Control Volume Entropy Balance Illustrating an Impossible Process

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ME 200 L28: Control Mass Entropy Balance andDirectionality of Processes

https://engineering.purdue.edu/ME200/

Spring 2014 MWF 0930-1020 AMProfessor Wassgren

Lecture by Robert Kapaku; slides adapted from Prof. Gore

TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edu

Control Volume Entropy Balance Illustrating an Impossible Process

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Given: Steam at 100oC, 1 bar is pressurized through a diffuser to 1.5 bars, 120oC 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 done

other than flow work, Steady state, Steady flow, Mass is conserved.

Equations: Starting with basic conservation equations from the equation sheet, we arrive at:

2

22000 2711 4 2676 2 265 33

100 1 7 3549 2

120 1 5 7 2693 47 2693 7 3549 0 0856 0

ii e e i CV e i

i

oi g

oe

CV

Vm m m;h h ; m( s s );

V ( . . ) . m / s

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

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

process!Adiabatic diffuser with given pressure gain leads to decrease in entropy.In reality, this diffuser design will not function! The pressure gain will be less than what is assumed here.

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State 1: 1bar, 100 C

State 2: 1.5 bar, 120 C

Saturated State: 1.5b bar, 111 C

T-s Diagram and Diffuser Action (This diffuser will not work!)

State 2: 1.5 bar, h2>h1, s2<S1

State 1: 1bar, 100 C

State 2: 1.5 bar, 120 C

State 1: 1bar, 100 C

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

Entropy Balance Equation

►Control Mass equations result from recognizing that there can be no inflows and outflows of mass

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

2 1

2 1 1

CMi i e e j CM irr

i e j irr

final

CM j CM irrinitial j irr

j CMj

CM

dS m s m s Q Tdt

dS (Q T ) dt ( ) dt

m( s s ) (Q T )

s s ( / m )(Q / T ) / m

2 1 1 2 1 2

CVi i e e j g

i e j g

dE m h m h Q Wdt

e e q w

Control Mass Entropy Generation: Example 1

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Given: Saturated liquid water at 10 bar is heated in a piston-cylinder device while maintaining pressure until the volume increases by a factor of 10.Assume the boundary temperature is equal to the water temperature.Find: (a) Work done in a reversible process, (b) Heat transfer in areversible process, and (c) entropy production in kJ/kg-K,if the work done is (90% of theoretical value).Assumptions: Change in KE, PE neglected, Control mass.Equations:

2 1 1 2 CMs s ( q / T ) / m

2 1 1 2 1 2 1 2 2 1 1 2 2 1 Rev Rev

u u q w ; w p( v v ); q T( s s )

St P bar T, C v, m3/kg vf, m3/kg vg, m3/kg x

1 10 179.9 1.1273(10-3) 1.1273(10-3) 0.1944 0

2 10 179.9 1.1273(10-2) 1.1273(10-3) 0.1944 0.0525

2 32

2 3

1 1273 10 1 1273 10 0 0101 0 05250 1944 1 1273 10 0 1932

f

g f

v v . ( ) . ( ) .x .v v . . ( ) .

Important: Know why these equations are simplified this way!

Control Mass Entropy Generation: Example 1

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St P bar T, C u, kJ/kg uf, kJ/kg ug, kJ/kg x

1 10 179.9 761.68 761.68 2583.6 0

2 10 179.9 857.331 761.68 2583.6 0.0525

22 20 0525 0 0525 2583 6 761 68 761 68

95 6508 761 68 857 3308

f

g f

u ux . ;u . ( . . ) .

u u

. . . kJ / kg

St P bar T, C h, kJ/kg hf, kJ/kg hg, kJ/kg x

1 10 179.9 762.81 762.81 2778.1 0

2 10 179.9 868.613 762.81 2778.1 0.0525

22 20 0525 0 0525 2778 1 761 68 762 81

105 803 762 81 868 613

f

g f

h hx . ;h . ( . . ) .

h h

. . . kJ / kg

Control Mass Entropy Generation: Example 1

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St P bar T, C s, kJ/kgK sf, kJ/kgK sg, kJ/kgK x

1 10 179.9 2.1387 2.1387 6.5863 0

2 10 179.9 2.3722 2.1387 6.5863 0.0525

22 20 0525 0 0525 6 5863 2 1387 2 1387 2 3722f

g f

s sx . ; s . ( . . ) . . kJ / kgK

s s

2 1 1 2 CMs s ( q / T ) / m

1 2 2 1

2 31 2 2 1

273 179 9 2 3722 2 1387 105 75

1000 1 1273 10 1 1273 10 10 15

Rev

Rev

q T( s s ) ( . )( . . ) . kJ / kg

w p( v v ) ( . ( ) . ( )) . kJ / kg

1 2

1 2 2 1 1 2

0 9 10 15 9 13595 6508 9 135 104 7858

w . ( . ) . kJ / kgq u u w . . . kJ / kg

2 1 1 2 2 3722 2 1387 104 790 2335 0 2313 0 0022

CM / m ( s s ) ( q / T ) ( . . ) ( . /. . . kJ / kg K

179.9+273¿

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

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10

1 2

On the p-v diagram drawn to scale State 1 and State 2

September 17th, 2010 ME 200 11

In-Class Example