Control Volume Entropy Balance Illustrating an Impossible Process
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• ME 200 L28: Control Mass Entropy Balance and Directionality of Processes https://engineering.purdue.edu/ME200/ Spring 2014 MWF 0930-1020 AM Professor Wassgren Lecture by Robert Kapaku; slides adapted from Prof. Gore TAs: Robert Kapaku [email protected] Dong Han [email protected]
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ME 200 L28: Control Mass Entropy Balance and Directionality of Processes https://engineering.purdue.edu/ME200/ Spring 2014 MWF 0930-1020 AM Professor Wassgren Lecture by Robert Kapaku ; slides adapted from Prof. Gore TAs: Robert Kapaku [email protected] Dong Han [email protected]. - PowerPoint PPT Presentation
Transcript of 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 [email protected] Dong Han [email protected]
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:
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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
