SOLUTIONS MANUAL FOR PRINCIPLES OF TURBOMACHINERY By Seppo A. Korpela || Updated Version 2024 | A+
SOLUTIONS MANUAL PRINCIPLES OF TURBOMACHINERY by Seppo A. Korpela Department of Mechanical and Aerospace Engineering A+
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PRINCIPLES OF TURBOMACHINERY
SOLUTIONS MANUAL
by
Seppo A. Korpela
Department of Mechanical and Aerospace Engineering
January 2012
Copyright ⃝2011-2012,
c Seppo A. Korpela
, Chapter 2
Exercise 2.1 Steam flows through a bank of nozzles shown in the figure below,
with wall thickness t2 = 2 mm, spacing s = 4 cm, blade height b = 2.5 cm, and
exit angle α2 = 68◦ . The exit velocity V2 = 400 m/s, pressure is p2 = 1.5 bar,
and temperature is T2 = 200 C. Find the mass flow rate.
Given:
b = 2.5 C s = 4 cm t2 = 0.2 cm α2 = 68◦ V2 = 400 m/s
T2 = 200 C = 473.15 K p1 = 1.50 bar = 150 kPa v2 = 1.4443 m3 /kg
Find: Mass flow rate.
Solution:
A2 = b(2 cos α2 − t2 ) = 2.5(4 cos(68◦ ) − 0.2) = 3.25 cm
A2 V2 3.25 · 400
ṁ = = 4 = 0.09 kg/s ⇐
v2 10 · 1.4443
Note that the specific volume could also be approximated as
,Exercise 2.2 Air enters a compressor from atmosphere at pressure 102 kPa and
temperature 42 C. Assuming that its density remains constant determine the spe-
cific compression work required to raise its pressure to 140 kPa in a reversible
adiabatic process, if the exit velocity is 50 m/s.
Given: Since the air is stagnant in the atmosphere, its conditions are the stagnation
conditions. The flow is isentropic and
T2 = 42 C = 315.15 K p1 = 102 kPa p2 = 140 kPa V2 = 50 m/s
Find: Specific work done.
Solution: For isentropic flow T ds = dh − vdp leads to dh = dp/ρ. In addition
h0 = h + V 2 /2 is constant. The flow is assumed incompressible because the
pressure changes only slightly and the exit velocity is small.
p 2 − p1 1 2
w= + V2
ρ 2
The density is
p1 102
ρ= = = 1.128 kg/m2
RT1 0.287 · 315.156
so that
140 − 102 502
w= + = 43.95 kJ/kg ⇐
1.128 2 · 1000
Exercise 2.3 Steam flows through a turbine at the rate of ṁ = 9000 kg/h. The
rate at which power is delivered by the turbine is Ẇ = 440 hp. The inlet total
pressure is p01 = 70 bar and total temperature is T01 = 420 C. For a reversible
and adiabatic process find the total pressure and temperature. leaving the turbine.
Given: The flow is isentropic and
T01 = 420 C p01 = 70 bar ṁ = 9000 kg/h Ẇ = 440 hp
Find: T02 and p02 .
Solution:
Ẇ 440 · 0.7457 · 3600
w= = = 131.2 kJ/kg
ṁ 9000
From steam tables h)1 = 3209.8 kJ/kg, s1 = 6.5270 kJ/kg K. At the exit
h02 = h01 − w = 3209.8 − 131.2 = 3078.6 kJ/kg s2 = 6.5270 kJ/kg K
From the superheated steam tables, or using EES,
p02 = 43.58 bar T02 = 348.5 C ⇐
3
, Exercise 2.4 Water enters a pump as saturated liquid at total pressure of p01 =
0.08 bar and leaves it at p02 = 30 bar. If the mass flow rate is ṁ = 10, 000 kg/h
and the process can be assumed to take place reversibly and adiabatically, deter-
mine the power required.
Given: The flow is isentropic and
p01 = 0.08 bar p02 = 30 bar ṁ = 10.000 kg/h
Find: Ẇ .
Solution: For isentropic flow
p02 − p01 30 − 0.08) · 105
w= = = 2.998 kJ/kg
ρ 998
Therefore
10, 000 · 2.998
Ẇ = ṁw = = 8.33 kW ⇐
3600
Exercise 2.5 Liquid water at 700 kPa and temperature 20 C flows at velocity
15 m/s. Find the stagnation temperature and stagnation pressure.
Given: The flow is isentropic and
p = 700 kPa T = 20 K V = 15 m/s
Find: T0 and p0 .
Solution: Since water is incompressible
V2 152
T0 = T + = 20 + = 20.027 = 20.027 C ⇐
2cp 2 · 4187
1 998 · 152
p0 = p + ρV 2 = 700 + = 700 + 122.75 = 812.3 kPa ⇐
2 2 · 1000
Exercise 2.6 Water at temperature T1 = 20 C flows through a turbine with inlet
velocity V1 = 3 m/s, static pressure p1 = 780 kPa and elevation z1 = 2 m. At
the exit the conditions are V2 = 6 m/s, p2 = 100 kPa and z2 = 1.2 m. Find the
specific work delivered by the turbine.
Given: Assuming that the process is isentropic and given that T1 = 20 C, and
p1 = 780 kPa z1 = 2 m V1 = 3 m/s
4
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