam at 1 bar, find the decrease in work and increase in entropy in kW/K.Assumptions: Change in PE neglected, No heat transfer, work done on Turbine shaft, Steady state, Steady flow, Mass is conserved.

2

2

2

2

CVCV CV i i

H ,L B,S I

CVe e i e

E I E

CVS i i es CVA i i e

CVCV i i e e CV

H ,L I I

dE VQ W m ( h gZ )

dt

dmVm ( h gZ ) ; m m

dt

W / m h h ,W / m h h

dSQ / T m s m s

dt

St P,T h s x

i 20,280 2976.4 6.6828 -

es 1,99.63 2423.2 6.6828 0.8883

e 1,99.63 2675.5 7.3594 1.0000

x=(s-sf)/(sg-sf)=(6.6828-1.3026)/(7.3594-1.3026)=5.3802/6.0568=0.8883; hes= 417.46 +0.8883 (2258) = 2423.22

CVS iW / m 2976.4 2423.2 553.18 kJ / kg 00CVa iW / m 2976.4 2675.5 3 .9 kJ / kg

150 150 276 59 54 23CVa TW kW ; / . . % ( low ) 276 59CVsW . kW ;

0 5 7 3594 6 6828 0 3383CV e e i iI I

m s m s . ( . . ) . kW / K

State 2a: Saturated

State 1: 20 bar, 280 C

On the T-s diagram drawn to scale State 1 and State 2 are close to eachother as illustrated below.

State 2s: Mixture

Entropy Generation: Example 3Entropy Generation: Example 3

9

Given: Consider R134 throttled from p3 =120 lbf/in2 to p4 =40 lbf/in2.Find: Find the change in entropy of R134.Assumptions: Change in KE, PE neglected, No heat transfer, No work doneother than flow work, Steady state, Steady flow, Mass is conserved.

3 4

4

23

24 4

4

40 91

40 91

40 91 20 57 85 31 0 2384

120 0 0839 10

40 0 0452 0 2197

1

f4 4 g4 f4 4 f4 fg4

of

o of f g

h h . Btu / lbmTable A10E

40.91=h +x (h -h ); x ( . h ) h

x ( . . ) / . .

s s ( lbf / in ) . Btu / lbm R(Table A E )

s s ( lbf / in ) . Btu / lbm R ;s . Btu / lbm R

s (

0 2384 0 0452 0 2384 0 2197 0 03442 0 05238

0 08

0 0868 0 0839 0

68

0029

o

oCV

. ) . . ( . ) . .

. Btu

/ m . . . Bt

/ lb

u / lbm

m

R

R

Adiabatic throttle with a pressure loss and phase change lead to increase inEntropy while keeping Enthalpy constant. Entropy is generated by fluid frictionOr viscosity in this case, in spite of the process being (externally) adiabatic.

10

State 4

State 3

T-s Diagram and Demonstration of Throttle Action; h-s diagram

State 3

State 4

Entropy Generation: Example 4Entropy Generation: Example 4

11

Given: Consider R134 condensed from saturated vapor (state 2) to saturated liquid at p3 =120 lbf/in2.Find: Find the change in entropy generation rate in the process of condensing R134.Assumptions: Change in KE, PE neglected, Heat transfer to a sink at 90.54oF-δ and heat transfer to sink at 80.54oF. No work done other than flow work, Steady state, Steady flow, Mass is conserved.

3

23 2

2 3

3 2

10 40 91

120 0 0839 0 2165

113 82 40 91 72 91

72 91 72 83 459 67 0 0839 0 2165

0 1369 0 1326 0

f g

of

CV

CV

CV

Table A E : h h . ; h Btu / lbm

s s ( lbf / in ) . ;s . Btu / lbm R

q h h . . . Btu / lbm

q / T ( s s )

. / ( . . ) ( . . )

. .

/ m

.I

2h = =113.82

3 2

62 8

72 83

72 91 62 83 459 67 0 1326

0 148

3

0

o

CV

o

CV

o

f receiving heat sin k is at . F

q / T ( s s ) . / ( . . ) .

. Btu / lbm Entropy generationbecauseof heat transfer

through finite delta T if receiving heat sin k is

R.

/ m

at . F.

12

State 2State 3

T-s Diagram and Demonstration of Condenser Action; h-s diagram

State 2

State 3

Sink Temperature