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Chapter

# Homework 7 SolutionsExam

Department
Mechanical Engineering
Course Code
ME 3333
Professor
All

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Borgnakke and Sonntag
6.25
Superheated vapor ammonia enters an insulated nozzle at 20°C, 800 kPa, shown
in Fig. P6.25, with a low velocity and at the steady rate of 0.01 kg/s. The
ammonia exits at 300 kPa with a velocity of 450 m/s. Determine the temperature
(or quality, if saturated) and the exit area of the nozzle.
Solution:
C.V. Nozzle, steady state, 1 inlet and 1 exit flow, insulated so no heat transfer.
Energy Eq.6.13: q + hi + V2
i/2 = he + V2
e/2,
Process: q = 0, Vi = 0
Table B.2.2: hi = 1464.9 = he + 4502/(2×1000) he = 1363.6 kJ/kg
Table B.2.1: Pe = 300 kPa Sat. state at 9.2°C :
he = 1363.6 = 138.0 + xe × 1293.8,
=> xe = 0.947, ve = 0.001536 + xe × 0.4064 = 0.3864 m3/kg
A
e = m
.
eve/Ve = 0.01 × 0.3864 / 450 = 8.56 × 10-6 m2
Inlet
Low V
Exi
t
Hi V
Hi P, A Low P, A
cb
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Borgnakke and Sonntag
6.28
A diffuser, shown in Fig. P6.28, has air entering at 100 kPa, 300 K, with a
velocity of 200 m/s. The inlet cross-sectional area of the diffuser is 100 mm2. At
the exit, the area is 860 mm2, and the exit velocity is 20 m/s. Determine the exit
pressure and temperature of the air.
Solution:
Continuity Eq.6.3: m
.
i = AiVi/vi = m
.
e = AeVe/ve,
Energy Eq.(per unit mass flow)6.13: hi + 1
2Vi2 = he + 1
2Ve2
h
e - hi = 1
2 ×2002/1000 1
2 ×202/1000 = 19.8 kJ/kg
T
e = Ti + (he - hi)/Cp = 300 + 19.8/1.004 = 319.72 K
Now use the continuity equation and the ideal gas law
v
e = vi
AeVe
AiVi = (RTi/Pi)
AeVe
AiVi = RTe/Pe
P
e = Pi
Te
Ti
AiVi
AeVe = 100
319.72
300
100 × 200
860 × 20 = 123.92 kPa
Inlet
Low V
Exi
t
Hi V
Hi P, A
Low P, A
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Borgnakke and Sonntag
6.46
A small turbine, shown in Fig. P 6.46, is operated at part load by throttling a 0.25
kg/s steam supply at 1.4 MPa, 250°C down to 1.1 MPa before it enters the turbine
and the exhaust is at 10 kPa. If the turbine produces 110 kW, find the exhaust
temperature (and quality if saturated).
Solution:
C.V. Throttle, Steady, q = 0 and w = 0. No change in kinetic or potential
energy. The energy equation then reduces to
Energy Eq.6.13: h1 = h2 = 2927.2 kJ/kg from Table B.1.3
C.V. Turbine, Steady, no heat transfer, specific work: w = 110
0.25 = 440 kJ/kg
Energy Eq.: h1 = h2 = h3 + w = 2927.2 kJ/kg
h3 = 2927.2 - 440 = 2487.2 kJ/kg
State 3: (P, h) Table B.1.2 h < hg
2487.2 = 191.83 + x3 × 2392.8
T
v
1
2
3
T = 45.8°C , x3 = 0.959
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for
testing or instructional purposes only to students enrolled in courses for which this textbook has been
adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108
of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.