PRINCIPLES OF TURBOMACHINERY
SOLUTIONS MANUAL
by
Seppo A. Korpela
Department of Mechanical and Aerospace Engineering
January 2012
c
Copyright ⃝2011-2012,
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 = 200 C = 473.15 K
t2 = 0.2 cm
α2 = 68◦
V2 = 400 m/s
v2 = 1.4443 m3 /kg
p1 = 1.50 bar = 150 kPa
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
= 4
= 0.09 kg/s
⇐
v2
10 · 1.4443
Note that the specific volume could also be approximated as
ṁ =
v2 =
R̄T2
8.314 · 473.15
=
= 1.457 m3 /kg
Mp2
18 · 150
V1
t1
s
1
t2
2
α2
V2
2
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 specific 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
502
140 − 102
+
= 43.95 kJ/kg
⇐
w=
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
=
= 131.2 kJ/kg
ṁ
9000
From steam tables h)1 = 3209.8 kJ/kg, s1 = 6.5270 kJ/kg K. At the exit
w=
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, determine 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
w=
p02 − p01
30 − 0.08) · 105
=
= 2.998 kJ/kg
ρ
998
Therefore
Ẇ = ṁw =
10, 000 · 2.998
= 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
T0 = T +
V2
152
= 20 +
= 20.027 = 20.027 C
2cp
2 · 4187
⇐
998 · 152
1
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
4
V1 = 3 m/s
p2 = 100 kPa
z1 = 1.2 m
V1 = 6 m/s
Find: w
Solution: Since water is incompressible
w = h01 − h02s
w=
p01 − p02 V12 − V22
=
+
+ g(z1 − z2 )
ρ
2
780 − 100) 103 32 − 62
+
+ 9.81 (2 − 2.12) = 681.36 − 13.50 + 7.85
998
2
w = 675.71 J/kg
⇐
Note the small contributions from potential and kinetic energy.
Exercise 2.7 Air at static pressure of 2 bar and static temperature of 300 K flows
with velocity 60 m/s. Find total temperature and pressure.
Given:
p1 = 2 bar
T1 = 300 K
V1 = 60 m/s
Find: T01 and p01
Solution: The total temperature is
T01 = T1 +
V12
602
= 300 +
= 301.8 K
2 cp
2 · 1004.5
⇐
and since the velocity is quite small, air may be taken to be incompressible with
density
p1
2 · 105
ρ=
=
= 2.323 kg/m3
RT1
287 · 300
and the stagnation pressure is
1
2.323 · 602
p01 = p1 + ρV12 = 2 · 105 +
= 204.2 kPa
2
2
⇐
Exercise 2.8 Air at static temperature of 300 K and static pressure of 140 kPa
flow with velocity 60 m/s. Evaluate the total temperature and total pressure of air.
Repeat the calculation if the air speed is 300 m/s.
Given:
p1 = 140 kPa
T1 = 300 K
V1 = 60 m/s
5
Find: T01 and p01
Solution:
T01 = T1 +
V12
602
= 301.8 K
= 300 +
2 cp
2 · 1004.5
⇐
and since the velocity is quite small, air may be taken to be incompressible with
density
p1
2 · 105
ρ=
=
= 2.323 kg/m3
RT1
287 · 300
and the stagnation pressure is
1
2.323 · 602
p01 = p1 + ρV12 = 140 · 103 +
= 142.93 kPa
2
2
⇐
and if V1 = 300 m/s then
T01 = T1 +
and
(
p01 = p1
T01
T1
V12
3002
= 300 +
= 344.8 K
2 cp
2 · 1004.5
)γ/(γ−1)
(
= 140
344.8
300
⇐
)3/5
= 227.9 kPa
⇐
Exercise 2.9 Air undergoes an increase of 1.75 kPa in total pressure through a
blower. The inlet total pressure is one atmosphere and the inlet total temperature
is 21 C. Evaluate the exit total temperature if the process is reversible adiabatic.
Evaluate the energy added to the air per unit mass flow.
Given:
p01 = 101.325 kPa
T01 = 294.15 K
∆p0 = 1.74 kPa
Find: T02
Solution: The stagnation density is
ρ01 =
p01
101.325 · 103
=
= 1.200 kg/m3
RT01
287 · 294.15
Since the process is reversible and adiabatic and the pressure increase is quite
small, the work done by the blower may be calculated from
w=
1.75
p02 − p01
=
= 1.458 kJ/kg
ρ01
1.200
6
⇐
and the stagnation temperature leaving the blower is
T02 = T01 +
w
1458
= 294.15 +
= 295.6 K = 22.4 C
cp
1004.5
⇐
Exercise 2.10 Air enters a blower from the atmosphere where pressure is 101.3 kPa
and temperature is 27 C. Its velocity at the inlet is and with velocity 46 m/s. At
the exit the total temperature is 28 C and the velocity is 123 m/s. Assuming that
the flow is reversible and adiabatic, determine (a) the change in total pressure in
millimeters of water and (b) the change in static pressure, also in millimeters of
water.
Given: The inlet stagnation pressure is p01 = 101.3 kPa and stagnation temperature and velocity are
T01 = 300.15 K
V1 = 46 m/s
T02 = 301.15 K
V2 = 123 m/s
At the exit
Find: ∆p0 in millimeters of water.
Solution: Work done by the blower is
w = cp (T02 − T01 ) = 1004.5(28 − 27) = 1004.5 kJ/kg
The stagnation density at the inlet is
ρ01 =
p01
101.3 · 103
=
= 1.176 kg/m3
RT01
287 · 300.15
The flow is isentropic and the pressure rise is small. Therefore
ws =
∆p0
ρ01
∆p0 = ρ01 w = 1.176 · 1.0045 = 1.181 kPa
In millimeters of water this is
∆H =
∆p0
1182
=
= 0.121 mm
ρw g
998 · 9.81
⇐ (a)
To calculate the difference in the static pressure, first calculate
1
462
p1 = p01 − ρ01 V12 = 300.15 −
= 100.06 kPa
2
2 · 1004.5
7
The static temperatures are
T1 = T01 −
V12
462
= 300.15 −
= 299.10 K
2cp
2 · 1004.5
T2 = T02 −
V22
1232
= 301.15 −
= 293.62 K
2cp
2 · 1004.5
The exit stagnation pressure is p02 = 101.300 + 1.182 = 102.482 kPa. and the
static pressure is
(
p2 = p02
T2
T02
)γ/(γ−1)
(
= 102.48
293.62
301.15
)3.5
= 93.789 kPa
and thus
∆Hs =
p2 − p1
93.789 − 100.06
=
= −641 mm
ρw g
998 · 9.81
⇐
(b)
Exercise 2.11 The total pressure, static pressure and the total temperature of air
at a certain point in a flow are 700 kPa, 350 kPa, and 450 K, respectively. Find the
velocity at that point.
Given: The properties are
T0 = 450 K
p = 350 kPa
p0 = 700 kPa
Find: The velocity
Solution:
(
T = T0
p
p0
)(γ−1)/γ
(
= 450
350
700
)1/3.5
= 369.15 K
and
√
√
V = 2cp (T0 − T ) = 2 · 1004.5(450 − 369.15) = 403.0 m/s
Exercise 2.12 Air has a static pressure of 2 bar and static temperature 300 K
while it flows at speed 1000 m/s. (a) Assuming air obeys the ideal gas law with
constant specific heats, determine its stagnation temperature and stagnation pressure. (b) Repeat part (a) using the air tables.
8
Given: The properties are
T1 = 300 K
p1 = 2 bar
V1 = 1000 m/s
Find: p01 and T01
Solution:
1
10002
2c
=
300
+
= 797.8 K
⇐ (a)
p
V12
2 · 1004.5
(
)γ/(γ−1)
(
)3/5
T01
797.8
= p1
= 200
= 61.33 bar
⇐ (a)
T1
300
T01 = T1 +
p01
Using air tables: At the inlet pr1 = 1.386
h01
V12
10002
=h+
= 300.19 +
= 800.19 kJ/kg
2
2cdot1000
⇐
(b)
Thus with the same entropy and this stagnation enthalpy pr0 = 43.38, and T01 =
780.15 K. Therefore
p01 = p1
43.38
pr0
=2
= 6.26 bar
pr
1.386
⇐
(b)
Exercise 2.13 At a certain location the velocity of air flowing in a duct is 321.5 m/s.
At that location the stagnation pressure is 700 kPa and stagnation temperature is
450 K. What is the static density at this location.
Given: The properties are
T01 = 450 K
Find: ρ1 .
Solution:
p1 = 700 kPa
V1 = 321.5 m/s
V12
321.52
T1 = T01 −
= 450 −
= 398.55 kPa
2cp
2 · 1004.5
(
(
)γ/(γ−1)
)3.5
389.55
T1
= 700
p1 = p01
= 457.66 kPa
T01
430
The density is then
ρ1 =
398550
p1
=
= 4.00 kg/m3
RT1
287 · 457.66
9
⇐
Exercise 2.14 Air flows in a circular duct of diameter 4 cm at the rate of 0.5 kg/s.
The flow is adiabatic with stagnation temperature 288 K. At certain location the
static pressure is 110 kPa. Find the velocity at this location.
Given: The properties are
T0 = 288 K
p = 110 kPa
ṁ = 0.5 kg/s
D = 0.04 m
Find: The velocity
Solution: Solving ṁ = ρAV = pAV /RT for temperature gives
T =
pAV
ṁR
Since T = T0 − V 2 /2cp a quadratic equation for velocity if found
V2+
2cp pA
V − 2cp T0 = 0
ṁR
the solution of which is
cp pA
V =−
ṁR
√
(
1−
2T0 ṁ2 R2
1+
cp p2 A2
)
The terms are
2cp ṁ2 R2
2 · 288 · 0.52 · 2872
=
= 0.618
cp p2 A2
1004.5 · 1100002 (π · 0.022 )2
cp RA
10004.5 · 110000 · π · 0.02
=
= 967.61
ṁR
0.5 · 287
Therefore
V = −967.61(1 −
√
1 + 0.6180) = 263.2 m/s
⇐
Exercise 2.15 Saturated steam enters a nozzle at static pressure 14 bar at velocity
52 m/s. It expands isentropically to pressure 8.2 bar. Mass flow rate is ṁ =
0.7 kg/s. Find the exit area if, (a) steam is assumed to behave as an ideal gas with
γ = 1.135, and cp = 2731 J/kg K; (b) the end state is calculated with properties
obtained from the steam tables.
10
Given: The properties are
T1 = 468.2 K
p1 = 14 bar
V1 = 52 m/s
ṁ = 0.7 kg/s
γ = 1.135
Find: A2
Solution: From steam tables the value of enthalpy is h1 = 2789.4 kJ/kg and the
specific volume is V1 = 0.1408 m3 /kg. Then assuming ideal gas behavior with
constant specific heats, gives
T01 = T1 +
V12
522
= 468.2 +
= 468.7 K
2cp
2 · 2731
and the stagnation pressure is
1
522
p01 = p1 + ρ1 V12 = 14 · 105 +
= 14.096 bar
2
2 · 0.1408
State 2 has p2 = 820 bar, so that
(
)0.135
( )(γ−1)/γ
820
p2
T2 = T1
= 468.2
1.135 = 439.3 K
p1
1400
Since no work is done T02 = T01 and
√
√
V2 = 2cp (T02 − T2 ) = 2 · 2731(468.7 − 439.3) = 400.5 m/s
The specific volume is
(
v2 = v1
p2
p1
)1/γ
(
= 0.1408
1400
820
)
1/1.315 = 0.2256 m3 /kg
Hence
A = ṁv2 /V2 = 0.7 · 0.2256/400.5 = 3.944 cm2
⇐
Using steam tables
x2 =
6.4684 − 2.0559
s2 − sf
=
= 0.9596
sg − Sf
6.6546 − 2.0559
h2 = hf + x2 hf g = 725.6 + 0.9596(2770.0 − 725.36) = 2687.0 kJ/kg
v2 = vf + x2 vf g = 1.116 · 10−3 + 0.9596(0.2354 − 1.116 · 10−3) = 0.2259 m3 /kg
11
Therefore
V2
√
√
2(h02 − h2 ) = 2(2791.0 − 2687.0 = 454.9 m2 /s
so that
A = ṁv2 /V2 = 0.7 · 0.2259/54509 = 0.3469 cm2
⇐
Exercise 2.16 A fluid enters a turbine with total temperature of 330 K and total
pressure of 700 kPa. The outlet total pressure is 100 kPa. If the expansion process
through the turbine is isentropic, evaluate (a) the work per unit mass flow if the
fluid is incompressible and having a density equal to 1000 kg/m3 , (b) the work
per unit mass flow if the fluid is air
Given: The properties are
T01 = 330 K
p01 = 700 kPa
p02 = 100 kPa
Find: w for water and air.
Solution: Since the process is isentropic, for water
ws =
p01 − p02
700 − 100
=
= 0.6 kJ/kg
ρ
998
For air
(
T02 = T01
p02
p01
)(γ−1)/γ
(
= 330
100
700
⇐
)1/3/5
= 189.3 K
so that
ws = h01 − h02 = cp (T01 − T02 ) = 1.0045(330 − 189.3) = 141.4 kJ/kg
⇐
Note that the air temperature leaving the turbine is very low, and operation such as
this would be unusual. It might happen in an expander in cryogenic applications.
Exercise 2.17 Air flows through a turbine which has a total pressure ratio 5 to 1.
The total-to-total efficiency is 80% and the flow rate is 1.5 kg/s. The desired output power is to be 250 hp(186.4 kW). Determine: (a) The inlet total temperature;
(b) The outlet total temperature; (c) the outlet static temperature if the exit velocity
is 90 m/s; (d) draw the process on a T S-diagram and determine the total-to-static
efficiency of the turbine.
12
Given: The properties of air are
p01 /p02 = 5
ηtt = 0.8
ṁ = 1.5 kg/s
V2 = 90 m/s
Ẇ = 186.4 kW
Find: T01 , T02 , T2 , the TS-diagram and ηts .
Solution:
w=
Ẇ
186.4
=
= 124.27 kJ/kg
ṁ
1.5
ws =
124.27
w
=
= 155.33 kJ/kg
ηtt
0.8
ws = cp (T01 − T02s )
T01 =
ws
155.33
=
= 419.51 K
(γ−1)/γ
cp [1 − (p02 /p01 )
]
1.0045(1 − 5−1/3.5 )
T02 = T01 −
T2 = T02 −
w
124.27
= 419.51 −
= 295.80 K
cp
1.0045
⇐
V22
902
= 295.80 −
= 291.76 K
2cp
2 · 1.0045
T02s = T01 −
⇐
⇐
w
155.33
= 419.51 −
= 264.9 K
2cp
2 · 1.0045
Assuming that V2s = V2
T2s = T02s −
Thus
ηts =
V2s2
902
= 264.9 −
= 260.8 K
2cp
2 · 1.0045
T01 − T02
419.51 − 295.80
=
= 0.78
T01 − T2s
419.51 − 260.8
⇐
Exercise 2.18 A blower has a change in total enthalpy of 6000 J/kg, an inlet
total temperature 288 K, and inlet total pressure 101.3 kPa. Find: (a) the exit
total temperature if the working fluid is air; (b) the total pressure ratio across the
machine if the total-to-total efficiency is 75%. Given: The properties of air are
h02 − h01 = 6000 J/kg
T01 = 288 K
Find: T02 , p02 /p01 .
13
p01 = 101.3 kPa
ηtt = 0.75
Solution:
T02 = T01 +
h02 − h01
6000
= 288 +
= 294.0 K
cp
1004.5
Next
ηtt =
⇐
T02s − T01
T02 − T01
T02s = T01 + ηtt (T02 − T01 ) = 288 + 0.75 · 6 = 292.5 K
so that
p02
=
p01
(
T02s
T01
)γ/(γ−1)
(
=
292.5
288
)3/5
= 1.0555
⇐
Exercise 2.19 A multi-stage turbine has a total pressure ratio 2.5 across each of
four stages. The inlet total temperature is T01 = 1200 K and the total-to-total
efficiency of each stage is 0.87. Evaluate the overall total-to-total efficiency of the
turbine if steam is flowing through it. Steam can be assumed to behave as a perfect
gas with γ = 1.3. Why is the overall efficiency higher than the stage efficiency?
Given: The properties of air are
p02
= 2.5
p01
T01 = 1200 K
ηtt = 0.87
Find: the total-to-total efficiency for the multistage turbine.
Solution:
p01
p01 p02 p03 p04
=
= 2.54 = 39
p05
p02 p03 p04 p05
Also
[
( )γ/(γ−1) ]
p02
T02 = T01 1 − ηtt
p01
and since the efficiency of each stage is the same
[
( )γ/(γ−1) ]
p02
T03 = T02 1 − ηtt
p01
[
(
T03 = T01 1 − ηtt
14
p02
p01
)γ/(γ−1) ]2
and so forth to
[
so that
T05
(
)γ/(γ−1) ]4
T05 = T01 1 − ηtt
p02
p01
[
(
= 1200 1 − 0.87 1 −
1
2.50.3/1.3
)]4
= 581.1 K
The overall efficiency is therefore
η0 =
T01 − T05
1 − T05 /T01
=
T01 − T05s
1 − (p05 /p01 )γ/(γ−1)
η0 =
1 − 581.1/1200
= 0.904
1 − (1/2.54 )0.3/1.3
The overall efficiency is larger than the stage efficiency because the internal energy at the exit of each stage is higher because of internal heating and this becomes
available during the next expansion process.
Exercise 2.20 Gases from a combustion chamber enter a gas turbine at total pressure of 700 kPa and total temperature of 1100 K. The total pressure and total
temperature at the exit of the turbine are 140 kPa and 780 K. If γ = 4/3 is used
for the mixture of combustion gases, which has a molecular mass of 28.97, find
the total-to-total efficiency and the total-to-static efficiency of the turbine, if the
exit velocity is 210 m/s.
Given: The inlet and exit conditions are and V2 = 210 m/s.
p01 = 700 kPa
p02 = 140 kPa
T01 = 1100 K
T02 = 780 K
Combustion gases with γ = 4/3 and R = 287 J/kg.
Find: the total-to-total efficiency.
Solution:
γR
= 4 · 287 = 1148 J/kg
cp =
γ−1
( )(γ−1)/γ
(
)1/4
p02
140
T02s = T01
= 1100
= 735.6 K
p01
700
ηtt =
T01 − T02
1100 − 780
=
= 0.878
T01 − T02s
1100 − 735.6
15
⇐
T2s = T02s −
ηts =
V2s2
2102
= 735.6 −
= 716.4 K
2cp
2 · 1148
T01 − T02
1100 − 780
=
= 0.834
T01 − T2s
1100 − 716.4
⇐
Exercise 2.21 Air enters a compressor from atmosphere at 101.3 kPa, 288 K. It
is compressed to static pressure of 420 kPa and at the exit its velocity is 300 m/s.
The compressor total-to-total efficiency is 0.82. (a) Find the exit static temperature, without making the assumption that V2s = V2 . (b) Find the exit static
temperature by assuming that V2s ̸= V2 .
Given: The exit velocity of air is V2 = 300 m and the properties of air are
p01 = 101.3 kPa
T01 = 288 K
p2 = 420 kPa
ηtt = 0.82
Find: T2 is two different ways.
Solution: Assume first that V2s = V2 , then
(
T2s = T01
T02s
p2
p01
)(γ−1)/γ
(
= 288
420
1013.3
)1/3.5
= 432.4 K
V22
3002
= T2s +
= 432.4 +
= 477.1 K
2cp
2 · 1004.5
As a result
T02 = T01 +
1
1
(T02s − T01 ) = 432.4 +
(477.1 − 288) = 518.7 K
ηtt
0.82
and
T2 = T02 −
V22
3002
= 518.7 −
= 473.9 K
2cp
2 · 1004.5
If V2 ̸= V2s , then
T02s = T2s +
⇐ (a)
V2s2
V 2 T2s
= T2s + 2
2cp
2cp T2
and
1
1
V22 T2s
V22
T2 = T01 + (T02s − T01 ) = T01 + (T2s − T01 +
)−
ηtt
ηtt
2cp T2
2cp
16
from which
T22
1
V22
V22 T2s
− [T01 + (T2s − T01 ) −
] T2 −
=0
ηtt
2cp
2ηtt cp
which when solved for T2 gives
√
419.22
419.2
T2 = −
+
+ 23620 = 469.5 K
2
4
⇐
(b)
Exercise 2.22 Liquid water issues at velocity V1 = 20 m/s from a bank of five
oblique nozzles shown in Figure 1. The nozzles with wall thickness t = 0.2 cm
are spaced s = 4 cm apart. The nozzle angle is α1 = 70 deg. Using the mass and
momentum balance: (a) Find the downstream velocity V2 . (b) Find the pressure
increase in the flow. (c) Show how to deduce this result from the final equation in
the example on mixing from the text. (d). If the thickness of the wall is vanishingly
small, what is the change in pressure.
Given:
t = 0.2 cm
s = 4 cm
V1 = 20 m/s
α1 = 70◦
Find: the pressure increase if δp = p2 − p1 , and also the pressure increase is
t = 0.
Solution: From the mass balance
V2 V1 (cos α1 − t/s)
From the momentum balance
ρV2 (V2 − V1 cos α1 ) = p1 − p2
and after eliminating V2 between this and the mass balance lead to
(
)
(
)
t
0.2
2 0.2
◦
2t
cos α1 −
= 998·20
cos(70 ) −
= 5.84 kPa
p2 −p1 = ρV1
s
s
4
4
⇐
Since Va = V1 cos α1 ) from
p2 −p1 = ρV (Va −V )
V =
t
t
Va A a
= V1 cos α1 (1−
) = V1 (cos α1 − )
Aa + Ab
s cos α1
s
17
V1
V2
α1
Figure 1: Nozzles with an oblique discharge.
and
Va − V = V1 cos α1 − V1 cos α1 + V1
so that
p2 − p1 =
t
ρV12
s
(
t
cos α1 −
s
t
s
)
⇐
Exercise 2.23 Consider the flow shown in Figure 1. Prove that the kinetic energy
lost in the flow as it moves to the downstream section is equal to that associated
with the transverse component of the velocity.
Given: The channel width s and angle α1 , and V .
Find: the expression for the loss in terms of the transverse component.
Solution: From continuity, after canceling s
V2 = V1 cos α1
From x-component of momentum equation, after canceling s
V2 (V2 − V1 cos α1 ) = p1 − p2
Substituting from continuity gives p1 = p2 . From energy equation
p1 V12
p2 V22
V2
+
=
+
+ζ 1
ρ
2
ρ
2
2
Canceling the pressure terms and from continuity V2 = V1 cos α1 yields
1 − cos2 α1 = ζ
V12 = V22 + ζV12
so that
ζ
V 2 sin2 α1
V2
V12
= 1
= t
2
2
2
18
⇐
Chapter 3
Exercise 3.1 Conditions in an air reservoir are 680 kPa and 560 K. From there
the air flows isentropically though a convergent nozzle to a back pressure of
101.3 kPa. Find the velocity at the exit plane of the nozzle.
Given: The properties are
T01 = 560 K
p01 = 630 kPa
p2 = 101.3 kPa
Find: The velocity V2 .
Solution: Since
101.3
p2
=
= 0.149
p01
630
the flow is choked and the velocity at the exit is the sonic velocity. To find it first
calculate
2 · 560
2T01
=
= 466.7 K
T∗ =
γ+1
2.4
and then
√
√
V ∗ = γRT ∗ = 1.4 · 287 · 466.7 = 433 m/s
⇐
Exercise 3.2 Air flows in a converging duct. At a certain location, where the area
is A1 = 6.5 cm2 , pressure is p1 = 140 kPa and Mach number is M1 = 0.6. The
mass flow rate is ṁ = 0.25 kg/s. (a) Find the stagnation temperature. (b) If the
flow is choked find the size of the throat area. (c) Give the percent reduction in
area from station 1 to the throat. (d) Find the pressure at the throat.
Given: At the location A1 = 6.5 cm2 and the properties are
M1 = 0.6
p1 = 140 kPa
ṁ = 0.25 kg/m
Find: The velocity T01 , A∗ , percent reduction in area, and p∗ .
Solution: Solve the equation
ṁ = ρ1 A1 V1 =
p 1 A1 √
M1 γRT1
RT1
for T1 . It yields
(
(
)2
)2
p1 A1 M1
140000 · 0.00065 · 0.6
γ
1.4
T1 =
=
= 232.7 K
ṁ
R
0.25
287
19
and then
(
T01 = T1
)
γ−1 2
1+
M1 = 232.7(1 + 0.2 · 0.62 ) = 249.4 K
2
(
)(γ+1)/2(γ+1)
1
2
γ−1 2
A∗
=
+
M
A1
M1 γ + 1 γ + 1 1
so that
⇐
)−(γ+1)/2(γ+1)
γ−1 2
2
A = A1 M 1
+
M
γ+1 γ+1 1
2
0.4
= 6.5 · 0.6(
+
· 0.62 )−3 = 5.47 cm2
⇐
2.4 2.4
Percent reduction is
6.5 − 5.47
A1 − A∗
=
= 0.1584 15.84 %
⇐
A1
6.5
(
∗
To calculate the exit pressure, first
)γ/(γ−1)
(
γ−1 2
p01 = p1 1 +
M1
= 140(10.2 · 0.62 )3.5 = 178.6 kPa
2
and then
(
∗
p = p01
2
γ−1
)γ/(γ−1)
(
= 178.6
2
2.4
)3/5
= 94.3 kPa
⇐
Exercise 3.3 Air flows in a convergent nozzle. At a certain location, where the
area is A1 = 5 cm2 , pressure is p1 = 240 kPa and temperature is T1 = 360 K.
Mach number at this location is M1 = 0.4. Find the mass flow rate.
Given: At the location A1 = 5 cm2 and the properties are
T1 = 360 K
p1 = 140 kPa
M1 = 0.4
Find: Find the mass flow rate.
Solution:
√
√
V1 = M1 γRT1 = 0.4 1.4 · 287 · 360 = 152.1 m/s
ρ1 =
p1
240000
=
= 2.323 kg/m3
RT1
287 · 360
so that
ṁ = ρ1 A1 V1 = 2.323 · 5 · 10−4 · 152.1 = 0.177 kg/s
20
⇐
Exercise 3.4 The area of a throat in a circular nozzle is At = 1 cm2 . For a choked
flow find the diameter where M1 = 0.5. Determine the value of Mach number at a
location where the diameter is D2 = 1.941 cm. Assume the flow to be isentropic
and γ = 1.4.
Given: At the throat A∗ = 1.0 cm2 and D2 = 1.941 cm.
Find: The D1 and M2
Solution: From
(
)(γ+1)/2(γ+1)
(
)3
1
1
A∗
2
γ−1 2
2
0.4 2
=
+
M
=
+
0.5 ) = 1.34
A1
M1 γ + 1 γ + 1 1
0.5 2.4 2.4
so that
√
A1 = 1.34A∗ = 1.34 · 1 = 1.34 cm2
D1 =
4A1
= 1.31 cm
π
⇐
For D2 = 1.941 cm, the area is A2 = πD22 /4 = 2.96 cm2 , then solve
(
)(γ+1)/2(γ+1)
2
2.96
1
γ−1 2
A2
=
= 2.96 =
+
M
A∗
1
M2 γ + 1 γ + 1 2
for M2 . Carrying out the solution by iteration gives M2 = 0.2
⇐
Exercise 3.5 In a location in a circular nozzle where the area is A1 = 4 , Mach
number has the value M1 = 0.2. Find the diameter at a location where M = 0.6.
Given: At the location A1 = 4 cm2 and M1 = 0.2
Find: The area at which M2 = 0.6.
Solution: From
(
)(γ+1)/2(γ+1)
(
)3
A1
2
2
1
γ−1 2
1
0.4 2
=
+
M
=
+
0.2
= 2.964
A∗
M1 γ + 1 γ + 1 1
0.2 2.4 2.4
therefore A∗ = 4/2.964 = 1.350 cm2 and then from
(
)(γ+1)/2(γ+1)
)3
(
1
1
2
γ−1 2
0.4 2
A2
2
=
= 1.188
=
+
M
+
0.6
A∗
M2 γ + 1 γ + 1 2
0.6 2.4 2.4
so that A2 = 1.188 · 1.350 = 1.604 cm2 and
√
√
D2 = 4A2 /π = 4 · 1.604/π = 1.43 cm
21
⇐
Exercise 3.6 Air flows through a circular duct 15 cm in diameter with a flow rate
2.25 kg/s. The total temperature and static pressure at a certain location in the
duct are 30 C and 106 kPa, respectively. Evaluate (a) the flow velocity, (b) the
static temperature, (c) the total pressure, and (d) the density at this location.
Given: At the location where D1 = 15.0 cm2 and the flow rate is ṁ = 2.25 kg/s,
the properties are
T01 = 303.15 K
p1 = 106 kPa
Find: The velocity, the static temperature, the total pressure, the static density.
Solution:
πD12
A1 =
= 0.01767 m2
4
From the mass balance
(
)1/2
√
γ
p1 A 1 M 1 √
T01
ṁ = ρ1 A1 V1 =
γRT1 = p1 A1 M1
RT1
RT01 T1
or
M14
2
2
+
M12 −
γ−1
γ−1
(
ṁ
p1 A1
)2
RT01
=0
γ
Substituting the numerical values into this gives
√
M14 + 5M12 − 0.4483 = 0
M12 = −2.5 + 2.52 + 0.4483
M1 = 0.297
Therefore
(
T1 = T01
and
V1 = M 1
γ−1 2
M1
1+
2
√
)−1
=
303.15
= 297.9 K
1 + 0.2 · 0.2972
√
γRT1 = 0.297 1.4 · 287 · 297.9 = 102.8 m/s
The stagnation pressure is
(
)γ/(γ−1)
)3.5
(
303.15
T01
= 106
= 112.7kPa
p01 = p1
T1
297.9
and the static density is
ρ1 =
106
p1
=
= 1.24 kg/m3
RT1
0.287 · 297.9
22
⇐
⇐
⇐
Exercise 3.7 Conditions in an air reservoir are 380 kPa and 460 K. From there
the air flows though a convergent nozzle to a back pressure of 101.3 kPa. The
polytropic efficiency of the nozzle is ηp = 0.98. Find: (a) Exit plane pressure, (b)
Exit plane temperature, (c) Velocity at the exit plane of the nozzle.
Given: The properties of air are.
T01 = 460 K
p01 = 380 kPa
pb = 101.3 kPa
ηp = 0.98
Find: pe , Te , Ve .
Solution: Since pb /p01 = 101.3/380 = 0.266, the flow is choked. The polytropic
exponent is
n=
Then
pe = p01
(
2
γ+1
γ
1/4
=
= 1.389
ηp + γ(1 − ηp )
0.98 + 1.4 · 0.02
)n/(n−1)
(
= 380
2
2.4
)
1/389/0.389 = 198.1 kPa
⇐
and
2
2
T01 =
460 = 383.3 K
⇐
γ+1
2.4
√
√
and the Mach number is Me = (n − 1)/(γ − 1) = 0.389/0.4 = 0.986 so
that
√
√
Ve = Me γRTe = 0.986 1.4 · 287 · 383.3 = 387.0 m/s ⇐
Te =
Exercise 3.8 Air issues from a reservoir at conditions 260 kPa and 540 K into a
converging nozzle. The nozzle efficiency is estimated to be ηN = 0.986. The
back pressure is pb = 101.3kPa. Find: (a) The exit Mach number, (b) Exit plane
temperature, (c) Exit plane pressure, (d) Exit velocity.
Given: The properties of air are.
T01 = 540 K
p01 = 260 kPa
pb = 101.3 kPa
ηN = 0.986
Find: Me , pe , Te , Ve .
Solution: Since pb /p01 = 101.3/260 = 0.389, the flow is choked. The Mach
number can be calculated from
]1/2
[
√
(γ − 1)(3ηN − 1) − 2γ + [(γ − 1)(3ηN − 1) − 2γ]2 + 8ηN (1 − ηN )(γ − 1)2
Me =
2(γ − 1)2 (1 − ηN )
23
[
Me =
0.4 · 1.958 − 2.8 +
√
(0.4 · 1.958 − 2.8)2 + 8 · 0.989 · 0.014 · 0.42
2 · 0.16 · 0.014
]1/2
= 0.9883
The exit temperature is
Te =
2T01
2 · 540
=
= 450 K
γ+1
2.4
⇐
The polytropic exponent is n = 1 + (γ − 1)Me2 = 1.3907 and the exit pressure is
(
pe = p01
2
γ+1
)n/(n−1)
(
= 260
2
2.4
)1.3907
0.3907 = 135.9 kPa
⇐
and the velocity is
Ve = Me
√
√
γRTe = 0.9882 1.4 · 287 · 450 = 420.2 m/s
⇐
Exercise 3.9 At the inlet to a nozzle the conditions are M1 = 0.3, p01 = 320 kPa,
and T01 = 430 K. The flow is irreversible with polytropic exponent n = 1.396.
Show that
( )γ/(γ−1) ( )(n−1)/n
p01
p1
T02
=
T2
p1
p2
Find the Mach number at a location where p2 = 210 kPa.
Given: T01 = 430 K, p01 = 320 kPa, M1 = 0.3, and n = 1.396.
Find: Show that the expression in the statement of the exercise is correct. Also
find the Mach number at the location where p2 = 210 kPa.
Solution: From
( )(n−1)/n
p1
T1
T1 T01
=
=
p2
T2
T01 T2
Therefore
T01
T01
=
T2
T1
(
p1
p2
)(n−1)/n
and since T01 − T02
T02
=
T2
(
p01
p1
)(γ−1)/γ (
24
p1
p2
)(n−1)/n
⇐
From
p01
=
p1
(
γ−1 2
1+
M1
2
so that
p1 =
and
)γ/(γ−1)
= (1 + 0.2 · 0.32 )3/5 = 1.0644
320
= 300.63 kPa
1.0644
(
) ( )n/(n−1)
γ−1 2
p1
1−
M1
2
p2
(
)0.396
T02
300.62
= (1 + 0.2 · 0.32 )
1.396 = 1.127
T2
210
T01 T1
T02
=
=
T2
T1 T2
and also
T02
γ−1 2
=1+
M2
T2
2
so that
√
M2 =
2
γ−1
(
) √
T02
2
−1 =
(1.127 − 1) = 0.797
T2
0.4
⇐
Exercise 3.10 Flow from a reservoir with p01 = 260 kPa and T01 = 530 K flows
through a nozzle. It is estimated that the static enthalpy loss coefficient is ζ =
0.020. The exit pressure is p2 = 180 kPa. (a) Find the exit Mach number. (b)
Find the polytropic efficiency of the nozzle.
Given: The conditions are
T01 = 530 K
p01 = 260 kPa
p2 = 180 kPa
ζ = 0.020
Find: ηp .
Solution: The static temperature for an isentropic process is
(
( )(γ−1)/γ
)1/3.5
180
p2
= 530
T2s = T01
= 4771 K
p01
260
The static temperature is obtained from
ζ=
T2 − T2s
T02 − T2
T2 =
ζT02 + T 2s
0.02 · 530 + 477.1
=
= 478.2 K
1+ζ
1.002
25
From
T02
γ−1 2
= 1+
M2 M2 =
T2
2
√
2 T02
(
− 1) =
γ − 1 T2
√
2 530
(
− 1) = 0.736 ⇐
0.4 478.2
The polytropic exponent is obtained from
T2
=
T01
(
p2
p01
)(n−1)/n
therefore
n=
(
ηN =
n−1
n
)(
1
1−
γ
γ−1
ln(T2 /T01 )
ln(p2 /p01 )
)
=
=
1
1−
ln(478.2/530)
ln(180/260)
0.3885 1.4
= 0.9793
1.3885 0.4
= 1.3885
⇐
Exercise 3.11 A two dimensional nozzle has a shape
4 x(x − 1)
+
5 2(x + 2)
√
The
√ nozzle stretches from 0 < x/a < 2( 2 − 1). The throat is at xt /a =
2( 2 − 1). The scale factor a is chosen such that the half-width of the nozzle at
x = 0 is 4a/5. Assume that 4f¯ = 0.02 and the inlet Mach number is M1 = 0.5.
Calculate and plot p/p1 as a function of x. Calculate the Mach number along the
nozzle and graph it on the same plot.
Given: The shape of the flow channel and the friction factor.
Find: Plot p/p1 and the Mach number.
Solution:
y=
%Nozzle Calculation
%Nozzle Calculation
clear all; clf
k=1.4; f=0.02/4; w=1;
x=[0:0.005:2*(sqrt(2)-1)]; y=4/5+x.*(x-2)./(4+2*x);
n=length(x);
M(1)=0.5
mx(1)=M(1)^2;
d(1)=4*y(1)*w/(w+y(1));
A(1)=2*y(1)*w;
26
p0(1)=1;
p(1)=1;
for i=2:n
d(i)=4*y(i)*w/(w+y(i));
A(i)=2*y(i)*w;
mx(i)=mx(i-1)-2*mx(i-1)*(1+0.5*(k-1)* ...
mx(i-1))*(A(i)/A(i-1)-1)/(1-mx(i-1))+ ...
mx(i-1)*k*mx(i-1)*(1+0.5*(k-1)*mx(i-1))* ...
4*f*(x(i)-x(i-1))/(d(i)*(1-mx(i-1)));
M(i)=sqrt(mx(i));
p0(i)=p0(i-1)-p0(i-1)*k*mx(i-1)*2*f*(x(i)-x(i-1))/d(i);
r(i)=((2+(k-1)*mx(i))/(2+(k-1)*mx(1)))^(k/(k-1));
p(i)=p0(i)*p(1)/(p0(1)*r(i));
end
plot(x,p,x,M);grid;
Exercise 3.12 Steam enters a nozzle from a steam chest at saturated vapor state
at pressure p0 = 0.8 bar. It expands isentropically through a steam nozzle. Find
the degree of supersaturation when it crosses the Wilson line at x = 0.96.
Given: Saturated steam at p01 = 80 kPa.
Find: The degree of saturation if it isentropically expands to x2 = 0.96.
Solution: The temperature of saturated steam at this pressure is T01 = 366.6 K.
Assuming that the adiabatic index according to Zeuner’s formula is γ = 1.135,
then for an isentropic process
(
T2 = T01
p2
p01
)(γ−1)/γ
(
= 366.6
36.6
80
)0.135/1.135
= 333.7 K
The pressure of saturated steam at this temperature is pss = 20.43 kPa. Hence the
degree of supersaturation is
S=
p2
36.6
=
= 1.774
pss
20.43
⇐
Exercise 3.13 Consider a supersonic flow over a convex corner with angle θ2 =
5◦ , when the inflow moves in the direction of θ1 = 0◦ . The upstream Mach number
is M1 = 1.1, pressure is p1 = 130 kPa and T1 = 310 K. Find: (a) Mach number,
(b) temperature, and (c) pressure after the expansion is complete.
27
Given: The conditions at the inlet are
M1 = 1.1
p1 = 130 kPa
T1 = 310 K
and the wall makes an angle θ2 = 15◦ .
Find: M2 , T2 and p2 .
Solution: The Prandtl-Meyer function a the inlet is
(
)
)
(
1
1
−1
−1
√
√
µ1 = tan
= tan
= 65.38◦
1.12 − 1
M12 − 1
then
√
)
(√
√
γ+1
γ−1
−1
−1
2
ν1 =
tan
(M − 1) − tan
M12 − 1
γ−1
γ+1 1
(√
)
√
√
2.4
0.4
tan−1
(1.12 − 1) − tan−1 1.12 − 1 = 1.336◦
ν1 =
0.4
2.4
so that
ν2 = ν1 + θ2 = 1.336◦ + 5◦ = 6.336◦
Since
√
ν2 =
γ+1
tan−1
γ−1
(√
)
√
γ−1
−1
2
(M − 1) − tan
M22 − 1
γ+1 2
With γ = 1.4 and ν1 = 6.336◦ , this equation can be solved by iterations for M2 .
It gives M2 = 1.306
⇐ (a)
Next
T0 = T1 (1 +
and
T2 = T0 (1 +
and
(
p2 = p 1
T2
T1
γ−1 2
M1 ) = 310(1 + 0.2 · 1.12 ) = 385.0 K
2
γ − 1 2 −1
M1 ) = 385(1 + 0.2 · 1.12 )−1 = 287.1 K
2
(
)γ/(γ−1)
= 130
287.1
310
28
)3/5
= 99.36 kPa
⇐
Exercise 3.14 Consider the steam flow from a low pressure nozzle at an angle
α = 65◦ . At the inlet of the nozzle steam is saturated vapor at pressure p0 =
18 kPa. Steam exhausts into the inter-blade space, where pressure is 7 kPa. Find
the angle θ by which the flow turns on leaving the nozzle, the far downstream
velocity, and its direction.
Given: Steam, with γ = 1.135 and R = 8314/18 = 461.9 J/kg K at the inlet as
saturated steam at p0 = 18 kPa and leaves at the nozzle angle α1 = 65◦ . Its far
downstream pressure is p2 = 7 kPa.
Find: The turning angle θ, V2 and α2 .
Solution: The nozzle is choked because p2 /p0 = 0.389. The saturation temperature is T0 = 330.9 K. Then
(
T2 = T0
From
p2
p0
)(γ−1)/γ
(
= 330.9
7
18
)0.135/1.135
= 295.7 K
T0
γ−1 2
=1+
M2
T2
2
√
(
) √
(
)
2
T0
2
330.9
M2 =
−1 =
− 1 = 1.327
γ − 1 T2
0.135 295.37
so that
V2 = M2
√
√
γRT2 = 1.327 1.135 · 461.9 · 295/7 = 522.6 m/s
√
√
γ+1
γ−1
ϕ=
tan−1
(M 2 − 1)
γ−1
γ+1 2
√
√
3.135
0.135
−1
ϕ=
tan
(1.3282 − 1) = 49.2◦
0.135
2.135
Next
In addition
(
−1
µ2 = tan
√
therefore
θ2 = ϕ −
1
M22 − 1
)
(
−1
= tan
1
√
1.3282 − 1
)
π
+ µ2 = 49.36 − 90 + 4839 = 8.1◦
2
29
⇐
= 48.86◦
⇐
⇐
Exercise 3.15 Consider the steam flow from a low pressure nozzle at an angle
α = 65◦ . At the inlet of the nozzle steam is saturated vapor at pressure p0 =
18 kPa. Steam exhausts into the inter-blade space, where pressure is 7 kPa. Using
the continuity equation, find the angle θ by which the flow turns on leaving the
nozzle, the far downstream velocity, and its direction.
Given: Steam, with γ = 1.135 and R = 8314/18 = 461.9 J/kg K at the inlet as
saturated steam at p0 = 18 kPa and leaves at the nozzle angle α1 = 65◦ . Its far
downstream pressure is p2 = 7 kPa
Find: The turning angle θ, V2 and α2 .
Solution: The flow is choked because p2 /p0 = 0.389. The specific volume of
saturated steam at p0 = 18 kPa is v0 = 8.46 m3 /kg and the temperature is T0 =
330.2 K. The specific volume at the throat is
(
)1/0.135
γ + 1 1/(γ−1)
2.135
v1 = v0
= 8.46
= 13.7 m3 /kg
2
2
and the temperature is
T1 =
2T 0
2 · 330.2
=
= 309.3 K
γ+1
1.135
and the velocity is
V1 =
√
√
γRT1 = 1.135 · 461.9 · 309.3 = 403.1 m/s
The temperature far downstream is
( )0.135/1.135
( )(γ−1)/γ
7
p2
T2 = T0
= 330.9
= 295.7 K
p0
18
and the Mach number can be calculated from
T0
γ−1 2
=1+
M2
T2
2
√
) √
)
(
(
2
2
T0
330.9
−1 =
− 1 = 1.327
M2 =
γ − 1 T2
0.135 295.7
so that
V2 = M 2
√
√
γRT2 = 1.327 1.135 · 461.9 · 295.7 = 522.6 m/s
30
⇐
The specific volume is
v2 =
RT2
461.9 · 295.7
=
= 19.5 m3 /kg
p2
7000
and by continuity the angle is
)
V1 v2
cos α1
α2 = cos
V2 v1
(
)
403.1 · 19.5
−1
◦
α2 = cos
cos(65 ) = 62.5◦
522.6 · 13.77
(
−1
θ = α1 − α2 = 65◦ − 62.73◦ = 2.5◦
31
⇐
⇐
Chapter 4
Exercise 4.1 Steam enters a rotor of an axial turbine with an absolute velocity
V2 = 320 m/s at the angle α2 = 73◦ . The axial velocity remains constant. The
blade speed is U = 165 m/s. The rotor blades are equiangular so that β3 = −β2
and the magnitude of the relative velocity remains constant across the rotor. Draw
the velocity triangles. Find, (a) the relative flow angle β2 , (b) the magnitude of
the velocity V3 after the flow leaves the rotor, and (c) the flow angle α3 which V3
makes with the axial direction.
Given: The rotor blades are equiangular with β3 = −β2 and W3 = W2 . At the
inlet of the rotor
V2 320 m/s
α2 = 73◦
U = 165 m/s
Find: β2 , V3 , α3 .
Solution: The axial and tangential components of the velocity are
Vx2 = V2 cos α2 = 320 cos(73◦ ) = 93.56 m/s
Vu2 = V2 sin α2 = 320 sin(73◦ ) = 306.0 m/s
The the tangential component of the relative velocity is
Wu2 = Vu2 − U = 306.0 − 165.0 = 141.0 m/s
√
so that
W2 =
2
2
Wx2
+ Wu2
=
√
93.562 + 141.02 = 169.2 m/s
and
β2 = tan−1 (Wu2 /Wx2 ) = tan−1 (141.0/93.56) = 56.44◦
Since W3 = W2
β3 = −56.44◦
⇐ (a)
Wu3 = −W u2 = −141.0m/s
and
√
Vu3 = Wu3 +U = 141.0+165.0 = 24.0 m/s V3 =
2
2
Vx3
+ Vu3
=
√
93.562 + 24.02 = 96.6 m/s ⇐
so that
α3 = tan−1 (Vu3 /Vx3 ) = tan−1 (24.0/93.56) = 14.4◦
32
⇐
(c)
Exercise 4.2 Water with density 998 kg/m3 flows in centrifugal pump at the rate
of 22 liters per second. The impeller radius is r2 = 7.7 cm and the blade width
at the impeller exit is b2 = 0.8 cm. If the flow angles at the impeller exit are
α2 = 67◦ and β2 = −40◦ , find the rotational speed of the shaft in rpm.
Given: The flow rate, the impeller radius and the height of its blade. The flow
angles are also known at the exit.
Find: The rotational speed of the shaft.
Solution: From
Q = 2πr2 b2 Vr2
Then
Vr2 =
Q
0.022 · 104
=
= 5/68 m/s
2πr2 b2
2π · 7.7 · 0.8
Vu2 = Vr2 tan α2 = 5.68 tan(67◦ ) = 13.39 m/s
Wu2 = Wr2 tan β2 = 5.68 tan(−40◦ ) = −4.77 m/s
and
U = Vu2 − Wu2 = 13.39 + 4.77 = 18.16 m/s
So that
Ω=
U
18.16 · 100 · 30
=
= 2252 rpm
r2
7.7 · π
⇐
Exercise 4.3 In a velocity diagram at the inlet of a turbine the angle of the absolute velocity is 60◦ and the flow angle of the relative velocity is −51.7◦ . Draw
the velocity diagram and find the value of U/V and Vx /U . Given: The angles
α = 60◦ and β = −51.7◦ .
Find: U/V and Vx /U .
Solution: Let V = 1, then
Vx = V cos α = cos(60◦ ) = 0.5
and
Vu = V sin α = sin(60◦ ) = 0.866
Wu = Wx tan(β) = 0.5 tan(−51.7◦ ) = −0.633
so that
U = Vu − Wu = 0.866 + 0.633 = 1.5
and U/V = 1/5
⇐
The ratio Vx /U = 0.866/1.5 = 1/3
⇐
33
Exercise 4.4 A small axial-flow turbine has an output power of 37 kW when handling one kg of air per second with an inlet total temperature of 335 K. The totalto-total efficiency of the turbine is 80%. The rotor operates at 50, 000 rpm and
the mean blade diameter is 10 cm. Evaluate (a) the average driving force on the
turbine blades, (b) the change in the tangential component of the absolute velocity
across the rotor, and (c) the required total pressure ratio across the turbine.
Given: Ẇ = 37 kW, ṁ = 1 kg/s, T01 = 335 K, ηtt = 0.8, and r2 = 0.05 m
Ω = 55, 000 rpm.
Find: The force Fu , Vu2 − Vu3 and p03 /p01 .
Solution: The blade speed is
U = r2 Ω =
0.05 · 50, 000 · π
= 261.8 m/s
30
so that from Ẇ = Fu U , the force is Fu = Ẇ /U = 37, 000/261.8 = 141.3 N ⇐
The specific work is w = Ẇ /ṁ = 37 kJ/kg and the change in tangential
velocity is from
w = u(VU 2 − Vu3 )
Vu2 − Vu3 =
w
37, 000
=
= 141.3 m/s
U
261.8
⇐
The isentropic work is
ws =
w
37
ws
46.25
=
= 46.25 kJ/kg rmso that T02s = T01 −
= 335−
= 289 K
ηtt
0.8
cp
1.0045
and
p01
=
p03
(
T01
T02s
)γ/(γ−1)
(
=
335
289
)3.5
= 1.678
⇐
Exercise 4.5 The exit flow angle of stator in an axial steam turbine is 68◦ . The
flow angle of the relative velocity leaving the rotor is −67◦ . Steam leaves the
stator at V2 = 120 m/s and the axial velocity is Vx2 = 0.41U . At the exit of
the rotor blades the axial steam velocity is Vx3 = 0.42U . The mass flow rate is
ṁ = 2.2 kg/s. Find, (a) the flow angle entering the stator assuming it to be the
same as the absolute flow angle leaving the rotor, (b) the flow angle of the relative
velocity entering the rotor, (c) the reaction, and (d) the power delivered by the
stage.
Given: α2 = 68◦ , β3 = −67◦ , V2 = 120 m/s, Vx2 = 0.41U , Vx3 = 0.42U , and
ṁ = 2.2 kg/s.
34
Find: β2 , α3 , and R.
Solution: From
Vx2 = V2 cos α2 = 120 cos(68◦ ) = 44.95 m/s
Vu2 = V1 sin α2 = 120 sin(68◦ ) = 111.3 m/s
and U = Vx2 /0.41 = 44.95/0.41 = 109.64 m/s. Thus
Wu2 = Vu2 − U = 111.26 − 109.64 = 1.62 m/s
(
and
−1
β2 = tan
Wu2
Wx2
)
(
−1
= tan
1.62
44.95
)
= 2.1◦
⇐
(b)
Next
Vx3 = 0.42U = 0.42 · 109.64 = 46.05 m/s
Wu3 = Wx3 tan β3 = 46.05 tan(−67◦ ) = −108.5 m/s
so that
Vu3 = Wu3 + U = −108.5 + 109.64 = 1.16 m/s
(
and
−1
α3 = tan
Vu3
Vu2
)
(
−1
= tan
1.16
46.05
)
= 1.44◦
⇐
(a)
Next
w = U (Vu2 − Vu3 ) = 109.64(112.26 − 1.16) = 12072 J/kg
and Ẇ = ṁw = 202 · 12072 = 26.6 kW
The reaction is
R=
⇐
(d).
108.52 + 46.042 − 1.622 − 44.952
W32 − W22
=
= 0.492
2e
2 · 12072
⇐ (c)
Exercise 4.6 The axial component of air flow leaving a stator in an axial flow
turbine is Vx2 = 175 m/s and its flow angle is 64◦ . The axial velocity is constant,
the reaction of the stage is R = 0.5, and the blade speed is U = 140 m/s. Since the
reaction is fifty percent the relationships between the flow angles are β2 = −α3
and α2 = −β3 . Find the flow angle of the velocity entering the stator.
Given: Vx2 , U , α2 , U and R.
Find: α2 .
Solution: The tangential velocity is
Vu2 = Vx2 = tan α2 = 175 tan(64◦ ) = 358.8 m/s
35
and
Wu2 = Vu2 − U = 358.8 − 140 = 218.8 m/s
Therefore
(
−1
β2 = tan
Wu2
Wx2
(
)
−1
= tan
218.8
175
)
= 51.35◦ α3 = −54.3◦
⇐
Exercise 4.7 The air flow leaving the rotor of an axial flow turbine is Vx3 =
140 m/s and its flow angle is 0◦ . The axial velocity is constant and equal to the
blade speed. The inlet flow angle to the rotor is α2 = 60◦ . Find the reaction.
Given: Vx , U = Vx , α1 = 0◦ , and α2 = 60◦ .
Find: R
Solution: Since the flow at the inlet and exit of the stage is axial
√
√
√
W3 = Vx2 + U 2 = U 2 = 140 2 = 198 m/s
and
Vu2 = Vx2 tan α2 = 140 tan(60◦ ) = 242.5 m/s
so that
Wu2 = Vu2 − U = 242.5 − 140 = 102.58 m/s
and
√
√
2
W2 = Wx2 + Wu2
= 1402 + 102.52 = 173.5 m/s
The specific work is
w = U (Vu2 − Vu3 ) = U Vu2 = 140 · 242.5 = 33, 950 J/kg
so that
R+
W32 − W22 1982 − 173.52
= 0.134
2w
2 · 33, 950
⇐
Exercise 4.8 A large centrifugal pump operates at 6000 rpm and produces a head
of 800 m while the flow rate is 30, 000 liters per minute. (a) Find the value of the
specific speed. (b) Estimate the efficiency of the pump.
Given: H, Ω, Q.
Find: The specific speed and η.
Solution: The flow rate is standard units is
Q=
30, 000
= 0.5 m3 /s
1000 · 60
36
The specific speed is
√
√
Ω Q
6000π 0.5
Ωs =
=
= 0.533
(gH)3/4
30 · (9.81 · 800)3/4
From the figure for Q = 500 liters/s, the efficiency is 0.84
⇐
⇐
(a)
(b).
Exercise 4.9 A fan handles air at the rate of 500 liters per second when operating at 1800 rpm. (a) What is the flow rate if the same fan is operated at 3600
rpm? (b) What is the percentage increase in total pressure rise of the air assuming
incompressible flow? (c) What is the power input required at 3600 rpm relative
to that at 1800 rpm. Assume that the operating point of the fan in terms of the
dimensionless parameters is the same in both cases.
Given: Q1 , Ω1 , Ω2 , and D1 = D2 .
Find: Q2 , increase in total pressure, Ẇ2 .
Solution: From
Q1
Q2
Ω2
3600
=
Q2 = Q1
= 500
= 1000 l/s
3
3
Ω1 D1
Ω2 D2
Ω1
1800
(a)
Next
Ẇ2
Ẇ1
=
3 5
ρΩ1 D1
ρΩ32 D25
so that
Ẇ2 = Ẇ1
Then
Ẇ =
Ω32
= 8Ẇ1
Ω31
⇐ (c)
ṁ
(p02 − p01 )ηtt = Q∆p0 ηtt
ρ
and
Ẇ1 Q2 ηtt
(∆p0 )1
2
1
=
= =
(∆p0 )2
Q1 ηtt Ẇ2
8
4
so that the percent increase is 300 %
⇐
(b).
Exercise 4.10 An axial flow pump having a rotor diameter of 20 cm handles water
at the rate of 60 liters per second when operating at 3550 rpm. The corresponding
increase in total enthalpy of the water is 120 J/kg and the total-to-total efficiency
is 75%. Suppose a second pump in the same series is to be designed to handle
37
water having a rotor diameter of 30 cm and operating at 1750 rpm. For this second
pump what will be the predicted values for (a) the flow rate, (b) the change in the
total pressure of water, (c) the input power.
Given: Q1 , D1 , Ω1 , ηt t, (∆h0 )1 , D2 and Ω2 .
Find: Q2 , (∆p0 )2 , and Ẇ2 .
Solution: From
) ( )3
( ) ( )3
(
30
Q1
Q2
Ω2
D2
1750
= 99.8 l/s ⇐
=
Q2 = Q1
= 60
3
3
Ω1 D1
Ω2 D2
Ω1
D1
3550
20
Next
Ẇ1 = ṁ1 (∆h0 )1 = ρQ1 (∆h0 )1 =
998 · 60 · 120
= 7.18 kW
1000
then from
Ẇ2
Ẇ1
=
3 5
ρΩ1 D1
ρΩ32 D25
(
)3 ( )5
( )3 ( )5
D2
1750
30
Ω1
Ẇ2 = Ẇ1
= 7.18
= 6.54 kW Lef tarrow (b)
Ω2
D1
3550
20
Then
Ẇ2
6540
(∆h0 )2 =
=
= 65.6 J/kg
ρQ2
99.8
and
(∆p0 )2 = ρ (∆h0 )2 = 998 · 65.6 = 49.1 kPa ⇐ (c)
Exercise 4.11 A small centrifugal pump handles water at the rate of 6 liters per
second with input power of 5 hp and total-to-total efficiency of 70%. Suppose the
fluid being handled is changed to gasoline having specific gravity 0.70. What are
the predicted values for (a) flow rate, (b) input power, (c) total pressure rise of the
gasoline?
Given: Q1 , Ẇ1 , ηt t, ρ1 , Q2 = Q1 and the specific gravity of gasoline.
Find: Ẇ2 , (∆p0 )2 .
Solution: Since the diameter and the shaft speed are the same, then from
Q2
Q1
=
3
Ω 1 D1
Ω2 D23
Q2 = Q1 ⇐ (a)
and from
Ẇ1
Ẇ2
=
3 5
ρ 1 Ω 1 D1
ρ2 Ω32 D25
38
(a)
Ẇ2 = Ẇ1
ρ2
= 5 · 0.7 = 3.5 hp
ρ1
Next from
Ẇ2 = ṁ2 (∆h0 )2 =
⇐ (b)
Q2 (∆p0 )2
ηtt
so that
(∆p0 ) =
ηtt Ẇ2
3/5 · 0.7457 · 1000
=
= 304.5 kPa
Q2
6
⇐ (c)
Exercise 4.12 A blower handling air at the rate of 240 liters per second at the
inlet conditions of 103.1 kPa for total pressure and 288 K for total temperature. It
produces a pressure rise of air equal to 250 mm of water. If the blower is operated
at the same rotational speed, but with an inlet total pressure and total temperature
of 20 kPa and 253 K, find; (a) the predicted value for the mass flow rate, (b) the
total pressure rise.
Given: Ω2 = Ω1 , Q1 , p01 , T01 , (∆p0 )1 , p02 , and T02 .
Find: ṁ2 , (∆p0 )2 .
Solution: From
Q1
Q2
=
Q2 = Q1
3
Ω1 D1
Ω2 D23
The stagnation densities are
ρ01 =
p01
101, 300
=
= 1.226 kg/m3
RT01
287 · 288
p02
20, 000
=
= 0.275 kg/m3
RT02
287 · 253
Density across the blower is assumed small so that
ρ02 =
ṁ2 = ρ02 Q2 = 0.275 · 0.24 = 0.066 kg/s
Next
⇐ (a)
(∆p0 )2
(∆p0 )1
=
2 2
ρ01 Ω1 D1
ρ02 Ω22 D22
so that
(∆p0 )2 = (∆p0 )1
0.275
ρ02
= 250
= 56.2 mm H2 O
ρ01
1.226
39
⇐
(b)
Exercise 4.13 Consider a fan with a flow rate of 1500 cfm, (cubic feet per minute)
and has a shaft speed 3600 rpm. If a similar fan one half its size is to have the same
tip speed, what will the flow rate be at a dynamically similar operating condition?
What is the ratio of power consumption of the second fan compared to the first
one?
Given: D2 = D1 /2, U2 = U1 , Ω1 , Q1 .
Find: Q2 , Ẇ2 /Ẇ1 .
Solution: Since U2 = U1 , then Ω2 D2 = Ω1 D1 and Ω2 = 2Ω1 . Next from
Q1
Q2
1500 · 2
Ω2 D23
=
Q
=
Q
=
= 375 cfm ⇐ (a)
2
1
3
3
3
Ω1 D1
Ω2 D2
Ω 1 D1
8
and since Ω1 D1 = Ω2 D2 , from
dotW1
dotW2
D22
1
=
Ẇ
=
Ẇ
= Ẇ1
2
1
3 5
3 5
2
ρΩ1 D1
ρΩ2 D2
D1
4
⇐
(b)
Exercise 4.14 A fan operating at 1750 rpm at a volumetric flow of 4.25 m3 /s
develops a head of 153 mm of water. It is required to build a larger, geometrically
similar fan which will deliver the same head at the same efficiency as the existing
fan, but at the rotational speed of 1440 rpm. (a) Determine the volume flow rate of
the larger fan. (b) If the diameter of the original fan is 40 cm, what is the diameter
of the larger fan. (c) What are the specific speed of these fans.
Given: Ω1 , Ω2 , Q1 , D1 , and H2 = H1 .
Find: Q2 , D2 , and Ωs .
Solution: From
H1
H2
= 2 2
2 2
Ω 1 D1
Ω 2 D2
since H2 = H1 , then D2 = D1 Ω1 /Ω2 . Next from
Q1
Q2
=
3
Ω 1 D1
Ω2 D23
Q2 = Q1
Ω2 D23
Ω21
17502
=
Q
=
4.25
= 6.28 m3 /s (a)
1
Ω1 D13
Ω22
14402
Next convert the head to standard units.
(ws )1 ==
998 · 9.81 · 0.153
ρH2 O gH1
=
= 1246 J/kg
ρ
1.20
40
and the density is the standard density at T = 293 K and p = 101.3215 kPa. Then
√
√
Ω Q
1750 · π 4.25
Ωs = 3/4 =
= 1.80
⇐ (b)
30 12463/4
ws
Exercise 4.15 The impeller of a centrifugal pump, with an outlet radius r2 =
8.75 cm and a blade width b2 = 0.7 cm, operates at 3550 rpm and produces a
pressure rise of 522 kPa at the flow rate of 1.5 liters per minute. Assume that
the inlet flow is axial and that the pump efficiency is 0.63. Find: (a) The specific
2
speed. (b). Show that the expression (??) for work reduces to w = (Vu2
+ U22 −
2
Wu2
)/2 and calculate the work two ways and confirm that they are equal.
Given: r2 , b2 , Ω, ∆p, Q, α1 and η.
Find: Ωs , and the work in terms of kinetic energies.
Solution: The radial velocity component at the exit is
Q
1/45 · 104
=
=
= 6.28 m/s
2πr2 b2
60 · 2π · 0.7
Vr2
The head is
H=
∆p0
522, 000
=
= 53.37 m
ρg
998 · 9.81
and the specific speed is
√
√
3550 · π 1.45/60
Ω Q
Ωs =
=
= 0.528
(gH)3/4
30(9.81 · 53.37)3/4
⇐
(a)
Next the blade speed is
U2 = r2 Ω =
8.75 · 3550π
= 32.53 m/s
100 · 30
and the isentropic and actual work are
ws = gH = 981 · 53.37 = 523 J/kg
w=
w2
523
=
= 830 J/kg
η
0.63
Next the kinetic energy terms are
Vu2 =
w
830
=
= 25.52 m/s
U2
32.53
41
and
Wu2 = Vu2 − U2 = 25.52 − 32.53 = −7.0 m/s
and
√
√
2
2
W2 = Wr2
+ Wu2
= 6.282 + 7.02 = 9.41 m/s
√
√
2
= 6.282 + 25.522 = 26.28 m/s
V2 = Vr22 + Vu2
thus
1
1
w = (V22 + U22 − W22 ) = (26.283 + 32.532 − 9.412 ) = 830 J/kg
2
2
42
⇐ (b)
Chapter 5
Exercise 5.1 Steam leaves the nozzles of a de Laval turbine with the velocity
V2 = 1000 m/s. The flow angle from the nozzle is α2 = 70◦ . The blade velocity
is U = 360 m/s and the mass flow rate is 800 kg/h. Take the rotor velocity
coefficient to be cR = 0.8. The rotor blade is equiangular. Draw the velocity
diagrams and determine: (a) The flow angles of the relative velocity at the inlet
to the rotor, (b) the flow angle leaving the rotor, (c) the tangential force on the
blades, (d) the axial thrust on the blades, (e) the power developed, and (f) the rotor
efficiency.
Given: V2 , α2 , U , cR , and ṁ.
Find: β2 , α3 , Fu , Fx , and ηR .
Solution: The components of the velocities are
Vu2 = V2 sin α2 = 1000 sin(70◦ ) = 939.7 m/s
Wu2 = Vu2 − U = 939.7 − 360 = 579.7 m/s
Vx2 = V2 cos α2 = 1000 cos(70◦ ) = 342, m/s
√
and
W2 =
2
2
Wx2
+ Wu2
=
√
Wx2 = Vx2
3422 + 579.72 = 673.1 m/s
The relative flow angle is
(
)
(
)
Wu2
579.7
−1
−1
β2 = tan
= tan
= 59.46◦
Wx2
342.0
⇐
(a)
and
W2 = cR W2 = 0.8 · 673.1 = 538.5 m/s
so that
Wu3 = W3 sin β3 = 583.5 sin(−59.46◦ ) = −463.8 m/s
Vu3 = U + Wu3 = 360 − 463.8 = −103.8 m/s
The flow angle at the exit of the rotor is
)
(
)
(
−103.8
Vu3
−1
−1
= tan
α3 = tan
= −20.77◦
Vx3
273.6
⇐
(b)
Next the force on the blades is obtained. First
w = U (Vu2 − V u3) = 360(939.7 + 103.8) = 375.64 kJ/kg
43
so that
Ẇ = ṁw =
800 · 375.64
= 83.476 kW
3600
⇐
(e)
then
Fu =
83, 476
Ẇ
=
= 231.9 N
U
360
⇐
(c)
The axial thrust is
Fx = ṁ(Vx2 − Vx3 ) =
800
(342 − 273.6) = 15.2 N
3600
⇐
(d)
The rotor efficiency is
ηR =
w
2 · 375, 640
=
= 0.751
V2 /2
10002
⇐
(f )
Exercise 5.2 The diameter of a wheel of a single stage impulse turbine is 1060 mm
and its shaft speed is 3000 rpm. The nozzle angle is 72◦ and the ratio of the blade
speed to the speed at which steam issues from the nozzles is 0.42. The ratio of
the relative velocity leaving the blades is 0.84 of that entering the blades. The
outlet flow angle of the relative velocity is 3◦ more than the inlet flow angle. The
mass flow rate of steam is 7.23 kg/s. Draw the velocity diagram for the blades
and determine: (a) The axial thrust on the blades. (b) The tangential force on the
blades. (c) Power developed by the blade row. (d) Rotor efficiency.
Given: D, α2 , Ω, U/V2 , W3 /W2 , the difference between β2 and β3 , and ṁ.
Find: Fx , Fu , Ẇ , and ηR .
Solution: The blade speed is
U = r2 Ω =
0.53 · 3000 · π
= 166.5 m/s
30
and the nozzle velocity is V2 = U/0.42 = 166.5/0.42 = 396.4 m/s. Also W3 =
0.84W2 . Next the components of the velocities are
Vu2 = V2 sin α2 = 396.4 sin(72◦ ) = 377.0 m/s
Wu2 = Vu2 − U = 377.0 − 166.5 = 210.5 m/s
Vx2 = V2 cos α2 = 396.4 cos(72◦ ) = 122.5, m/s
√
and
W2 =
2
2
Wx2
+ Wu2
=
√
Wx2 = Vx2
122.52 + 210.52 = 243.6 m/s
44
The relative flow angle is
(
(
)
)
210.5
Wu2
−1
−1
β2 = tan
= tan
= 59.81◦ β3 = −59.81◦ −3◦ = −62.81◦
Wx2
122.5
and
W3 = cR W2 = 0.8 · 243.6 = 204.6 m/s
so that
Wu3 = W3 sin β3 = 204.6 sin(−62.81◦ ) = −182.0 m/s
Wx3 = W3 cos β3 = 204.6 cos(−62.81◦ ) = 93.5 m/s
and
Vu3 = Wu3 + U = −182 + 166.5 = −15.5 m/s
Vx3 = Wx3 = 93.5 m/s
The flow angle at the exit of the rotor is
(
)
(
)
Vu3
−15.5
−1
−1
α3 = tan
= tan
= −9.41◦
Vx3
93.5
Next the force on the blades is obtained.
Fx = ṁ(Vx2 − Vx3 ) = 7.23(93.5 − 122.5) = −209.7 N
⇐
(a)
Fu = ṁ(Vu2 − Vu3 ) = 7.23(−15.5 − 377.0) = 2838 N
⇐
(b)
The specific work done is
w = U (Vu2 − V u3) = 166.5(377.0 + 15.5) = 65.35 kJ/kg
so that
Ẇ = ṁw = 7.23 · 65.35 = 472.5 kW
then the rotor efficiency is
ηR =
w
2 · 65, 350
=
= 0.832
V2 /2
396.42
⇐
(c)
Exercise 5.3 The wheel diameter of a single stage impulse steam turbine is 40 cm
and the shaft speed is 3000 rpm. The steam issues from nozzles at velocity
275 m/s at the nozzle angle of 70◦ . The rotor blades are equiangular and friction reduces the relative velocity as the steam flows through the blades to 0.86
45
times the entering velocity. Find the power developed by the wheel when the
axial thrust is Fx = 120 N.
Given: D, α2 , Ω, V2 , cR , and Fx .
Find: Ẇ .
Solution: The blade speed is
U = r2 Ω =
0.2 · 3000 · π
= 62.83 m/s
30
Next the components of the velocities are
Vu2 = V2 sin α2 = 275 sin(70◦ ) = 258.4 m/s
Wu2 = Vu2 − U = 258.4 − 62.83 = 195.6 m/s
Vx2 = V2 cos α2 = 275 cos(70◦ ) = 94.06, m/s
Wx2 = Vx2
and
Wu3 = −cR Wu2 = −0.86 · 195.6 = −168.2 m/s
and
Vu3 = Wu3 + U = −168.2 + 62.83 = −105.4 m/s
so that
w = U (Vu2 − Vu3 ) = 62.83(258.54 + 105.4) = 22.86 kJ/kg
In addition V x3 = cN Vx2 = 0.86 · 94.06 = 80.89 m/s. From the expression for
the axial component of the blade force
Fx = ṁ(Vx2 − Vx3 )
ṁ =
Fx
12)
=
= 9.11 kg/s
Vx2 − Vx3
94.06 − 80.89
so that
Ẇ = ṁw = 9.11 · 22.86 = 208.3 kW
Exercise 5.4 Steam issues from the nozzles of a single stage impulse turbine with
velocity 400 m/s. The nozzle angle is at 74◦ . The absolute velocity at the exit is
94 m/s and its direction is −8.2◦ . If the blades are equiangular: (a) Find the power
developed by the blade row when the steam flow rate is 7.3 kg/s. (b) The the rate
of irreversible energy conversion per kg of steam flowing through the rotor.
Given: V2 , V3 , α2 , α3 , ṁ, and that β3 = −β2 .
46
Find: Ẇ , rate of loss of energy by irreversiblities.
Solution: The components of the velocities are
Vu2 = V2 sin α2 = 400.0 sin(74◦ ) = 384.5 m/s
Vx2 = V2 cos α2 = 400.0 cos(74◦ ) = 110.3, m/s
Also
Wx2 = Vx2
Vu3 = V3 sin α3 = 94.0 sin(−8.2◦ ) = −13.4 m/s
Vx3 = V3 cos α2 = 94.0 cos(−8.2◦ ) = 93.0, m/s
Wx3 = Vx3
From
Wu2 = W2 sin β2 = Vu2 − U
Wu3 = W3 sin β3 = Vu3 − U
so that
W2 sin β3 − W3 sin β2 = Vu2 − Vu3
from which
W2 (1 + cv ) sin β2 = Vu2 − Vu3
Since the velocity coefficient is the same in the nozzles and the rotor, it can be
written as
W3
Wx3
Vx3
93.0
cv =
=
=
=
= 0.843
W2
Wx2
Vx2
110.3
From the previous equation
W2 sin β2 =
Vu2 − Vu3
1 + cv
In addition
W2 cos β2 = Vx2
the ratio of these gives
tan β2 =
384.5 + 13.4
Vu2 − Vu3
=
= 1.957
Vx2 (1 + cR )
110.3(1 + 0.843)
Thus β2 = 62.93◦ and
W2 =
110.3
Vx2
+
= 242.4 m/s W3 = cv W2 = 0.843·242.4 = 204.5 m/s
cos β2 cos(62.94◦
47
From these
Wu2 = W2 sin β2 = 242.4 sin(62.94◦ ) = 215.8 m/s
Wu3 = W3 sin β3 = 242.4 sin(−62.94◦ ) = −182.1 m/s
and the blade speed is
U = Vu2 − Wu2 = 384.5 − 215.8 = 168.7 m/s
The specific work is
w = U (Vu2 − V u3) = U (Wu2 − Wu3 ) = 168.7(215.8 + 182.8) = 67.13 kJ/kg
so that
Ẇ = ṁw = 7.3 · 67.13 = 490.0 kW
The loss is
1
1
KEloss = (W22 − W32 ) = (242.42 − 204.62 ) = 8.454 kJ/kg
2
2
and the power loss is
Ẇloss = ṁKEloss = 7.3 · 8.454 = 61.7 kW
Exercise 5.5 Carry out the steps in the development of the expression for ratio of
the optimum blade speed to the steam velocity for a single stage impulse turbine
with equiangular blades. Note that this expression is independent of the velocity
coefficient. Carry out the algebra to obtain the expression for the rotor efficiency
at this condition. (a) Find numerical value for the velocity ratio when the nozzle
angle is 76◦ . (b) Find the rotor efficiency at this condition assuming that cR = 0.9.
(c) Find the flow angle of the relative velocity entering the blades at the optimum
condition.
Given: α2 , optimum conditions.
Find: ηR and β3 .
Solution: At optimum conditions
1
1
U
= sin α2 = sin(76◦ ) = 0.485
V2
2
2
48
⇐ (a)
then
sin2 α2
1
ηR = (1 + cv C)
= (1 + 0.9) sin(76◦ ) = 0.894
2
2
⇐
(b)
From
W2 sin β2 = V2 sin β2 − U
W2 cos β2 = V2 cos β2
by dividing gives
tan β2 =
sin α2 − 12 sin α2
1
sin α2 − U/V2
=
= tan α2
cos α2
cos α2
2
(
so that
−1
β2 = tan
)
1
◦
tan(76 ) = 63.50◦
2
⇐
(c)
Exercise 5.6 Steam flows from a set of nozzles of a single stage impulse turbine
at α2 = 78◦ with velocity V2 = 305 m/s and the blade speed is U = 146 m/s.
The outlet flow angle of the relative velocity is 3◦ greater than its inlet angle and
the velocity coefficient is cR = 0.84. The nozzle velocity coefficient is cN = 1.
The power delivered by the wheel is 1000 kW. Draw the velocity diagrams at the
inlet and outlet of the blades. Calculate the mass flow rate of steam.
Given: V2 , U , α2 , β3 = −β2 − 3◦ , cR = 0.84, CN = 1.0 and Ẇ .
Find: ṁ.
Solution: The components of the velocities are
Vu2 = V2 sin α2 = 305 sin(78◦ ) = 290.1 m/s
Vx2 = V2 cos α2 = 305 cos(78◦ ) = 94.3, m/s
Wx2 = Vx2
Wu2 = Vu2 − U = 290.1 − 146 = 144.1 m/s
so that
W2 sin β3 − W3 sin β2 = Vu2 − Vu3
√
√
2
2
= 94.32 + 144.12 = 172.7 m/s
W2 = Wx2
+ Wu2
and the flow angle of the relative velocity is
)
(
(
)
Wu2
144.1
−1
−1
β2 = tan
= tan
= 56.80◦
Wx2
94.3
49
At the exit of the rotor
W3 = CR W2 = 0.84 · 172.2 = 144.6 m/s
and β3 = −(56.8◦ + 3◦ ) = −59.8◦ . Then
Wu3 = W3 sin β3 = 144.6 sin(−59.8◦ ) = −125.0 m/s
Wx3 = W3 cos β3 = 144.6 cos(−59.8◦ ) = 72.7 m/s
Then
Vu3 = Wu3 + U = −125.0 + 146 = 21.0 m/s
(
therefore
α3 = tan
−1
Vu3
Vx3
)
(
−1
= tan
21.0
72.7
Vx3 = 72.7 m/s
)
= 16.11◦
The specific work is
w = U (Vu2 − V u3) = 146(290.1 − 21.0) = 39.29 kJ/kg
so that
ṁ =
Ẇ
1000
=
= 25.5 kg/s
w
39.29
Exercise 5.7 Steam flows from a set of nozzles of a single stage impulse turbine
at angle α2 = 70◦ . (a). Find the maximum total-to-static efficiency if the velocity
coefficients are cR = 0.83 and cN = 0.98. (b). If the rotor efficiency is 90 % of its
maximum value, find the possible outlet flow angles for the relative velocity.
Given: α2 , cR = 0.83, CN = 0.83 and speed ratio U/V2 = 0.9(U/V2 )opt .
Find: ηts and ṁ.
Solution: At optimal condition
U
sin α2
sin(70◦ )
=
=
= 0.470
V2
2
2
and
1
1.83 2 ◦
ηRmax = (1 + CR ) sin2 α2 =
sin (70 ) = 0.808
2
2
Also
0.982 (1 + 0.83) 2 ◦
1
ηts = c2N (1 + cR ) sin2 α2 =
sin (70 ) = 0.776
2
2
50
⇐
(a)
The actual rotor efficiency is
ηR = f ηRmax = 0.9 · 0.808 = 0.727
and therefore
ηR = 2(1 + cR )λ(sin α2 − λ) =
which reduces to
f
(1 + cR ) sin2 α2
2
1
λ2 − sin α2 λ + f sin2 α2 = 0
4
The solution of this is
One root is
λa =
√
1
λ sin α2 (1 ± (1 − f ))
2
√
U
sin(70◦ )
(1 + (1 − 0.9) = 0.6184 =
2
V2
and the other root is
√
sin(70◦ )
U
λa =
(1 − (1 − 0.9) = 0.3123 =
2
V2
The flow angles are
(
−1
β2a = tan
(
−1
β2b = tan
sin α2 − λa
cos α2
sin α2 − λb
cos α2
)
= 43.21◦
)
= 61.06◦
Exercise 5.8 The nozzles of a single stage impulse turbine have a wall thickness
t = 0.3 cm and height b = 15 cm. The mean diameter of the wheel is D = 116 cm
and the nozzle angle is α2 = 72◦ . The number of nozzles in a ring is 72. The
specific volume of steam at the exit of the nozzles is 15.3 m3 /kg and the velocity
there is V2 = 366m/s. (a) Find the mass flow rate of steam through the steam
nozzle ring. (b) Find the power developed by the blades for an impulse wheel of
equiangular blades, given that the velocity coefficient cR = 0.86 and cN = 1.0.
The shaft turns at 3000 rpm.
Given: t, b, D, α2 , Z, v1 , cR , cN , and V2 .
Find: ṁ and Ẇ .
51
Solution:
Vu2 = V2 sin α2 = 366 sin(72◦ ) = 348.1 m/s
Vx2 = V2 cos α2 = 366 cos(72◦ ) = 113.1, m/s
The exit area is
A = (2πr cos α2 − Zt)b = (2π · 0.58 cos(72◦ ) − 72 · 0.003)0.15 = 0.1365, m2
Mass flow rate is
ṁ =
366 · 0.1365
V2 A
=
= 3.266 kg/s ⇐ (a)
v2
15.3
Next
Wu2 = Vu2 − U = 348.14 − 182.2 = 165.9 m/s
Wx2 = 113.1 m/s
and the magnitude of the relative velocity leaving the nozzles is
√
√
2
2
W2 = Wx2
+ Wu2
= 113.12 + 165.92 = 200.8 m/s
and its flow angle is
β2 = tan
(
−1
Wu2
Wx2
)
(
−1
= tan
165.9
113.1
)
= 55.71◦
Therefore after the rotor
W3 = cR W2 = 0.86 · 200.8 = 172.7 m/s
Then
β3 = −55.71◦
Wu3 = W3 sin β3 = 172.7 sin(−55.71◦ ) = −142.7 m/s
Wx3 = W3 cos β3 = 172.7 cos(−55.71◦ ) = 97.3, m/s
The tangential component of the absolute velocity at the exit of the rotor is
Vu3 = U + Wu3 = 182.2 − 142.7 = 39.5 m/s
and the specific work is
w = U (Vu2 − Vu3 ) = 182.2(348.1 − 39.5) = 56.23 kJ/kg
and the power is
Ẇ = ṁw = 3.266 · 56.23 = 183.6 kW
52
⇐ (b)
Exercise 5.9 The isentropic static enthalpy change across a stage of a single stage
impulse turbine is ∆hs = 22 kJ/kg. The nozzle exit angle is α2 = 74◦ . The mean
diameter of the wheel is 148 cm and the shaft turns as 1500 rpm. The blades are
equiangular with a velocity coefficient cR = 0.87. The nozzle velocity coefficient
is cN = 0.98. (a) Find the steam velocity at the exit from the nozzles. (b) Find the
flow angles of the relative velocity at the inlet and exit of the wheel. (c) Find the
overall efficiency of the stage.
Given: r2 , α2 , Ω, cR , cN , and ∆hs .
Find: V2 , β2 , β3 and ηtt .
Solution: The blade speed is
U = r2 Ω =
0.74 · 1500 · π
= 116.2 m/s
30
For isentropic flow assuming that the exit velocity is the same as at the inlet to the
stage
ws = ∆hs = 2U (V2s sin α2 − U )
therefore
(
V2s =
∆hs
+U
2U
)
1
=
sin α2
(
)
22, 000
1
+ 116.2
= 219.4 m/s
2 · 116.2
sin(74◦ )
Then from
CN =
V2
V2s
V2 = CN V2s = 0.98 · 219.4 = 215.0 m/s
⇐ (a)
and
U
116.2
=
= 0.540
V2
215.0
To find the flow angles, first calculate
λ=
Vu2 = V2 sin α2 = 215 sin(74◦ ) = 206.7 m/s
Vx2 = V2 cos α2 = 215 cos(74◦ ) = 59.3, m/s
then
Wu2 = Vu2 − U = 206.7 − 116.2 = 90.5 m/s
53
Wx2 = 59.3 m/s
and the angle at which the relative velocity enters the rotor is
(
(
)
)
90.5
Wu2
−1
−1
β2 = tan
= tan
= 56.76◦
⇐
Wx2
59.3
(b)
and the exit angle from the rotor is β3 = −56.76◦ .
The efficiency is
ηts = 2λCn2 (1 + CR )(sin α2 − λ)
= 2 · 0.540 · 0.982 (1 + 0.85)(sin(74◦ ) − 0.534) = 0.808
⇐
(c)
Exercise 5.10 An impulse turbine has a nozzle angle α2 = 72◦ and steam velocity
V2 = 244 m/s. The velocity coefficient for the rotor blades is cR = 0.85 and the
nozzle efficiency is ηn = 0.92. The output power generated by the wheel is
Ẇ = 562 kW when the mass flow rate is ṁ = 23 kg/s. Find the efficiency of the
turbine.
Given: α2 , V2 , cR , cN , ṁ, and Ẇ .
Find: ηts .
Solution: The specific work is
w=
Ẇ
562
=
= 24.43 kJ/kg
ṁ
23
and from
w = U (Vu2 − Vu3 ) = U (Wu2 − Wu3 ) = U (1 + cR )WU 2
But
Wu2 = V2 sin α2 − U
so that
(1 + cR )U 2 − (1 + cR )V2 sin α2 U +
The solution of this is
1
U = V2 sin α2 ±
2
and
244 sin(72◦ )
U1 =
+
2
√
√
w
=0
1 + cR
w
1 2 2
V2 sin α2 −
4
1 + cR
2442 sin2 (72◦ ) 24, 430
−
= 132 m/s
4
1.85
54
and the second root is
244 sin(72◦ )
U2 =
−
2
√
2442 sin2 (72◦ ) 24, 430
−
= 100.1 m/s
4
1.85
From these
U2
U1
= 0.541 lambda2 =
= 0.410
V2
V2
The efficiency using either root becomes
λ1 =
ηts = 2λc2N (1 + cR )(sin α2 − λ)
ηts = 2 · 0.541 · 0.922 (1 + 0.85)(sin(72◦ ) − 0.541) = 0.755
⇐
Exercise 5.11 A two row velocity compounded impulse wheel is part of a steam
turbine with many other stages. The steam velocity from the nozzles is V2 =
580 m/s and the means speed of the blades is U = 116 m/s. The flow angle
leaving the nozzle is α2 = 74◦ and the flow angle of the relative velocity leaving
the first set of rotor blades is β3 = −72◦ . The absolute velocity of the flow as it
leaves the stator vanes between the two rotors is α4 = 68◦ and the outlet angle of
the relative velocity leaving the second rotor is β5 = −54◦ . The steam flow rate is
ṁ = 2.4 kg/s. The velocity coefficient is cv = 0.84 for both the stator and rotor
rows. (a) Find the axial thrust from each wheel. (b) Find the tangential thrust from
each wheel. (c) Find the efficiency of the rotors defined as the work out divided
by the kinetic energy available from the nozzles.
Given: V2 , U , α2 , β3 , α4 , β5 , cN = CR , and ṁ.
Find: Fx , Fu , ηR .
Solution: The velocity components are
Vu2 = V2 sin α2 = 580 sin(74◦ ) = 557.5 m/s
Vx2 = V2 cos α2 = 580 cos(74◦ ) = 159.9 m/s
and
Wu2 = Vu2 − U = 557.5 − 116 = 441.5 m/s
Wx2 = 159.9 m/s
and the magnitude of the relative velocity leaving the nozzles is
√
2
2
+ Wu2
= 469.6 m/s W3 = 0.84 · 469.6 = 394.4 m/s
W2 = Wx2
55
Next
Wu3 = W3 sin β3 = 394.4 sin(−72◦ ) = −375.1 m/s
Wx3 = W3 cos β3 = 394.4 cos(−72◦ ) = 121.9 m/s
The components of the absolute velocity are
Vu3 = Wu3 + U = −375.1 + 116 = −259.1 m/s
and
√
2
2
V3 = Vx3
+ Vu3
= 286.3 m/s
Next
Vx3 = 121.9 m/s
V4 = 0.84 · 286.3 = 240.5 m/s
Vu4 = V4 sin α4 = 240.5 sin(68◦ ) = 223.0 m/s
Vx4 = V4 cos α4 = 240.5 cos(68◦ ) = 90.1 m/s
and the components of the relative velocity become
Wu4 = Vu4 − U = 223.0 − 116 = 107.0 m/s
and its magnitude is
√
2
2
W4 = Wx4
+ Wu4
= 139.9 m/s
Wx4 = V x4 = 90.1 m/s
W5 = 0.84 · 139.9 = 117.5 m/s
The components after the second rotor become
Wu5 = W5 sin β5 = 117.5 sin(−54◦ ) = −95.1 m/s
Wx5 = W5 cos β5 = 117.5 cos(−54◦ ) = 69.1 m/s
and finally
Vu5 = Wu5 + U = −95.1 + 116 = 20.9 m/s
Vx5 = Wx5 = 69.1 m/s
The force components are then
Fx1 = ṁ(Vx2 − Vx3 ) = 2.4(159.9 − 121.9) = 91.2 N
Fu1 = ṁ(Vu2 − Vu3 ) = 2.4(557.5 − 259.1) = 1960.0 N
For the second rotor
Fx2 = ṁ(Vx4 − Vx5 ) = 2.4(90.1 − 69.1) = 50.4 N
56
Fu2 = ṁ(Vu4 − Vu5 ) = 2.4(223 − 20.9) = 485.0 N
The power output is
Ẇ1 = Fu1 U = 1960 · 116 = 227.4 kW
Ẇ2 = Fu2 U = 485 · 116 = 52.26 kW
an the efficiency is
η=
w1 + w2
2U
= 2 (Vu2 − Vu3 + Vu4 − Vu5 )
2
V2 /2
V2
with a numerical value
η=
2 · 116
(555705259.1 + 223 − 20.9) = 0.703
5802
⇐
Exercise 5.12 A velocity-compounded impulse wheel has two rows of moving
blades with a mean diameter of D = 72 cm. The shaft rotates at 3000 rpm. Steam
issues from the nozzles at an angle α2 = 74◦ with velocity V2 = 555 m/s. The
mass flow rate is ṁ = 5.1 kg/s. The energy loss through each of the moving
blades is 24 percent of the kinetic energy entering the blades based on the relative
velocity. Steam leaves the first set of moving blades at β3 = −72◦ the guide vanes
between the rows at α4 = 68◦ and the second set of moving blades at β5 = −52◦ .
(a) Draw the velocity diagrams and find the flow angles at the blade inlets both
for absolute and relative velocities. (b) Find the power developed by each row of
blades. (c) Find the rotor efficiency as a whole.
Given: V2 , D, Ω, α2 , β3 , α4 , β5 , cN = CR , and ṁ.
Find: Fx , Fu , ηR .
Solution: The blade speed is
0.36 · 3000 · π
= 113.1 m/s
30
√
and the velocity coefficients are cR = cN = 0.76 = 0.872.
The velocity components are
U = r2 Ω =
Vu2 = V2 sin α2 = 555 sin(74◦ ) = 533.5 m/s
Vx2 = V2 cos α2 = 555 cos(74◦ ) = 153.0 m/s
57
and
Wu2 = Vu2 − U = 533.5 − 113.1 = 420.4 m/s
Wx2 = 153.0 m/s
and the magnitude of the relative velocity leaving the nozzles is
√
2
2
W2 = Wx2
+ Wu2
= 447.4 m/s W3 = 0.872 · 447.4 = 390.0 m/s
Next
Wu3 = W3 sin β3 = 390.0 sin(−72◦ ) = −370.9 m/s
Wx3 = W3 cos β3 = 390.0 cos(−72◦ ) = 120.5 m/s
The components of the absolute velocity are
Vu3 = Wu3 + U = −370.9 + 113.1 = −257.8 m/s
and
√
V3 =
Next
Vx3 = 120.5 m/s
V4 = 0.872 · 284.6 = 248.1 m/s
2
2
Vx3
+ Vu3
= 284.6 m/s
Vu4 = V4 sin α4 = 248.1 sin(68◦ ) = 230.0 m/s
Vx4 = V4 cos α4 = 248.1 cos(68◦ ) = 92.9 m/s
and the components of the relative velocity become
Wu4 = Vu4 − U = 230.0 − 113.1 = 116.9 m/s
and its magnitude is
√
2
2
W4 = Wx4
+ Wu4
= 149.3 m/s
Wx4 = V x4 = 92.9 m/s
W5 = 0.872 · 149.3 = 130.2 m/s
The components after the second rotor become
Wu5 = W5 sin β5 = 130.2 sin(−52◦ ) = −102.6 m/s
Wx5 = W5 cos β5 = 130.2 cos(−52◦ ) = 80.2 m/s
and finally
Vu5 = Wu5 + U = −102.6 + 113.1 = 10.5 m/s
58
Vx5 = Wx5 = 80.2 m/s
The work done is
w1 = U (Vu2 − Vu3 ) = 113.1(533.5 + 257.8) = 89, 496 J/kg
w2 = U (Vu4 − Vu5 ) = 113.1(230.0 − 10.5) = 24, 825 J/kg
and the power is
Ẇ1 = ṁw1 = 5.1·89.496 = 456.4 kW
Ẇ2 = ṁw2 = 5.1·24.825 = 126.6 kW
so the efficiency is
η=
w1 + w2
2(89496 + 24825)
=
= 0.742
2
V2 /2
5552
⇐ (c)
The flow angles are
(
)
(
)
Wu2
420.4
−1
β2 = tan
= tan
= 70◦
Wx2
153.0
(
)
(
)
Vu3
−257.8
−1
−1
α3 = tan
= tan
= −64.9◦
Vx3
120.5
)
(
)
(
116.9
Wu4
−1
−1
= tan
= 51.5◦
β4 = tan
Wx4
92.9
−1
Exercise 5.13 Steam flows from the nozzles of a zero percent repeating stage at
an angle α2 = 69◦ and speed V2 = 450 m/s and enters the rotor with blade speed
moving at U = 200 m/s. (a) Find its efficiency when the loss coefficients are
calculated from Soderberg’s correlation. (b) Find the work delivered by the stage.
Given: V2 , U , and α2 .
Find: η, w.
Solution: The velocity components are
Vu2 = V2 sin α2 = 450 sin(69◦ ) = 420.1 m/s
Vx2 = V2 cos α2 = 450 cos(69◦ ) = 161.3 m/s
so that
Wu2 = Vu2 − U = 420.1 − 200 = 220.1 m/s
59
Wx2 = Vx2 = 161.3 m/s
√
2
2
Wx2
+ Wu2
= 272.9 m/s. Next
(
)
(
)
220.1
Wu2
−1
−1
= tan
β2 = tan
= 53.76◦
Wx2
161.3
and W2 =
β3 = −53.73◦
The amount of turning is
∆α = 68◦
∆β = 107/5◦
The loss coefficients are
(
ζN = 0.04 + 0.06
(
ζR = 0.04 + 0.06
69
100
)2
107.5
100
= 0.0686
)2
= 0.1094
The velocity coefficients are
cN = √
1
= 0.9674
1 + ζN
cR = √
1
= 0.9494
1 + ζR
Now
W3 = CR W2 = 0/9494 · 272.9 = 259.1 m/s
and
Wu3 = W3 sin β3 = 259.1 sin(−53.76◦ ) = −209.0 m/s
Wx3 = W3 cos β3 = 2559.1 cos ∗(−53.76◦ ) = 153.1 m/s
and the components of the absolute velocity after the rotor are
Vu3]=U +Wu3 =300−209.0=−9.0 m/s
V[ x3
= 153.1 m/s
so that the flow angle is
(
α3 = tan
and V3 =
−1
Vu3
Vx3
)
(
−1
= tan
−9.0
153.1
)
= −3.35◦
√
2
2
Vx3
+ Vu3
= 153.3 m/s. The value of the speed ratio is
λ=
200
U
=
= 0.444
V2
450
60
and the word delivered is
w = U (Vu2 − Vu3 ) = 200(420.1 + 9.0) = 85.818 kJ/kg
⇐
(b)
⇐
(a)
The efficiency is
ηtt =
(ζR −
4)λ2
4λ(sin α2 − λ)
= 0.888
+ (4 − 2ζR ) sin α2 + ζR + ζN
Exercise 5.14 Show that for a repeating stage the efficiency of a zero percent
reaction is
4λ(sin α2 − λ)
ηtt =
2
(ζR − 4)λ + (4 − 2ζR )λ sin α2 + ζR + ζN
and this maximum is at the condition λ = U/V2 given by
√
ζR + ζS − (ζR + ζN )(ζR + ζS − ζR sin2 α2 )
λ=
ζR sin α2
Given: A stage with zero reaction.
Find: ηtt , and λ for maximum efficiency.
Solution: Examination of the figure for a stage with zero reaction shows that
ηtt =
h01 − h03
h01 − h03ss
which can be written as
1
h01 − h03ss
h03 − h03ss
−1=
−1=
ηtt
h01 − h03
w
which can be written as
2
h3 − h3ss + 12 (V32 − V3ss
)
1
−1=
ηtt
w
Since
T2
T3s
=
T2s
T3ss
then
h3s − h3ss =
T3ss
T3s
(h2 − h2s ) =
(h2 − h2s )
T3s
T2
61
and the expression for efficiency becomes
h3 − h3s +
1
−1=
ηtt
T3s
(h2
T2
− h2s ) + (1 −
T3ss
)V32
T3
w
or
ζR W32 + TT3s2 ζN V22 + (1 −
1
−1=
ηtt
2w
Neglecting the temperature factors leads to
T3ss
)V33
T3
1
ζR W32 + ζN V22
−1=
ηtt
2w
Since W3 = W2 and
W22 = V22 − 2U V2 sin α2 + U 2
w = 2U W2 sin β2 = 2U (V2 sin α2 − U )
Substituting these into the expression for efficiency and introducing the parameter
λ = U/V2 gives
1
ζR (λ2 − 2λ sin α2 + 1) + ζN
−1=
ηtt
4λ(sin α2 − λ)
This can be now written as
ηtt =
(ζR −
4)λ2
4λ(sin α2 − λ)
+ (4 − 2ζR )λ sin α2 + ζR + ζN
⇐
(a)
Differentiating this with respect to λ and setting the derivative to zero gives
ζr sin α2 λ2 − 2(ζR + ζN )λ + (ζR + ζN ) sin α2 = 0
and the solution of this is
√
ζR + ζS − (ζR + ζN )(ζR + ζS − ζR sin2 α2 )
λ=
ζR sin α2
62
⇐
(b)
Chapter 6
Exercise 6.1 At inlet to the rotor in a single stage axial-flow turbine the magnitude of the absolute velocity of fluid is 610 m/s. Its direction is 61◦ as measured
from the cascade front in the direction of the blade motion. At exit of this rotor the
absolute velocity of the fluid is 305 m/s directed such that its tangential component is negative. The axial velocity is constant, the blade speed is 305 m/s, and the
flow rate through the rotor is 5 kg/s. (a) Construct the rotor inlet and exit velocity
diagrams showing the axial and tangential components of the absolute velocities.
(b) Evaluate the change in total enthalpy across the rotor. (c) Evaluate the power
delivered by the rotor. (d) Evaluate the average driving force exerted on the blades.
(e) Evaluate the change in static and stagnation temperature of the fluid across the
rotor, assuming the fluid to be a perfect gas with cp = 1148 J/kg K. (f) Calculate
the flow coefficient and the blade loading coefficient. Are they reasonable?
Given: V2 , α2 , V1 , U , and ṁ.
Find: ∆T , ∆T0 , ϕ, ψ.
Solution: For a The axial and tangential velocity components at the inlet to the
rotor are
Vx = V2 cos α2 = 610 cos(61◦ ) = 295.7 m/s
Vu2 = V2 sin α2 = 610 sin(61◦ ) = 533.5 m/s
At the exit of the rotor cos α1 = Vx /V3 so that
(
)
295.7
−1
α3 = cos
= −14.16◦
305
and
Vu3 = V3 sin α2 = 610 sin(−14.16◦ ) = −74.6 m/s
To construct the velocity diagrams the relative flow angles are needed. First the
relative velocity components are
Wu2 = Vu2 − U = 533.5 − 305 = 228.5 m/s
(
so that
−1
β2 = tan
Wu2
Wx2
)
(
−1
= tan
228.5
295.7
Wx2 = Vx2
)
= 37.7◦
The relative velocity component after the rotor are
Wu3 = Vu3 − U = −74.5 − 305 = −379.5 m/s
63
Wx3 = Vx3 =
so that
(
−1
β3 = tan
Wu3
Wx3
(
)
−1
= tan
−379.5
295.7
)
= −52.08◦
⇐ (a)
The specific work delivered is
w = U (vu2 − Vu3 ) = 305(533.5 + 74.6) = 185, 480 J/kg
which gives a drop in stagnation enthalpy of
h02 − h03 = 185.480 kJ/kg
⇐
(b)
Ẇ = ṁ = w = 5 · 185.48 = 92.7 MW
⇐
The power delivered is
(c)
and the force on the blades is
Fu =
Ẇ
185, 480
=
= 3040 N
U
305
⇐
(d)
The drop in the stagnation temperature of
T02 − T03 =
w
185, 4801
=
= 161.6 K
cp
1148
(e)
and the drop in static temperature is
T2 − T3 = T02 − T03 −
1
6102 − 3052
(V22 − V32 ) = 161.6 +
= 40.0 K ⇐ (e)
2cp
2 · 1148
The flow coefficient and the blade loading coefficients are (reasonable)
ϕ=
Vx
295.7
=
= 0.969 ψ = 2.00
U
305
⇐ (f )
Exercise 6.2 A small axial-flow turbine must have an output power of 37 kW
when handling one half of a kilogram of combustion gases per second with an
inlet total temperature of 410 K. The value of the gas constant is 287 J/kg K and
γ = 4/3. The total-to-total efficiency of the turbine is 80%. The rotor operates
at 50, 000 rpm and the mean blade diameter is 10 cm. Evaluate (a) the average
driving force on the turbine blades, (b) the change in the tangential component
64
of the absolute velocity across the rotor, and (c) the required total pressure ratio
across the turbine.
Given: Ẇ , T01 , Ω, D, ηtt and ṁ.
Find: Fu , Vu2 − Vu3 , p01 /p03 .
Solution: The blade speed is
U = rΩ =
0.05 · 50, 000 · π
= 261.8 m/s
30
and the average blade force is
Fu =
Ẇ
37, 000
=
= 141.3 N
U
261.8
⇐ (a)
The specific work is
w=
Ẇ
37, 000
=
= 74 kJ/kg
ṁ
0.5
and therefore the change in the tangential velocity components is
Vu2 − Vu3 =
w
74000
=
= 282.7 m/s
U
261.8
⇐
(b)
The isentropic work is
ws =
74
w
=
= 92.5 kJ/kg
ηtt
0.8
so that the exit stagnation temperature for an isentropic expansion is
T03s = T01 −
ws
925, 000
= 410 −
= 329.4 K
cp
1148
and the pressure ratio is
p01
=
p03
(
T01
T03s
(
)γ/(γ−1)
=
410
329.4
)4
= 2.4
⇐
(c)
Exercise 6.3 A turbine stage of a multi-stage axial turbine is shown in Figure ??.
The value of the gas constant is 287 J/kg K and γ = 4/3. The inlet gas angle
to the stator is α1 = −36.8◦ and the outlet angle from the stator is α2 = 60.3◦ .
65
The flow angle of the relative velocity at the inlet to the rotor is β2 = 36.8◦ and
the flow leaves at β3 = −60.3◦ . (a) If the blade speed is U = 220 m/s, find the
axial velocity, which is assumed constant throughout the turbine. (b) Find the
work done by the fluid on the rotor blades for one stage. (c) The inlet stagnation
temperature to the turbine is 950 K and the mass flow rate is ṁ = 400 kg/s. If
this turbine produces a power output of 145 MW, find the number of stages. (d)
Find the overall stagnation pressure ratio if its isentropic adiabatic efficiency is
ηt = 0.85. (e) Why does the static pressure fall across the stator and the rotor?
Given: α1 , α2 , β2 , β3 , ηtt , U , Ẇ , ṁ, and T01 .
Find: Vx , w, number of stages, p01 /p0e .
Solution: From Vu2 = Wu2 + U follows
Vx tan α2 = Vx tan β2 + U
so that
Vx =
U
220
=
= 218.9 m/s
tan α2 − tan β2
tan(60.3◦ ) − tan(36.8◦ )
⇐
(a)
Work delivered by a repeating stage is
w = U (Vu2 − Vu3 ) = U Vx (tan α2 − tan α1 )
w = 220 · 218.9(tan(60.3◦ ) − tan(−36.8◦ )) = 120.5 kJ/kg
⇐
(b)
⇐
(c).
To work done by all the stages is
wt =
Ẇ
145 · 106
=
= 362.5 kJ/kg
ṁ
400
Hence the number of stages is N = wt /w = 362.5/120.5 = 3.00
The temperature drop per stage is
T01 − T03 =
w
120, 500
=
= 104.9 K
cp
1148
and the pressure drop across the entire machine is
T01] − T0e = 3 · 104.9 = 314.7 K
and the exit temperature for an isentropic process is
T03s = T01 −
314.7
T01 − T0e
= 950 −
= 579.7 K
ηtt
0.85
66
and the pressure ratio is
(
)γ/(γ−1) (
)4
p01
T01
950
= 7.21
=
=
p0e
T0es
579.7
⇐ (d)
The static pressure drops because velocity increases. Static pressure drops across
the rotor because the relative stagnation enthalpy remains constant and the relative
velocity increases across the rotor.
Exercise 6.4 A single stage axial turbine has a total pressure ratio of 1.5 to 1, with
an inlet total pressure 300 kPa and temperature of 600 K. The absolute velocity
at the inlet to the stator row is in the axial direction. The adiabatic total-to-total
efficiency is 80%. The relative velocity is at an angle of 30◦ at the inlet of the rotor
and it exits at −35◦ . If the flow coefficient is ϕ = 0.9, find the blade velocity.
Use compressible flow analysis with cp = 1148 J/kg K, γ = 4/3, and R =
287 J/kg K.
Given: p01 /p03 , p01 , T01 , α1 , β2 , β3 , ηtt , ϕ,
Find: U .
Solution: From
w = U (Vu2 − Vu3 ) = U (Wu2 − Wu3 ) = U Vx (tan β2 − tan β3 )
Dividing by U 2 gives
ψ = ϕ(tan β2 − tan β3 ) = 0.9(tan(30◦ ) − tan(−35◦ )) = 1.15
Then from
T01
=
T03s
(
p01
p03
)γ/(γ−1)
= 1.50.25 = 1.1067
so that T03s = 600/1.1067 = 542.2 K, and
T03 = T01 − ηtt (T01 − T03s ) = 600 − 0.8(600 − 542.2) = 553.7 K
The work is now
w = cp (T01 − T03 ) = 1148(600 − 553.7) = 53.15 kJ/kg
From w = ψU 2 the blade speed is
√
√
w
53150
U=
=
= 215 m/s
ψ
1.15
67
⇐
Exercise 6.5 An axial turbine has a total pressure ratio of 4 to 1, with an inlet
total pressure 650 kPa and total temperature of 800 K. The combustion gases that
pass through the turbine have γ = 4/3, and R = 287 J/kg K. (a) Justify the
choice of two stages for this turbine? Each stage is normal stage and they are
designed the same way, with the blade loading coefficient equal to 1.1 and the
flow coefficient equal to 0.6. The absolute velocity at the inlet to the stator row
is at angle 5◦ from the axial direction. The overall total-to-total efficiency of the
turbine is 91.0%. Find: (b) The angle at which the the absolute velocity leaves the
stator,(c) the angle of the relative velocity at the inlet of the rotor, (d) the angle at
which the relative velocity leaves the rotor. (e) Draw the velocity diagrams at the
inlet and outlet of the rotor. (f) What are the blade speed and the axial velocity?
(g) A consequence of the design is that each stage has the same work output and
efficiency. Find (g) the stage efficiency, and (h) the pressure ratio for each stage.
Given: p01 /p0e , p01 , T01 , psi, ϕ, and ηtt .
Find: Justify that two stages is appropriate, α2 , β2 , β3 , U , Vx , ηs , p01 /p03 .
Solution: The stagnation temperature at the exit of the turbine is
(
T0es = T01
p0e
p01
)(γ−1)/γ
= 800 · 0.250.25 = 565.7 K
and the specific work delivered is
w = ηtt cp (T01 − T0es ) = 0.91 · 1148(800 − 565.7) = 244.8 kJ/kg
The temperature drop is
T02 − T0e =
w
244, 800
=
= 213.2 K
cp
1148
⇐ (a)
A typical temperature drop across a stage is about 120 K. Hence two stages are
justified and ws = w/2 = 122.39 kJ/kg. Each stage has the same flow angles.
Hence their efficiency is the same.
Solving from
1 − R − ψ/2
tan α1 =
ϕ
for the reaction R gives
R=1−
ψ
− ϕ tan α1 = 1 − 0.555 − 0.6 tan(5◦ ) = 0.3975
2
68
and
tan β1 = −
0.3975 + 0.55
R + ψ/2
=−
= −1.5792 β1 = −57.65◦
ϕ
0.6
tan α2 =
⇐
(c)
1 − R + ψ/2
1 − 0.3975 + 0.55
=
= 1.9208 α2 = 62.50◦
ϕ
0.6
R − ψ/2
0.3975 − 0.55
=−
= 02542 β2 = 14.26◦
ϕ
0.6
The blade speed is
√
√
w
122, 390
=
= 333.6 m/s ⇐ (e)
U=
ψ
1.1
tan β2 = −
⇐
(b)
and the axial velocity is
Vx = ϕU = 0.6 · 333.6 = 200.1 m/s
⇐
(d)
The stage efficiency is obtained as follows. For each stage
so that
ηs =
T01 − T02
1 − T02 /T 01
=
T01 − T02s
1 − (p02 /p01 )γ/(γ−1 )
ηs =
T02 − T0e
1 − T0e /T 02
=
T02 − T0es
1 − (p0e /p02 )γ/(γ−1 )
[
p02
= 1−
p01
[
p0e
= 1−
p02
1
ηs
1
ηs
(
(
T02
1−
T01
T0e
1−
T02
)]γ/(γ−1)
)]γ/(γ−1)
and
[
(
)]γ/(γ−1) [
(
)]γ/(γ−1)
p0e
p02 p0e
1
T02
1
T0e
=
= 1−
1−
1−
1−
p01
p01 p02
ηs
T01
ηs
T02
Hence
(
)(
)
(
)
( )(γ−1)/γ
T02
T0e 1
T02
T0e 1
p0e
1−
1−
− 1−
+1−
+1−
T01
T02 ηs
T01
T02 ηs
p01
69
Which when solved for the stage efficiency 1/ηs gives
1
2 − T02 /T01 − T0e /T02
=
+
ηs
2(1 − T02 /T01 )(1 − T0e /T02 )
√
(2 − T02 /T01 − T0e /T02 )2 − 4(1 − T02 /T01 )(1 − T0e /T02 )(1 − p0e /p01 )(γ−1)/γ
2(1 − T02 /T01 )(1 − T0e /T02 )
and
and
T02
693.4
=
= 0.827
T01
800
0.287 −
1
=
ηs
√
586.8
T0e
=
= 0.846
T02
693.4
0.0824 − 4 · 0.133 · 0.154 · 0.293
= 1.108
2 · 0.133 · 0.154
so that
ηs = 0.9024
and
ηs =
1 − T02 /T01
1 − (p02 /p01 )(γ−1)/γ
so that
[
(
)]γ/(γ−1) (
)4
p02
1
T02
0.133
= 1−
1−
= 1−
= 0.528
p01
ηs
T01
0.9024
and p02 = 0.528 · 0.528 = 343 kPa. Similarly
ηs =
1 − T0e /T02
1 − (p0e /p02 )(γ−1)/γ
[
(
)]γ/(γ−1) (
)4
p0e
1
T02
0.154
= 1−
1−
= 1−
= 0.474
p02
ηs
T01
0.9024
or
p01
= 1.895
p02
p02
= 2.111
p0e
⇐
(h)
Exercise 6.6 For a steam turbine rotor the blade speed at the casing is U =
300 m/s and at the hub its speed is 240 m/s. The absolute velocity at the casing section at the inlet to the rotor is V2c = 540 m/s and at the hub section it is
V2h = 667 m/s. The angle of the absolute and relative velocities at the inlet and
exit of the casing and hub sections are αc2 = 65◦ , βc3 = −60◦ , αh2 = 70◦ , and
70
βh3 = −50◦ . The exit relative velocity at the casing is Wc3 = 456 m/s and at the
hub it is Wh3 = 355 m/s. Evaluate for the tip section: (a) The axial velocity at
the inlet and exit; (b) the change in total enthalpy of the steam across the rotor; (c)
the outlet total and static temperatures at the hub and casing sections, if the inlet
static temperature is 540 C and inlet total pressure is 7 MPa and they are the same
at all radii. Assume that the process is adiabatic and steam can be considered a
perfect gas with γ = 1.3. The static pressure at the exit of the rotor is the same
for all radii and is equal to the static pressure at inlet of the hub section. Repeat
the calculations for the hub section. (d) Find the stagnation pressure at the outlet
at the casing and the hub.
Given: Uc , Uh , V2c , V2h , α2c , α2h , β3c , β3h , W3c , W3h , T2 , p02 .
Find: Vx3c , Vx2c , ∆h0 , T03 , T3 , and similarly for the hub.
Solution: The axial velocities are
Vx2c
Vx2h
Vx3c
Vx3h
V2c cos α2c = 540 cos(65◦ ) = 228.2 m/s
⇐ (a)
◦
V2h cos α2h = 667 cos(70 ) = 228.1 m/s
⇐ (a)
◦
V3c cos α3c = 456 cos(−60 ) = 228.0 m/s
⇐ (a)
V3h cos α3h = 355 cos(−50◦ ) = 228.2 m/s
⇐ (a)
=
=
=
=
and the flow coefficients are
ϕc =
Vx
228.1
=
= 0.76
Uc
300
ϕh =
Vx
228.0
=
= 0.95
Uh
240
The tangential components of the velocities are
Vu2c = V2c sin α2c = 540 sin(65◦ ) = 489.4 m/s
Vu2h = V2h sin α2h = 667 sin(70◦ ) = 626.8 m/s
and
Vu3c = Wu1c + Uc = W3c sin β3c + Uc = 456 sin(−60◦ ) + 300 = −94.6 m/s
Vu3h = Wu1h + Uh = W3h sin β3h + Uh = 667 sin(−50◦ ) + 240 = −31.9 m/s
The stagnation enthalpy change is therefore
∆h0c = wc = Uc (Vu2c − Vu3c ) = 300(489.34 + 94.9) = 175, 290 J/kg ⇐ (b)
∆h0h = wh = Uh (Vu2h − Vu3h ) = 240(626.8 + 31.9) = 158, 088 J/kg ⇐ (b)
71
Note that if rVu = C then w is independent of the radius. Here
rc Vu2c
489.4
= 1.25
= 0.976
rh Vu2h
626.8
so that the condition for a free vortex distribution does not quite hold. Take
1
1
w = (wc + wh ) = (175, 290 + 158, 088) = 166, 690 J/kg
2
2
The blade loading coefficients are
ψc =
wc
175, 290
=
= 1.947
Uc2
3002
ψh =
wh
158, 088
=
= 2.744
2
Uh
2402
Using the average value U = 0.5(Uc + Uh ) = 270 m/s, the average value for the
blade loading coefficient is ψ = w/U 2 = 166, 690/2702 = 2.286. The average
force on the blades is therefore
Fu = ψU = 2.286 · 270 = 617.2 N
The reactions at the casing and the hub are
Rc = −ϕc tan β3c −
ψ
= −0.76 tan(−60◦ )−0.5·1.947 = 0.343
2
Rh = −ϕh tan β3h −
The negative reaction at the hub is undesirable.
The static temperature at the in let to the rotor is uniform T2 = 813.15 K. and
the gas constant is R = R̄/M = 8314/18 = 461.9 J/kg K. The specific heat is
cp = γR/(γ −1) = 1.3·416.9/0.3
= 2001.2001.5 J/kg K, and the speed of sound
√
is c2 = sqrtγRT2 = 1.3 · 461.9 · 813.15 = 698.8 m/s. So the Mach number is
M2c =
V2c
540
=
= 0.773
c2
689.8
and the stagnation temperature at the casing is
T2c = T2 (1 +
γ−1 2
M2c ) = 813.15(1 + 0.15 · 0.7732 ) = 886.0 K
2
From the work done on the casing end of the blade the exit stagnation temperature
is
wc = cp (T02c − T03c )
T03c = T02c −
72
175, 290
wc
= 886.0 −
= 798.4 K
cp
2001.6
ψ
= −0.95 tan
2
and the absolute velocity is
√
√
2
2
V3c = Vx3c
+ Vu3c
= 228.22 + 94.62 = 247.0 m/s
so that
T3c = T03c −
V3c2
2472
= 798.4 −
= 783.2 K
2cp
2 · 2001.6
At the hub
V2h
667
=
= 0.995
c2
689.8
and the stagnation temperature at the hub is
M2h =
T2h = T2 (1 +
γ−1 2
M2h ) = 813.15(1 + 0.15 · 0.9952 ) = 924.3 K
2
The exit stagnation temperature is
T03h = T02h −
wh
158, 088
= 924.3 −
= 845.3 K
cp
2001.6
and the absolute velocity is
√
√
2
2
V3h = Vx3h
+ Vu3h
= 228.22 + 31.92 = 230.4 m/s
so that
T3h = T03h −
2
230.4
V3h
= 845.3 −
= 832.0 K
2cp
2 · 2001.6
The stagnation pressure is a uniform p02 = 7 MPa at the inlet to the rotor. so that
the static pressure at the casing is
(
)−(γ−1)/γ
γ−1
p2c = p02 1 +
= 7000(1 + 0.15 · 0.7732 )−1.3/0.3 = 4826 kPa
2
and a similar calculation at the hub give
(
)−(γ−1)/γ
γ−1
= 7000(1 + 0.15 · 0.9952 )−1.3/0.3 = 4018 kPa
p2h = p02 1 +
2
At the exit p3 = 4016 kPa and it is uniform. The mach number at the casing is
M3c =
V3c
V3c
247
=√
=√
= 0.360
c3c
γRT3c
1.3 · 2001.6 · 783.2
73
and
(
p03c = p3c
γ−1 2
1+
M3c
2
)γ/(γ−1)
= (1 + 0.15 · 0.3602 )1.3/0.3 = 4368 kPa
At the hub
M3h =
V3h
V3h
230.2
=√
=√
= 0.326
c3h
γRT3h
1.3 · 2001.6 · 832.0
and
(
p03h = p3h
γ−1 2
1+
M3h
2
)γ/(γ−1)
= (1 + 0.15 · 0.3262 )1.3/0.3 = 4303 kPa
Exercise 6.7 Combustion gases, with γ = 4/3 and R = 287 kJ/kg K, flow thorough a turbine stage. The inlet flow angle for a normal stage is α1 = 0◦ . The
flow coefficient is ϕ = 0.52 and the blade loading coefficient is ψ = 1.4. (a)
Draw the velocity diagrams for the stage. (b) Determine the angle at which relative velocity leaves the rotor. (c) Find the flow angle at the exit of the stator.
(d) A two-stage turbine has an inlet stagnation temperature of T01 = 1250 K and
blade speed U = 320 m/s. If the adiabatic efficiency of the turbine is ηt = 0.89,
find the stagnation temperature of the gas at the exit of the turbine and the stagnation pressure ratio for the turbine. (e) Assuming that the density ratio across the
turbine based on static temperature and pressure ratios is the same as that based
on the stagnation temperature and stagnation pressure ratios, find the ratio of the
cross-sectional areas across the two-stage turbine.
Given: ϕ, ψ, α1 , T01 , U , ηtt and two stages.
Find: β3 , α2 , T03 , Ae /A1 .
Solution: With the flow coefficient and blade loading coefficient known, as well
as the inlet flow angle α1 = 0, the reaction can be calculated from
tan α1 =
1 − R − ψ/2
ϕ
R=1−
ψ
1.4
=1−
= 0.3
2
2
and the flow angles are
1 − R + ψ/2
tan α2 =
ϕ
(
−1
α2 = tan
74
1 − 0.3 + 0.7
0.52
)
= 69.62◦
⇐ (c)
(
)
0.3 + 0.7
β1 = tan
−
= −62.53◦ ⇐ (b)
0.52
(
)
R − ψ/2
0.3 − 0.7
−1
tan β2 = −
α2 = tan
−
= 37.59◦ ⇐ (b)
ϕ
0.53
R + ψ/2
tan β1 = −
ϕ
−1
With the flow angles fixed and axial velocity equal for both stages, the work is the
same. Hence
w = ψU 2 − 1.4 · 3202 = 143, 360 J/kg
therefore
wts =
wt = 2w = 286, 720 K/kg
286, 720
wt
=
= 322.2 kJ/kg
ηtt
0.89
and
T0e = T01 −
wt
286, 720
= 1250 −
= 1000.0 K
cp
1148
T0es = T01 −
322, 200
wts
= 1250 −
= 969.4 K
cp
1148
Therefore
p01
=
p0e
(
T01
T0es
With p1 = ρ1 RT1 ]
)γ/(γ−1)
(
=
1250
969.4
)4
= 2.76
⇐
(d)
p e = ρ e T Te ,
ρ1
p1
p01 T05
1000
= |f racTe T1 ≈
= 2.76
= 2.208
ρe
pe
p0e T01
1250
From ρ1 Vx1 A1 = ρe Ve Ae the area ratio is Ae /A1 = 2.208
The detailed calculations show that
Vx = ϕU = 0.52 · 320 = 166.4 m/s
and
Ve5
166.42
Te = T0e −
= 1000 −
= 988.2 K
2cp
2 · 1148
so that
pe
=
p0e
(
Te
T0e
(
)γ/(γ−1)
=
75
988.2
1000
)4
= 0.9526
and
T1 = T01 −
so that
p1
=
p01
Therefore
(
166.42
V15
= 1450 −
= 1237.9 K
2cp
2 · 1148
T1
T01
)γ/(γ−1)
(
=
1237.9
1250
)4
= 0.962
p5
p5 p05 p01
0.9526
=
=
= 0.3582
p1
p05 p01 p1
2.76 · 0.962
so that
ρ1
p1 Te
988.2
=
=
= 2.229
ρe
pe T1
0.362 · 1250
and the approximation is reasonably good.
⇐
(e)
Exercise 6.8 Steam enters a 10-stage fifty percent reaction turbine at pressure
0.8 MPa and 200 C and leaves at pressure 5 kPa and with quality equal to 0.86.
(a) If the steam flow rate is 7 kg/s, find the power output and the overall efficiency
of the turbine. (b) The steam enters each stator stage axially with velocity of
75 m/s. The mean rotor diameter for all stages is 1.4 m and the axial velocity is
constant through the machine. Find the rotational speed of the shaft. (c) Find the
inlet and exit flow angles at the mean blade height assuming equal enthalpy drops
for each stage.
Given: R, p01 , T01 , α1 , p0e , x = 0.86, Vx , D, ṁ and the number of stages is 10
and equal stagnation enthalpy drops across each stage.
Find: Ẇ , ηtt , α2 , and β1 .
Solution: From the steam tables with p01 = 80, 00 kPa and T01 , enthalpy is h01 =
2839.3 kJ/kg and entropy is s1 = 6.8158 kJ/kg K. For an isentropic process to
exit pressure of p0e = 5 kPa, entropy is ses = s1 , the quality is
xe =
ses − sf
6.1858 − 0.4764
=
= 0.808
sg − sf
7.9187
and
h0es = hf + xes hf g = 137.82 + 0.808 · 2424.4 = 2077.3 kJ/kg
h0e = hf + xe hf g = 137.82 + 0.86 · 2424.4 = 2222.8 kJ/kg
Therefore
ηtt =
2839.3 − 2222.8
h01 − h0e
=
2839.3 − 2077.3 = 0.811
h01 − h0es
|
76
⇐
(a)
Since equal work is done by each stage
w=
2839.3 − 2222.8
= 61.74 kJ/kg
10
Since the inlet and exit are axial and the reaction is 50 percent, the velocity triangles are symmetric and β2 = 0 so that Vu2 = u. Then the expression for work can
be solved for the blade speed
√
w = U (Vu2 − Vu3 ) = U Vu2 = U 2 U = sqrtw = 61740 = 248.5 m/s
Therefore
U
248.5 · 30
=
= 3390 rpm ⇐ (b)
r
0.7 · π
The flow angle to the rotor is
)
( )
(
U
248.5
−1
−1
α2 = −β1 = tan
= tan
= 73.2◦
Vx
75
Ω=
⇐
(c)
Exercise 6.9 Combustion gases enter axially into a normal stage at stagnation
temperature T01 = 1200 K and stagnation pressure p01 = 1500 kPa. The flow
coefficient is ϕ = 0.8 and the reaction is R = 0.4 and the inlet Mach number to
the stator is M1 = 0.4. Find; (a) the blade speed, (b) the Mach number leaving
the stator and the relative Mach number leaving the rotor. (c) Using the Soderberg
loss coefficients find the efficiency of the stage. (d) Repeat the calculations with
inlet Mach number M1 = 0.52.
Given: ϕ, α1 , R, T01 and p01 .
Find: α2 , β1 , R, and plot results for α1 = 10◦ to 30◦ .
Solution: With α1 = 0 and R = 0.4, then ψ = 2(1 − R) = 1.2. As a consequence
tan α2 =
1 − R + ψ/2
0.6 + 0.6
=
= 1.5 α2 = 56.31◦
ϕ
0.8
−(R + ψ/2)
0.4 + 0.6
=−
= 1.25 β1 = −51.34◦
ϕ
0.8
0.4 − 0.6
−(R − ψ/2)
=−
= 0.25 β2 = 14.04◦
tan β2 =
ϕ
0.8
With M1 = 0.4 and the inlet flow axial
(
)−1
γ−1 2
0.42
T1 = T01 1 +
= 1200(1 +
M1
) = 1168.8 K
2
6
tan β1 =
77
and
V1 = Vx = M1
√
√
γRT1 = 0.4 1.333 · 287 · 1168.8 = 267.5 m/s
Then U = Vx /ϕ = 267.5/0.8 = 334.4 m/s and the work done is
w = U Vx (tan α2 − tan α3 ) = 334.4 · 267.5 tan(56.31◦ ) = 134.2 kJ/kg
The velocity after the stator is
V2 =
Vx
267.5
=
= 482.2 m/s
cos α2
cos(56.31◦ )
The static temperature is
T2 = T02 −
V22
482.22
= 1200 −
= 1098.7 K
2cp
2 · 1.148
and
V2
482.2
=√
= 0.744
γRT2
1.333 · 287 · 1098.7
The stagnation temperature after the rotor is
M2 = √
T03 = T02 −
⇐
(a)
w
134.2
= 1200 −
= 1083.1 K
cp
1.148
Next the relative velocity is
W3 =
Wx
267.5
=
= 428.3 m/s
cos β3
cos(−51.34◦ )
and
T3 = T03 −
and
V32
267.52
= 1083.1 −
= 1051.9 K
2cp
2 · 1148
W3
428.3
=√
= 0.675
γRT3
1.333 · 287 · 1051.9
The deflections are
∆α = α2 − α1 = 56.31◦
M3R = √
∆β = β2 − β3 = 14.0451.34 = 65.38◦
78
⇐ (b)
so that
(
ζS = 0.04 + 0.06
(
ζR = 0.04 + 0.06
∆α
100
∆β
100
)2
(
= 0.04 + 0.06
)2
(
= 0.04 + 0.06
56.31
100
65.38
100
)2
= 0.0590
)2
= 0.0656
(
)
ϕ2
ζS
ζR
1
−1=
+
ηtt
2ψ cos2 β3 cos2 α2
(
)
0.82
0.0590
0.0656 2
◦
=
(51.34 ) +
= 0.09596
2 · 1.2
cos
cos2 (56.312 )
so that
and the total-to-total efficiency is ηtt = 0.9124 ⇐ (c).
If the inlet Mach number is increased to M1 = 0.52, then M2 = 0.9889 and
M3R = 0.9131. The efficiency remains the same.
Exercise 6.10 For a normal turbine stage fluid enters the stator at the angle 10◦ .
The relative velocity has an angle −40◦ as it leaves the rotor. The blade loading
factor is 1.6. (a) Determine the exit angle of the flow leaving the stator and the angle of the relative velocity as it enters the rotor. Determine the degree of reaction.
(b) For the conditions of part (a) calculate the flow exit angle and the angle of the
relative velocity entering the rotor, as well as the degree of reaction and plot them
as functions of α1 , from α1 = 10◦ to 30◦ .
Given: ψ, α1 , β3 .
Find: α2 , β1 , R, and plot results for α1 = 10◦ to 30◦ .
Solution:
1
= tan α1 − tan β1 = tan(10◦ ) − tan(−40◦ ) = 1.015 ϕ = 0.985
ϕ
Reaction is
R=1−
ψ
− ϕ tan α1 = 1 − 0.8 − 0.985 tan(10◦ ) = 0.026
2
⇐ (b)
α2 = tan−1 (ψ/ϕ + tan α1 ) = tan−1 (1.6/0985 + tan(10◦ )) = 60.96◦ ⇐ (a)
From
β2 = tan−1 (tan α2 − 1/ϕ) = tan−1 (tan(60.96◦ ) − 1.015) = 38.15◦ ⇐ (a)
79
%Hw 6.10
alpha3deg=10; alpha3=alpha3deg*pi/180;
beta3deg=-40; beta3=beta3deg*pi/180;
phi=1/(tan(alpha3)-tan(beta3)); psi=1.6
alpha2=atan(psi/phi+tan(alpha3));
alpha2deg=alpha2*180/pi;
R=1-0.5*phi*(tan(alpha2)+tan(alpha3))
% Computer part
alpha3adeg=linspace(-20,10);
alpha3a=alpha3adeg*pi/180;
n=length(alpha3a);
for i=1:n
phia(i)=1/(tan(alpha3a(i))-tan(beta3));
alpha2a(i)=atan(psi/phia(i)+tan(alpha3a(i)));
beta2a(i)=atan(tan(alpha2a(i)-1/phia(i)));
Ra(i)=1-0.5*phia(i)*(tan(alpha2a(i))+tan(alpha3a(i)));
end
plot(alpha3adeg,Ra); grid;
Exercise 6.11 For Example 6.6 plot the variation of the reaction from the hub to
the casing.
%HW=6.11
clear all
Rgas=287; cp=1148; k=4/3; T01=1100; p01=420000;
rho01=p01/(Rgas*T01); c01=sqrt(k*Rgas*T01);
rm=0.17; kappa2=0.7; rc2=2*rm/(1+kappa2); rh2=kappa2*rc2;
kappa3=0.65; rc3=2*rm/(1+kappa3); rh3=kappa3*rc3 ;
alpha2mdeg=60.08; alpha2m=alpha2mdeg*pi/180;
alpha3mdeg=-21.20; alpha3m=alpha3mdeg*pi/180;
beta2mdeg=25.98;
beta2m=beta2mdeg*pi/180;
beta3mdeg=-58.59; beta3m=beta3mdeg*pi/180;
Rinv=1-(tan(alpha2m)+tan(alpha3m))/(tan(beta2m)+tan(beta3m));
Rm=1/Rinv; phi1=-2*Rm/(tan(beta2m)+tan(beta3m))
phi=0.8; psim=phi*(tan(beta2m)-tan(beta3m));
Rm=-0.5*phi*(tan(beta3m)+tan(beta2m));
psi=2*(1-Rm-phi*tan(alpha3m)); kp=k/(k-1);
Vx=231; U=Vx/phi; Vum2=Vx*tan(alpha2m); Vum3=Vx*tan(alpha3m);
80
n=100;
for i=1:n+1;
zeta(i)=(i-1)*0.01;
r2(i)=2*((1-kappa2)/(1+kappa2))*zeta(i)+2*kappa2/(1+kappa2);
r3(i)=2*((1-kappa3)/(1+kappa3))*zeta(i)+2*kappa3/(1+kappa3);
alpha2(i)=atan(tan(alpha2m)/r2(i));
alpha3(i)=atan(tan(alpha3m)/r3(i));
beta2(i)=atan(tan(alpha2m)/r2(i)-r2(i)/phi);
beta3(i)=atan(tan(alpha3m)/r3(i)-r3(i)/phi);
R(i)=1-(1-Rm)/r2(i)^2;
end
alpha2deg=alpha2*180/pi; alpha3deg=alpha3*180/pi;
beta2deg=beta2*180/pi; beta3deg=beta3*180/pi;
plot(zeta,alpha2deg,zeta,alpha3deg,zeta,beta2deg,zeta,beta3deg)
figure(2); plot(zeta,R)
m=100;
%Numerical check
a=(2*pi*Vx*rho01);
sum=0; dr=(rc2-rh2)/m;
Vxr=Vx^2/(2*cp*T01);
Vur=Vum2^2/(2*cp*T01);
rhom=rho01*(1-Vxr-Vur)^(1/(k-1));
b=(2*pi*Vx*rhom);
for i=1:m+1
r(i)=rh2+(i-1)*dr;
rho(i)=(1-Vxr-Vur*rm^2/r(i)^2)^(1/(k-1));
f(i)=r(i)*rho(i);
end
mdota=0.5*a*(rc2^2-rh2^2)
mdotb=0.5*b*(rc2^2-rh2^2)
mdot=a*trapz(r,f)
Rh = 0.204
Rc = 0.610
81
Exercise 6.12 For a normal turbine stage the exit blade angle of the stator at 70◦
and relative velocity has an angle −60◦ as it leaves the rotor. For a range of
flow coefficients ϕ = 0.2 − 0.8 calculate and plot the gas exit angle from the
rotor, the angle the relative velocity makes as it leaves the stator, the blade loading
coefficient and the degree of reaction. Comment on what is a good operating range
and what are the deleterious effects in the flow over the blades if the mass flow
rate is reduced too much or if it is increased far beyond this range.
Given: ψ, α1 , β3 .
Find: α2 , β1 , R, and plot results for α1 = 10◦ to 30◦ .
Solution:
%HW6.12
clear all
phi=0.2:0.05:0.8;
alpha2d=70; alpha2=alpha2d*pi/180;
beta1d=-60; beta1=beta1d*pi/180;
psi=-1+phi.*(tan(alpha2)-tan(beta1));
alpha1=atan(tan(alpha2)-psi./phi);
alpha1d=alpha1*180/pi;
R=1-0.5.*phi.*(tan(alpha2)+tan(alpha1));
beta2=atan(tan(alpha2)-1./phi);
beta2d=beta2*180/pi;
subplot(2,2,1); plot(phi,alpha1d); grid;
xlabel(’Flow Coefficient \phi’);
ylabel(’Air rotor outlet angle \alpha_1’);
subplot(2,2,2); plot(phi,beta2d); grid;
xlabel(’Flow Coefficient, \phi’);
ylabel(’Flow angle, \beta_2’);
subplot(2,2,3); plot(phi, psi); grid;
xlabel(’Flow Coefficient \phi’);
ylabel(’Blade Loading Factor, \psi’);
subplot(2,2,4); plot(phi,R); grid;
xlabel(’Flow Coefficient \phi’);
ylabel(’Degree of Reaction, R’);
Exercise 6.13 Given: T01 , p01 , α1 , β3 , ψ, and V2 .
Find: α2 , β2 , R, and DpLS .
82
Solution: The flow coefficient is
ϕ=
1
1
=
= 0.985
◦
tan α1 − tan β1
tan(10 ) − tan(−40◦ )
The flow angles after the stator are
α2 = tan−1 (ψ/ϕ + tan α1 ) = tan−1 (1.6/0.985 + tan(10◦ )) = 60.96◦
⇐ (a)
β2 = tan−1 (tan α2 − 1/ϕ) = tan−1 (tan(60.96◦ ) − 1/0.985) = 38.15◦
⇐
The reaction is
1
1
R = 1 − ϕ(tan α2 + tan α1 ) = 1 − (tan(60.96◦ ) + tan(10◦ )) = 0.026
2
2
The Ainley-Mathieson correlations give
]( )
[
(α )
( α )2
s 2
2
2
+ 0.821
− 0.129
Ypa = −0.627
100
100
c
]( )
[
( α )2
(α )
s
2
2
+ 0.242
+ 1.489
− 1.676
100
100
c
( α )2
(α )
2
2
− 0.356
+ 0.399
+ 0.007 = 0.0268
100
100
Ype
[
]( )
( α )2
s 2
α2
2
= −1.56
+ 1.554
− 0.0639
100
100
c
[
]( )
( α )2
s
α2
2
+ 3.73
− 3.435
+ 0.289
100
100
c
( α )2
α
2
2
+ −0.82
+ 0.7806
+ 0.078 = 0.1058
100
100
and then
[
] ( )|α1 /α2 |
( α )2
t/c
1
Yp = Ypa +
(Ype − Ypa )
= 0.029
α2
0.2
The static temperature after the stator is
T2 = T02 −
4202
V2
= 700 −
= 699.8 K
2cp
2 · 1148
83
(b)
so that the Mach number is
M2 = √
420
V2
=
= 0.8116
1.333 · 287 · 699.38
γRT2
and the ratio of the stagnation pressure to the static pressure is
p02
=
p2
(
γ−1 2
1+
M2
2
)γ/(γ−1)
= (1 +
0.792 4
) = 1.517
6
and from the definition of the stagnation pressure loss coefficient
(
)
p01
p2
= 1 + Yp 1 −
= 1.010
p02
p02
Therefore
p02 = p01
380
p02
=
= 376.3 kPa
p01
1.01
and
376.3
p2
=
= 248.0 kPa
p02
1.517
and the loss of stagnation pressure is
p2 = p02
dp0LS = p01 − p02 = 380.0 − 376.2 = 3.8 kPa
The static enthalpy loss coefficient is obtained as follows. First
(
T2s = T02
p2
p02
)(γ−1)/γ
(
= 700
248.0
376.3
)0.25
The stagnation density is
ρ02 =
and therefore
ζN = (
p02
= 1.873 kg/m3
RT02
2dp0LS /ρ02
)
= 0.0205
1 + γ−1
M22 V22
2
84
= 630.8 K
Chapter 7
Exercise 7.1 The inlet and exit total pressures of a air flowing through a compressor are 100 kPa and 1000 kPa. The inlet total temperature is 281 K. What is
the work of compression if the adiabatic total-to-total efficiency is 0.75
Given: Inlet p01 and T01 , and exit p0e as well as ηtt .
Find: Work to compress the air.
Solution: From
ηtt =
ws
h02s − h01
=
w
h02 − h01
the work done is
w=
so that
1
cp T01
(h02s − h01 ) =
ηtt
ηtt
1004.5 · 281
w=
0.75
[(
1000
100
[(
)1/3.5
p02
p01
]
)(γ−1)/γ
−1
]
− 1 = 350 kJ/kg
Exercise 7.2 Air flows through an axial fan rotor at mean radius of 15 cm. The
tangential component of the absolute velocity is increased by 15 m/s through the
rotor. (a) Evaluate the torque exerted on the air by the rotor, if the flow rate is
0.472 m3 s and the pressure and temperature of the air are 100 kPa and 300 K. (b)
What is the rate of energy transfer to the air, if the rotational speed is 3000 rpm.
Given: Inlet p1 , T1 , and the change in the swirl velocity; the flow rate Q, rotational
speed Ω, and r.
Find: Torque and power.
Solution: Density at the inlet is
ρ1 =
p1
100
=
= 1.61 kg/m3
RT1
0.287 · 300
Mass flow rate is
ṁ = ρ1 Q = 1.161 · 0.472 = 0.5575 kg/s
85
The blade speed is
U = rΩ =
0.15 · 3000 · π
= 47.12 kg/s
30
The power needed to compress the air is
Ẇ = ṁU (Vu2 − vu1 ) = 0.5575 · 47.12 · 15 = 394 W
⇐
(b)
and the torque is
T =
Ẇ
394 · 30
=
= 1.23 N m
Ω
3000·
⇐
(a)
Exercise 7.3 The blade speed of a compressor rotor is U = 280 m/s and the
total enthalpy change across a normal stage is 31.6 kJ/kg. If the flow coefficient
ϕ = 0.5 and the inlet to the rotor is axial, find the gas angle leaving the rotor.
Given: ϕ and w, and U .
Find: Find the value of the flow angle at the exit of the rotor.
Solution: Since the flow at the inlet is axial, the equation for work reduces to
w = U (Vu2 − Vu1 ) = U Vu2 = U Vx tan α2
or
ψ = ϕ tan α2
With
31600
w
=
= 0.403
U2
2802
( )
(
)
ψ
0.403
−1
−1
α2 = tan
= tan
= 38.9◦
ϕ
0.5
ψ=
then
⇐
Also
Vx = ϕU = 0.5 · 280 = 140 m/s
and
Wu2 = Vu2 − U = Vx tan α2 − U = 140 tan(38.9◦ ) − 280 = −167.1 m/s
(
and
β2 = tan
−1
Wu2
Vx
)
(
−1
= tan
86
−167.1
140
)
− 50.05◦
⇐
Exercise 7.4 Air flows through an axial flow fan, with an axial velocity of 40 m/s.
The absolute velocities at the inlet and the outlet of the stator are at angles of 60◦
and 30◦ , respectively. The relative velocity at the outlet of the rotor is at an angle
−25◦ . Assume reversible adiabatic flow and a normal stage. (a) Draw the velocity
diagrams at the inlet and outlet of the rotor. (b) Determine the flow coefficient. (c)
Determine the blade loading coefficient. (d) Determine at what angle the relative
velocity enters the rotor. (e) Determine the static pressure increase across the
rotor in Pascals. The inlet total temperature is 300 K and the inlet total pressure is
101.3 kPa. (f) Determine the degree of reaction.
Given: Inlet p01 , T01 , Vx , α1 , β2 , α2 . Reversible adiabatic flow.
Find: ϕ, ψ, R and p2 − p1 .
Solution: The blade speed
U = Vu2 −Wu2 = Vx (tan α2 −tan β2 ) = 40(tan(60◦ )−tan(−15◦ )) = 87.93 m/s
and the flow coefficient is
ϕ=
Vx
40
=
= 0.455
U
87.93
⇐
(b)
The work done is
w = U Vx (tan α2 − tan α1 ) = 87.63 · 40(tan(60◦ ) − tan(30◦ )) = 4061.5 J/kg
so that
w
4061.5
=
= 0.525
U2
87.932
Writing the equation for work as
ψ=
w = U (Wu2 − Wu1 ) = Vx (tan β1 − tan β2 )
so that
(
−1
β1 = tan
ψ
tan β2 −
ϕ
)
(
−1
= tan
⇐
(c)
ψ = ϕ(tan β1 − tan β2 )
0.525
tan(−25 ) −
0.455
◦
)
= −58.3◦
⇐ (d)
The reaction is
0.455
1
[tan(60◦ ) + tan(30◦ )] = 0.475 ⇐ (f )
R = 1− ϕ(tan α2 +tan α1 ) = 1−
2
2
87
Since the velocities are low, the flow may be assumed incompressible and since it
is reversible and adiabatic
∆p0
w=
ρ
The density is
ρ01 =
p01
101.3
=
= 1.1765 kg/m3
RT01
0.287 · 300
and it may be assumed that ρ1 = ρ01 . Thus
p02 = p01 + ρw = 101.3 + 1.1765 · 4.0615 = 106.1 kPa
The velocities are
V1 =
40
Vx
=
= 46.19 m/s
cos α1
cos(30◦ )
V2 =
Vx
40
=
= 80.00 m/s
cos α2
cos(60◦ )
The static pressure at the inlet is
1
1.1765 · 46.192
p1 = p01 − ρV12 = 101.3 −
= 100.05kPa
2
2 · 1000
1
1.1765 · 80.02
p2 = p02 − ρV22 = 106.1 −
= 102.31kPa
2
2 · 1000
so that p2 − p1 = 102.31 − 100.05 = 2.26 kPa
⇐ (e)
Exercise 7.5 Air flows through an axial flow compressor. The axial velocity is
60 percent of the blade speed at the mean radius. The reaction ratio is 0.4. The
absolute velocity enters the stator at an angle of 55 deg from the axial direction.
Assume a normal stage. (a) Draw the velocity diagrams at the inlet and outlet
of the rotor. (b) Determine the flow coefficient. (c) Determine the blade loading
coefficient. (d) Determine at what angle the relative velocity enters the rotor. (e)
Determine at what angle the relative velocity leaves the rotor. (f) Determine at
what angle the absolute velocity leaves the stator.
Given: Vx = 0.U , R = 04, and α2 = 55◦ .
Find: α1 , β1 , β2 and ψ.
88
Solution: Since Vx = 0.6U , the flow coefficient is ϕ = 0.6. The flow angle α1
can be obtained from
(
)
(
)
2 · 0.6
2(1 − R)
−1
−1
◦
α1 = tan
− tan α2 = tan
− tan(55 ) = 29.76◦
ϕ
0.6
Then the blade loading coefficient is
ψ = 2(1 − R tan α1 ) = 2(1 − 0.4 − 0.6 tan(29.76◦ )) = 0.514
and then
R + ψ/2
tan β1 = −
ϕ
R − ψ/2
tan β2 = −
ϕ
(
−1
β1 = tan
(
−1
β2 = tan
0.5 + 0.514/2
−
0.6
0.5 − 0.514/2
−
0.6
)
= −47.59◦
⇐
(e)
= −13.42◦
⇐
(e)
)
Exercise 7.6 The angle at which the absolute velocity enters the rotor of single
stage axial compressor is α1 = 40◦ and the relative velocity at the inlet of the rotor
is β1 = −60◦ . These angles at the inlet of the stator are α2 = 60◦ and β2 = −40◦ .
The mean radius of the rotor is 30 cm and the hub to tip radius is 0.8. The axial
velocity is constant and has a value Vx = 125 m/s. The inlet air is atmospheric
at pressure 101.325 kPa and temperature 293 K. (a) Find the mass flow rate. (b)
What is the rotational speed of the shaft under these conditions? (c) What is the
power requirement of the compressor?
Given: α1 , α2 , β1 , r, rh /rt and Vx . The inlet conditions are p01 and T01 .
Find: ṁ, Ω, Ẇ .
Solution: To calculate the mass flow rate from
ṁ = ρ1 AVx
the area and density need to be determined. The area can be written as
A = π(rc2 − rh2 ) = πrc2 (1 − κ2 )
in which κ = rh /rc = 0.8 and from
1
rm = (rh + rc )
2
89
rc
(1 + κ)
2
thus
rc =
2rm
2 · 0.3
1
=
= m
1+κ
1 + 0.8
3
A=
π
(1 − 0.64) = 0.1257 m2
9
and
The velocity at the inlet is
V1 =
Vx
125
=
= 163.18 m/s
cos α1
cos(40◦ )
The stagnation Mach number is
V1
163.18
=√
= 0.4756
γRT01
(1.4 · 287 · 293
M01 =
and the Mach number is
0.4756
M01
=√
M1 = √
= 0.4876
2
1 − 0.2 · 0.27562
M
1 − γ−1
01
2
Then
T1 =
and
(
p1 = p01
T01
1+
T1
T01
γ−1
M12
2
=
293
= 279.75 K
1 − 0.2 · 0.48762
)γ/(γ−1)
(
= 101.325
279.75
293
)3.5
= 86.17 kPa
and the density is
ρ1 =
p1
81.17
=
= 1.073 kg/m3
RT1
0.287 · 279.75
The mass flow rate is
ṁ = rho1 AVX = 1.073 · 0.1257 · 125 = 16.86 kg/s
⇐
(a)
From
U = Vu1 −Wu1 = Vx (tan α1 −tan β1 ) = 125 [tan(40◦ ) − tan(60◦ )] = 321.4 m/s
so that
Ω=
U
321.4 · 30
=
= 10230 rpm
rm
0.3 · π
90
⇐
(b)
The work done is
w = U Vx (tan α2 − tan α1 ) = 321.4 · 125 [tan(60◦ ) − tan(40◦ )] = 35.87 kJ/kg
and the power is
Ẇ = ṁw = 16.86 · 35.87 = 604.8 kW
⇐
(c)
Exercise 7.7 The angle at which the absolute velocity enters the rotor a compressor stage is α1 = 35◦ and the relative velocity makes an angle β1 = −60◦ . The
corresponding angles at the inlet to the stator are α2 = 60◦ and β2 = −35◦ . The
stage is normal and the axial velocity is constant through the compressor. (a) Why
does the static pressure rise across both the rotor and the stator. (b) Draw the velocity triangles. (c) If the blade speed is U = 290ms, find the axial velocity. (d)
Find the work done per unit mass flow for a stage and the increase in stagnation
temperature across it. (e) The stagnation temperature at the inlet is 300 K. The
overall adiabatic efficiency of the compressor is ηc = 0.9 and the overall stagnation pressure ratio is 17.5. Determine the number of stages in the compressor. (f)
How many axial turbine stages will it take to power this compressor?
Given: Inlet T01 , U , α1 , α2 , β1 , β2 and p0e /p01 as well as ηc .
Find: Vx , w and ∆T0 across as stage, the number of stages in the compressor and
suggest how many turbine stages are needed to power the compressor.
Solution: The first part asks why pressure increases across the stator and the rotor. Because the flow is turned toward the axis in the stator and the axil velocity
remains constant, its kinetic energy decreases. This means that the static enthalpy
increases, since the stagnation remains constant. With an increase in static enthalpy pressure increases, as seen from
T ds = dh − vdp
for isentropic flow and there is no qualitative change as irreversibilities are introduced. The same arguments hold for the rotor, but now the relative velocity is
turned toward the axis and since the relative stagnation enthalpy remains constant,
the static enthalpy and thus also the pressure increased across the rotor. Part (b)
asks for the velocity diagrams. Since the angles are given, and the velocity U is
also given, they can be drawn. From
1
= tan α1 − tan β1 = tan(35◦ ) − tan(−60◦ ) = 2.4323
ϕ
91
so that phi = 0.411 and
Vx = ϕU = 0.4112̇90 = 119.2ms
⇐
(c)
Next
ϕ
(tan α2 + tan α1 ) = 0.5
2
as was expected as the velocity diagrams are symmetric. Then
R=1−
ψ = 2(1 − R − ϕ tan α1 ) = 1 − 2 · 0.411 tan(35◦ ) = 0.424
and
w = ψU 2 = 0.424 · 2902 = 35, 678 K/kg
DT0 =
w
= 35.5 K
cp
⇐
(d)
With the pressure ratio known, the exit temperature can be determined from
( )(γ−1)/γ
1 p0e
1
f racT0e T01 = 1 +
=1+
(17.51/3.5 ) = 2.401
ηc p01
0.9
so that T0e = 2.401·300 = 721.8 K. The number of stages can now be determined
from
T0e − T01
421.8
N=
=
= 11.88 N = 12 ⇐ (e)
T03 − T01
35.5
Typically one turbine stage is needed to power 5 or 6 compressor stages, so 2
turbine stages are needed. This is the answer to part (f).
Exercise 7.8 The blade speed of a rotor of an axial air compressor is U = 150 m/s.
The axial velocity is constant and equal to Vx = 75 m/s. The tangential component of the relative velocity leaving the rotor is Wu2 = −30 m/s, the tangential
component of the absolute velocity entering the rotor is Vu1 = 55m/s. The stagnation temperature and pressure at the inlet to the rotor are 340 K and 185 kPa. The
stage efficiency is 0.9 and one-half of the loss in stagnation pressure takes place
through the rotor. (a) Draw the velocity diagrams at the inlet and exit of the rotor.
(b) Find the work done per unit mass flow through the compressor. (c) Draw the
states in an hs-diagram. (d) The stagnation and static temperatures between the
rotor and the stator. (e) The stagnation pressure between the rotor and the stator.
Given: Inlet p01 and T01 , Vu1 , Vx , Wu2 and ηtt .
Find: Work to compress the air, T02 and T2 , p02 .
92
Solution: The remaining velocity components are
Vu2 = U + Wu2 = 150 − 30 = 20mm/s
Wu1 = Vu1 − U = 65 − 150 = 95 m/s
and the work done is
w = U (Vu2 − Vu1 ) = 150(120 − 55) = 9750 J/kg
and
T02 = T01 +
w
9750
= 340 +
= 349.7 K
cp
1004.5
⇐
⇐
(b)
(d)
The isentropic work is
ws = ηtt w = 0.9 · 9750 = 8775 J/kg
and
T03s = T01 +
so that
(
p03 = p01
T03s
T01
8775
= 348.7 K
1004.5
)γ/(γ−1)
(
= 185 +
349.7
340
)3.5
= 202.18 kPa
The ideal pressure reached after the rotor is there were no losses and the same
amount of work is done as in the actual case is
(
)γ/(γ−1)
(
)3.5
T02
349.7
p02i = p01
= 185
= 204.15 kPa
T01
340
and the stagnation pressure loss is
Dp0L = p02i − p03 = 204.15 − 202.18 = 1.97 kPa
The stagnation pressure after the rotor is
⇐
p02 = p03 + 0.5Dp0L = 202.18 + 0.985 = 203.17 kPa
(e)
The velocity after the rotor is
√
√
2
= 752 + 1202 = 141.51 m/s
V2 = Vx2 + Vu2
and the static temperature is
T2 = T02 −
141.512
V22
= 349.7 −
= 339.7 K
2cp
2 · 1004.5
93
⇐
(d)
Exercise 7.9 Air from ambient at 101.325 kPa and temperature 20 C enters the
first stage of multistage axial flow compressor with velocity 61 m/s. The blade tip
radius is 60 cm and the hub radius is is 42 cm. The shaft speed is 1800 rpm. The
air enters a stage axially and leaves it axially at the same speed. The rotor turns the
relative velocity 18.7◦ toward the direction of the blade movement. The total-tototal stage efficiency is 0.87. (a) Draw the inlet and exit velocity diagrams for the
rotor. (b) Draw the blade shapes. (c) Determine the flow coefficient and the blade
loading factor. (d) Determine the mass flow rate. (e) What is the work required
per unit mass. (f) What is the total pressure ratio for the stage? (g) Determine the
degree of reaction.
Given: p01 , T01 , Vx , Dβ, rc , rh , Ω, and ηtt .
Find: ψ, ϕ, R, ṁ, w and p0e /p01 .
Solution: Since the inlet velocity is axial and it is quite small the density at the
inlet is the same as the stagnation density
ρ01 =
p01
101.325
=
= 1.205 kg/m3
RT01
287 · 293
and the mass flow rate is
ṁ = ρAVx = ρπ(rc2 − rh2 )Vx = 1.205 · π(0.62 − 0.422 ) = 42.40 kg/s
⇐ (d)
The mean radius and the blade speed are
1
0.6 + 0.42
0.51 · 1800 · π
r = (rc + rh )
= 0.51 m U = rm Ω =
= 96.13 m/s
2
2
30
Now the flow angles can be determined. The relative velocity is at the angle β1
entering the rotor. It is
(
)
(
)
−U
−96.13
−1
−1
β1 = tan
= tan
= −57.6◦
Vx
61
and
β2 = β1 + 18.7◦ = −57.6◦ + 18.7◦ = −38.9◦
The tangential component of the velocity entering the stator is
Vu2 = Wu2 + U = Vx tan β2 + U = 62 tan(−38.9◦ ) + 96.13◦ = 46.91 m/s
94
(
and
−1
α2 = tan
Vu2
Vx
(
)
= tan
−1
46.91
61
)
= 37.56◦
The specific work done is
w = U (Vu2 − Vu1 ) = 96.1(46.91 − 0) = 4509 J/kg
⇐
(e)
and
ws = ηtt w = 0.87 · 4509 = 3922.8 J/kg
so that
ψ=
4509
Vx
w
61
=
= 0.488 ϕ =
=
= 0.635
2
2
U
96.13
U
96.13
⇐ (c)
The stagnation temperature after the rotor is
T02 = T01 +
w
4509
= 293 +
= 297.5 K
cp
1004.5
T02s = T01 +
ws
3922.8
= 293 +
= 296.9 K
cp
1004.5
and
The pressure ratio is therefore
p02
=
p01
(
T02s
T01
)γ/(γ−1)
(
=
296.9
293
)3.5
= 1.047
The reaction is
R=1−
ϕ
0.635
(α2 + tan α1 ) = 1 −
(tan(37.56◦ ) − 0) = 0.756
2
2
⇐
(g)
Exercise 7.10 Air from ambient at 101.325 kPa and temperature 300 K enters an
axial flow compressor stage axially with velocity 122 m/s. The blade tip radius
is 35 cm and the hub radius is 30 cm. The shaft speed is 6000 rpm. At the exit
relative velocity is at an angle −45◦ . The total-to-total stage efficiency is 0.86. (a)
Draw the inlet and exit velocity diagrams for the rotor. (b) Draw the blade shapes.
(c) Determine the flow coefficient and the blade loading factor. (d) Determine the
mass flow rate. (e) What is the total pressure ratio for the stage? (f) Determine the
degree of reaction.
95
Given: Inlet p01 , T01 , Vx , rc , rh , Ω β1 and ηtt .
Find: ψ, ϕ, R, and p03 /p01 .
Solution: The mean radius and the blade speed are
1
0.35 + 0.30
0.325 · 6000 · π
r = (rc +rh )
= 0.325 m U = rm Ω =
= 204.2 m/s
2
2
30
Now the flow angles can be determined. The relative velocity is at the angle β1
entering the rotor. It is
(
)
(
)
−U
−204.2
−1
−1
= tan
= −59.1◦
β1 = tan
Vx
122
From the expressoion for the tangential component of the velocity
Vx tan α2 = U + Vx tan β2
which gives
(
α2 = tan
−1
U
+ tan β2
Vx
)
(
−1
= tan
)
204.2
◦
+ tan(−45 ) = 34.0◦
122
The specific work done is
w = U (Vu2 − Vu1 ) = 204.2(82.3 − 0) = 16, 806 J/kg
and
ws = ηtt w = 0.86 · 16806 = 14, 453 J/kg
so that
ψ=
w
16806
Vx
122
=
= 0.40 ϕ =
=
= 0.60
2
2
U
204.2
U
204.2
The stagnation temperature after the rotor is
T02 = T01 +
w
16, 806
= 300 +
= 316.7 K
cp
1004.5
T02s = T01 +
ws
14, 453
= 300 +
= 314.4 K
cp
1004.5
and
96
⇐ (c)
The pressure ratio is therefore
(
)γ/(γ−1) (
)3.5
p02
T02s
314.4
=
=
= 1.178
p01
T01
300
⇐ (e)
The reaction is
R=1−
ϕ
0.60
(α2 + tan α1 ) = 1 −
(tan(34.0◦ ) − 0) = 0.80
2
2
⇐
(f )
The stagnation density at the inlet is
ρ01 =
p01
101.325
=
= 1.205 kg/m3
RT01
287 · 300
The static temperature at the inlet is
T1 = T01 −
V12
1222
= 300 −
= 292.6 K
2cp
2 · 1004.5
and the static pressure is
(
)γ/(γ−1)
(
)3.5
T1
292.6
p1 = p01
= 101.325
= 92.83 kPa
T01
300
and the static density is
ρ1 =
p1
92.83
=
= 1.1055 kg/m3
RT1
287 · 292.6
and the mass flow rate is
ṁ = ρAVx = ρπ(rc2 − rh2 )Vx = 1.1055 · π(0.352 − 0.302 ) = 13.77 kg/s
⇐ (d)
Exercise 7.11 Carry out design calculations for a compressor stage with blade
loading factor in the range ψ = 0.25 to ψ = 0.55, flow coefficient ϕ = 0.7 and
reaction R = 0.6. Keep the diffusion factor equal to 0.45. Calculate and plot 1−η,
solidity, and the static pressure rise 1 − V22 /V12 for the rotor and stator (including
the de Haller criterion). What are the stagnation losses across the stator and rotor?
Given: The range of blade loading factors and R = 0.6
Find: Solidity, diffusion factor, loss, diffusion.
Solution:
97
% Compressor calculations
clear all; clf;
% Specified conditions
R=0.4; phi=0.6; DF=0.45;
% Blade angles
alpha1=atan((1-R-0.5*psi)./phi);
alpha2=atan((1-R+0.5*psi)./phi);
beta1=-atan((R+0.5*psi)./phi);
beta2=-atan((R-0.5*psi)./phi);
% Solidities using Lieblein’s diffusion factor
sigmar=0.5*sin(beta2-beta1)./(cos(beta1)-(1-DF).*cos(beta2));
sigmas=0.5*sin(alpha2-alpha1)./(cos(alpha2)-(1-DF).*cos(alpha1));
% Loss coefficients
omegar=0.014*sigmar./cos(beta2);
omegas=0.014*sigmas./cos(alpha1);
% Pressure rise and de Haller Conditions
w1sq=phi.^2+(0.5*psi+R).^2;
w2sq=phi.^2.*(1+tan(beta2).^2);
v2sq=phi.^2+(0.5*psi+1-R).^2;
v1sq=phi^2.*(1+tan(alpha1).^2);
hallerr=1-w2sq./w1sq; hallers=1-v1sq./v2sq;
% Stagantion pressure loss and efficiency
Dp0LR=(0.5*phi^2./psi).*omegar./cos(beta1).^2;
Dp0LS=(0.5*phi^2./psi).*omegas./cos(alpha2).^2;
Dp0L=Dp0LR+Dp0LS;
% Plots
n=length(hallers);
dehaller=0.44*ones(1,n);
subplot(2,2,1), plot(psi,hallerr,’.-’,psi,hallers,psi,dehaller);
grid on; axis([0.25 0.52 0.2 0.7]);
xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12)
ylabel(’1-V_2^2/V_1^2’,’Fontsize’,12)
text(0.26, 0.47,’de Haller’)
legend(’Rotor’,’Stator’)
set(gca,’Fontsize’,12); hold on;
subplot(2,2,2), plot(psi,Dp0LR,’.-’,psi,Dp0LS,psi,Dp0L);
grid on; axis([0.25 0.52 0.00 0.08]);
xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12)
98
ylabel(’Loss, 1-\eta’,’Fontsize’,12)
legend(’Rotor’,’Stator’,’Total’)
subplot(2,2,3), plot(psi,sigmar,’.-’,psi,sigmas);
grid on; axis([0.25 0.52 0.0 3.0]);
xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12)
ylabel(’Solidity’,’Fontsize’,12)
legend(’Rotor’,’Stator’)
s=[1.0 1.4 2.0];
subplot(2,2,4);
for i=1:3
ddf(i,:)=1-cos(beta1)./cos(beta2) + ...
0.5*cos(beta1).*(tan(beta2)-tan(beta1))./s(i);
plot(psi,ddf(i,:)); hold on;
end
axis([0.25 0.52 0.2 0.6]); grid;
xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12)
ylabel(’Diffusion Factor’,’Fontsize’,12)
text(0.405,0.55,’\sigma=1.0’), text(0.405,0.35,’\sigma=2.0’)
text(0.405,0.482,’\sigma=1.4’);
% End of script
Exercise 7.12 A compressor stage with reaction ratio R = 0.54, and stator outlet
metal angle χ3 = 14.5◦ . The camber angle is θ = 32◦ , pitch chord ratio is
s/c = 0.82, and the position of maximum camber a/c = 0.45. The ratio of the
blade height to the chord is b/c = 1.7. (a) Find the deviation of the flow leaving
the stator. (b) Find the deflection. (c) Find the flow coefficient and blade loading
coefficient if the inflow is at zero incidence.
Given: R, χ2 , θ, s/c, a/c, b/c.
Find: Deviation, deflection, ϕ, ψ.
Solution: The flow angle at the exit of the stator is
√
( a )2 √ s
∗
α
s
3
θ
+
θ
α3∗ = χ3 + δ ∗ = χ3 + 0.902
c
c 500
c
and solving this for α3∗ gives
α3∗
√
χ3 + 0.92(a/c)2 θ s/c
√
=
= 21.12◦
θ
1 − 500 s/c
99
and the deviation at the exit is
δ ∗ = α3∗ − χ3 = 21.12◦ − 14.5◦ = 6.62◦
⇐
(a)
From the tangent difference formula
(
∗
−1
α2 = tan
tan α3∗ +
)
1.55
1 + 1.5s/c
)
(
1.55
−1
◦
= 47.24◦
= tan
tan(21.12 ) +
1 + 1.5 · 0.82
and the deflection is
ϵ∗ = α2∗ − α3∗ = 47.24◦ − 21.12◦ = 26.12◦
⇐
(b)
The flow coefficient is
1
= tan α3∗ + tan α2∗ = tan(21.12◦ ) + tan(47.24◦ ) = 1.429
ϕ
ϕ = 0.7
⇐ (c)
and
ψ = ϕ(tan α2∗ − tan α3∗ ) = 0.7[tan(47.24◦ ) − tan(21.12◦ )] = 0.474 ⇐
(c)
Exercise 7.13 The circular arc blades of a compressor cascade have camber θ =
30◦ and the maximum thickness at a/c = 0.5. The spacing to chord ratio is
s/c = 1.0. The nominal outflow angle is α∗ = 25◦ . (a) Determine the nominal
incidence. (b) Determine that lift coefficient at the nominal incidence if the drag
coefficient is CD = 0.016.
Given: θ, s/c, a/c, α2∗ , and CD .
Find: The nominal incidence and the lift coefficient.
Solution: From the tangent difference formula
(
)
1.55
∗
−1
∗
α2 = tan
tan α3 +
1 + 1.5s/c
(
)
1.55
−1
◦
∗
tan(25 ) +
= 47.37◦
α2 = tan
2.5
100
The deviation is obtained from
√
δ ∗ = mθ
in which
m = 0.92
so that
( a )2
c
+
s
c
α3∗
25
= 0.92 · 0.25 +
= 0.28
500
500
δ ∗ = 0.28 · 30◦ · 1 = 8.4◦
⇐
(a)
The metal angle at the exit is
χ2 = χ3 + θ = 16.6◦ + 30◦ = 46.6◦
and the incidence at the inlet is
i∗ = α2∗ − δ ∗ = 47.37◦ − 46.6◦ = 0.77◦
The lift coefficient is
CL = 2
(s)
c
cos αm (tan α2∗ − tan∗3 ) − CD tan αm
the mean flow direction is
(
)
[
]
1
−1
∗
∗
−1 1
◦
◦ ∗
αm = tan
(tan α2 + tan α3 ) = tan
(tan(47.37 ) + tan(35 )3 ) = 37.83◦
2
2
so that
CL = 2·1·cos(37.82◦ )(tan(47.37◦ )−tan(25◦ ))−0.018 tan(37.82◦ ) = 0.966
⇐ (b)
Exercise 7.14 Air with density 1.21 kg/m3 flows into a compressor stator with
velocity V2 = 120 m/s and leaves at the angle α3 = 30◦ . If the Leiblein’s diffusion
factor is to be held at 0.5 and the the stagnation pressure loss across a compressor
stator is 0.165 kPa, find the static pressure increase across the stage assuming that
the flow is incompressible.
Given: ρ1 , V2 , Dp0 LS, and DF .
Find: p3 − p2 .
101
Solution: The definition of the diffusion factor
DF =
V2 − V3 Vu2 − Vu3 s
+
V2
2V2
c
can be written as
DF = 1 −
cos α2
Vx s
+
(tan α2 tan α3 )
cos α3 2V2 c
or
cos α2 cos α2 s
+
(tan α2 tan α3 )
cos α3
2 c
Letting x = cosα2 then inserting the numerical values gives
(√
)
1
2
1
4x
1 − x2
=1− √ x+
−√
2
10
x
3
3
DF = 1 −
which reduces to
√
9
13x2 − 5 3x +
=0
16
and the solution of this is
√
√
5 3 + 75 − 13 · 9/4
x=
26
x = 0.593236 α2 = 53.61◦
Assuming incompressible flow the difference is the stagnation pressures is
1
p02 − p03 = p2 − p3 + ρ(V22 − V32 )
2
and
[
]
1 2
cos2 α2
p3 − p2 = ρV2 1 −
− ∆p0
2
cos2 α3
(
)
1.21
cos2 (53.61◦ )
2
=
· 120 1 −
− 125, 000 = 4.46 kPa
2
cos2 (30◦ )
Exercise 7.15 Air at temperature 288 K and at pressure 101.325 kPa flows into
a compressor with 10 stages. The efficiency of the first stage is 0.920 and the
second stage 0.916. For the rest of the stages the stage efficiency drops by 0.004
successively so that the last stage has an efficiency 0.884. The axial velocity is
constant and the flow angles are the same at the inlet and exit of each of the stages.
Hence the work done by each stage is the same. (a) Find the overall efficiency of
102
the compressor. (b) Find the overall efficiency by using the theory for a polytropic
compression with small stage efficiency ηp = 0.902.
Given: Inlet p01 and T01 , and the stage efficiencies of ten stages.
Find: Find the overall efficiency.
Solution: The following Matlab script gives the overall effieciency.
%HW 7.15
clear all
p0(1)=101325; T0(1)=288; T0s(1)=T0(1);
n=10; r=4.3;
k=1.4; kp=k/(k-1);
etapg=0.902;
etag=(r^(1/kp)-1)/(r^(1/(kp*etapg))-1);
T0eg=T0(1)*r^(1/kp)
T0e=T0(1)+(T0eg-T0(1))/etag;
dT0a=(T0e-T0(1))/n
dT0=16.8323
etap=[0.92, 0.916, 0.912, 0.908, 0.904, ...
0.900, 0.896, 0.892, 0.888, 0.884]
for i=2:n+1
T0s(i)=T0(i-1)+etap(i-1)*dT0;
T0(i)=T0(i-1)+dT0
p0(i)=p0(i-1)*(T0s(i)/T0(i-1))^kp
end
rt=p0(n+1)/p0(1)
T0ss=T0(1)*r^(1/kp)
eta=(T0ss-T0(1))/(T0(n+1)-T0(1))
etaa=(r^(1/kp)-1)/(r^(1/(etapg*kp))-1)
The exact value for the overall efficiency is ηc = 0.8846 and the approximate
value is ηc = 0.8804. Hence the use of the average small stage efficiency does not
change the overall efficiency by much.
Exercise 7.16 Air flows in a axial flow fan of free vortex design, with hub radius
rh = 7.5 cm and casing radius 17 cm. The fan operates at 2400 rpm. The volumetric flow rate Q = 1.1 m3 /s and the stagnation pressure rise is 3 cm H2 O. The
fan efficiency is ηtt = 0.86. (a) Find the axial velocity. (b) Find the work done on
103
the fluid. (c) Find the absolute and relative flow angles at the inlet and exit of the
rotor when the inlet is axial.
Given: rh and rc , Ω, Q, ∆p, ηtt .
Find: w α1 , α − 2, β1 , β2 .
Solution: Free vortex design. Then mean radius is
1
1
rm = (rh + rc ) = (7.5 + 17.0) = 12.25 cm
2
2
and the blade speed at the mean radius is
Um = rm Ω =
0.1225 · 2400 · π
= 30.79 m/s
30
The blade speed at the hub and the casing are
Uh =
rh
Um
rm
Uc =
rc
Um
rm
The density is
ρ01 =
p01
101325
=
= 1.20 kgm2
RT01
287 · 293
and the flow area is
A = π(rc2 − rh2 ) = π(172 − 7.52 ) = 731.2 cm2
and the axial velocity is
Vx =
Q
1.1
=
= 15.0m/s
A
0.07312
The ideal work is
ws =
∆p0
0.03 cdot9.81 · 998
=
= 243.8 J/kg
ρ
1.2
and the actual work is w = ws /eta = 243.8/0.86 = 283.4 J/kg. Thus since the
inlet is axial
w
243.8
Vu2 =
=
= 9.21 m/s
Um
30.79
Now
(
)
(
)
Um
30.79
−1
−1
β1m = tan
−
= tan
−
= −64◦
⇐ (a)
Vx
15
104
and
(
−1
α2m = tan
Vu2m
Vx
)
(
−1
= tan
9.217
−
15
)
= 31.6◦
⇐ (a)
From Vu2m = Um + Wu2m which is also Vx tan α2m = Um + Vx tan β2m from
which
)
(
)
(
30.79
Um
−1
◦
−1
= tan
tan(31.6 ) −
= −55.2◦ ⇐ (a)
β2m = tan
tan α2m −
Vx
15
Similarly at the hub and the casing
α1h = 0◦
α2h = 45◦
α1c = 0◦
α2c = 23.8◦
β1h = −51.4◦
β2h = −14.2◦
β1c = −70.6◦
β2c = −67.4◦
Exercise 7.17 Consider an axial flow compressor in which flow leaves the stator
with a tangential velocity distributed as a free vortex. The hub radius is 20 cm and
the static pressure at the hub is 94 kPa and the static temperature there is 292 K.
The radius of the casing is 30 cm and the static pressure at the casing is 97 kPa.
The total pressure at this location is 101.3 kPa. Find the exit flow angles at the
hub and the casing.
Given: pc , p01h , ph Th , rh , rc .
Find: αc and αh .
Solution: The density at the hub is
ρh =
From
ph
94
=
= 1.12 kg/m3
RTh
0.287 · 292
1
p0h = ph + ρh Vh2
2
the velocity at the hub is
√
√
2(p0h − ph )
2(101.3 − 94)1000
=
= 114.1 m/s
Vh =
ρh
1.122
so that
∫
pt − p h =
rc
rh
Vu
ρ dr = ρK 2
r
∫
105
rc
rh
dr
ρK 2
=
r3
2
(
1
1
− 2
2
rh rc
)
so that
√
K=
2(pc − ph )
=
ρ(1/rh2 − 1/rc2 )
√
2(97000 − 94000)
= 9.81 m2 /s
1.12(1/0.102 − 1/0.152 )
Also, assuming that the density is the same
1
1
pc + ρVc2 = ph + ρVh2
2
2
The velocity at the casing is
√
√
2(p0c − pc )
2(101300 − 97000)
Vc =
=
= 87.56 m/s
ρ
1.12
and
Vuc =
K
9.81
=
= 65.4 m/s
rc
0.15
Vuh =
9.81
= 98.1 m/s
0.1
and
Vx =
and
√
Vc2
−
Vuc2
√
√
2
= 87.552 − 65.42 = 58.21 m/s
= Vh2 − Vuh
(
)
65.4
αc = tan
= tan
= 48.3◦
58.21
)
(
)
(
98.1
Vuh
−1
−1
= tan
= 59.3◦
αh = tan
Vx
58.21
−1
Vuc
Vx
)
(
−1
106
Chapter 8
Exercise 8.1 An industrial air compressor has 29 backward swept blades with
blade angle −21◦ . The tip speed of the blades is 440 m/s and the radial component of the velocity is 110 m/s. Air is inducted from atmospheric conditions at
101.3 kPa and 298 K with an axial velocity equal to 95 m/s. The hub to tip ratio
at the inlet is 0.4. The total-to-total efficiency of the compressor is 0.83, and the
mass flow rate is 2.4 kg/s. (a) Find the total pressure ratio using the Stodola slip
factor. (b) Find the tip radius of the impeller.
Given: p01 , T01 , V1 , r1h /r1s , ṁ, Z , χ2 , U2 , Vr2 .
Find: p03 /p01 and U2 .
Solution: The flow coefficient is
ϕ=
Vr2
110
=
= .35
U2
44)
The slip coefficient is
σ =1−
π cos χ2
π cos(−21◦ )
=1−
= 0.899
Z
29
and the tangential velocity is
Vu2 = σU + Vr2 tan χ2 = 0.899 · 440 + 110 tan(−21◦ ) = 353.3 ms
The work done is
w = U Vu2 = 440 · 353.3 = 155.44 kJ/kg
and the isentropic work is
ws = ηw = 0.83 · 155.44 = 129.02 kJ/kg
The total temperature at the end of the isentropic process is
T02s = T01 +
ws
129.02
= 298 +
= 426.4 K
cp
1.0045
and the pressure ratio is
(
)γ/(γ−1) (
)3.5
426.4
p02
T02s
=
=
= 3.505
p01
T01
298
107
⇐
(a)
At the inlet
T1 = T01 −
M1 = √
952
V12
= 298 −
= 293.5 K
2cp
2 · 1004.5
V1
95
=√
= 0.277
γRT1
1.4 · 287 · 293.5
The stagnation density is
ρ01 =
p01
101300
=
= 1.184 kg/m3
RT01
287 · 298
and the static density is
(
ρ1 = ρ01
T1
T01
)1/(γ−1)
(
= 1.184
293/5
298
)2.5
= 1.140 kg/m3
From the mass flow rate
ṁ = ρ1 A1 V1
A1 =
ṁ
2.4
=
= 0.02216 m2
ρ1 V1
1.1.4 · 95
and from the expression for the area
A1 = π(rs2 − rh2 ) = πrs2 (1 − κ2 )
the shroud radius is
√
rs =
A1
=
π(1 − κ2 )
√
0.002216
= 0.0916 m ⇐ (b)
π(1 − 0.16)
Exercise 8.2 A centrifugal compressor has 23 radial vanes and an exit area equal
to 0.12 m2 . The radial velocity is 27 m/s and the tip speed of the impeller is
350 m/s. The total-to-total efficiency is 0.83. (a) Find the mass flow rate of air, if
at the inlet the total pressure and temperature are 101.3 kPa and 298 K. (b) Find
the exit Mach number. (c) If the blade height at the exit is b = 3 cm and there is
no leakage flow, find the tip radius of the impeller. (d) Find the rotational speed of
the compressor wheel. (e) Find the required power neglecting mechanical losses.
Given: p01 , T01 , Z , χ2 , U2 , Vr2 , and ηtt .
Find: ṁ, M2 , r2 , and Ω.
108
Solution: The slip factor is
σ =1−
π cos χ2
π
=1−
= 0.863
Z
23
The tangential velocity is
Vu2 = σU = 0.863 · 350 = 302.2 ms
and the specific work is
w = U2 Vu2 = 302.2 · 350 = 105.8 kJ/kg
and the temperature at the exit is
T02 = T01 +
w
105.8
= 298 +
= 403.3 K
cp
1.0045
The isentropic process leads to T02s
)]
[
(
)]
[
(
403.3
T02
T02s = T01 1 + ηtt
−1
= 298 1 + 0.83
−1
= 385.4 K
T01
298
The exit pressure is
(
p02 = p01
T02s
T01
)γ/(γ−1)
(
= 101.3
)3.5
385.4
−1
= 249.2 kPa
298
The velocity at the exit is
√
√
2
V2 = Vr22 + Vu2
= 272 + 302.22 = 303.3 m/s
and the static temperature at the exit is
T2 = T02 −
V22
303.32
= 403.3 −
= 357.5 K
2cp
2 · 1004.5
The Mach number at the exit is
√
) √
)
(
(
2
T02
2 403.3
−1 =
− 1 = 0.8
M2 =
γ − 1 T2
0.4 357.5
109
The static pressure at the exit is
(
p2 = p02
T2
T02
)γ/(γ−1)
(
= 249.2
357.5
403.3
)3.5
= 163.4 kPa
The static density is
ρ2 =
163.4
p2
=
= 1.593 kg/m3
RT2
0.287 · 357.5
The mass flow rate is
ṁ = ρ2 A2 Vr2 = 1.593 · 0.12 · 27 = 5.16 kg/s
The exit radius is therefore
r2 =
A2
0.12
=
= 0.637 m
2πb2
2π · 0.03
Ω=
U
35) · 30
=
= 5250 rpm
r2
0.637 · π
and the shaft speed is
The power to the compressor is
Ẇ = ṁw = 5.16 · 105.7 = 545.7 kW
⇐
(e)
Exercise 8.3 A centrifugal compressor has an axial inlet and the outlet blades at
an angle such that the tangential component of the exit velocity has a value equal
to 0.9 times the blade speed. The outlet radius is 30 cm and the desired pressure
ratio is 3.5. The inlet stagnation temperature is T0 = 298 K. If the total-tototal efficiency of the compressor is 0.8, at what angular speed does it need to be
operated.
Given: Axial inlet with T01 . At the exit Vu2 = 0.9U2 , r2 = 0.3 m, p03 /p01 = 3.5,
ηtt = 0.8.
Find: Ω.
Solution: Work done is
w = Vu2 U2 = 0.9U22
110
so that the isentropic work is
[(
ws =
ηtt0.9U22
Solving this for U2 gives
√
U2 =
and
Ω=
= cp (T02s − T01 ) = cp T01
)]
p03
−1
p01
1004.5 · 298(3.51/3.5 − 1)
= 423.0m/s
0.8 · 0.9
423.0 · 30
U2
=
= 13, 464 rpm Lef tarrow
r2
0.3 · π
Exercise 8.4 A small centrifugal compressor as a part of a turbocharger operates
at 55,000 rpm. It draws air from atmosphere at temperature 288 K and pressure
101.325 kPa. The inlet Mach number is M1 = 0.4 and the flow angle of the
relative velocity is β1s = −60◦ at the shroud. The radius ratio at the inlet is
κ = r1h /r1s = 0.43. Find, (a) the tip blade speed at the inlet, (b) the mass flow
rate and (c) the throat are if the inducer is choked.
Given: Axial inlet T01 , p01 , M1 , β1s , and r1s /r2 . The shaft speed is Ω.
Find: U1s , ṁ, At .
Solution: The inlet static temperature is
)−1
(
γ−1 2
T1 = T01 1 +
M1
= 288(1 + 0.2 · 0.42 )−1 = 279.1 K
2
and the stagnation density is
ρ01 =
101325
p01
=
= 1.226 kg/m3
RT01
287 · 288
The static density is therefore
(
)1/(γ−1)
(
)1/(γ−1)
279.1
T1
= 1.226
= 1.333 kg/m3
ρ1 = ρ01
T01
288
The relative Mach number at the inlet shroud is
M1Rs =
M1
0.4
=
= 0.8
cos β1s
cos(−60◦ )
111
The sound speed and stagnation sound speed at the inlet are
√
√
c1 = γRT1 = 1.4 · 287 · 279.1 = 334.9 m/s
√
√
c01 = γRT01 = 1.4 · 287 · 288.0 = 340.2 m/s
and the blade speed at the shroud inlet is
√
√
√
2
2
− M12 = 334.9 0.82 − 0.42 = 232.0 m/s ⇐ (a)
− V12 = c1s M1Rs
U1s = W1s
and the inlet radius of the shroud is
r1s =
U1s
232.0 · 30
=
= 0.0403 m
Ω
55, 000 · π
The mean radius at the inlet is
1
r1s
0.0403
r1 = (r1h + r1s ) =
(1 + κ) =
(1 + 0.62) = 0.0288 m
2
2
2
The inlet area is
2
2
2
A1 = π(r1s
− r1h
) = pir1s
(1 − κ2 ) = π · 0.4032 (1 − 0.622 ) = 0.004155 m2
The inlet velocity is
V1 = M1 c1 = 0.4 · 334.9 = 133.96 m/s
and the mass flow rate is
ṁ = ρ1 A1 V1 = 1.133 · 0.00415 · 133.9 = 0.6305 kg/s
⇐ (b)
and the mass balance gives
(
A1 cos β1s M1Rs
γ−1 2
M1Rs
1+
2
)−(γ+1)/2(γ−1)
(
= At Mt
γ−1 2
1+
Mt
2
)−(γ+1)/2(γ−1)
and when the throat is choked
At = A1 cos1s (
2
γ+1
M1RS
2
)−(γ+1)/2(γ−1) = 0.0020 m
2
+ γ−1
M1Rs
γ+1
112
⇐ (c)
Exercise 8.5 A centrifugal compressor in a turbocharger operates at 40,000 rpm
and inlet Mach number M1 = 0.35. It draws air from atmosphere at temperature
293 K and pressure 101.325 kPa. The radius ratio is r1s /r2 = 0.71 and the diffusion ratio is W1s /W2 = 1.8. The inlet angle of the relative velocity at the shroud is
β1s = −63◦ . The slip factor is σ = 0.85 and the flow angle at the exit is α2 = 69◦ .
Find, (a) the tip speed of the blade at the inlet, (b) the tip speed of the blade at the
outlet, (c) the loading coefficient, and (d) the metal angle at the exit.
Given: Axial inlet and T01 , p01 , M1 , β1s , r1s /r2 , W1s /W2 , σ and α2 .
Find: U1s , U2 , ψ, χ2 .
Solution: The inlet static temperature is
(
)−1
γ−1 2
T1 = T01 1 +
M1
= 293(1 + 0.2 · 0.352 )−1 = 286.0 K
2
and the stagnation density is
ρ01 =
p01
101325
=
= 1.205 kg/m3
RT01
287 · 293
The static density is therefore
(
ρ1 = ρ01
T1
T01
)1/(γ−1)
(
= 1.205
286.0
293
)1/(γ−1)
= 1.134 kg/m3
The sound speed at the inlet is
√
√
c1 = γRT1 = 1.4 · 287 · 286 = 339.0 m/s
and the velocity is
V1 = M1 c1 = 0.35 · 339 = 118.6 m/s
The relative Mach number at the inlet shroud is
M1Rs =
0.35
M1
=
= 0.771
cos β1s
cos(−63◦ )
and the blade speed at the shroud inlet is
√
√
√
2
2
U1s = W1s
− V12 = c1s M1Rs
− M12 = 339.0 0.7712 − 0.352 = 232.8 m/s ⇐ (a)
113
The radius of the inlet shroud is
r1s =
U1s
232.8 · 30
=
= 0.0556 m
Ω
40, 000 · π
and the exit radius is
r2 = r1s
0.0556
r2
=
= 0.0783 m
r1s
0.71
⇐ (b)
and the blade speed there is
U2 = U1s
With
0.0783
r2
= 232.8
= 328.0 m/s
r1s
0.0556
W1s
r1s /r2
√
=−
W2
sin β1s 1 − 2ψ + ψ 2 / sin2 α2
which can be written as
1 − 2ψ +
where
(r1s /r2 )2
c=
=
(W1s /W2 )2 sin2 β1s
ψ2
=c
sin2 α2
(
0.71
1.8 sin(−65◦ )
)2
= 0.196
Hence the loading factor can be found by solving
ψ 2 − 2 sin2 α2 ψ + (1 − c) sin2 α2 = 0
or
√
ψ = sin α2 − sin4 α2 − (1 − c) sin2 α2 = 0.629
2
⇐ (c)
The choice of the minus sign gives backward swept blades.
Then
(
)
σ
tan χ2 = 1 −
tan α2
ψ
gives
χ2 = tan−1 [tan(60◦ )(1 − 0.85/0.629)] = −42.48◦
114
⇐ (d)
Exercise 8.6 A small centrifugal compressor draws atmospheric air at 293 K and
101.325 kPa, At the inlet r1h = 3.2 cm and r1s = 5 cm. The total-to-total efficiency of the compressor is 0.88. The relative Mach number at the inlet shroud
is 0.9 and the corresponding relative flow angle is β1s = −62◦ . At the outlet the
absolute velocity is at the angle α2 = 69◦ . The diffusion ratio is W1s /W2 = 1.8
and the radius ratio is r1s /r2 = 0.72. (a) Find the rotational speed of the shaft. (b).
Find the blade loading coefficient w/U22 . (c) Find the flow coefficient ϕ = Vr2 /U2
(d) Find the blade width at the exit.
Given: Axial inlet and T01 , p01 , Ω, ηtt , M1Rs , β1s , r1s /r2 , W1s /W2 , and α2 .
Find: Ω, ψ = w/U22 , ϕ = Vr2 /U2 , b2 .
Solution: The inlet static temperature is
(
)−1
γ−1 2
T1 = T01 1 +
M1
= 293(1 + 0.2 · 0.52 )−1 = 279.0 K
2
and the sound speed at the inlet is
√
√
c1 = γRT1 = 1.4 · 287 · 279 = 334.8 m/s
and the velocity is
V1 = M1 c1 = 0.5 · 334.8 = 167.7 m/s
The blade speed at the shroud inlet is
√
√
√
2
2
2
U1s = W1s − V1 = c1s M1Rs
− M12 = 334.8 0.812 − 0.52 = 213.1 m/s
The shaft speed is
Ω=
U1s
213.1 · 30
=
= 40, 700 rpm
r1s
0.05 · π
⇐
and the exit radius is
r2 = r1s
r2
0.05
=
= 0.0694 m
r1s
0.72
and the blade speed there is
U2 = U1s
213.1
r2
=
= 296.0 m/s
r1s
0.72
115
(a)
By squaring and adding the equations
V2 cos α2 = W2 cos β2
V2 sin α2 − U2 = W2 sin β2
leads to
V22 − 2U2 V2 sin α2 + U22 − W22 = 0
which when solved for V2 gives
√
V2 = U2 sin α2 − U22 sin2 α2 − U22 + W22
= 296 cos(69◦ ) −
and
√
2962 cos2 (69◦ ) − 2962 + 150.672 = 169.35 m/s
Vu2 = V2 sin α2 = 169.35 sin(69◦ ) = 158.1 m/s
The work done is
w = U2 Vu2 = 158.1 · 296 = 46.795 kJ/kg
and the blade loading factor is
ψ=
w
Vu2
158.1
=
=
= 0.534
2
U2
U2
296
⇐
(b)
The radial velocity is
Vr2 = V2 cos α2 = 169.35 cos(69◦ ) = 60.69 m/s
and the flow coefficient is
ϕ=
Vr2
60.69
=
= 0.205
U2
296
⇐
(c)
The stagnation temperature at the exit is
T02 = T01 +
46, 795
w
= 293 +
= 339.6 K
cp
1004.5
The isentropic work is
ws = ηtt w = 0.88 · 46, 795 = 41, 180 J/kg
116
and the stagnation temperature for an isentropic process is
T02s = T01 +
w2
41, 180
= 334.0 K
= 293 +
cp
1004.5
The stagnation pressure at the exit is
(
)γ/(γ−1)
(
)3.5
T02s
334.0
p02 = p01
= 101325
= 160.24 kPa
T01
293
The flow function at the inlet is
(
)−(γ+1)/2(γ−1)
γ
γ−1 2
1.4
F1 = √
M1 1 +
M1
= √ ·0.5(1+0.2·0.52 )−3 = 0.957
2
γ−1
0.4
The stagnation Mach number at the exit is
M02 =
169.35
V2
=√
= 0.458
c02
1.4 · 287 · 339.6
and the Mach number is
M2 = √
M02
1−
γ−1
2
M02
2
=√
0.458
1 = 0.2 · 0.4582
= 0.468
The flow function is
)−(γ+1)/2(γ−1)
(
γ
γ−1 2
F2 = √
M2 1 +
M2
2
γ−1
From
1.4
=√
· 0.468(1 + 0.2 · 0.4682 )−3 = 0.9115
0.4
√
√
A2
T02 p01 F1
339.6 101325 0.957
=
=
= 0.7149
A1
T01 p02 F2
293 160240 0.9115
With
2
2
A1 = π(r1s
− r1h
) = π(0.052 − 0.0322 ) = 0.004637 m2
then
A2 = 0.7149 · 0.004637 = 0.003315 m2
and the blade width is
A2
0.003315
b2 =
=
= 0.0076 m = 7.6 mm
2πr2
2]pi · 0.0694
117
⇐ (d)
Exercise 8.7 Show that in the incompressible limit the angle of the relative velocity at the inlet is optimum at 54.7◦ .
Given: The expression which yields the correct angle.
Find: β1s in the incompressible limit.
Solution: The general equation is
(
cos2 β1s =
2
3 + γM1R
2
2M1R
√
)(
1− 1−
2
4M1R
2 2
(2 + γM1R
)
)
For M1R ≪ 1, the radical becomes
√
√
2
2 2
4M1R
4 1
1−
= 1 − M1R
= 1 − M1R
2 2
(2 + γM1R )
9
9
Therefore
(
2
cos β1s =
So that cos β1s =
2
3 + γM1R
2
2M1R
√1
3
)
2
2 2
2(3 + γM1R
1
)
(1 − 1 + M1R
)=
≈=
9
2·9
3
and thus β1s = 54.7◦ .
Exercise 8.8 Show that the expression for the dimensionless mass flow rate for a
compressor with pre-swirl at angle α1 is
3
M1R
(tan α1 − tan β1s )2 cos3 β1s
(
)
Φf =
γ − 1 2 cos2 β1s
1+
M1R
2
sin2 α1
Plot the results for α1 = 30◦ , with β1s on the abscissa and Φf on the ordinate, for
relative Mach numbers 0.6, 0.7, 0.8. For a given mass flow rate does the pre-swirl
increase or decrease the allowable relative Mach number and does the absolute
value of the relative flow angle increase or decrease with pre-swirl.
Solution: The mass balance can be written as
2
(1 − κ2 )W1s cos β1s
ṁ = ρ1 A1 V1 = ρ1 A1 W1s cos β1s = ρ1 πr1s
which in the non-dimensional form is
Φ=
2
2
W1s
ρ1 U1s
W1s
ρ1 πr1s
2
(1
−
κ
)
cos
β
=
(1 − κ2 ) cos β1s
1s
2
2
ρ01 πr2 c01
ρ01 U2 c01
118
Since
U1s = Vu1 − Wu1
Vx = W1s cos β1s
this can be written as
Φ=
ρ1 Vx2 c31 W1s c201
(1 − κ2 )(tan α1 − tan β1s )2 cos β1s
ρ01 c21 c301 c01 U22
or as
ρ1
Φ=
ρ01
With
Φf =
so that
(
T1
T01
)3
2
ΦM0u
1−κ
3
M1R
(1 − κ2 )(tan α1 − tan β1s )2 cos3 β1s
3
M0u
M12 =
2
W1s
cos2 β1 s
V12
Vx2
=
=
c21
c21 cos2 α1
c21 cos2 α1
M 2 (tan α1 − tan β1s )2 cos3 β1s
ϕf = ( 1R
)(3γ−1)/(2γ−2)
2 cos2 β1 s
1 + γ−1
M
1R cos2 α1
2
Exercise 8.9 Water with density 998 kg/m3 flows through the inlet pipe of a centrifugal pump at a velocity of 6 m/s. The inlet shroud radius is 6.5 cm and water
flow has an axial entry. The relative velocity at the exit of the impeller is 15 m/s
and is directed by backward curved impeller blades such that the exit angle of the
absolute velocity is α2 = 65◦ . The impeller rotates at 1800 rpm and has a tip radius of 15 cm. Assume that the adiabatic efficiency of the pump is 75 %. Evaluate
(a) the power into the pump, (b) the increase in total pressure pressure of the water
across the impeller, (c) the change in static pressure of the water between the inlet
and outlet of the impeller. (d) If the inlet mean inlet radius is 5.0 cm, what is the
ratio of the change in kinetic energy of the water across the impeller to the total
enthalpy of the water across the pump. (d) What is the ratio of the change in the
absolute kinetic energy, the change in the relative kinetic energy and the change
in the kinetic energy owing to the centrifugal effect as a fraction of work done.
(f) If the velocity at the exit of the volute is 6 m/s, what is the ratio of the change
in static pressure across the rotor to the change in static pressure across the entire
pump?
, Given: V1 , r1s , R1h , Ω, W2 , α2 , r2 .
Find: Ẇ , ∆p0 , ∆p, ∆KE/∆h0 , ∆KEU /∆h0 , ∆KEW /∆h0 , (∆p)R /(∆p)p .
Solution: The inlet area is
2
2
) = 0.00942 m2
− r1h
A1 = π(r1s
119
and the mass flow rate is
ṁ = ρ1 A1 V1 = 998 · 0.00942 · 6 = 56.4 kg/s
The blade speed is
U2 = r2 Ω =
0.15 · 1800 · π
= 28.2 m/s
30
By squaring and adding the equations
V2 cos α2 = W2 cos β2
V2 sin α2 − U2 = W2 sin β2
leads to
V22 − 2U2 V2 sin α2 + U22 − W22 = 0
which when solved for V2 gives
√
V2 = U2 sin α2 − U22 sin2 α2 − U22 + W22
√
= 28.27 cos(65 ) − 28.272 sin2 (65◦ ) − 28.272 + 15.02 = 16.56 m/s
◦
in which the minus sign was chosen for backward swept blades.
Next
Vu2 = V2 sin α2 = 16.56 sin(65◦ ) = 15.00 m/s
Vr2 = V2 cos α2 = 16.56 cos(65◦ ) = 7.00 m/s
so that
Wu2 = Vu2 − U2 = 15.00 − 28.27 = 13.29 m/s
Wr2 = 7.00 m/s
and the flow angle of the relative velocity is
(
)
(
)
Wu2
13.27
−1
−1
β2 = tan
= tan
−
= −62.19◦
Wr2
7.00
and the specific work is
w = Vu2 U2 = 15.0 · 28.27 = 318.2 J/kg
and the power is
Ẇ = ṁw = 56.4 · 318.2 = 23.95 kW
120
⇐
(a)
The isentropic work is
ws = ηtt w = 0.75 · 424.3 = 318.2 J/kg
The stagnation pressure increase is therefore
∆p0 = ρws = 998 · 317.6 = J/kg
and the static pressure increase is
998
ρ
p2 − p1 = p02 − p01 − (v22 − V12 ) = 317.6 −
(16.562 − 62 ) = 198.75 kPa
2
2
The kinetic energies are obtained from
1
1
1800 · π
U1 = r1 Ω = (r1h + r1s )Ω = (0.065 + 0.035)
= 9.425 m/s
2
2
30
√
and
W1 =
so that
V12 + U12 =
√
62 + 9.4252 = 11.17 m/s
1
1
∆KEV = (V22 − V12 ) = (16.562 − 62 ) = 119.1 J/kg
2
2
1
1
∆KEU = (U22 − U12 ) = (28.272 − 9.4252 ) = 355.3 J/kg
2
2
1
1
∆KEW = (W12 − W22 ) = (11.172 − 192 ) = 50.17 J/kg
2
2
so that
∆KEV
119.1
=
= 0.281
w
424.3
∆KEU
355.3
=
= 0.827
w
424.3
∆KEW
50.1
=−
= −0.118
w
424.3
For the stator ∆p0 = 0, so that
p3 − p2 =
so that
) 988 (
)
ρ( 2
V2 − V32 =
16.562 − 62 = 118.8 kPa
2
2
∆pR
198.75
=
= 0.626
∆pR + ∆pV
198.75 + 118.8
121
⇐ (d)
Exercise 8.10 Centrifugal pump handling water operates with backward curving
blades. The angle between the relative velocity and the tip section is 45◦ . The
radial velocity at the tip section is 4.5 m/s, the flow at the inlet is axial, and the
impeller rotational speed is 1800 rpm. Assume that there is no leakage and that
the mechanical friction may be neglected, and the total-to-total efficiency is 70%.
(a) Construct the velocity diagram at the impeller exit. (b). Evaluate the required
tip radius for a water pressure rise of 600 kPa, (c) For the total pressure rise of
600 kPa, evaluate the difference between the total and static pressure of water at
the impeller tip section.
Given: Vr2 , Wu2 , β2 , p3 − p1 .
Find: r2 , p03 − p3 .
Solution: From
w = Vu2 U2 = U2 (U2 + Wu 2)
and
∆p0 = ρηtt w = ρηtt U2 (U2 + Wu2 )
and
∆p0
600, 000
=
= 858.86
ρηtt
998 · 0.7
so that
U22 + Wu2 U2 −
∆p0
=0
ρηtt
the solution of which is
√
√
1 2
1
∆p0
U2 = − Wu2 +
Wu2 +
= 2.25 + 2.252 + 858.86 = 31.64mm/s
2
4
ρηtt
Hence
r2 =
U2
31.64 · 30
=
= 0.168 m
Ω
1800 · π
⇐
(b)
and
Vu2 = U2 + Wu2 = 31.64 − 4.5 = 27.145 m/s
and
V2 = sqrtVu2 + Vr2 =
√
4.52 + 27.142 = 27.51 m/s
so that the difference in the stagnation and static pressure is
998
1
p02 − p2 = ρV22 =
· 27.512 = 377.6 kPa
2
2
122
⇐
(c)
Exercise 8.11 A centrifugal water pump has an impeller diameter D2 = 27 cm
and when its shaft speed is 1750 rpm it produces a head H = 33 m. (a) Find the
volumetric flow rate. (b) Find the blade height at the exit of the impeller. (c) Find
the blade angle at the exit of the impeller if it has 11 blades.
Given: D2 , Ω, H, Z.
Find: Q, b2 , χ2 .
Solution: The blade tip speed is
U2 = r2 Ω =
0.135 · 1750 · π
= 24.74 m/s
30
The blade loading coefficient is
ψs =
gH
9.81 · 33
=
= 0.5289
2
U
24.742
From the expression
ψs =
0.383
1/3
Ωs
the specific speed becomes
(
)3 (
)3
0.383
0.383
Ωs =
=
= 0.380
ψs
0.5289
√
Ω Q
Ωs =
(gh)3/4
Then from
(
)2 (
)2
Ωs (gh)3/4
0.380 · 30(9.81 · 33)3/4
Q=
=
= 0.0250 m3 /s
Ω
1750 · π
The flow coefficient is
√
√
ϕ = 0.1715 Ωs = 0.1715 0.380 = 0.1057
so that the radial component of the velocity is
Vr2 = ϕU2 = 0.1057 · 24.74 = 2.615 m/s
and the blade width is
b2 =
0.1057
ϕ
=
= 0.0113 m
2πr2 Vr2
2π · 0.135 · 2.615
123
⇐ (b)
⇐ (a)
The hydraulic efficiency is
ηh = 1 −
0.4
0.4
= 1/4 = 0.821
1/4
Q
25
and the work done is
w=
ws
gH
9.81 · 33
=
=
= 394.3 J/kg
ηh
ηh
0.821
The tangential component of velocity is
w
394.3
=
= 15.94 m/s
U2
24.74
Vu2 =
The flow angle is therefore
(
α2 = tan
−1
Vu2
Vr2
)
(
−1
= tan
15.94
2.615
)
= 80.7◦
and
Wu2 = Vu2 − U2 = 15.94 − 24.74 = −8.80 m/s
(
so that
−1
β2 = tan
Wu2
Wr2
)
(
−1
= tan
8.80
−
2.615
)
= −73.46◦
The slip coefficient is
σ =1−
0.63 · π
0.63π
=1−
= 0.82
Z
11
so that the metal angle can be obtained by solving
Vu2 = σU2 − Vr2 tan χ2
for χ2 , which is
(
)
(
)
Vu2 − σU2
15.94 − 0.82 · 24.74
−1
−1
χ2 = tan
= tan
−
= −59.0◦
Vr2
2.615
⇐ (c)
Exercise 8.12 A centrifugal water pump has an impeller diameter of D2 = 25 cm
and when its shaft speed is 1750 rpm, it delivers 20 liters per second of water. (a)
Find the head of water delivered by the pump. (b) Find the power to drive the
124
pump. (c) The impeller has 9 blades. Use the Stanitz slip factor to find the blade
angle at the exit of the impeller. (d) Use the Wiesner slip factor to find the blade
exit blade angle.
Given: D2 , Ω, Q, Z.
Find: H, ṁ, χ2 .
Solution: The blade speed is
U2 = r2 Ω =
The specific speed is
0.125 · 1750 · π
= 22.91 m/s
30
√
Ω Q
Ωs =
(gH)3/4
and the blade loading coefficient is
ϕs =
gH
U22
In addition, the blade loading is related to the specific speed by the empirical
equation
0.383
ψs = 1/3
Ωs
Equating these expressions for ψs and solving for gH gives
(
gH =
0.383U22
√
(Ω Q)1/3
)4/3
Then
H=
(
=
0.383 · 22.912
√
(1750 · π 0.020/30)1/3
gH
277.1
=
= 28.24 m
g
9.81
⇐
)4/3
= 277.1 J/kg
(a)
and the mass flow rate is
ṁ = ρQ = 998 · 0.020 = 19.96 kg/m3
The hydraulic efficiency is
ηh = 1 −
0.4
0.4
=
1
−
= 0.811
Q1/4
201/4
125
and the actual work is
w=
ws
gH
277.1
=
=
= 341.7 J/kg
ηh
ηh
0.811
and the power to the pump is
Ẇ = ṁw = 19.96 · 341.7 = 6.82 kW
⇐
(b)
The specific speed is
√
√
1750 · π 0.02
Ω Q
=
= 0.3816
Ωs =
(gh)3/4
30 277.13/4
and the flow coefficient is
√
√
ϕ = 0.1715 Ωs = 0.1715 0.3816 = 0.1059
and the radial component of the exit velocity is
Vr2 = ϕU2 = 0.1059 · 22.91 = 2.43 m/s
The tangential component is
Vu2 =
w
341.7
=
= 14.9 m/s
U2
22.91
and the flow angle at the exit is
(
)
(
)
Vu2
14.9
−1
−1
α2 = tan
= tan
= 80.76◦
Vr2
2.43
The tangential component of the relative velocity is
Wu2 = Vu2 − U2 = 14.9 − 22.91 = −7.99 m/s
(
so that
−1
β2 = tan
Wu2
Wr2
)
(
= tan
−1
7.99
−
2.43
)
= −73.1◦
The first guess of the slip coefficient is obtained from the Stanitz formula
σ =1−
0.63π
0.63 · π
=1−
= 0.78
Z
9
126
and this is used to obtain the metal angle
(
)
(
)
Vu2 − σU2
14.9 − 0.78 · 22.91
−1
−1
χ2 = tan
= tan
= −50.71◦
Vr2
2.43
Using this in the Wiesner’s formula
√
√
cos χ2
cos(50.71◦ )
σ =1−
=
= 0.829
Z 0.7
90.7
Iterating gives
(
−1
χ2 = tan
Vu2 − σU2
W Vr2
)
(
= tan
−1
14.9 − 0.829 · 22.91
2.43
)
= −59.3◦
and one more iteration leads to χ2 = −62.4◦ ⇐ (c)
Exercise 8.13 A centrifugal pump delivers water at 0.075 m3 /s with a head of
20 m while operating at 880 rpm. The hub to shroud radius ratio at the inlet
is 0.35 and the relative velocity makes an angle −52◦ at the inlet. (a) Find the
reversible work done by the pump. (b) What is the work done by the impeller. (c)
Find the radius of impeller and the inlet radius ratio of the shroud to the hub. (d)
Determine the blade width at the exit of the impeller. (e) Assume a reasonable
number of blades and calculate the blade angle at the exit. Use the Pfleiderer
equation to determine more accurately the number of blades and recalculate the
blade angle at the exit if needed. (f) What is the power to drive the pump?
Given: Q, Ω, H, β1s , κ.
Find: ws , w, r2 , r1s /r2 , b2 , χ2 .
Solution: The isentropic work is
ws = gH = 9.81 · 20 = 196.2 J/kg
⇐
(a)
Since the inlet flow is axis, the flow rate can be written as
Q = A1 V1 =
2
(1 − κ2 )U1s
πr1s
πr3 (1 − κ2 )U Ω
= 1s
tan(−β1s )
tan(−β1s )
from which
(
)1/3 (
)1/3
75 tan(52◦ )30
Q tan(−β1s )
=
= 0.0723 m
r1s =
π(1 − κ2 )Ω
1000π(1 − 0.352 ) · 880 · π
127
⇐ (c)
and the tip speed of the blade at the inlet shroud is
U1s = r1s Ω =
0.0723 · 880 · π
= 6.66 m/s
30
The hydraulic efficiency is
ηh = 1 −
0.4
0.4
= 1 − 1/4 = 0.864
1/4
Q
75
so that the specific work done by the rotor is
w = f racws ηh =
196.2
= 227.1 J/kg
0.864
⇐
(b)
The specific speed is
√
√
Ω Q
880π 0.075
Ωs =
=
= 0.481
(gH)3/4
30196/23/4
1/3
(Note that ψs = 0.383/(Ωs ) = 0.489, which is close).
The blade speed is
√
√
w
227.1
U2 =
=
= 21.56 m/s
ψs
0.481
The flow coefficient is
√
√
ϕ = 0.1715 Ωs = 0.1715 0.481 = 0.119
Next the velocity components at the exit are
Vu2 =
w
227.1
=
= 10.53 m/s
U2
21.56
Vr2 = ϕU2 = 0.119 · 21.56 = 2.565 m/s
The volumetric efficiency is obtained by using the correlation in the text. It is
ηv = 1 −
Thus
QR =
0.0988
C
=1−
= 0.977
n
Q
0.0750.3387
Q
0.075
=
= 0.0768 m3 /s
ηv
0.977
128
and the blade width is
b2 =
QR
0.0768
=
= 0.0204 m
2πr2 V r2
2π · 0.234 · 2.565
⇐
(d)
The overall efficiency can be obtained from the correlation in the text. It is
η = 0.8393
The exit flow angle is
(
α2 = tan
−1
Vu2
Vr2
)
(
−1
= tan
10.53
2.565
)
= 76.3◦
The tangential component of the relative velocity is
Wu2 = Vu2 − U2 = 10.53 − 21.56 = −11.02 m/s
(
so that
−1
β2 = tan
Wu2
Wr2
)
(
−1
= tan
11.02
−
2.565
)
= −76.9◦
Assume that Z = 6 and assume that the metal angle is χ2 = −70◦ , then
√
√
χ2
cos(−70◦ )
σ = 1 − 0.7 = 1 −
= 0.833
Z
60.7
and this is used to obtain the metal angle
(
)
(
)
Vu2 − σU2
10.53 − 0.833 · 21.56
−1
−1
χ2 = tan
= tan
= −70.94◦
Vr2
2.565
After one iteration the metal angle is χ2 = −71.19◦ . Using the Pfleider’s formula
gives
(
)
(
)
1 + r1s /r2
1
Z = 6.5
cos
(β1s + χ2 )
1 − r1s /r2
2
)
(
)
(
1 + 0.3091
1
◦
◦
(−52.0 − 71.9 ) = 6.4 ⇐ (e)
= 6.5
cos
1 − 0.3091
2
Thus Z = 6 is an appropriate number. Finally the power to the pup is
Ẇ = ρQ
998 · 75 · 96.2
ws
=
= 17.5 kW
ηh
1000 · 0.839
129
⇐
(f )
Exercise 8.14 A fan draws in atmospheric air at 0.4 m3 /s at pressure 101.32, kPa
and temperature 288 K. The total pressure rise across the fan, which has 30 radial
blades, is 2.8 cm of water. The inner radius is 14.8 cm and outer radius is 17.0 cm.
The rotational speed of the fan is 980 rpm. Use the Stanitz slip factor and a fan
efficiency of 78 %. (a) If the velocity into the fan is radially outward, find the
angle of the relative velocity at the inlet. (b) Determine the power to the fan if 4
% is lost to mechanical friction. (c) Find the angle blade angle at the exit.
Given: p01 , T01 , Ω, Q, r1s , r2 , b2 , η, ηm , total pressure rise in h of water.
Find: β1 , Ẇ , β2 .
Solution: The stagnation density at the inlet is
ρ01 =
p01
101325
=
= 1.226 kg/m3
RT01
287 · 288
The pressure rise is small and this can be taken to be incompressible flow. The
stagnation pressure rise is
∆p0 = ρH2 O gh = 998 · 9.81 · 0.028 = 274.1 Pa
and the isentropic work is
ws =
∆p0
274.1
=
= 223.6 J/kg
ρ
1.226
and the actual work is
w=
ws
223.6
=
= 286.7 J/kg
ηh
0.78
The power to the fan is therefore
Ẇ =
ρQw
1.226 · 0.4 · 286.7
=
= 146.4 W
ηm
0.96
The blade speed at the inlet is
U1 = r2 Ω =
0.148 · 980 · π
= 15.19 m/s
30
and at the exit it is
U2 = U1
17
r2
= 15.19
= 17.45 m/s
r1
14.8
130
and the inlet radial velocity is
Vr1 =
Q
0.4
=
= 2.76 m/s
2πr1 b
2π · 0.148 · 0.156
The tangential component of the relative velocity at the inlet is
wu1 = Vu1 − U1 = 0 − 15.19 = −15.19 m/s
The flow angle is therefore
(
)
)
(
Wu1
15.19
−1
−1
β1 = tan
= −79.7◦
= tan
−
Wr11
2.76
At the exit
14.8
r1
= 2.76
= 2.40 m/s
r2
17.0
and the tangential component of the velocity is
Vr2 = Vr1
Vu2 =
W
286.7
=
= 16.43 m/s
U2
17.45
The slip factor is
σ =1−
0.63π
0.63π
=1−
= 0.934
Z
30
so that the metal angle is
(
)
(
)
Vu2 − σU2
16.43 − 0.934 · 17.45
−1
−1
χ2 = tan
= tan
= 3.3◦
Vr2
2.4
Exercise 8.15 A pump draws water at the rate of 75 liters per second from a
large tank with the air pressure above the free surface at 98.00 kPa. The pump is
z = 2 m above the water level in the tank. The pipe diameter is 14.0 cm and the
suction pipe is 20 m in length. The entrance loss coefficient is Ki = 0.2 and the
loss coefficient of the elbow is Ke = 0.6 and the pipe roughness is 55 µ m. Find
the suction specific speed if the shaft speed is 1800 rpm. The viscosity of water is
1.08 · 10−3 kg/m s.
Given: Q, pa , z1 , Dp , Ki , Ke , ϵ, Ω.
Find: Ωss .
131
Solution: From the Bernoulli equation
[ ( )
] 2
Vp
pa 1 2
pt
L
+ V1 + gz1 = + f
Ki + Ke
ρ
2
ρ
D +
2
The velocity in the pipe is
Vp =
4Q
4 · 0.075
Q
=
=
= 4.87 m/s
2
Ap
πD
π · 0.142
Reynolds number is
Re =
ρVp Dp
998 · 4.87 · 0.14
=
= 630, 300
µ
1080 · 10−6
The friction factor is obtained from the Colebrook formula in Chapter 3. For a
pipe with roughness ϵ = 55 µm, it gives f¯ = 0.01677. Thus the head loss is
[
(
)
]
20
4.872
hl = 0.01677
+ 0.2 + 0.6
= 3.867 m
0.14
2 · 9.81
The vapor pressure is pv = 37.8 kPa and the pressure at the free surface is pa =
98 kPa. Thus the NPSH is
NPSH =
pa
pv
98000
3780
− z1 − hl −
=
− 2 − 3.863 −
= 3.75 m
ρg
ρg
998 · 9.81
998 · 9.81
The suction specific speed is therefore
√
√
Ω Q
1800 · π 0.075
Ωss =
=
= 3.455
(gNPSH)3/4
30 · (9.81 · 3.75)3/4
So cavitation is likely to take place.
Exercise 8.16 Consider a volute consisting only a circular section. (a) If the tangential velocity varies as Vu = K/r, show that the value of K in terms of the
volumetric flow rate Q, the radius of the section R and the radius to the center of
the section a is given by
K=
Q
√
2πR(λ − λ2 − 1)
132
in which λ = a/R. (b) For R = 0.5 m, a = 2 m and Q = 1.5 m3 /s, find the
pressure difference p2 − p1 at the centerline, between the outside and inside edge
of the section.
Given: R, a, Q.
Find: K and p2 − p1 .
Solution: With the coordinate origin at the
√ center of the pipe its circumference is
2
2
2
given by x + y = a . From this y = a2 − x2 for the upper half of the pipe.
Then flow rate is given by
∫
∫
K
Q=
V dA =
dA
A
A r
but dA = 2y(x)dx and r = R + x, so that
∫ a √ 2
K a − x2
Q=2
dx
R+x
a
Let z = x/a and λ = R/a, and then dx = adz and the equation becomes
∫ 1 √
1 − z2
Q = 2aK
dz
−1 λ + z
Since R > a it follows that λ > 1. It remains to evaluate the integral, which is an
exercise in calculus. First write
∫ 1 √
1 − z2
I=
dz
−1 λ + z
as
∫
1
I=
−1
(√
1 − z2
λ−z
−√
λ+z
1 − z2
∫
This is
I = (1 − λ )
1
2
−1
)
∫
dz +
−1
dz
√
+
(λ + z) 1 − z 2
or considering the two integrals separately
∫ 1
dz
2
√
I1 = (1 − λ )
2
−1 (λ + z) 1 − z
133
1
∫
λ−z
√
dz
1 − z2
1
−1
∫
I2 =
λ−z
√
dz
1 − z2
1
−1
λ−z
√
dz
1 − z2
In evaluating I1 let z = tanh u so that dz = du/ cosh2 u and
∫
∫
du
du
2
2
I1 = (1 − λ )
= (1 − λ )
2
λ cosh u + sinh u
cosh u(λ + tanh u)
But
λ ( + −u ) 1 ( − −u ) λ + 1 u λ − 1 −u
e e
+
e e
=
e +
e
2
2
2
2
λ cosh u + sinh u =
Thus
∫
du
+ 2(1 − λ2 )
u
(λ + 1)e + (λ − 1)e−u
I1 = 2(1 − λ )
2
Next let v = eu and dv = eu du, then
∫
eu du
(λ + 1)e2u + λ − 1
∫
dv
(λ + 1)v 2 + λ − 1
√
√
√
Now write λ + 1v = λ − 1 tan θ, so thatdv = (λ − 1)/(λ + 1) tan θdθ.
Hence
(√
)
∫
√
√
λ
+
1
dθ
I1 = −2 λ2 − 1
= −2 λ2 − 1 tan−1
eu
cos2 θ(tan2 θ + 1)
λ−1
I1 = 2(1 − λ )
2
Now from z = tanh u = (eu − e−u )/(eu + e−u ) it follows that ]
√
1+z
u
e =
1−z
and
√
I1 = −2 λ2 − 1 tan−1
(√
so that
λ+1
λ−1
√
1+z
1−z
)
(π
)
√
|1−1 = −2 λ2 − 1
−0
2
√
I1 = −π λ2 − 1
The second integral
∫
I2 =
1
−1
λ−z
√
dz =
1 − z2
134
∫
λ sin θ
cos θdθ
cos θ
or
∫
I2 =
√
π
π
(λ−sin θ)dθ = λθ−cos θ = z sin−1 z|1−1 + 1 − z 2 |1−1 = λ −λ(− ) = λπ
2
2
Hence
I = π(λ −
and
K=
√
λ2 − 1
Q
√
2πR(λ − λ2 − 1)
135
Chapter 9
Exercise 9.1 Combustion gases with γ = 4/3 and cp = 1148 J/kg K, at T01 =
1050 K and p01 = 310 kPa enter a radial inflow turbine. At the exit of the stator
M2 = 0.9. As the flow leaves the turbine it is diffused to atmospheric pressure at
p4 = 101.325 kPa. The total-to-total efficiency of the turbine is ηtt = 0.89. Find
the stator exit angle.
Given: T01 , p01 , M2 , p4 , ηts .
Find: α2
Solution: Since T02 = T01 , the temperature at the inlet the the rotor can be obtained from
γ − 1 2 −1
1050
T2 = T 02(1 +
M2 ) =
= 925.1 K
2
1 + 0.81/6
and the velocity is
V2 = M2
√
√
γRT2 = 0.8 1.333 · 287 · 925.1 = 535.49 m/s
From
T01 − T03
T01 − T4s
The stagnation temperature at the exit is
[
(
( )(γ−1)/γ )]
p4
T03 = T01 1 − ηts 1 −
p01
))]
[
(
(
101.325
= 822.1 K
T03 = 1050 1 − 0.89 1 −
310
The word delivered is
ηts =
w = cp (T01 − T03 ) = 1148(1050 − 822.1) = 261.65 kJ/kg
Since w = U 2 , the blade speed is
√
√
U2 = w = 261, 640 = 511.5 m/s
and the flow angle is
α2 = sin
(
−1
U2
V2
)
(
−1
= sin
136
511.5
535.49
)
= 72.8◦
⇐
Exercise 9.2 During a test air runs through a radial inflow turbine at the rate of
ṁ = 0.323 kg/s when the shaft speed is 55, 000 rpm. The inlet stagnation temperature is T01 = 1000 K and the pressure ratio is p01 /p3 = 2.1. The blade radius
at the inlet is r2 = 6.35 cm. The relative velocity entering the blade is radial and
the flow leaves the blade without swirl. Find, (a) the spouting velocity, (b) the
total-to-static efficiency, and (c) the power delivered.
Given: T01 , p01 /p3 , ṁ, Ω, and r2 .
Find: V0 , ηts and Ẇ .
Solution: The blase speed is
U2 = r2 Ω =
6.35 · 55, 000 · π
= 365.73 m/s
100 · 30
[
and
(
ws = cp (T01 − T3ss ) = cp T01 1 −
[
so that
(
V0 = 2cp T01
V0 =
(
1−
p3
p01
p3
p01
)(γ−1)/γ ]
1
= V02
2
)(γ−1)/γ )]
√
2 · 1148 · 980 (1 − (1/2.1)0 .35) = 617.20 m/s
⇐ (a)
and the work and isentropic work delivered are
w = U22 = 365.752 = 133.7 kJ/kg
so that
ηts =
1
617.22
ws = V02 =
= 190.5 kJ/kg
2
2
w
133.7
=
= 0.702
ws
90.5
⇐
(b)
and the power delivered is
Ẇ = ṁw = 0.35 · 133.7 = 46.8 kW
⇐
(c)
Exercise 9.3 A radial turbine delivers Ẇ = 80 kW as its shaft turns at 44, 000
rpm. Combustion gases with γ = 4/3 and cp = 1148 J/kg K enter the rotor with
relative velocity radially inward at radius r2 = 8.10 cm. At the exit the shroud
radius is r3s = 6.00 cm and at this location M3Rs = 0.59. The exit pressure is
137
p3 = 101.325 kPa and exit temperature is T3 = 650 K. Find the hub to shroud
ratio κ = r3h /r3s at the exit.
Given: T3 , p3 , Ẇ , Ω, r2 , and r3s .
Find: b3
Solution: The blade speed is
0.081 · 44, 000 · π
= 373.22 m/s
30
0.060 · 44, 000 · π
U3s = r3s Ω =
= 276.46 m/s
30
and the relative velocity at the shroud is
√
√
W3s = M3s γRT3 = 0.59 1.333 · 287 · 650 = 294.25 m, /s
U2 = r2 Ω =
The absolute velocity at the exit is
√
√
2
2
V3 = W3s
− U3s
= 294.252 − 276.462 = 100.8 m/s
The specific work is
w = U22 = 373.222 = 139.3 kJ/kg
and the mass flow rate is therefore
Ẇ
80, 000
=
= 0.574 kg/s
w
139.3
The static density at the exit is
ṁ =
ρ3 =
p3
101325
=
= 0.543 kg/m3
RT3
287 · 650
and the exit area is
A3 =
ṁ
0.574
=
= 0.01049 m2
ρ3 V 3
0.543 · 100.8
√
so that
r3h =
2
r3s
−
A3
= 0.0161 m
π
and the radius ratio is
κ=
0.0161
r3h
=
= 0.268
r3s
0.060
138
⇐
Exercise 9.4 A radial inflow turbine rotor, with rotor inlet radius r2 = 9.3 cm and
blade height b2 = 1.8 cm, turns at 42, 000 rpm. Its working fluid is a gas mixture
with cp = 1148 J/kg K and γ = 4/3. The exhaust pressure is p3 = 101.325 kPa
and the total-to-static efficiency is ηts = 0.82. The nozzle (stator) angle is α2 =
67◦ and the velocity coefficient for the flow through the stator is cv = 0.96 and the
Mach number at the inlet to the rotor is M2 = 0.9. Find, (a) the inlet stagnation
pressure to the stator, and (b) the stagnation pressure loss across the stator.
Given: r2 , b2 , Ω, p3 , α2 , and ηts .
Find: T01 and ∆p0LS .
Solution: The blade speed is
U2 = r2 Ω =
9.3 · 42, 000 · π
= 409.03 m/s
100 · 30
and the absolute velocity is
V2 =
U2
409.03
=
= 444.36 m/s
sin α2
sin(67◦ )
The radial component is
Vr2 = Wr2 = V2 cos α2 = 444.36 cos(67◦ ) = 173.62 m/s
Since the M2 is known, the inlet static temperature is
T2 =
V22
444.362
=
= 637.0 K
γRT M22
1.333 · 287 · 0.92
and the stagnation temperature T02 = T01 is
T02 = T2 (1 +
γ−1 2
0.81
M2 ) = 637(1 +
) = 723.0 K
2
6
The velocity for isentropic flow is
V2s =
V2
444.36
=
= 462.88 m/s
cv
0.96
and the static temperature for the isentropic process is
T2s = T02 −
V2s2
462.882
= 637 −
= 629.7 K
2cp
2 · 1148
139
The specific work is
w = U 2 = 409.032 = 167.31 kJ/kg
The exit stagnation temperature is therefore
T03 = T02 −
w
167, 310
= 723 −
= 577.3 K
cp
1148
From the total-to-static efficiency
ηts =
T01 − T03
T01 − T3ss
the static temperature T3ss is
T3ss = T01 −
1
1
(T01 − T03 ) = 723 −
(723 − 573.3)545.3 K
ηts
0.822
The stagnation pressure at the inlet is therefore
)γ/(γ−1)
(
)4
(
723
T01
p01 = p3
= 101.325
= 313.25 kPa
T3ss
545.3
⇐
(a)
The stagnation pressure p02 is obtained from
(
)4
( )γ/(γ−1)
629.7
T2s
p02 = p01
= 313.2
= 299.1 kPa
T2
637
and the stagnation pressure loss is
∆p0LS = p01 − p02 = 313.2 − 299.1 = 14.1 kPa
⇐
(b)
Exercise 9.5 Gas with γ = 4/3 and cp = 1148 K/kg K flows in a radial inflow
turbine, in which the inlet stagnation temperature is T01 = 980 K and the inlet
stagnation pressure is p01 = 205.00 kPa. The exit pressure is p3 = 101.325 kPa
and the exit temperature is T3 = 831.5 K. The stagnation temperature at the exit
is T03 = 836.7 K. The pressure at the inlet to the rotor is T2 = 901.6 K and the
pressure is p2 = 142.340 kPa. The shaft speed is 160,000 rpm and the radius ratio
is r3 /r2 = 0.57. Assume that the relative velocity is radial at the inlet and that
there is no exit swirl. Find, (a) the total-to-static efficiency, (b) the flow angles α2
and β3 , and (c) ηS and ηR .
140
Given: T01 , p01 , p3 , T3 , p2 , T2 , Ω, r3 /r2 .
Find: ηts , β3 , ηN , and ηR .
Solution: The velocity at the inlet to the rotor is
√
√
V2 = 2cp (T02 − T2 ) = 2 · 1148(980 − 901.6) = 424.3 m/s
and the Mach number is
M2 = √
424.3
V2
=√
= 0.722
γRT2
1.333 · 287 · 901.6
The stagnation pressure at the inlet to the rotor is
(
)γ/(γ−1)
(
)4
γ−1 2
0./722
p02 = p2 1 +
M2
= 142, 340 1 +
= 198.69 kPa
2
6
The temperature T2s can be calculated from
( )(γ−1)/γ
(
)0.25
p02
198.69
T2s = T2
= 901.6
= 894.6 K
p01
205.00
The static enthalpy loss coefficient for the stator is
ζS =
2cp (T2 − T2s )
2 · 1148(901.6 − 894.6)
=
= 0.0895 ⇐ (c)
2
V2
424.32
and the total-to-static efficiency is
ηts =
1 − T03 /T01
1 − 836.7/908
=
= 0.905
(γ−1)/γ
1 − (p3 /p01 )
1 − (101.325/205.00)0.25
⇐
The specific work delivered is
w = cp (T01 − T03 ) = 1148(980 − 836.7) = 164.51 kJ/kg
The blade speed is
U2 =
√
w=
√
164, 510 = 405.6 m/s
The flow angle at the inlet to the rotor is
( )
(
)
U2
405.6
−1
−1
α2 = sin
= sin
= 72.94◦
V2
421.90
141
⇐
(b)
(a)
The blade velocity at the exit at the mean radius is
U3 = U2
r3
= 405.6 · 0.57 = 231.2 m/s
r2
The velocity at the exit is
√
√
V3 = 2cp (T03 − T3 ) = 2 · 1148(836.7 − 831.5) = 109.2 m/s
and the flow angle of the relative velocity is
( )
(
)
U3
231.2
−1
−1
= − tan
= −64.7◦
β3 = − tan
V3
109.2
⇐
(c)
The calculation of the static enthalpy loss coefficient of the rotor flow is carried
out by
]
[
( )2
r3
2 sin2 β3
1
1
ζS Tr
ζR =
−1−
−
ηts
2 tan2 β3 r2
2 sin2 α2 (r3 /r2 )2
[
]
1
0.572
0.0895
2 sin2 (64.72 )
=
−1−
−
= 0.0974
0.905
2 tan2 (64.72 ) 2 sin2 (73◦ )
0.325
The temperature ratio Tr = T2s /T2 was taken to be unity in the previous forgoing
calculation. It can now determined to be
[( )
]
2
r3
T2s
γ−1 2 2
1 + ζS
1
=1=
M2 sin α2
= 0.919
−1 +1−
2
T2
2
r2
tan2 α2
sin β3
and when this is used for the value of Tr the static enthalpy loss coefficient for the
rotor can be recalculated. The result is
ζR = 0.117
⇐
(c)
Exercise 9.6 For an exit with no swirl show that
√
W3 (r)
sin(α2 − β2 ) r3 r2
=
+ cot2 β3
2
W2
cos α2
r2 r3
in which r is the radius at an arbitrary location of the exit plane of the blade and
r3 and β3 are the mean values of the exit radius and angle. Show further that at
the mean radius
sin(α2 − β2 ) r3
W3
=−
W2
cos α2 sin β3 r2
142
and plot the angle β3 for the range 0.53 < r3 /r2 < 0.65 when W3 /W2 = 2 and
α2 = 70◦ and β2 = −40◦ .
Given: A range of r3 /r2 , and W3 /W2 = 2 as well as α2 = 70c irc and β2 = −40◦ .
Find: β3
Solution: From the exit velocity diagram
W32 (r) = U32 (r) + V32
and U3 (r) = U3 r/r3 in which U3 is the blade speed at the mean radius r3 , this
expression can be written as
W32 (r) = U32
(
or
W32 (r)
=
U32
r2
+ cot2 β3 U32
r32
r2
+ cos2 β3
r32
)
=
r2
U22 32
r2
(
r2
+ cos2 β3
r32
The tangential component of the velocity give
U2 = V2 sin α2 − W2 sin β2
and dividing by W2 leads to
U2
V2
=
sin α2 − sin β2
W2
W2
From the radial components
V2 cos α2 = W2 cos β2
the equation
f racV2 W2 =
is obtained. Thus
or
cos β2
cos α2
U2
= tan α2 cos β2 − sin β2
W2
U2
sin α2 cos β2 − cos α − 2 sin β2
=
W2
cos α2
143
)
which is also
sin(α2 − β2
U2
=
W2
cos α2
Therefore
sin(α2 − β2 ) r3
W3 (r)
=
W2
cos α2
r2
At the shroud
W3s
sin(α2 − β2 ) r3
=
W2
cos α2
r2
and at the hub
W3h
sin(α2 − β2 ) r3
=
W2
cos α2
r2
√
so that
W3s
=
W3h
√
r2
+ cot2 β3
r32
√
r2
+ cot2 β3
2
r3s
√
r2
+ cot2 β3
2
r3h
2
r3s
+ r32 cos2 β3
2
r3h
+ r32 cot2 β3
at the mean radius
sin(α2 − β2 ) r3 √
W3
=
1 + cot2 β3
W2
cos α2
r2
or
W3
sin(α2 − β2 ) r3
=−
W2
cos α2 sin β3 r2
For β2 − 40◦ and α2 = 70◦ the relationship
−
r3
sin(110◦ )
sin β3 =
= 1.373739
r2
2 cos(70◦ )
so that
sin β3 = −1.373739
r3
r2
and the results are shown in the table below.
r3 /r2
β3
0.53
−46.7◦
0.56
−50.3◦
0.59
−54.1◦
144
0.62
−58.4◦
0.65
−63.2◦
Exercise 9.7 An inexpensive radial inflow turbine has flat radial blades both at
the inlet and the exit of the rotor. The shaft speed is 20,000 rpm. The radius of
the inlet to the rotor is 10 cm and the mean radius at the exit is 6 cm. The ratio
of blade widths is b3 /b2 = 1.8. The inlet stagnation temperature is T01 = 420 K
and the exhaust flows into the atmospheric pressure 101.325 kPa. Assume that
the gases which flow through the turbine have γ = 4/3 and cp = 1148 J/kg K.
If the power delivered by the turbine is 10 kW, find (a) the mass flow rate, (b)
the static temperature at the exit of the stator, (c) the static temperature at the exit
of the turbine, (d) the blade height at the inlet and the exit of the turbine, (e) the
total-to-total efficiency, and the total-to-static efficiency.
Given: r3 , r2 , α2 , cN , ζR , T01 , Ω, p3 , and b3 /b2 .
Find: ṁ, T3 , p01 , b3 , b2 , ηtt and ηts .
Solution: The blade speeds are
U2 = r2 Ω =
and
U3 = U2
0.1 · 20, 000 · π
= 209.4 m/s
30
r3
0.06
= 209.4
= 125.7 m/s
r2
0.1
The specific work delivered is
w = U2 − U3 = 209.42 − 125.72 = 28.07 kJ/kg
and the mass flow rate is therefore
ṁ =
Ẇ
10, 000
=
= 0.356 kg/s
w
28, 070
⇐ (a)
The inlet velocity to the rotor is
V2 =
U2
209.4
=
= 216.8 m/s
sin α2
sin(75◦ )
and the radial component of the velocity is
Vr2 = V2 cos α2 = 216.8 cos(75◦ ) = 56.12 m/s
W2 = 56.12 m/s
The static temperature leaving the stator is
T2 = T02 −
216.82
V22
= 420 −
= 399.5 K
2cp
2 · 1148
145
⇐ (b)
The isentropic velocity is
V2
216.8
=
= 223.5 m/s
cN
0.97
V2s =
and the isentropic static temperature is therefore
T2s = T02 −
223.52
V2s2
= 420 −
= 398.2 K
2cp
2 · 1148
The ratio of stagnation pressures at the inlet and exit of the stator are therefore
p01
=
p02
(
T2
T2s
)γ/(γ−1)
(
=
399.5
398.2
)4
= 1.013
The stagnation temperature after the rotor is
T03 = T02 −
w
28.07
= 420 −
= 395.5 K
cp
1.148
Mass balance gives
ρ2 r2 b2 W2 = ρ3 r3 b3 W3
This equation contains three unknowns ρ2 , ρ3 and W3 . One may proceed by trial
by assuming a trial value for the flow angle at the exit. As the first attempt the
value α3 = 50◦ is chosen. Then
V3 =
U3
125.7
=
= 164.0 m/s
sin α3
sin(50◦ )
and the axial velocity component at the exit is
Vx3 = V3 cos α3 = 164.0 cos(50◦ ) = 105.4 m/s
W3 = 105.4 m/s
The static temperature is
T3 = T03 −
V32
164.02
= 395.5 −
= 383.3 K
2cp
2 · 1148
Next the temperature for an isentropic expansion is calculated from
T3s = T3 − ζR
105.42
W32
= 383.8 − 0.5
= 381.4 K
2cp
2 · 1148
146
The static pressure at the inlet can now be obtained from
( )γ/(γ−1)
(
)4
T2
399.5
p 2 = p3
= 122.0 kPa
= 101.325
T3s
381.4
The densities can now be calculated. They are
ρ2 =
p2
122, 000
=
= 1.064 kg/m3
RT2
287 · 399.5
and
p3
101, 325
=
= 0.920 kg/m3
RT3
287 · 383.8
The value of density ρ̄3 may now be calculated from
ρ3 =
ρ̄3 = ρ2
0.1 · 56.12
r2 W2 b2
= 0.920
= 0.524 kg/m3
r3 W3 b3
0.06 · 105.4 · 1.8
This is in error by the amount r = ρ3 − ρ̄3 = 0.396 kg/m3 . By writing a Matlab
scrip to carry out the calculations, and changing the trial value of α3 , the error is
made nearly zero when α3 = 65.414◦ . The calculations with this value give
V3 =
U3
125.7
=
= 138.2 m/s
sin α3
sin(65.414◦ )
and the axial velocity component at the exit is
Vx3 = V3 cos α3 = 138.2 cos(65.414◦ ) = 57.5 m/s
W3 = 57.5 m/s
The static temperature is
T3 = T03 −
V32
138.22
= 395.5 −
= 387.2 K
2cp
2 · 1148
⇐ (c)
Next the temperature for an isentropic expansion is calculated from
T3s = T3 − ζR
W32
57.52
= 387.2 − 0.5
= 386.5 K
2cp
2 · 1148
The static pressure at the inlet can now be obtained from
(
( )γ/(γ−1)
)4
399.5
T2
= 101.325
p2 = p 3
) = 115.68 kPa
T3s
386.5
147
The densities can now be calculated. They are
ρ2 =
p2
115, 680
=
= 1.0088 kg/m3
RT2
287 · 399.5
and
p3
101, 325
=
= 0.9117, kg/m3
RT3
287 · 387.2
The value of density ρ̄3 may now be calculated from
ρ3 =
ρ̄3 = ρ2
0.1 · 56.12
r2 W2 b2
= 0.9117
= 0.9117 kg/m3
r3 W3 b3
0.06 · 57.50 · 1.8
and the error is now only e = −0.000014 kg/m3 .
The static temperature
)0.25
( )(γ−1)/γ
(
p02
1
= 385.3 K
T3ss = T3s
= 386.5
p01
1.013
and the stagnation pressure at the inlet to the stator is
)γ/(γ−1)
(
)4
(
420
T01
p01 = p3
= 101.325
= 143.11 kPa
T3ss
385.26
The blade widths can be now determined. They are
b2 =
ṁ
0.3562
=
= 0.010 m
2πr2 ρ2 W2
2 · π · 0.10 · 1.0088 · 56.12
⇐
(d)
b3 =
0.3562
ṁ
=
= 0.018 m
2πr3 ρ3 W3
2 · π · 0.06 · 0.9117 · 57.50
⇐
(d)
and
The total-to-static efficiency is
ηts =
T01 − T03
420 − 395.5
=
= 0.704
T01 − T3ss
420 − 385.3
⇐
(e)
To determine the total-to-total efficiency, the stagnation temperature T03ss is needed.
It is
(
( )(γ−1)/γ
)0.25
110.320
p03
= 385.3
= 393.5 K
T03ss = T3ss
p3
101.325
so that
420 − 395.5
T01 − T03
=
= 0.924 ⇐ (d)
ηtt =
T01 − T03ss
420 − 393.5
148
Exercise 9.8 Combustion gases with γ = 4/3 and cp = 1148 J/kg K enter a
stator of a radial flow turbine with T01 = 1150 K, p01 = 1300 kPa, and M1 = 0.5,
and with a flow rate of ṁ = 5.2 kg/s. The radius of the inlet is r1 = 17.4 cm,
the exit from the stator is at r2 = 15.8 cm and the inlet to the rotor is at r2 =
15.2 cm. The chord of the stator is ct = 4.8 cm and the width of the channel
is b = 1.6 cm. The rotational speed of the rotor is 31, 000 rpm. The exit static
pressure is p3 = 320 kPa. The trailing edge thickness of the 17 stator vanes can
be ignored. Find, (a) the total-to-static efficiency of the turbine, (b) the stagnation
pressure loss across the stator, and (c) the stagnation pressure loss across the gap.
Given: T3 , p3 , Ẇ , Ω, r2 , and r3s .
Find: b3
Solution:
: (a) 0.811, (b) 10.6 kPa, (c) 3.67 kPa.
Exercise 9.9 Combustion gases enter the stator of a radial inflow turbine at the
stagnation pressure p01 = 346 kPa and stagnation temperature T01 = 980 K. They
enter the rotor at the speed V2 = 481.4 m/s with the relative flow making an angle
β2 = −35◦ and exhaust into the atmosphere at 101.3325 kPa. The total-to-static
efficiency of the turbine is ηts = 0.83. Find, (a) the angle at which the flow enters
the rotor and (b) the relative Mach number at the inlet.
Given: T3 , p3 , Ẇ , Ω, r2 , and r3s .
Find: b3
Solution:
: α2 = 61.41◦ ,(b) M2R = 0.67.
149
Chapter 10
Exercise 10.1 A Pelton wheel operates from an effective head of He = 300 m and
at a flow rate of 4.2 m3 /s. The wheel radius is r2 = 0.75 m and its rotational speed
is 450 rpm. The water which leaves the penstock is divided into 5 streams. The
nozzle coefficient is cN = 0.98 for each of the nozzles. Impulse blades turn the
flow into the direction β3 = −65◦ and as a result of friction the relative velocity
reduces by an amount which gives a velocity coefficient cv = 0.90. Find, (a) the
efficiency of the turbine, (b) the power specific speed, and (c) the nozzle diameter
and the number of buckets in the wheel.
Given: He , Q, Ω, r2 , cN , cv , β3
Find: ηts , Ωsp , d, Z
Solution: The exit velocity from the nozzle is
√
√
V2 = cN 2gHe = 0.98 2 · 9.81 · 300 = 75.19 m/s
and the blade speed is
U = r2 Ω =
0.75 · 450 · π
= 35.34 m/s
30
The relative velocity entering the wheel is
W2 = V2 − U = 75.19 − 35.34 = 39.85 m/s
and the relative velocity at the exit is
W3 = cv W2 = 0.90 · 39.85 = 35.5, m/s
The tangential component of the relative velocity at the exit is
Vu3 = W3 sin β3 = 35.86 sin(−65◦ ) = −32.50 m/s
The tangential component of the absolute velocity at the exit is
Vu3 = U + Wu3 = 35.34 − 32.50 = 2.83 m/s
and the specific work is
w = U (Vu2 − Vu3 ) = 35.34(39.85 − 2.83) = 2556.8 J/kg
150
or
w = U (Vu2 − U )(1 − cv sin β3 ) = 2556.8J/kg
The power is therefore
Ẇ = ρQw = 998 · 4.2 · 255.8 = 10.7 MW
and the efficiency is
w
Ẇ
=
= 0.869
ρQgHe
ηHe
η=
⇐
(a)
The specific speed is
√
√
Ω Q
450 · π 4.2
Ωs =
=
= 0.24
(gHe )3/4
30(9.81 · 400)3/4
and the power specific speed is
√
√
Ωsp = ηΩs = 0.869 0.24 = 0.22
⇐ (b)
The flow rate in each jet is
Qj =
Q
4.2
=
= 0.84 m3 /s
N
5
and the jet area is
Aj =
Qj
0.84
=
= 0.0111 m2
V2
75.19
so that the jet diameter is
√
d=
4Aj
=
π
√
4 · 0.0111
= 0.119 m
π
The ratio of the wheel diameter to the jet diameter is
2 · 0.75
D
=
= 12.6
d
0.119
⇐ (c)
and the number of buckets is
Z=
D
+ 15 = 21.3
2d
151
Z = 21
⇐
(c)
Exercise 10.2 The flow rate through a small Francis turbine is 4.5 m3 /s, its head
is 150 m and the the rotational speed is 450 rpm. The inlet radius is r2 = 0.6 m and
the water leaves the guide vanes at the angle α2 = 72◦ and velocity V2 = 53.3 m/s.
It leaves the turbine without swirl. (a) Find the velocity coefficient of the stator
(inlet spiral and gates). (b) Find the inlet angle of the relative velocity, β2 . (c)
What is the output power? (d) What is the torque on the shaft? (d) Determine the
power specific speed and comment if the runner shape in the figure is appropriate.
Given: He , Q, V2 , r2 , α2 , cv , β3
Find: cN , β2 Ẇ , torque, Ωs , Ωsp .
Solution: The nozzle coefficient is
cN = √
V2
53.7
=√
= 0.9825 m/s
2gHe
2 · 9.81 · 150
⇐
(a)
The tangential component of the velocity is
Vu2 = V2 sin α2 = 53.3 sin(72◦ ) = 50.69 m/s
and the radial component is
Wr2 = V2 cos α2 = 53.3 cos(72◦ ) = 16.44 m/s
The blade speed is
U2 = r2 Ω =
0.6 · 450 · π
= 28.27 m/s
30
and the tangential component of the relative velocity is
Wu2 = Vu2 − U2 = 50.69 − 28.27 = 22.42 ms
The angle of the relative velocity is therefore
β2 =
Wu2
22.42
=
= 54.6◦
Wr2
16.44
⇐
(b)
The specific work is
w = U2 Vu2 = 28.27 · 50.69 = 1433.2 J/kg
and the power is
Ẇ = ρQw = 998 · 4.5 · 1433.3 = 6437 kW
152
⇐
(c)
and the efficiency is
η=
w
1433.3
=
= 0.974
gHe
9.81 · 150
The shaft torque is
T =
Ẇ
6, 437, 000 · 30
=
= 136.6 kN
Ω
450 · π
⇐
The specific speed and power specific speed are
√
√
450 · π 4.5
Ω Q
=
= 0.421
Ωs =
(gHe )3/4
30(9.81 · 150)3/4
and
Ωsp =
√
ηΩs =
√
0.9740.421 = 0.415
⇐
(d)
⇐
(e)
(e)
Exercise 10.3 The pressure at the entrance of a Francis turbine runner is 189.5 kPa
and at the exit it is 22.6 kPa. The shaft turns at 210 rpm. At the exit the flow
leaves without swirl. The inlet radius is r2 = 910 mm, and the exit radius is
r3 = 760 mm. The relative velocity entering the runner is W2 = 10.2 m/s, and
the flow angle of the relative velocity leaving the runner is β3 = −72◦ . The blade
height at the inlet is b2 = 600 mm. (a) Compute the stagnation pressure loss in
the runner. (b) Find the power delivered by the turbine.
Given: p2 , p3 , Ω, r2 , r3 , β3 , W2 .
Find: dp0LR and Ẇ .
Solution: The blade speed at the inlet is
U2 = r2 Ω =
0.91 · 210 · π
= 20.0 m/s
30
and at the exit it is
U3 = U2
0.76
r3
= 20.0
= 16.7 m/s
r2
0.91
The radial velocity at the exit is
Vr3 = −
16.7
U3
=−
= 5.43 m/s
tan β3
tan(−72◦ )
153
The mass balance gives
Vr2 =
r3
0.76
5.43 = 4.54 m/s
Vr3 =
r2
0.91
and thus the relative flow angle into the rotor is
)
)
(
(
Vr2
5.45
−1
−1
β2 = cos
= cos
= 63.6◦
W2
10.2
and the tangential component of the velocity is
Vu2 = U2 + Vr2 tan β2 = 20.0 + 4.54 tan(63.6◦ ) = 29.15 m/s
The specific work is
w = U2 Vu2 = 20/0 · 29.15 = 583.31 J/kg
The isentropic work is ws = (p02 − p03 )/ρ. The stagnation pressure at the inlet is
p02 = p2 + ρ
V22
29.52
= 189, 600 + 998
= 623.8 kPa
2
2
and at the exit it is
p03 = P3 + ρ
V32
5.432
= 226, 000 + 998
= 37.2 kPa
2
2
and
623, 800 − 373, 000
p02 − p03
=
= 587.68 J/kg
ρ
998
and the stagnation pressure loss is
ws =
dp0LR = ρ(ws − w) = 998(587.68 − 583.31) = 4.36 kPa
⇐
The flow rate is
Q = 2πr2 b2 Vr2 = 2π · 0.19 · 0.6 · 4.54 = 15.56 m3 /kg
and the power delivered is
Ẇ = ṁw = 998 · 15.56 · 583.3 = 9.06 MW
154
⇐
(b)
(a)
Exercise 10.4 A Francis turbine has an inlet radius r2 = 1450 mm and outlet
radius r3 = 1220 mm. The blade width is constant b = 370 mm. The shaft speed
is 360 rpm and the volumetric flow rate is Q = 16.7 m3 /s. The flow enters the
runner at α2 = 78◦ . Water leaves the the turbine without swirl and the outlet
pressure p3 = 35 kPa. The loss through the runner is 0.20W23 /2g m. Find the
pressure p2 at the inlet and the head loss through the runner.
Given: r2 , r3 , b, Q, α2 , p3 and given the head loss across the runner as 0.02W32 /2g.
Find: p2 .
Solution: The blade speed is
U2 = r2 Ω =
1.45 · 360 · π
= 54.66 m/s
30
The flow area at the inlet to the runner is
A2 = 2πr2 b = 2π · 1.11 · 0.37 = 3.37 m2
The radial component of the velocity is
Vr2 =
Q
16.7
=
= 4.95 m/s
A2
3.37
and the tangential component is
Vu2 = Vr2 tan α2 = 4.95 tan(78◦ ) = 23.31 m/s
and the absolute velocity is
√
√
2 +V2 =
V2 = Vrw
4.952 + 23.312 = 23.83 m/s′
u2
Hence the specific work is
w = U2 Vu2 = 54.66 · 23.31 = 1374 J/kg
The blade speed at the exit is
U3 = U2
r3
1.22
= 54.66
= 46.00 m/s
r2
1.45
The exit area is
A3 = 2πrr b = 2π · 1.22 · 0.37 = 2.384 m2
155
and the relative velocity at the exit is
√
√
W3 = U32 + Vr32 = 46.002 + 5.892 = 46.37 m/s
Hence the head loss is
HR = 0.2
W32
0.2 · 46.372
=
= 21.92 m
2g
2 · 9.81
The stagnation pressure at the exit is
p03 = p3 + ρ
V32
5.892
= 35000 + 998
= 52.30 kPa
2
2
The stagnation pressure loss is
∆p0LR = ρgHr = 998 · 9.81 · 21.92 = 214.6 kPa
so that the stagnation pressure at the inlet is
p02 = p03 + ∆p0LR + ρw = 52.3 + 214.6 +
998 · 1274
= 1538.3 kPa
1000
and the static pressure is
V22
998 · 23.832
p2 = p02 − ρ
= 1, 538, 300 −
= 1255, kPa
2
2
Exercise 10.5 The relative velocities at the inlet and the exit of a Francis turbine
are W2 = 10.0 m/s and W3 = 33.7 m/s. The shaft speed is Ω = 200 rpm. The
inlet radius of the runner is r2 = 1880 mm and its outlet radius is r3 = 1500 mm.
The runner blade width is constant b = 855 mm. Find, (a) the flow rate through
the turbine and (b) the torque assuming the flow leaves without exit swirl.
Given: W2 , W3 , Ω, r2 , r3 , b.
Find: Q and torque.
Solution: The blade speeds are
U2 = r2 Ω =
and
U3 = U2
1.88 · 200 · π
= 39.37 m/s
30
r3
1.88
= 39.37
= 31.42 m/s
r2
1.50
156
and the redial velocity at the exit is
√
√
Vr3 = W32 − U32 = 33.72 − 31.422 = 12.20 m/s
The flow rate is
Q = 2πr3 bV r3 = 2π · 1.50 · 0.855 · 12.20 = 98.3 m3 /s
⇐
(a)
and the radial velocity at the inlet is
Vr2 = Vr3
r3
1.50
= 12.20
= 9.74 m/s
r2
1.88
The angle of the relative flow at the inlet is
)
(
)
(
Vr2
9.734
−1
−1
β2 = cos
= cos
= 13.3◦
V2
10.0
Now the tangential component of the relative velocity is
Wu2 = Wr2 tan β2 = 9.74 tan(1.3◦ ) = 2.31 m/s
and the tangential component of the absolute velocity is therefore
Vu2 = U2 + Wu2 = 39.37 + 2.31 = 41.68 m/s
and the specific work is
w = U2 Vu2 = 39.37 · 41.68 = 1641 J/kg
and the power delivered is
Ẇ = ρQw = 998 · 98.3 · 1641 = 161 MW
The torque is therefore
T =
Ẇ
161 · 106 · 30
=
= 7, 685, 000 N m
Ω
200 · π
⇐
(b)
Exercise 10.6 A Francis runner is to be designed for effective head of He =
140 m and flow rate Q = 20 m3 /s. Assume that the efficiency is η = 0.9 and
that there is no exit swirl. Use the formula of Lugaresi and Massa to calculate
157
the specific speed. Use other the formulas in the text to obtain br and U2 /V0 from
Ωsp . Assume that the mechanical and volumetric losses are negligible. Find the
specific diameter on this basis. (a) Find the diameter at the inlet. (b) Find the blade
speed at the inlet. (c) Find the flow angles of the absolute and relative velocities
at the inlet.
Given: He , Q, η.
Find: D2 , U2 , α2 and β2 .
Solution: The power delivered is
Ẇ = ηρQgHe = 0.9 · 998 · 20 · 9.81 · 140 = 24, 672 kW
and the specific work is
w=
Ẇ
24, 670, 000
=
= 1236 J/kg
ρQ
998 · 20
The specific speed is
(
Ωs = 1.14
and
Ωsp =
Hr
He
√
)0.512
ηΩs =
(
= 1.14
√
100
140
)0.512
= 0.9596
0.90 · 0.9596 = 0.9104
The shaft speed can now be determined from
Ωs (gHe )3/4
0.9596 · 30 · (9.81 · 140)3/4
√
√
Ω=
=
= 462 rpm
Q
20 · π
The blade width to diameter ratio is
br = 0.0505Ω2sp + 0.26Ωsp + 0.018 = 0.213
and the U2 /V0 ratio is
U2
= 0.724
= 0.74Ω0.238
sp
V0
Thus
√
8U2
Ds =
=
Ωs V0
√
8 · 0.724
= 2.13
0.9596
158
Hence
√
√
Ds Q
2.13 20
D2 =
=
= 1.567 m
(gHe )1/4
(9.81 · 140)1/4
and
b2 br D2 = 0.213 · 1.567 = 0.334 m
and
D2
1.567 · 462 · π
Ω=
= 37.93 m/s
2
2 · 30
Now the radial velocity at the inlet is
U2 =
Vr2 =
Q
20
=
= 12.12 m/s
2πr2 b
2π0.7835 · 0.335
and the tangential velocity component can be obtained from
Vu2 =
so that
(
α2 = tan
−1
Vr2
Vu2
w
1236.1
=
= 32.59 m/s
U2
37.93
)
(
−1
= tan
12.12
32.59
)
= 69.5◦
⇐
(a)
and the tangential component of the relative flow is
Wu2 = Vu2 − U2 = 32.59 − 37.93 = −5.33 ms
and it makes the angle
(
)
(
)
Wr2
12.12
−1
−1
β2 = tan
= tan
= −23.6◦
Wu2
−5.33
⇐
(b)
Exercise 10.7 Water enters the runner of a Francis turbine with a relative velocity
at angle −12◦ . The inlet radius is 2.29 m and the mean radius at the exit is 1.37 m.
The rotational speed is 200 rpm. The blade height at the inlet is b2 = 1.22 m and
at the exit the inclined width of the blade is b3 = 1.55 m. The radial velocity at the
runner inlet is 10.0 m/s and the flow leaves the runner without swirl. Evaluate (a)
the change in total enthalpy of the water across the runner, (b) the torque exerted
by the water on the runner normalized to a metric tonne per second, (c) the power
developed, (d) the flow rate of water, (e) the change in total pressure across the
runner if the total-to-total efficiency is 95%, and the volumetric and mechanical
159
losses can be neglected. (f) What is the change in static pressure across the runner,
and (h) the total-to-static efficiency.
Given: r2 , r3m , Ω, β2 , b2 , η, Vr2 , b3 .
Find: h02 − h03 , T , Ẇ , Q, p02 − p03 , p2 − p3 , ηts .
Solution: The blade speeds are
U2 = r2 Ω =
2.29 · 200 · π
= 47.96 m/s
30
and the blade speed at the mean radius at the exit is
U3m = U2
1.37
r3m
= 47.96
= 28.69 m/s
r2
2.29
The flow areas are
A2 = 2πr2 b2 = 2π · 2.29 · 1.22 = 17.55 m/s
A3 = 2πr3m b3 = 2π · 1.37 · 1.55 = 13.34 m/s
and the flow rate is
Q = Vr2 A2 = 10 · 17.55 = 175.5 m3 /s
⇐ (d)
and the mass flow rate is
ṁ = ρQ = 998 · 175.5 = 175.2 tonne/s
The mean radial velocity leaving the runner is
V3m =
Q
175.54
=
= 13.16 m/s
A3
12.342
The angle of the relative flow at the exit is
)
(
(
)
U3
28.69
−1
−1
β3 = − tan
= − tan
= −65.4◦
Wm3
13.16
The tangential component of the relative velocity at the inlet is
Wu2 = Vr2 tan β2 = 10 tan(−12◦ ) = −2.12 m/s
160
and the corresponding value of the absolute velocity is
Vu2 = U2 + Wu2 = 47.96 − 2.12 = 45.84 m/s
so that
V2 =
√
√
2
Vr22 + Vu2
= 102 + 45.842 = 46.91 m/s
and the flow angle is
(
−1
α2 = tan
Vu2
Vr2
)
(
−1
= − tan
45.84
10.0
)
= 77.7◦
The work done is
w = U2 Vu2 = 47.96 · 45.384 = 2198.4 J/kg
so this is also the total enthalpy difference across the rotor
h02 − h03 = 2198.4 J/kg
⇐ (a)
The power delivered is
Ẇ = ρQw = 998 · 175.54 · 2198.4 = 385.1 MW
⇐ (c)
and the torque is therefore
T =
Ẇ
385, 100, 000 · 30
=
= 18, 39, 000 N m
Ω
200 · pi
Hence per ton of water flow it is
Ts =
T
1, 839, 100
=
= 105 kN m
175.2
ṁ
⇐ (b)
The isentropic work is
ηHe =
w
2198.4
=
= 2314.1 J/kg
η
0.95
The total pressure loss is
∆p0LR = ρ(gHe − w) = 998(2314.1 − 2198.4) = 115.5 kPa
161
and the change in stagnation pressure is
p02 − p03 = ρgHe − ∆p0LR =
998 · 2314.1
− 115.5 = 2194 kPa ⇐
1000
(e)
The change in static pressure is
998
1
2
) = 2, 194, 000−
p2 −p3 = p02 −p03 − ρ(V22 −V3m
(46.912 −13.162 ) = 1182 kPa ⇐
2
2
:
Exercise 10.8 The Otari number 2 power plant in Japan delivers 89.5 MW of
power when the flow rate is 207 m3 /s and the head is 48.1 m. The diameter of
the Kaplan turbine is D2t = 5.1 m and hub-to-tip ratio κ = 0.56. The generator
has 36 poles and delivers power at line frequency is 50 Hz. (a) Find the efficiency
of the turbine. (b) Calculate and flow angles entering and leaving the rotor and
construct a graph to show their variation across the span.
Solution: (a) The efficiency is
η=
Ẇ0
89.5 · 106
=
= 0.918
ρQgHe
998 · 207 · 9.81 · 48.1
(b) With a 50 Hz line frequency and 36 poles, the shaft speed is Ω = 120·50/36 =
166.67 rpm. The tip speed of the runner blade is
D2t
2.55 · 166.67 · π
Ω=
= 44.5 m/s
2
30
The axial velocity is uniform and it is given by
U2t =
Vx2 =
4Q
4 · 207
=
= 14.76 m/s
2
−κ )
π · 5.12 (1 − 0.562 )
2
πD2t
(1
Since each blade element delivers the same amount of work, the tangential velocity at the tip can be determined from
Vu2t =
0.91 · 9.81 · 51
ηgHe
=
= 9.73 m/s
U2t
39.9
The flow angles for the absolute and relative velocities can now be determined.
The calculations, at the mean radius r2m = 1.99 m the absolute velocity makes an
angle
(
)
(
)
Vu2t rt
9.73 · 2.55
−1
−1
α2m = tan
= tan
= 40.2◦
Vx2 rm
14.76 · 1.99
162
(f )
The tangential component of the relative velocity at this location is
Wu2m = Vx2 tan α2m − U2t
rm
44.5 · 1.99
= −22.25 m/s
= 14.76 tan(40.2◦ ) −
rt
2.55
and the flow angle is
(
−1
β2m = tan
Wu2m
Wx2
)
(
−1
= tan
−22.25
14.76
)
= −56.4◦
The flow leaves the runner axially. Therefore the tangential component of the
relative velocity is Wu3m = −U2t rm /rt and with the axial velocity constant, the
flow angle is
)
(
)
(
Wu3m
−44.5 · 1.99
−1
−1
β3m = tan
= tan
= −67.0◦
Wx3
14.76 · 2.55
163
Chapter 11
Exercise 11.1 A fluid coupling operates with oil flowing in a closed circuit. The
device consists of two elements, the primary and secondary, each making up onehalf of a torus, as shown in the figure in the text. The input power is 100 hp and
input rotational speed is 1800 rpm. The output rotational speed is 1200 rpm. (a)
Evaluate the efficiency of and output power of this device. (b) At what rate must
energy as heat be transferred to the cooling system, to prevent a temperature rise
of the oil in the coupling?
Given: Ẇp , Ωp , Ωs .
Find: η, Q̇ = Ẇp − Ẇs .
Solution: Since the torque in the primary and the secondary are the same
Ẇp = T Ωp
and
Ẇs = Ẇp
Ẇp = T Ωp
Ωs
1200
= 100
= 66.7 HP
Ωp
1800
The efficiency is
η=
Ẇs
66.7
=
= 0.667
100
Ẇp
and the rate of heat transfer from the converter fluid to the hardware and the surroundings is
Q̇ = Ẇp − Ẇs = 33.3 · 745.7 = 24.86 kW
After the hardware has reached its steady state temperature all the heat is transferred to the surroundings.
Exercise 11.2 (a) Carry out the algebraic details to show that the expression for
the flow rate through a fluid coupling is given by
√ √(
)(
)
Ω2s
r12
Dh
Q = AΩp r2
1− 2
1− 2
(1)
fL
Ωp
r2
and if for a low values of slip the friction factor is related to the flow rate by an
expression
cµA
c
=
f=
Re
ρQD
164
express the flow rate in terms of the slip for small values of s. (b) Carry out the
algebraic details to show that the expression for the torque of a fluid coupling is
given by
√ √(
)(
)(
)
Ω2s
r12
r12 Ωs
Dh
2 3
T = ρAΩp r2
1− 2
1− 2
1− 2
(2)
fL
ΩP
r2
r2 Ωp
(c) What is the appropriate form for this equation for low values of slip.
Given: The form of the equation for the flow rate and torque.
Find: Derive the equations.
Solution: The details are shown in the text. For small values of slip
1−
Ω2s
Ωs
Ωs
)(1 +
) = s(2 − s)
= (1 −
2
Ωp
Ωp
Ωp
and with
c
cµA
D
ReD
=
so that
=
Re
ρQD
fL
cL
and the flow rate becomes
√
√
ρD2 Q
Q = AΩp r2
s(2 − s)(1 − r12 /r22 )
cµAL
f=
Squaring and solving for Q gives
(
Q=
AΩ2p r22 s(2
which can be reduces to
Q=
(
2AΩ2p r22 s
r2
− s) 1 − 12
r2
r12
1− 2
r2
)
ρD2
cµL
)
ρD2
cµL
⇐ (b)
or the flow rate is directly proportional to slip.
The expression for torque coefficient is obtained from
)
(
r12 ρD2
2
2
2 2
T = ρQ(Ωp r2 − Ωs r1 ) = ρQΩp r2 s(2 − s) 1 − 2
r2 cµL
165
and
(
)(
)
r12
r12 ρAD2 Ωp
T
CT =
= s(2 − s) 1 − 2
1 − (1 − s) 2
ρΩ2p r25
r2
r2
cµLr2
and for a small s this becomes
(
r2
CT = 2s 1 − 12
r2
)2
ρAD2 Ωp
cµLr2
and this is also linear in s for small values.
Exercise 11.3 A fluid coupling operates with an input power of 200 hp, 5% slip
and a circulatory flow rate of 1500 l/s. (a) What is the rate at which energy as
heat must be transferred from the coupling in order for its temperature remain
constant? (b) What would be the temperature rise of the coupling over a period of
one-half hour assuming that no heat is transferred form the device and that it has a
mass of 45 kg, consisting of 70% metal with a specific heat 840 J/kg K, and 30%
oil with a specific heat 2000 J/kg K.
Given: Ẇp , s and a time duration of 30 minutes.
Find: η, Q̇ = Ẇp − Ẇs .
Solution: The output power is
Ẇs = ˙˙Wp
ΩS
= (1 − s)Ẇp = 0.95 · 200 = 190 hp
Ωp
The rate at which heat is transferred out from the coupling oil is
Q̇ = Ẇp − Ẇs = 10 · 745.7 = 7457 W
⇐
(a)
The change in internal energy is
U2 − U1 = Q̇∆t = 7457 · 30 · 60 = 13, 422, 600 J
The change in internal energy is also
U2 − U1 = ms cs (T2 − T1 ) + mf cf (T2 − T1 ) = (ms cs + mf cf )(T2 − T1 )
which can be written as
U2 − U1 = m(fs cs + ff cf )(T2 − T1 )
so that
T2 − T1 =
13, 422, 600
= 251 C
45(0.7 · 840 + 0.3 · 2000)
166
⇐ (b)
Exercise 11.4 In a fluid coupling the input and output shafts rotate at 2000 and
1800 rpm, respectively. The fluid is an oil having a specific gravity of 0.88 and
viscosity 0.25 kg/m s. The outer mean radius of the torus is r2 = 15 cm the inner
mean radius is r1 = 7.5 cm. The radial height is b = 2r2 /15. The axial flow
area around the torus is the same as the flow area at the outer clearance between
the primary and secondary rotors. Given that the relative roughness of the flow
conduit is 0.01, find the volumetric flow rate and the axial velocity.
Given: Ωp , Ωs , r2 , r1 , b, SG, µ, ϵ
Find: Q
Solution: The cross sectional area of the flow is
[
]
A = π (r2 = b/2)2 − (r2 − b/2)2 = 2πr2 b
and the circumference is
C = 2π(r2 − b/2) + 2π(r1 + b/2) + 4πr2
so that the hydraulic diameter is
D=
The flow rate is
4A
4 · 2πr2 b
=
= 2b
C
4πr2
√
Q = AΩp r2
Dh
fL
√(
√
or
Q=
2πr22 bΩp
b
2πf
Ω2
1 − 2s
Ωp
√(
)(
Ω2
1 − 2s
Ωp
r2
1 − 12
r2
)(
With
r1 = 7.5 cm
r2 = 15 cm
b=
)
r2
1 − 12
r2
)
2
R2 = 2 cm
15
and the flow area is
A = 2πr2 b = 2π0.15 · 0.02 = 0.0118 m2
Assume that the flow is laminar and f = 64/Re. If Reynolds number is Re =
3200, then f = 64/3200 = 0.02 and
Q = 0.3256 m3 /s
167
so that the axial velocity is
Vx =
Q
0.3256
=
= 17.28 m/s
A
0.0118
and the Reynolds number is
Re =
ρVx D
0.88 · 998 · 17.28 · 0.04
=
= 2428
µ
0.25
And the friction factor is f = 64/Re = 0.0264. Iterating gives finally f = 0.0346
and the flow rate is
Q = 0.2476 m3 /s = 247.6 liters/s
⇐
(a)
and the axial flow rate is
Vx =
Q
0.02476
=
= 13.l1 m/s
A
2π · 0.15 · 0.02
⇐
(b)
Exercise 11.5 Show that the kinetic energy loss model at the inlet to the turbine
given by
1 2
r2 (Ωp − Ωs )2
2
is based on the conversion of the change in the one half of the tangential component of the velocity squared, irreversibly into internal energy. To show this, note
that the incidence of the relative velocity at the inlet to the turbine is β2 since
the blades are radial. This leads to a leading edge separation, after which the
flow reattaches to the blade. After this reattachement the radial component of the
relative velocity is the same as in the flow incident on the blade.
Given: The velocity triangles
Find: The loss
Solution: Inspection of the velocity diagrams shows that
Vus2 = Us2 + Wus2
′
Vus2
= Us2
and therefore the expression
′
= Wus2 = Up2 − Us2 = r2 (Ωp − Ω)
Vus2 − Vus2
so that
1
1
′
)2 = r22 (Ωp − Ωs )2
(Vus2 − Vus2
2
2
168
Exercise 11.6 For a fluid coupling for which r1 /r2 = 0.7, find the value of the
slip at which the power is maximum.
Given: The expressions for power delivered by the secondary
Find: For what value of the ratio ω = Ωs /Ωp is the power maximum.
Solution: From
)
(
√
Ẇs
r12
2
= ω 1 − ω 1 − 2ω
Tm Ωp
r2
in which ω = ΩS /Ωp . Let w̄ = Ẇs /Tm Ωp , and k = r12 /r22 the following equation
is obtained
√
w̄ = ω 1 − ω 2 (1 − kω)
Differentiating with respect to ω and setting the derivative to zero gives
√
ω
1ω 2 (1 − 2kω) − √
(ω − kω 2 ) = 0
2
1−ω
which simplifies to
2kω 3 − 2ω 2 − 2kω + 1 = 0
With k = 0.72 = 0.49 the correct root of this cubic equation gives
ω = 0.6053
and the slip is
s = 1 − ω = 1 − 0.6053 = 0.3947
⇐
Exercise 11.7 A torque converter operates with oil flowing in a closed circuit. It
consists of a torus is made of a pump, a turbine, and a stator. The input and output
rotational speeds are 4000 rpm and 1200 rpm, respectively. At this operating condition the torque exerted on the stator is twice that exerted on the pump. Evaluate
(a) the output to input torque ratio and (b) the efficiency.
Given: Ωp , Ωs , and Tf = 2Tp .
Find: Ts /Tp , Ẇs /Ẇp .
Solution: From
Tp = ṁ(r1 Vu2 − r1 Vu1 )
Ts = ṁ(r2 Vu2 − r3 Vu3 )
Tf = ṁ(r1 Vu1 − r3 Vu3 )
169
it is seen that Ts = Tp + Tf . Therefore
Ts = Tp + 2Tp = 2Tp
and
or
Ts
=3
Tp
3 · 1200
Ts Ωs
Ẇs
=
=
= 0.9
Tp Ωp
4000
Ẇp
⇐
⇐
(a)
(b)
Exercise 11.8 A torque converter multiplies the torque by is designed to have to
provide a torque multiplication ratio of 3.3 to 1. The circulating oil flow rate is
500 kg/s. The oil enters the fixed vanes in the axial direction at 10 m/s, and leaves
at an angle 60 deg in the direction of the blade motion. The axial flow area is
constant. Find the torque which the primary exerts on the fluid and the torque by
the fluid on the blades of the secondary. The inlet and outlet radii of fixed vanes
are 15 cm.
Given: r2 , α1 = 0, α2 = 60c irc, Vx1 = Vx2 = 10 m/s, Ts = 3.3Tp , and ṁ =
500 kg/s
Find: Tf , Tp , Ts .
Solution: The tangential component of the velocity leaving the fixed member is
Vu2 = Vx2 tan α2 = 10 tan(60◦ ) = 17.32 m/s
and V u1 = 0. Thus
Tf = ṁr2 Vu2
Tp = ṁ(r3 Vu3 − r2 Vu2 )
Tf = ṁr3 Vu3
From
Tp + Tf = Ts
Ts = 3.3Tp
Tf = 2.3Tp
and
Tf = 500 · 0.15 · 17.32 = 1299 N m
and thus Tp = 565 N m and Ts = 1864 N m.
170
⇐
Exercise 11.9 Develop the equations for the torque ratio and efficiency for a
torque converter given in the text. (a) At what ratio of the rotational speeds is
the efficiency maximum. From this and the experimental curves shown in the text
estimate (b) the ratio r2 /r3 , and the value of Qr1 tan α1 /Ar32 Ωp .
Given: The graphs in the text.
Find: r2 /r3 , and Qr1 tan α1 /Ar32 Ωp
Solution: From
Ts = ρQ(r22 Ωp − r32 Ωs )
and
Tp = ρQ(r22 Ωp − r1 Vu1 )
Defining the ratios
T =
Ts
Tp
ω=
Ωs
Ωp
R=
r2
r3
a=
r1Vu1
r32 Ωp
the ratio of the first two equations can be written as
T =
The efficiency is
η=
R2 − ω
R2 − a
Ts Ωs
ω(R2 − ω)
=
Tp Ωp
R2 − a
The graph shown that efficiency and torque ratio are zero when ω = R2 = 0.85.
The maximum efficiency is obtained by differentiating
dη
R2 − 2ω
= 2
=0
dω
R −a
so that ωmax = R2 /2 = 0.425 ⇐ (a). At this value the graph shows that
T = 1.6. Then from the equation for the torque ratio may be written as
(R2 − a)T = R2 − ω
and solving this for a gives
a = R2 −
0.85 − 0.425
R2 − ω
= 0.85 −
= 0.584
T
1.6
171
⇐
(b)
Chapter 12
Exercise 12.1 Reconsider the ducted windmill but now let the duct be a cylindrical control volume Ae in cross sectional area. Show that the axial force on the
blades is
Fd = ρAb Vb (V − Vb )
which is in agreement with the equation for wind turbine drag.
Given: The flow variables and the control volume.
Find: The drag force.
Solution: From a cylindrical control volume
ρAe V = ṁe + ρ(Ae − Ab ) + ρAb Vb
in which Ae is the cross sectional area of the duct, Ab is the cross sectional area
of the slipstream where the velocity is Vb and ṁe is the rate at which mass leaves
the control volume across its lateral boundary. Solving this for ṁe gives
ṁe = ρAb (V − Vb )
The x-momentum balance gives
ρAb Vb2 + ρ(Ae − Ab )V 2 + ṁV − ρAe V 2 = −Fd
which reduces to
ρAb (V − Vb )[V − (V + Vb )] = −Fd
so that
Fd = ρAb Vb (V − Vb )
⇐
Exercise 12.2 The non-dimensional pressure difference can be expressed as
p + − pa
1
ρV 2
2
in which the pressure difference is between is that just before the disc and the free
stream. (a) Use the momentum theory for a wind turbine to express this in terms
of the interference factor a. For which value of a is this maximum? Interpret this
physically. (b) Develop the expression for
p a − p−
1
ρV 2
2
172
For which value of a is this the maximum?
Given: The formulas for the pressure difference across the disk.
Find: Express the pressure difference in terms of the interference factor a and for
which value of a is the pressure difference the maximum.
Solution: From
p0+
p a 1 2 p+ 1
=
+ ρV =
+ (1 − a)V 2
ρ
ρ
2
ρ
2
so that
p+ − p a
1
= ρV 2 = 1 − (1 − a)2 = a(2 − a)
ρ
2
Differentiating to obtain the maximum of this gives
2 − 2a = 0
a=1
which means that p+ is the stagnation pressure.
Downstream
p0−
p− 1
pa 1
=
+ (1 − a)2 ρV 2 =
+ (1 − 2a)V 2
ρ
ρ
2
ρ
2
Therefore
p a − p−
= (1 − a)2 − (1 − 2a)2 = a(2 − 3a)
1
2
ρV
2
Differentiating this with respect to a and setting the derivative to zero gives
a=
1
3
This value of a corresponds to a maximum power from the turbine. Hence also
maximum for the difference p+ − p− . or the lowest value of p− .
Exercise 12.3 A wind turbine operates at wind speed of V = 12 m/s. Its blade
radius is R = 20 m and its tip speed radius RΩ/V = 4. It operates at the condition Cp = 0.3. a. Find the rate of rotation of the blades. b. Find the power
developed by the turbine. c. Find the value of the interference factor a based on
the momentum theory. d. Find the pressure on the front of the actuator disc, if the
free stream pressure is 101.30 kPa.
Given: V , ΩR/V , Cp , and R.
173
Find: Ω, Ẇ , a, p+ .
Solution: The shaft speed is
Ω=
5V
5 · 12 · 30
=
= 28.6 rpm
R
20 · π
From the expression for the power coefficient
Cp = 4a(1 − a)2
and expanding the right hand side and setting it equal to 0.4 gives
4a3 − 8a2 + 4a − 0.4 = 0
This has the solutions a = (0.1330, 0.5874, 1.2749). The value in the range 0 <
a < 1/2 is the appropriate one, or a = 0.1339 Lef tarrow (c).
The power delivered is
1
1
Ẇ Cp AV 3 = 0.4 · 1.2 · π202 · 123 = 521 kW
2
2
⇐
(b)
The velocity at the actuator disk is
Vd = (1 − a)V = 0.867 · 12 = 10.4 m/s
⇐
(c)
The pressure upstream of the disk is
1
1
p+ = pa + V 2 − ρ(1 − a)2 − V 2
2
2
or
1
p+ = pa + ρa(2−a)V 2 = 101, 325+0.5·1.2·0.1330·1.867·122 = 101, 325+21 = 101, 346 Pa
2
Exercise 12.4 Using the axial momentum theory calculate the ratio of the slipstream radius to that of the disc radius in terms of the interference factor a. If
the wind turbine blades are 80 m long, what is the radius of the slip stream far
downstream. What is the radius of the streamtube far upstream.
Given: a = 1/3, R.
Find:
Rb
,
R
Ra .
174
⇐ (
Solution: From continuity
πRa2 V = πR2 (1 − a)2 V = πRb2 (1 − 2a)V
it follows that
Ra √
= 1−a
R
so that for a = 1/3 these give
Ra √
= 2
R
√
Rb
2
=
R
3
Rb
=
R
Ra =
√
√
Ra =
√
1−a
1 − 2a
280 = 113 m
2
80 = 65.3 m
3
⇐
⇐
Exercise 12.5 In a wind with speed V = 8.7 m/s and air density ρ = 1.2 kg/m3 ,
a wind turbine operates at a condition with Cp = 0.31. Find the blade length, if
the power delivered to the turbine is to be Ẇ = 250 kW.
Given: V , Cp , and Ẇ .
Find: R.
Solution: From
Cp = 4a(1 − a)2
with Cp = 0.31 the value of a can be obtained from
4a3 − 8a2 + 4a − 0.31 = 0
The roos are a = (1.2491, 0.6564, 0.0945). The only one in the the range of wind
turbines is a = 0.0945 ⇐ (a). From
1
Ẇ = ρAV 3
2
the area is
A=
2Ẇ
2 · 250, 000
=
= 2041.1 m2
3
3
ρV Cp
1.2 · 8.7 · 0.31
so that the radius is
R=
A
2041.1
=
= 25.5 m
π
π
175
Lef tarrow (b)
Exercise 12.6 Consider a 3-bladed wind turbine with blade radius of R = 35 m
and constant cord of c = 80 cm, which operates with rotational speed of Ω = 10
rpm. The wind speed is V = 12 m/s. Find the axial and tangential induction
factors at r = 10 m assuming that angle of attack is 6◦ and CD = 0.01CL .
Given: Z, R, r, c, V , and Ω.
Find: a′ and a.
Solution: Using the Matlab script on page 425 of the text gives
a′ = 0.09390423
a = 0.06017493
]
Exercise 12.7 For the Example 12.4 calculate the force Fx and torque T by using
the mean values at r = 0.6 and compare with the numerical solution given in the
text.
Given: Z, R, X, c, α, ϵ.
Find: Fx and T .
Solution: At r/R = 0.6 the values for a′ and a are a′ = 0.0055 and a = 0.0341,
therefore
Fx = 2a(1−a)πρV 2 R2 (1−0.22 ) = 2·0.046·0.954·π·1.2·122 ·402 ·(1−0.22 ) = 14, 638 N
and the torque is
T = πρΩV a′ (1 − a)R4 (1 − 0.24 ) = π · 1.2 ·
10π
· 354 · (1 − 0.24 ) = 23, 8670 N m
30
%HW 12.6
clear all;
R=35; V=12; c=0.80; alphad=6;
omega=10*pi/30; X=R*omega/V;
alpha=alphad*pi/180;
CL=2*pi*sin(alpha);
Z=3; rho=1.2;
dr=0.01; Fx=0; T=0; T2=0;
r=[0.2:dr:1.00]*R;
x=r*X/R;
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n=length(r); kmax=20;
fid=fopen(’span’,’w’);
for i=1:n
k=0; a(i) = 0; ap(i) = 0; temp=a(i);
phi(i) = atan((1-a(i))/(x(i)*(1+ap(i))));
theta(i)=phi(i)-alpha;
sigma(i)=Z*c/(2*pi*r(i)); sg(i)=sigma(i)/4;
Cx(i)=CL*cos(phi(i)); Cy(i)=CL*sin(phi(i));
while k < kmax
a(i)=sg(i)*Cx(i)/(sin(phi(i))^2+sg(i)*Cx(i));
ap(i)=sg(i)*Cy(i)/(cos(phi(i))*sin(phi(i))-sg(i)*Cy(i));
phi(i)=atan((1-a(i))/(x(i)*(1+ap(i))));
Cx(i)=CL*cos(phi(i)); Cy(i)=CL*sin(phi(i));
theta(i)=phi(i)-alpha;
res=abs(1-temp/a(i)); temp= a(i);
if res < 0.0001
break
end
k=k+1;
end
thetad(i)=theta(i)*180/pi; phid(i)=phi(i)*180/pi;
fprintf(fid,’%12.4f%12.4f%12.4f%12.4f%12.4f\n’,r(i)/R, ...
phid(i),thetad(i),a(i),ap(i));
g(i)=4*pi*rho*R*V^2*a(i)*(1-a(i))*r(i)*dr;
gg(i)=4*pi*rho*X*V^2*ap(i)*(1-a(i))*r(i)^3*dr;
Fx=Fx+g(i);
T=T+gg(i);
end
Fx=Fx-0.5*(g(1)+g(n));
T=T-0.5*(gg(1)+gg(n)); % Put these !!!!
P=T*X*V/R;
fclose(fid);
Fx1=2*rho*pi*V^2*a(41)*(1-a(41))*R^2*(1-0.2^2);
T1=rho*pi*omega*V*ap(41)*(1-a(41))*R^4*(1-0.2^4);
177