ME 200 L31: Review for Examination 3
Thu 4/10/14 Examination 3 (L22 – L30) 6:30 – 7:30 PM
WTHR 200, CL50 224, PHY 112
Kim See’s Office ME Gatewood Wing Room 2172
Please check HW and Examination Grades on Blackboard
Please pick up all graded Home Work and Examinations
from Class or Room 2172 after class today!
https://engineering.purdue.edu/ME200/
ThermoMentor© Program
Spring 2014 MWF 1030-1120 AM
J. P. Gore
gore@purdue.edu
Gatewood Wing 3166, 765 494 0061
Office Hours: MWF 1130-1230
TAs: Robert Kapaku rkapaku@purdue.edu
Dong Han han193@purdue.edu
Examination 3
• Extra Time Students 6-9 pm ME2063
• Civil Engineering Banquet Students 6-7 pm
ME1051 Can accommodate ~50 students. If
you are not a CE banquet student go to your
regular room
• Division 4 (Gore Students) WTHR 200
2
Isentropic Processes for Ideal Gases (Air) with Constant Specific Heat
v2 p2
2
s2
T2
T
v1
1
p1
s1
T1
State 1 and State 2 are defined in the Figure
by intersection of any two curves or lines
passing through the points on the graph. The
lines and curves passing through these points
define additional states where one of the
properties is constant. For example the vertical
line is the isentropic line along which s is
Constant.
p2
s2  s1  s 2  s  R ln
0
p1
o
s
v2 vr 2 ( RT2 / pr2 )


v1 vr1 ( RT1 / pr1 )
vr
Tabulated as a function
of T in Table A22 for air.
o
1
o
p2
exp(
s
o
o
2 / R)
 exp ( s 2  s 1 ) / R 
p1
exp( s o1 / R )

p2 pr 2

;
p1 pr1

pr
Tabulated as a function of T
in Table A22 for air.
3
Isentropic Processes for Ideal Gases with Constant Specific Heat
Constant specific heat assumption allows further simplification of the p-v
relation for the isentropic process for ideal gases:
T2
p2
s2  s1  c p ln  R ln
0
T1
p1
T2  p2 
 
T1  p1 
T2  p2 
 
T1  p1 
R / cp
; R / c p  ( c p  cv ) / c p  1  1 / k  ( k  1 ) / k
k 1
k
1
k
p2v2
v2  p2 

     p2v2 k  p1v1k
p1v1
v1  p1 
Isentropic processes for ideal gases can be analyzed using pv=mRT
and pvk = constant if specific heat is independent of temperature.
4
Isentropic Processes for Ideal Gases with Constant Specific Heat
p2 v 2  p v
n
T
n
1 1
p2 v 2 k  p1v1k
n=k
v = Constant; n = ±∞
n = -1; p and v both increase
and decrease together
p = const; n = 0
n=1
pv  RT  Constant 1  T  Constant 2
s
5
Polytropic Processes for Ideal Gases with Constant Specific Heat
p2 v 2  p v
n
n
1 1
p
v = Constant; n = ±∞
n = -1; p and v both increase
and decrease together
n = -1
p = const; n = 0
n=k
n=1
v
6
Maximum performance measures
(reversible processes)
• Power cycles
QC
TC
max  1 
1
QH
TH
Carnot
efficiency
• Refrigeration cycles
max
QC
TC


QH  QC TH  TC
• Heat pump cycles
QH
TH
 max 

QH  QC TH  TC
7
Example
A refrigeration cycle operating between two
reservoirs receives energy QC from the one at TC =
280 K and rejects energy QH to one at TH = 320 K.
For each of the following cases determine whether
the cycle operates reversibly, irreversibly, or is
impossible:
–
–
–
–
QC = 1500 kJ, Wcycle = 150 kJ. Impossible
QC = 1400 kJ, QH = 1600 kJ. Reversible
QH = 1600 kJ, Wcycle = 400 kJ. Irreversible
β = 5. Irreversible (βmax= 280/40 =7)
8
Example: Use of Clausius Inequality
QH=1000 kJ, TH=500 K and QC=600 kJ at TC:
(a) 200 K, (b) 300 K, (c) 400 K. Find if each
cycle is reversible, irreversible or ideal.
Solution: Use the given QH, QC values to find
work and ensure that the work produced does
not result in a negative value for scycle
 Q  


 T b
QH QC

 s cycle
TH TC
Example: Use of Clausius Inequality
(a)  s cycle 
1000 kJ 600 kJ

 1 kJ/K
500 K 200 K
scycle = +1 kJ/K > 0
Irreversibilities present within system
(b)  s cycle 
1000 kJ 600 kJ

 0 kJ/K
500 K 300 K
scycle = 0 kJ/K = 0
No irreversibilities present within system
1000 kJ 600 kJ

 0.5 kJ/K
(c)  s cycle 
500 K
400 K
scycle = –0.5 kJ/K < 0
Impossible
Entropy is a Property
► Entropy is easier to understand if thought of as a
property analogous to specific volume. It is defined as:
ds 
 QInt .Rev.
analogous to dv 
 WInt .Rev.
T
p
Int .Re v . not emphasized when defining work to avoid confusion.
 QRev
 Q 
ds  
;The second RHS is clearer.
 
T
 T  Re v
Subscript "Rev" relevant to path functions.
T is a point function. So TRev not written after integration
 Tds   Q
Rev
Entropy Change Calculations
QH=1000 kJ, TH=500 K and QC=600 kJ at TC:
(a) 200 K, (b) 300 K, (c) 400 K. Find if each heat
transfer is reversible find entropy change for
(the material that makes up) the reservoir.
Solution: Use the given QH, QC and TH, TC values
to find the entropy change for the reservoirs:
  QRev 

  S
T 

QC
QH
S H 
and SC 
TH
TC
Examples: Entropy Change for the Reservoirs
(Property of materials making up the reservoirs)
(a)
1000 kJ
600 kJ
S H 
 2 kJ/K; SC 
 3 kJ/K
500 K
200 K
Entropy can decrease and increase!
(b)
1000 kJ
600 kJ
S H 
 2 kJ/K; SC 
 2 kJ/K
500 K
300 K
Net entropy change can be zero!
(c)
1000 kJ
600 kJ
S H 
 2 kJ/K; SC 
 1.5 kJ/K
500 K
400 K
Net entropy change can be negative!
However, can’t continue that process in a cyclic manner!
Since by definition reservoir properties are fixed, these
changes must be compensated by reverse actions!
Entropy Change using Tables: Example 3
Given: 0.5 kg/s of steam at 280oC, 20 bar is expanded in a turbine to
1 bar in a constant entropy process.
Find: Find the work produced by the steam in kW and show the process
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.
dECV
V2
h
s
  QCV   WCV   mi ( h 
 gZ )i  St P,T
dt
2
H ,L
B ,S
I
i
20,280 2976.4 6.6828
dmCV
V2
es 1,99.63 2423.2 E me ( h  2  gZ )e ; dt  I mi  E me
e
1,99.63 2675.5 7.3594
W / m  h  h ,W / m  h  h
CVS
i
i
es
CVA
i
i
x
0.8883
-
e
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
WCVS / mi = 2976.4-2423.2=553.18 kJ/kg; WCVa / mi =2976.4-2675.5=300.9 kJ/kg
WCVa
= 150.45 kW
WCVs  276.59 kW
14
On the T-s diagram drawn to scale State 1 and State 2 are close to each
other as illustrated below.
State 1: 20 bar, 280 C
State 2s: Mixture
State 2a: Saturated
Entropy Change using Tables: Example 4
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 done
other than flow work, Steady state, Steady flow, Mass is conserved.
h3  h4  40.91Btu / lbmTable A10E
40.91=h f4 +x4 (hg4 -h f4 ); x4  ( 40.91  h f4 ) h fg4
x4  ( 40.91  20.57 ) / 85.31  0.2384
s3  s f ( 120lbf / in 2 )  0.0839 Btu / lbm o R ( Table A  10 E )
s f 4  s f ( 40lbf / in 2 )  0.0452 Btu / lbm o R ; s g 4  0.2197 Btu / lbm o R
s4  ( 1  0.2384 )0.0452  0.2384( 0.2197 )  0.03442  0.05238 
 0.0868 Btu / lbm o R
Adiabatic throttle with a pressure loss and phase change lead to increase in
Entropy while keeping Enthalpy constant.
16
T-s Diagram and Demonstration of Throttle Action; h-s diagram
State 3
State 3
State 4
State 4
17
Entropy Change for Solids: Example
2 ft3 of sand is heated from 32oF to 70oF find heat added
and change in entropy.
Assume: Sand is incompressible and has constant
specific heat given in Table A-19E ρ=94.9 lbm/ft3, cp = c=
0.191Btu/lbm-oF
Solution:
 QInt .Rev.   WInt .Re v.  dU
TdS  pdV  mcdT
70
QInt .Rev.   2( 94.9 )( 0.191 )dT  2( 94.9 )( 0.191 )( 70  32 )  1377.57 Btu
32
T2
dT
( 70  459.67 )
dS  mc
 S  mc ln  2( 94.9 )( 0.191 )ln
T
T1
( 32  459.67 )
 2( 94.9 )( 0.191 )ln
( 529.67 )
 2.6971 Btu / o R
( 491.67 )
Entropy Change for Liquids: Example
0.15 m3 of water at 350 K is heated to 400 K find heat
added and change in entropy considering constant
specific heat at 375 K from Table A-19.
Assume: Constant specific heat (and constant density)
at 375 K from Table A-19.
Solution:
 QInt .Rev.   WInt .Rev.  dU
TdS  pdV  mcdT
400
QInt .Rev. 
 0.15( 956.8 )4.22dT  0.15( 956.8 )4.22( 400  350 )
350
 30282.72 kJ
T2
dT
( 400 )
dS  mc
 S  mc ln  143.52( 4.22 )ln
T
T1
( 350 )
 80.874 kJ / K
Specific Entropy “s” (kJ/kg-K or BTU/lbm-oR)
Change for Ideal Gases, Constant cp, cv
► Remember Chapter 3 (Property Relations). Divide by “m” kg
Tds  du  pdv  dh  pdv  vdp  pdv
Tds  cv dT  pdv  c p dT  vdp
For Ideal Gases : c p  cv  R; pv  RT
dT
dv
dT
dv
 p
 cv
R
T
T
T
v
 T2
v2  
T2
v2 
dT
s    cv
 R ln
 R ln
   cv ln

T

T
v
T
v
1 

1
1  cv  const .
 1
dT
dp
dT
dp
also,ds  c p
v
 cp
R
T
T
T
p
ds  cv
 T2
p2  
T2
p2 
dT
s    c p
 R ln

c
ln

R
ln
  p

T

T
p
T
p
1 

1
1  c p  const .
 1
Specific Entropy “s” (kJ/kg-K or BTU/lbm-oR)
Change for Ideal Gases, Variable cp, cv
► Remember Chapter 3 (Property Relations). Divide by “m” kg
Tds  c p dT  vdp
For Ideal Gases : pv  RT
cp
dT
dp
dp
ds  c p
v

dT  R
T
T
T
p
 T2 c p
p2 
s   
dT  R ln

T T

p
1
 1

2
cp
 o
p2 
o
S  m  s  R ln  ; s   dT
p1 
T

T1
T
Air s o inTables A  22, A  22 E and Idea l gas s o A  23, A23E
s o  M s o ;M ( kg / kmol )  MolecularWeight
Entropy Change for Air: Example
0.15 kg of Air at 700 K, 10 atm pressure is heated to 1200 K in a piston
cylinder device at constant pressure, find heat added and change in
entropy considering variable specific heat and properties from Table A-22.
Solution:
 QInt .Rev.   WInt .Rev .  dU
T 1200

  QInt .Rev.  m  T 700 du  p( v2  v1 )
QInt .Rev .  0.15(( 933.33  512.33 ) 
T, K
u, kJ/kg
So
700
512.33
2.57277
1200
933.33
3.17888
1000( 0.287( 1200  900 ) / 1000 )
 63.15  12.91  76.06 kJ
 o
p2 
o
S  m  ( s 2  s 1 )  R ln 
p1 

 0.15(( 3.17888  2.57277 )  0 )
 0.0909 kJ / K ; Same order as( QInt .Rev. / T )
kJ/kg-K
Entropy Change for Ideal Gases: Example
2 lbm of N2 is heated at constant volume from 32oF to 70oF find
heat added and change in entropy.
Assume: Ideal Gas and variable specific heat with properties given
in Table A23E
Solution:
 QInt .Re v .   WInt .Re v .  dU ,T2  70o F  529.67 o R;T  32o F  491.67 o R
T  529.67
TdS  pdV  m

T  491.67
dU 

m
u 529.67  u 491.67
M

Interp.in A23E : QInt .Re v .  2( 2636.6  2437.89 ) / 28  14.193 Btu
S 
o
o
o
o
p
T
m
m
( s 2  s 1 )  R ln 2 
( s 2  s 1 )  R ln 2
M
p1
M
T1
2 
529.67 
Interp.in A23E :
( 45.699  45.128 )  1.986 ln
28 
491.67 
S  0.0302 Btu / o R  0 as it should be.
Comparable to QInt .Re v . / T  0.02783Btu / o R as it should be.
Entropy Generation using Tables: Example 1
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 done
other than flow work, Steady state, Steady flow, Mass is conserved.


dSCV
 m( si  se )  Q T  s CV
j
dt
dECV
V2
V2
  QCV   WCV   mi ( h 
 gZ )i   me ( h 
 gZ )e
dt
2
2
H ,L
B ,S
I
E
dmCV
Vi 2
  mi   me ;he  hi 
;Vi  2000( 2796.2  2719.6 )  390.89 m / s
dt
2
I
E
Table A  4 : sI  s( 120o C, 0.7bar )  7.6395 kJ / kg  K
sE  s( 160o C,1bar )  7.6597 kJ / kg  K
s CV / m  ( se  si )  7.6597  7.6395  0.0202 kJ / kg  K
24
T-s Diagram and Diffuser Action
State 2: 1 bar, 160 C
State 2: 1 bar, h2>h1, s2>S1
State 1: 0., 7bar, 100 C
State 1: 0.7bar, 100 C
25
On the T-s diagram drawn to scale State 1 and State 2
2
1
Entropy Generation Calculation: Example 2
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.
dECV
V2
  QCV   WCV   mi ( h 
 gZ )i 
dt
2
H ,L
B ,S
I
St
P,T
h
s
x
dmCV
V2
m
(
h


gZ
)
;
  mi   me
E e
e
2
dt
I
E
i
20,280
2976.4
6.6828 -
es
1,99.63 2423.2
6.6828 0.8883
WCVS / mi  hi  hes ,WCVA / mi  hi  he
e
1,99.63 2675.5
7.3594 1.0000
dSCV
  QCV / T   mi si   me se  s CV
dt
H ,L
I
I
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
WCVS / mi  2976.4  2423.2  553.18 kJ / kg WCVa / mi  2976.4  2675.5  300.9 kJ / kg
WCVs  276.59kW ; WCVa  150kW ;T  150 / 276.59  54.23% ( low )
s CV   me se   mi si  0.5( 7.3594  6.6828 )  0.3383kW / K
I
I
27
On the T-s diagram drawn to scale State 1 and State 2 are close to each
other as illustrated below.
State 1: 20 bar, 280 C
State 2s: Mixture
State 2a: Saturated
Entropy Generation: Example 3
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 done
other than flow work, Steady state, Steady flow, Mass is conserved.
h3  h4  40.91Btu / lbmTable A10E
40.91=h f4 +x4 (hg4 -h f4 ); x4  ( 40.91  h f4 ) h fg4
x4  ( 40.91  20.57 ) / 85.31  0.2384
s3  s f ( 120lbf / in 2 )  0.0839 Btu / lbm o R ( Table A  10 E )
s f 4  s f ( 40lbf / in 2 )  0.0452 Btu / lbm o R ; s g 4  0.2197 Btu / lbm o R
s4  ( 1  0.2384 )0.0452  0.2384( 0.2197 )  0.03442  0.05238 
 0.0868 Btu / lbm o R
s CV / m  0.0868  0.0839  0.0029 Btu / lbm o R
Adiabatic throttle with a pressure loss and phase change lead to increase in
Entropy while keeping Enthalpy constant. Entropy is generated by fluid friction
Or viscosity in this case, in spite of the process being (externally) adiabatic.
29
T-s Diagram and Demonstration of Throttle Action; h-s diagram
State 3
State 3
State 4
State 4
30
Entropy Generation: Example 4
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 90.54oFδ and heat transfer to sink at 80.54oF. No work done other than flow work,
Steady state, Steady flow, Mass is conserved.
Table A  10 E : h3  h f  40.91; h2 = hg =113.82 Btu / lbm
s3  s f ( 120lbf / in 2 )  0.0839 ; s2  0.2165Btu / lbm o R
qCV  h2  h3  113.82  40.91  72.91 Btu / lbm
s CV / m  qCV / T  ( s3  s2 )
 72.91 / ( 72.83  459.67 )  ( 0.0839  0.2165 )
 0.1369  0.1326  0.If receiving heat sin k is at 72.83o F
s CV / m  qCV / T  ( s3  s2 )  72.91 / ( 62.83  459.67 )  0.1326
 0.1480 Btu / lbm o R. Entropy generation because of heat transfer
through finite delta  T if receiving heat sin k is at 62.83o F .
31
T-s Diagram and Demonstration of Condenser Action; h-s diagram
State 3
State 2
State 2
Sink Temperature
32
State 3
Conservation Laws
33
Property Relations and Efficiency Definitions
Cycle Efficiency Definitions