4-1-10 [duct-100cm2] Air flows steadily through a long insulated duct with a crosssectional area of 100 cm2. At the inlet, the conditions are 300 kPa, 300 K, and 10 m/s. At
the exit, the pressure drops to 100 kPa due to frictional losses in the duct. Determine (a)
the exit temperature and (b) the exit velocity. Use the PG model for air. (c) Explain why
the temperature does not increase despite the presence of friction. (d) What-if scenario:
Would the answers change significantly if the IG model was used for air? Why?
SOLUTION
Assumptions
PG Model, adiabatic, no work transfer
Analysis
The specific heat ratio of air k=1.4. The constant pressure specific heat of air is Cp=1.005
kJ/kg-K.
Let state-1 represent the inlet and state-2 the exit state.
State-1:
p1  300 MPa
T1  300 K
State-2: p2  100 kPa
The energy equation, 1st law, for the system can be simplified as follows:
0
0
dE
 m  ji  je   Q  Wext  m  hi  he   Q  Wext
dt

0
hi  he  c p Ti  Te   0;  Te  Ti  300 K;
Of course if pe and ke are accounted for, there would be a slight change in temperature.
The mass flow equation, mi  me  m ,
100  104  10  300 

AV
AV
kg
1 1
1 1 p1
m


 0.34844
;
v1
RT1
s
 0.28699  300 
 V2 
mv2 mRT2  0.34844  0.28699  300 
m


 30 ;
4
A2
A2 p2
s
100  10  100 
Temperature does not increase due to the increased velocity and pressure decreased.
What-if Scenario: Use TEST solution to verify this answer and carrying out the what-if
scenario. Same results obtained in IG model. Cp is function of temperature and
temperature is barely changed.
4-1-29 [nozzle-1MPa] Air (use the PG model) expands in an isentropic horizontal nozzle
from inlet conditions of 1.0 MPa, 850 K, 100 m/s to an exhaust pressure of 100 kPa. (a)
Determine the exit velocity. What-if Scenario: What would the exit velocity be (b) if the
inlet kinetic energy were neglected in the energy equation? (c) if the nozzle were vertical
with the exhaust plane 4.0 m above the intake plane?
SOLUTION
Assumptions
PG Model, adiabatic, no work transfer, isentropic
Analysis
The specific heat ratio of air k=1.4. The constant pressure specific heat of air is Cp=1.005
kJ/kg-K.
Isentropic relationship,
k
p2  T2  k 1
 
p1  T1 
p 
T2  T1  2 
 p1 
k 1
k
1.4 1
 100  1.4
 850 
 440 K

 1000 
The energy equation, 1st law, for the system can be simplified as follows:
dE
dt
0
 m  ji  je   Q  Wext  m  hi  he   Q  Wext
hi  kei  pei
0
0
 he  kee  pee
0
0
hi  he  kee  kei
 c p Ti  Te  
Ve 2  Vi 2
2000
Ve  2000C p (Ti  Te )  Vi 2  2000  1.005(850  440)  1002  913.3
If ke is neglected,
Ve  2000C p (Ti  Te )  2000  1.005(850  440)  907.8
If nozzle is vertical with exhaust 4m above intake,
m
s
m
s
hi  he  kee  kei  gze
 c p Ti  Te  
Ve 2  Vi 2
gze

2000
1000
gze 
9.81  4 
m


Ve  2000 C p (Ti  Te ) 
 Vi 2  2000  1.005(850  440) 
 1002  913.25


1000 
1000 
s


4-1-34 [diffuser-100kPa] Air enters an insulated diffuser operating at steady state at 100
kPa, -5oC and 250 m/s and exits with a velocity of 125 m/s. Neglecting any change in pe
and thermodynamic friction, determine: (a) The temperature of air at the exit. (b) The
pressure at the exit. (c) The exit-to-inlet area ratio.
SOLUTION
Assumptions
PG Model, adiabatic, no work transfer, isentropic
Analysis
The specific heat ratio of air k=1.4. The constant pressure specific heat of air is Cp=1.005
kJ/kg-K.
The energy equation, 1st law, for the system can be simplified as follows:
dE
dt
0
 m  ji  je   Q  Wext  m  hi  he   Q  Wext
0
hi  kei  pei
0
 he  kee  pee
0
hi  he  kee  kei
Ve 2  Vi 2
2000
1252  2502
1.005  268  Te  
2000
Te  268  23.3  291 K or 18C
 c p Ti  Te  
Using the isentropic relationship,
p2  T2 
 
p1  T1 
k
k 1
k
1.4
 T  k 1
 291 1.41
p2  p1  2   100 
 133.4 kPa

 268 
 T1 
Exit to inlet area ratio, mi  me  m
 e AeVe  i AV
i i
Ae
V
p V RTe
pTV
 i i  i i
 i e i
Ai  eVe
RTi peVe
peTV
i e
Ae
100  291  250

 1.63
Ai 133.4  268  125
0
4-1-35 [diffuser-85pct] Repeat problem 4-1-34 assuming thermodynamic friction in the
diffuser causes its efficiency to go down to 85%.
SOLUTION
h
P2
3
2
2
P1
1
s
Assumptions
PG Model, adiabatic, no work transfer
Analysis
The specific heat ratio of air k=1.4. The constant pressure specific heat of air is Cp=1.005
kJ/kg-K.
The energy equation, 1st law, for the system can be simplified as follows:
dE
dt
0
 m  ji  je   Q  Wext  m  hi  he   Q  Wext
hi  kei  pei
0
0
 he  kee  pee
hi  he  kee  kei
Ve 2  Vi 2
 c p Ti  Te  
2000
1252  2502
1.005  268  Te  
2000
Te  268  23.3  291 K
0
0
diffuser 
0.85 
h2  h1 T2  T1

h3  h1 T3  T1
T2  T1 291  268

T3  T1
T3  268
0.85 T3  268   23 K
T3 = 295 K or 22C
Using the isentropic relationship,
k
p3  T3  k 1
 
p1  T1 
k
1.4
 T  k 1
 295 1.41
p3  p1  3   100 
 140 kPa

T
268


 1
Exit to inlet area ratio, mi  me  m
 e AeVe  i AV
i i
Ae
V
p V RTe
pTV
 i i  i i
 i e i
Ai  eVe
RTi peVe
peTV
i e
Ae 100  295  250

 1.57
Ai 140  268  125
8-2-10 [air-50kgs] A turbojet aircraft is flying with a velocity of 250 m/s at an altitude
where the ambient conditions are 20 kPa and -25oC. The pressure ratio across the
compressor is 12, and the temperature at the turbine inlet is 1000oC . Air enters the
compressor at a rate of 50 kg/s. Determine (a) the temperature and pressure of the gases
at the turbine exit, (b) the velocity of the gases at the nozzle exit and (c) the propulsive
efficiency of the cycle. Use the PG model. (d) What-if Scenario: What would the exit
velocity be if the aircraft velocity were 300 m/s?
SOLUTION
T
Turbojet Cycle
4
pc
5
6
3
2
1
pc
s
Analysis With reference to the accompanying T  s diagram the principal states are
evaluated as follows.
State-1. Diffuser (given p1  20 kpa, T1  25°C and V1  250
m
)
s
State-2 (given V2  0 ): The energy and entropy equation for the steady diffuser produces.
dE
 m1 j1  m2 j2  Q  Wext ;
dt
kJ
kJ
kJ
j  h  ke  pe, R  0.287
, c p  1.005
, cv  0.718
, k  1.4;
kgK
kgK
kgK
 j1  j2  h1  ke1  h2  ke2  Assumtion V2 is significantly smaller than V1 .
 h2  h1 
V12
V2
V12
 c p (T2  T1 )  1  T2  T1 
;
2000
2000
2000 c p
2
m

 250

s 
 T2  248 K  
 279 K;
2000  1.005
ds
Q
 m1 s1  m2 s2 
 sgen ;
dt
TB
 s1  s2 ;
T p 
 2  2 
T1  p1 
 279 K 
 p2  20 kPa 

 248 K 
0.4
( k 1)
1.4
T2
;  p2  p1 

 T1 
k
( k 1)
k
;
 30.2 kPa
State-3 (given p3  rp p2 , s3  s2 )
T3  p3 
 
T2  p2 
( k 1)
k
;  T3  T2 rp
( k 1)
k
;
0.4
T3  (279 K )(12) 1.4  567 K
p3  (30.2 kPa )(12)  362.4 kPa
State-4: Given p4  p3 ; T4  1500 K
State-5 (given s5  s4 ): The power output from the turbine can be equated to the power
input to the compressor. Alternatively, the two systems can be treated as a combined
system with two inlets and two exits with no heat or external work transfer.
m( j4  j5 )  m( j3  j2 );  h4  h5  h3  h2 ;
 c p (T4  T5 )  c p (T3  T2 );  T5  T2  T3  T4 ;
T 5  1273K  567 K  279 K  985 K
From isentropic relation, T4  T5  p4 p5 
k 1
k
T 
;  p5  p4  5 
 T4 
k
k 1
;
 985 K 
p5  362.4 kPa 

 1273 K 
1.4
0.4
 147.7 kPa
State-6 (given p6 ): The entropy and energy equation for the steady nozzle produces.
T p 
s6  s5 ;  6   6 
T5  p5 
k 1
k
0.4
 20 kPa  1.4
;  T6  985 K 
  556 K
 148 kPa 
The velocity of exhaust gases:
dE
 m5 j5  m6 j6  Q  Wext ;
dt
 j5  j6  h5  ke5  h6  ke6  h5  h6 
V62
;
2000
V62
 c p (T5  T6 ) 
 V6  2000 c p (985 K  556 K) ;
2000
m
Ans : V6  928.6
s
The propulsive efficiency,
m V6  V1 
m(V6  V1 )V1
WP 
; KE 
;
1000
2000
WP
1
1
2
p 



;
V6  V1
V6
WP  KEout 1  KEout WP
1
1
2V1
V1
2
p 
2
2

V
928.6
1 6 1
250
V1
 p  42.4%
What-if Scenario: Use TEST solution to verify this answer and carrying out the what-if
scenario.
15-1-4 [steam-200kPa] Saturated steam at 200 kPa is flowing with a velocity of 500 m/s.
A pitot tube brings the flow to stagnation and the stagnation pressure is measured as 350
kPa. Determine (a) the total pressure, (b) the total temperature and (c) the stagnation
temperature. Use the PC model.
SOLUTION
Assumptions
PC Model, isentropic
Analysis

State 1 given p1 = 200 kPa, V1 = 500 m , x1 =1.0
s
T1  120.2C
h1  2706.6
kJ
kgK

j1  2831.6
State 2  p2  350 kPa, V2  0 m/s, j2  j1 
T2  185.2C
h2  2831.6
kJ
kgK
State 3  s3  s1 , j3  j1 , V3  0 m/s 
h3  2831.6
kJ
kgK
pt  381.5 kPa
p3  381.5 kPa
Tt  186.1C
T3  186.1C
Tstag  185.2C
kJ
kgK
s1  7.127
kJ
kgK
15-1-23 [aircraft-1000kmh] An aircraft is cruising at a velocity of 1000 km/h at an
altitude of 10 km, where the static temperature and pressure are -50oC and 26.5 kPa,
respectively. Determine (a) the Mach number of the aircraft. Also determine (b) the
pressure and (c) temperature of the air brought to rest isentropically by a diffuser. (d)
What-if Scenario: How would the answer in part (c) change if the diffuser were adiabatic
but irreversible?
SOLUTION
Assumptions
PG Model, isentropic
Analysis:

State 1 given p1 = 26.5 kPa, V1 = 1000 km , T1 =-50C , z = 10 km
h

According to TEST Table H-3 High Speed Flows: Properties of Atmospheric Air, the
speed of sound at an altitude of 10 km is 299.5 m/s, or:
c1  kRT1  1.4  0.287   273  50   299.5
m
s
km  1000m  1hr 
m

V1  1000


  277.8
hr  1km  3600s 
s

and
M1 
V1 277.8

 0.9275
c1 299.5
State 2  V2 = 0, s2 = s1

The energy equation, 1st law, for the system can be simplified as follows:
dE
dt
0
 m  ji  je   Q  Wext  m  hi  he   Q  Wext
hi  kei  pei
0
0
0
 he  kee  pee
0
hi  he  kei
 c p Ti  Te  
Vi 2
2000
Vi 2
277.7i 2
Te 
 T1 
  50   11.6C
C p 2000
1.005  2000
0
Using the isentropic relationship,
k
p2  T2  k 1
 
p1  T1 
k
1.4
 T  k 1
 261.4 1.41
p2  p1  2   26.5 
 46.2 kPa

 223 
 T1 
According to TEST Table H-1 High Speed Flows: Isentropic Table for Air, the ratio T/Tt
at a mach number of 0.9275 is 0.853074. So:
T
Tt  T   t
T
1



  261.4 K
   50  273 
 0.853074 

According to TEST Table H-1 High Speed Flows: Isentropic Table for Air, the ratio p/pt
at a mach number of 0.9275 is 0.5739. So:
p 
 1 
pt   p   t    26.5 
  46.2 kPa
 0.5739 
 p
What-if Scenario Use TEST solution (posted) for performing any what-if scenario.
15-2-7 [nozzle-1200kPa] A convergent nozzle has an exit area of 4 cm2. Air enters the
nozzle with a total pressure of 1200 kPa, and a total temperature of 400 K. Assuming
isentropic flow, determine the mass flow rate for back pressure of (a) 900 kPa, (b) 634
kPa and (c) 400 kPa.
SOLUTION
Assumptions
PG Model, isentropic
Analysis:
Tr  400 K
Ae  4 cm 2
pr  1200 kPa
e
pb Let us first determine the critical pressure.
pt  pr and Tt  Tr since we assume isentropic
flow. Using a isentropic table for air gives us
p
 0.5283 meaning we
that at M=1 we have
pt
have p  0.5283 pt  633.9 kPa
cp
kJ
kair 
 1.4 and Rair  0.287
cv
kg K
Let us assume one dimensional isentropic flow.
(a) pb  900 kPa gives us a flow that is not choked. Where we can use:
m  e AeTe
p
Let’s use the relation e  0.75 where we know that it is a subsonic flow since there is a
pt
convergent nozzle and there by the isentropic flow table gives us:
T
M  0.65454  e  0.92113
Tt
pe
Let us now rewrite eq. 1 with Ve  M ece  M e kair RairTe and  e 
where
RairTe
T e 0.92113Tr gives us:
m  e AeVe 
pe Ae M e kair Rair 0.92113Tr
Rair 0.92113Tr
 0.857
kg
s
(b) & (c): Since both pb  634  633.936  p kPa and pb  400 kPa gives us a choked
flow will they both have the same mass flow rate. For choked flow we have M=1 at the
T
exit and there by Ae  A and M=1 isentropic flow    0.8333
Tt
m   AV 
p A M  kair RairT
RairT

p A kair
Rair 0.8333Tt
 0.970
kg
s
15-2-11 [tank-1MPa] Compressed air is discharged through a converging nozzle as
shown in the accompanying figure. The conditions in the tank are 1 MPa and 500 K
while the outside pressure is 100 kPa. The inlet area of the nozzle is 100 cm2 and the exit
area is 35 cm2. Determine (a) the exit velocity, (b) the exit temperature and (c) the force
of the air on the nozzle. Assume the conditions inside to remain unchanged during the
discharge.
SOLUTION
Assumptions
PG Model, isentropic
Analysis
(a) To find the exit velocity, first we must see if the flow is choked at the exit. For
p
k  1.4 , M  1 occurs at b  0.5283 , or rather, any outside pressure at or below
pr
pb   0.52831000 kPa   528.3 kPa results in a choked flow. Since patm  100 kPa ,
the exit flow is choked, and the exit velocity is simply the velocity of sound at this point.
Ve  ce  kRTe . The exit temperature, Te , can be found using the isentropic table for
Me  1.
Ve 
Te
 0.833 , or Te   0.833 500 K   416.65 K . Thus,
Tr
1.4 287 J/kg  K  416.65 K   409.16 m/s .
(b) The exit temperature is found above to be
Te   0.833500 K   416.65 K
Te
 0.833 , or
Tr
(c) For a cylindroid nozzle with a control volume drawn around the edges of the nozzle,
any radial forces sum to zero, so the net force on the nozzle by the air can be assumed to
be in the x-direction only. Thus, a momentum balance gives,
m Vi  Ve 
dM x
 T  pi Ai  pe Ae  pb  Ai  Ae  .
 mV
i x ,i  meVx ,e   Fx , or 0 
1000
dt
Since the exit area is choked at Mach 1, A*  Aexit  35 cm 2 , thus for the inlet,
Ai Ai 100 cm 2


 2.857 , which corresponds to M i  0.208 in the Isentropic table.
A* Ae 35 cm 2
p
p
 0.970 , or pi   0.9701000 kPa   970 kPa , and
Also from the table, i 
pr pt
Ti T
  0.9914 , or Ti   0.9914 500 K   495.7 K . Thus,
Tr Tt
pi Ai
pb ( Ai  Ae )
pe Ae
x
Vi  M i kRTi   0.208 

1.4  287 J/kg  K  495.7 K   92.83


m   AV  6.821 kg/m3 .01 m 2  92.83 m/s   6.33
T
m Ve  Vi 
1000
m Ve  Vi 
m
. Mass flow is
s
kg
. Finally, the force on the flow
s
 pi Ai  pe Ae  pb  Ai  Ae 
  pe  pb  Ae   pi  pb  Ai
1000
 6.33 409.16  92.83  528.3  100 0.0035  970  100 0.01  5.2 kN



 
 
1000

15-2-14 [rocket-700mps] A rocket motor is fired on a test stand. Hot exhaust gases leave
the exit with a velocity of 700 m/s at a mass flow rate of 10 kg/s. The exit area is 0.01 m2
and the exit pressure is 50 kPa. For an ambient pressure of 100 kPa, determine the rocket
motor thrust that is transmitted to the stand. Assume steady state and one-dimensional
flow.
SOLUTION
Assumptions
PG Model, isentropic
Analysis
The momentum equation produces
0
m  0  Ve 
1000
p0 ( A  Ae )
 T  p0 A  pe Ae  p0  A  Ae  .
mVe
T
 Ae  pe  p0 
1000
10  700 

 0.01 50  100 
1000
 6.5 kN
p0 A
pe Ae
x
15-2-26 [o2-M05] Oxygen flows at Mach 0.5 in a channel with a cross-sectional area of
0.16 m2. The temperature and pressure are 800 K and 800 kPa, respectively. (a) Calculate
the mass flow rate through the channel. The cross-sectional area is now reduced to 0.15
m2. Determine the (b) Mach number and (c) flow velocity at the reduced area. Assume
the flow to be isentropic.
SOLUTION
Assumptions
PG Model, isentropic
Analysis
State 1  given A1 = 0.16 m 2 , T1 = 800 K, p1 =800 kPa, M1 = 0.5

m   AV
1 
p1
800
kg

 3.85 3
RT1 0.260  800
m
c1  kRT1  1.395  .260  800  540
m
s
m
s
kg
m   AV  166
s
State 2 ( A2  0.15 m 2 , pt 2  pt1 , Tt 2  Tt1 , A )
V1  M 1c1  270
A2
since A remains
A
constant and we can find M2 and T2/Tt2 from high speed flow isentropic table for oxygen
From isentropic and constant mass flow rate relation we can find
M 2  0.55
V2  M 2 kRT2  294
m
s
15-3-1 [normal-600mps] An airstream with a velocity of 600 m/s, a pressure of 50 kPa
and a temperature of 250 K undergoes a normal shock. Determine (a) the velocity and (b)
the pressure at the exit. (c) Also determine the loss of total pressure due to the shock. (d)
What-if Scenario: What would answers be if the inlet velocity were 1200 m/s instead?
SOLUTION
Assumptions
PG Model, isentropic except across shock
Analysis:
State 1: p = 50 kPa, T = 250 K, V = 600 m/s
m
c1  kRT1  316.9
s
V
M1  1  1.89336 M = 1.89336
c1
pi
p1
50
 0.15079  Pt ,1 

 pt ,1  331.59 kPa
pt ,i
0.15079 0.15079
State 2: Normal shock occurs when M = 1.89336: M2 = 0.597
Ve
m
 0.39905  V2  V1  0.39905   600  0.39905  239.43
Vi
s
pe
 4.01534  p2   p1  4.01534    50  4.01534   200.77 kPa
pi
pt ,e
 0.77037  pt ,2   pt ,1   0.77037    331.59  0.77037   pt ,2  255.45 kPa
pt ,i
So,
pt ,1  pt ,2  331.59  255.45  Loss of Total Pressure  76.14 kPa
What-if Scenario Use TEST solution (posted) for performing any what-if scenario.
15-3-4 [normal-1MPa] Determine (a) the back pressure necessary for a normal shock to
appear at the exit of a converging-diverging nozzle, with an exit to throat area ratio of 2 if
the reservoir conditions are 1 MPa and 850 K. (b) What-if Scenario: What would the
back pressure be if the working gas were helium instead?
SOLUTION
Assumptions
PG Model, isentropic except across shock
Analysis:
From the isentropic table we obtain the following:
A3
p
T
 2, M 3  2.19697, 3  0.09397, 3  0.50901
Ath
pt
Tt
From the normal shock table we obtain the following:
M 4  0.54741,
p4
 5.46406
p3
pt1  pt 2  pt 3  1000 kPa
p3  0.09397 1000  93.97 kPa
p4  5.46406 p3  513.45 kPa
pb  513.45 kPa
What-if Scenario Use TEST solution (posted) for performing any what-if scenario.
15-3-10 [normal-60kPa] Air flowing steadily in a nozzle experiences a normal shock at a
Mach number of 2.5. If the pressure and temperature of air are 60 kPa and 273 K,
respectively, upstream of the shock, determine (a) the pressure, temperature, (b) velocity,
(c) Mach number and (d) the total pressure downstream of the shock. (e) What-if
Scenario: What would the velocity be for helium undergoing a normal shock under the
same conditions?
SOLUTION
Assumptions
PG Model, isentropic except across shock
Analysis:
Mach number M1 = 2.5. From the normal shock table we get M 2  0.513
Also,
p2
 7.125  p2  7.125  60kPa   427.5 kPa
p1
T2
 2.1375  T2  2.1375  273K   583.5 K
T1
Therefore, V2  M 2
And, pt ,2 
pt ,2
pt ,1
1000  kRT2   0.513 1000 1.40  0.287  583.5  248.3
pt ,1 
pt ,2 pt ,1
pt ,1 p1
p1   0.499 
m
s
1
 60   511.5 kPa
 0.0585
What-if Scenario Use TEST solution (posted) for performing any what-if scenario.
15-3-17 [normal-500K] For the converging-diverging nozzle shown in the accompanying
figure, (a) find the range of back pressure for which a normal shock appears in the
diverging section, and (b) the mass flow rate when the shock is present. The throat area is
10 cm2 and the exit area is 40 cm2.
SOLUTION
Assumptions
PG Model, isentropic except across shock
Analysis:
(a) Range of back pressure for which a normal shock appears in the diverging section
Use isentropic table and normal shock table:
For normal shock at the exit use supersonic flow (minimum pb ),
 p 
Ae
p
 4  i  0.0298, pi   i  pt ,i  (0.0298)(1000kPa)  29.8 kPa
p 
A*
pt ,i
 t ,i 
p 
pe
 9.91447  pb  pe   e  pi  (9.91447)(29.8kPa)  295.5 kPa
pi
 pi 
For subsonic flow (maximum pb ),
 p 
pi
 0.98511  pb  pi   i  pt ,i  (0.98511)(1000kPa)  985.11 kPa
p 
pt ,i
 t ,i 
295.5kPa  pb  985.11kPa
(b) The mass flow rate when the shock is present
Let use the exit mass flow right after the shock
p
Ae
T
p
T
 4  t ,e  0.34567, M e  0.47879, e  2.60755, e  9.91442, i  0.3667
A*
pt ,i
Ti
pi
Tt ,i
p 
pt ,e   t ,e  pt ,i  (0.34567)(1000kPa)  345.67kPa
p 
 t ,i 
T
Ti   i
T
 t ,i

 (Tt ,i )  (0.3667)(500K)  183.35K

T 
Te   e  (Ti )  (2.60755)(183.35K)  478.1K
 Ti 
pe  295.5kPa
mshock
mshock

pe
295.5kPa
1m 2
2 

Ae M e kRTe 
(40cm ) 
(0.479) (1.4)(287J/kgK)(478.1K)
2 
RTe
(0.287)(478.1K)
 10000cm 
kg
 1.8
s
Problems Added:
Refrigerant R-134a enters a compressor at 150 kPa, 10°C and leaves at 1000 kPa. The
refrigerant is then throttle through a capillary tube to a pressure of 200 kPa. Determine
the (a) quality and (b) temperature of the refrigerant at the capillary tube exit. Also find
the rate of (c) compressor power and (d) entropy generated per unit mass. Assume the
compression process to be isentropic. (e) What-if scenario: What would the answers be if
the exit pressure of the capillary tube were 400 kPa?
SOLUTION
Assumptions
PC Model, isentropic compression
Analysis:
State 1  given p1 and T1  :
x1  1.0
kJ
kgK
h1  261.61
kJ
kg
T2  73.1C
kJ
kgK
T3  62.0C
s1  1.02287
State 2  given p2 and s2  s1
x2  1.0
h2  306.58
State 3  given p3 and h3  h2 
x3  1.0
s3  1.14568
kJ
kg
Compressor power per unit mass,
dE
dt
0
 m  ji  je   Q  Wext  m  hi  he   Q  Wext
0
Wext
kW
  hi  he    h1  h2    261.61  306.58   45.0
m
kg
Entropy generated per unit mass,
dS
dt
0
 m  si  se   Q  S gen  m  si  se   Q  S gen
0
S gen
kW
  se  si    s3  s1   1.14567  1.02287   0.12281
m
kgK
What-if Scenario Use TEST solution (posted) for performing any what-if scenario.
TEST Codes:
States {
State-1: R-134a;
Given:
{ p1= 150.0 kPa; T1= 10.0 deg-C; Vel1= 0.0 m/s; z1= 0.0 m;
mdot1= 1.0 kg/s; }
State-2: R-134a;
Given:
{ p2= 1000.0 kPa; s2= "s1" kJ/kg.K; Vel2= 0.0 m/s; z2= 0.0 m;
mdot2= 1.0 kg/s; }
State-3: R-134a;
Given:
{ p3= 200.0 kPa; h3= "h2" kJ/kg; Vel3= 0.0 m/s; z3= 0.0 m; }
}
Analysis {
Device-A: i-State = State-1; e-State = State-2;
Given: { Qdot= 0.0 kW; T_B= 298.15 K; Sdot_gen= 0.0 kW/K; }
Device-B: i-State = State-2; e-State = State-3;
Given: { Wdot_ext= 0.0 kW; T_B= 298.15 K; }
}
A pump in a well 15m beneath ground level pumps water at 100 kPa, 20°C into a 10 m3
container at ground level with a mass flow rate of 5 kg/s. The water leaves the pump at
400 kPa through a pipe with diameter of 10cm and exits through a nozzle with an exit
area of 50 cm2. Assuming adiabatic and reversible flow throughout the whole system,
determine (a) the exit velocity and (b) the pumping power in hp. (c) If the tank was
initially half full and losing water at a rate of 0.003 m3/s, determine the amount of work it
takes to fill the tank. (e) What-if scenario: What would the answers be if the well were
30 m below ground level?
SOLUTION
Assumptions
SL Model
Analysis: