Mcconkey Chapter 11 solution

1. Assignment of Power Plant-I Designed by Sir Engr. Masood Khan
SOLVED PROBLEMS OF CHAPTER # 11
TITLE: ROTODYNAMIC MACHINERY

2. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
PROBLEM # 11.1
The velocity of steam at inlet to a simple impulse turbine is
1000 m/s, and the nozzle angle is 20o
. The blade speed is
400 m/s and the blades are symmetrical. Determine the
blade angle if the steam is to enter the blade without shock.
If the friction effects on the blade are negligible, calculate
the tangential force on the blades and the diagram power
for a mass flow of 0.75 kg/s. What is the axial thrust and
the diagram efficiency?
If the relative velocity at exit is reduced by friction to 80%
of that at inlet, what are then the diagram power and the
axial thrust? Calculate also the diagram efficiency in this
case.
GIVEN DATA:
Device = Simple Impulse Turbine
Velocity at Inlet = Cai = 1000 m/s.
Blade Velocity = Cb = 400 m/s.
Nozzle Angle = α = 20o
.
Mass Flow Rate = m° = 0.75 kg/s.
REQUIRED DATA:
Blade Angle at Inlet = βi.
Blade Angle at Exit = βe.
Tangential Force = Ft.
Diagram Power = Pd.
Axial Thrust, and
Diagram Efficiency = ηd.
Frictionless Case.
DIAGRAM:
SOLUTION:
For Frictionless Case; Cri = Cre, & βi = βe
From inlet Velocity Triangle; Using Cosine Law:
Cri = 638.38 m/s
Inlet Angle βi can be find from Cosine Law:
Ci
2
= Cri
2
+ Cb
2
– 2V1CbCos (180 – βi).
(1000) 2
= (638.38)2
+ (400)2
– 2(638.8)(400) Cos
(180 - βi).
βi = 31.91o
= βe & Cri = Cre
To find C2 from the exit Velocity Triangle, Using
Cosine Rule:
C2
2
= Cre
2
+ Cb
2
– 2CreVbCosβe = Cri
2
+ Cb
2
– 2CreCos
βi. = (638.8) 2
+ (400) 2
– 2(400) (638.8)
Cos31.9o
.
C2 = 362.72 m/s.
To Find Tangential Force:
Fd = mo
x Cw
Fd = mo
(Cri Cosβi + Cre Cos βe)
Fd = 2 x 0.75(Cri Cos βi) = 817.42 N.
To Find the Axial Force:
Fa = mo
(Cfi – Cfe) = 0
As for frictional case, Cfi = Cfe
To find the Diagram Power:
Pd = mo
x Cb ΔCw
Pd = m x Cb(Criosβi + Creosβe)
Pd = 2mo
CbCri Cosβi. = 325.394KW
To find Diagram Efficiency:
ηd = Power Output / Power Input
ηd = (325.393) / (1/2) (mo
/Ci
2
/ 1000)
ηd = 0.864 = 86.4 %
For Friction Case;

3. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
Inlet velocity factor is same while outlet velocity factor
changes.
Cre = KCri = 0.8(638) = 510.4 m/s
To find Cae using Cosine Law
Cae = (Cre)2
+ (Cb)2
– 2CreCb Cos 32o
.
Cae = 510.42
+ (400)2
– 2(510.4x400)Cos32o
Cae = 272.45 m/s.
Cfi = 638 Sin 31.91o
= 338 m/s
Cfe = 510.4 Sin 31.91o
= 270.47 m/s
Cwi = 638.38 Cos 31.91o
= 541 m/s
Cwf = 510.4 Cos 31.91o
= 432.84 m/s
To find Tangential Force:
Ft = mo
x ΔCw = mo
(CriCosβi + CreCosβe)
Ft=0.75(638.38Cos31.91o
+ 510Cos31.91o
=731.127 N
To find Axial Thrust:
Fa = mo
(Cfi – Cfe) = 0.79(338 – 270) = 52 N
To fin Output Power (Diagram Power):
Power, Pd = mo
Cb ΔCw = mo
Cb(Cwi + Cwe)
Power, Pa = 52 N
To find Input Power:
Pinput = mo
C1
2
/ 2 = (0.79 x 10002
) / 2 x 1000
Pinput = 325 KW
To find Diagram Efficiency:
ηd = Output Power / Input Power
ηd = 307 / 395 = 77.77 %
PROBLEM # 11.2
The steam from a nozzle of a single impulse turbine
discharges with a velocity of 600 m/s and at 20o
to the
plane of the wheel. The blade wheel rotates at 3000 rpm
and the mean blade radius is 590 mm. The axial velocity of
the steam at exit from the blade is 164 m/s and the blades
are symmetrical. Calculate:
(i) the blade angles;
(ii) the diagram work per unit mass flow rate of steam;
(iii) the diagram efficiency;
(iv) The blade velocity coefficient.
GIVEN DATA:
Velocity at Inlet, Cai = 600 m/s
Nozzle Angle, αi = 20o
N = 3000 rpm, r = 590 mm
Axial Flow of the steam at exit, Cfe = 164 m/s
Blade velocity, Cb = 2πrN/60 = 185.26 m/s
REQUIRED:
(i) Blade Angle = βI
(ii) Diagram Work / mo
(iii) Diagram Efficiency = ηd
(iv) The Blade Velocity Coefficient = K
DIAGRAM:
SOLUTION:
As Cb / Cai = 0.36 & Cos αi / 2 = 0.40
So Cb / Cai < Cos αi / 2
For inlet velocity diagram:
Cri
2
= Cb
2
+ Cai
2
-2CbCaiCosαi
Cri = 185.262
+ 6002
– 2(185.26 x 600)Cos 20o
=
430.25 m/s
To find βi
Cri
2
= Cb
2
+ Cai
2
– 2CbCaiCosαi.
6002
= 430.252
+ 185.262
– 2(430.25 x 185.26 Cos βi
βi = 28o
For Impulse Turbine: βi = βe; therefore βe = 28o
Cfi = 430.25 x Sin 28o
= 202.9 m/s

4. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
Also Tan βe = Cfe / Cwe
Cwe = Cfe / tan βe = 164 / tan 28o
= 306.629 m/s
Cos βe = Cwe / Cre → Cre = 347 m/s
To find Cae
Cae
2
= Cre
2
+ Cb
2
– 2CreCbCosβe
Cae
2
= 3472
+ 185.262
– 2(347 x 185.26) Cos 28o
Cae = 203.36 m/s
Now
(i) Blade Angle: Very clear from velocity diagram as: βi
= βe = 28o
(ii) Diagram Power / mo
Pd = Cb x ΔCw = Cb (Cwi + Cwe)
Pd = Cb(CriCosβi + CreCosβe) = 126.65 KW.
(iii) Diagram Efficiency:
ηd = (Cb x ΔCw) / (C1
2
/ 2)
ηd = 2 x 126650 x 1000 / (600)2
= 70 %
(iv) Blade velocity coefficient:
K = Cre / Cri = 347 / 430.25
K = 0.80
PROBLEM # 11.3
The nozzle of the impulse stage of a turbine receives steam
at 15 bar and 300 C and discharges at 10 bar. The nozzle
efficiency is 95% and the nozzle angle is 29o
. The blade
speed is that required for maximum work and the inlet
angle of the blades is that required for entry of the steam
without shock. The blade exit angle is 5o
less than the inlet
angle. The blade velocity coefficient is 0.9. Calculate for a
steam flow of 1350 kg/h:
(i) The diagram power;
(ii) The diagram efficiency.
GIVEN DATA:
Po = 15 bar P1 = 10 bar
T = 300 Co
α = 20o
ηnozzle = 95 %
Optimum condition → Blade angle (inlet) at optimum
condition.
Βe = outlet angle = βe – 5o
K = 0.9 mo
= 1350 kg/h.
REQUIRED:
1. Diagram Power.
2. Diagram Efficiency.
DIAGRAM:
SOLUTION:
By interpolation, first find ho & so from Super Heated
Steam table:
P h So
1.2 3040 6.9534
1.5 ? ?
1.6 3034.8 6.8844
1.5 – 1.2 / 1.6 – 1.2 = ho – 3040 / 3034.8 – 3040.
ho = 3036. and similarly So = 6.9016.
As the process through the Nozzle is Adiabatic i.e. So
= S1 therefore: at P1 = 10 bar So = S1 =
6.9016.
From S.S.T at P1 = 10 bar, Sg = 6.5865.
As Sg < S1, hence steam is super heated.
Now, from super heated steam table at P1, S1 & h1 =
2940.
Hence Enthalpy drops.
Theoretical Enthalpy drop = ho – h1
/
= 3036 – 2940 =
96

5. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
ηiso,Nozzle = Actual Enthalpy drop / Theoretical
Enthalpy drop.
ηiso,Nozzle = h0 – h1 / ho – h1
/
0.95 x 96 = ho – h1 = 91.2
Applying Steady State Steady Flow Equation across the
Nozzle to find Cai.
ho + Ce
2
/ 2 = h1 + Ci2
/ 2
Ce = 2 (ho – h1) x 1000 = 427.08 m/s.
As the Turbine is working at optimum conditions so,
Cb / Cai = Cos αi / 2
Cb = C1 (Cos αi / 2)
Cb = 200.62 m/s
Using Cosine law:
Cri2 = Cb
2
+ Cai
2
-2CbCaiCos 20o
Cri = 248.23 m/s
To find βi
Cai
2
= Cb
2
+ Cri
2
= 2CbVaiCos βi
β1 = 36o
To find Cae
Cre = K Cri = 223.40 m/s
To find Ce
Cae = 2232
– 200.622
Cae = Cf = 299.96
To find Power
Power = 0.375 x 200.62 x 427.08 x Cos 20o
Power = 30.19 KW
PROBLEM # 11.4
The following particulars apply to a tow row velocity
compounded impulse stage of a turbine: nozzle angle 17o
;
blade speed 125 m/s; exit angles of the first row moving
blades, the fixed blades, and the second row moving blades,
22, 26 and 30o
respectively. Take the blade velocity
coefficient for each row of blades as 0.9 and assume that
the absolute velocity of the steam leaving the stage in the
axial direction. Draw the velocity diagram for the stage and
obtain:
(i) the absolute velocity of the steam leaving the stage;
(ii) The diagram efficiency.
GIVEN DATA:
Device: 2-row Velocity Compounded Wheel Turbine.
Nozzle Angle, αi = 17o
Blade Velocity, Cb = 125 m/s
Exit angle of the first row moving blade, βe1 = 22o
Exit angle of the second row moving blade, βe2 = 30o
Blade velocity Coefficient, k = 0.9
REQUIRED:
i. Absolute Velocity of the Steam leaving the stage, Cae2.
ii. Diagram Efficiency, ηd.
DIAGRAM:
SOLUTION:
As Turbine is working at Optimum conditions,
Cb / Cai = Cos αi / 4. = 522.08 m/s
To find Cri1
Cri1
2
= Cb
2
+ Cai
2
-2CbCaiCos 20o
Cri1 = 404.19 m/s, Cre1 = 0.9Cri1
Cre1 = 0.9 x 404.19 = 363.77 m/s
To find Cae1
Cae1
2
= 363.772
+ 1252
– 2(363.77 x 125)Cos 22o
Cae1
2
= 252.256 m/s
Cai1 = 0.9 x 252.256 = 227.03 m/s
To find Cri2

6. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
Cri2
2
= 1252
+227.032
– 2(125 x 227.03)Cos 26o
Cri2 = 127.03 m/s
Cre2 = 0.9 x 127.03 = 114.38 m/s
Similarly Cae2 = 62.79 m/s
To find ηd
ηd = Power output / Input Power
ηd = mo
CbΔCw / mo
(Cai
2
/ 2)
ηd = 2x125(ΔCwi + ΔCwe) / Cai
2
ηd = 2x125[{Cri1Cosβi1 + Cre1Cosβe1}
+ {Cri2Cosβi2 + Cre2Cosβe2}] / Cai
2
Where βi1 & βi2 are the inlet angles of first and second row
of moving blades. From the figure above:
βi1 = 22.14o
βi2 = 50.8o
ηd = 2x125[{404.14cos22.14o
+ 363.77cos22o
}
+ {127.09cos50.8o
+ 114.38cos30o
}/ (55.08)2
ηd = 0.81 = 81%
PROBLEM # 11.5
The first stage of a turbine is a two-row velocity
compounded wheel. Steam at 40 bar and 400 C is expanded
in the nozzle to 15 bar, and has a velocity at discharge of
700 m/s. the inlet velocity the stage is negligible. The
relevant exit angles are: nozzle 18o
; first row blades 21o
;
fixed blades 26.5o
; second row blades 35o
. Take the blade
velocity coefficient for all blades as 0.9. The mean
diameter of the balding is 750 mm and the turbine shaft
speed is 3000 rpm. Draw the velocity diagram for this
wheel and calculate:
(i) the diagram efficiency;
(ii) The stage efficiency.
GIVEN DATA:
Device: 2-row velocity compounded wheel Turbine.
Po = 40 bar, P1 = 15 bar, T = 400 Co
Exit velocity, C1 = 700 m/s, Inlet velocity, Co = 0
Exit angle of Nozzle, αi1 = 18o
,
Exit angle of he 1st
row of moving blades, βe1 = 21o
Exit angle of the fixed blades, αi2 = 26.5o
Exit angle of the 2nd
row of moving blades, βe2 = 35o
Mean diameter of the blade, D = 750 mm
Turbine Shaft Speed, N = 300 rev. / sec
Velocity coefficient, K = 0.9
REQUIRED:
1. Diagram Efficiency, ηd
2. Stage Efficiency, ηstage
DIAGRAM:
SOLUTION:
Cb = 2πrN / 60 = 117.75
To find Cri1
Cri1 = √(Cb
2
+ Cai1
2
– 2Cai1CbCosαi1) = 589.13 m/s
Cre1 = 0.9 x 589.13 = 530.22 m/s
To find Cae1
Cae1 = √(Cb
2
+ Cre1
2
– 2CbCri1Cosβi2) = 422.44 m/s
Cai2 = 0.9 x 422.44 = 380.16 m/s
To find Cri2
Cri2 = √(Cb
2
+ Cai2
2
–2CbCai2Cosαi2) = 279.79 m/s
Cre2 = 0.9 x 279.76 = 251.789 m/s
To find Cae2
Cae2 = √(Cb
2
+ Cre2
2
– 2CbCre2Cosβe2) = 169.37 m/s
To find βi1
7002
= 117.752
+ 589.132
– 2x117.75x589.13xCosβi1
βi1 = 21.52
To find βi2

7. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
380.162
= 117.752
+ 279.762
– 2x117.75x279.76Cosβi2
Now ∆Cw = Cw1 + Cw2
∆Cw = {Cri1Cosβi1 + Cre1Cosβe1}+{Cri2Cosβi2 + Cre2Cosβe2}
∆Cw = {(589.13) Cos21.5o
+ (530.22) Cos21o
} + {(279.76)
Cos37.3o + (251.58) Cos35o
}
∆Cw = 1471.76
OUTPUT POWER:
Power Output = mo
Cb(∆Cw) = 173299.74 Watts
Diagram Efficiency, ηd = Output power / Input power
ηd = 173299.74 / (0.5 mo
Cai1
2
)
ηd = 173299.74 / 0.5(700)2
= 70.6 %
Stage Efficiency, ηstage,
By assuming that the process through the nozzle is Adiabatic,
then at inlet condition, at Po = 40 bar & T = 400 Co
From SHST, ho = 3213.6 & So = 6.7690
P1 = 1.5 MPa & S1 = 6.7690 (as the process is Adiabatic)
(6.769 – 6.746) = (h – 2927.2)
(6.953 – 6.746) (3040.4 – 2927.2)
h = 2939.6
at P1 = 1.6 MPa
(6.7690 – 6.6732) = (h – 2919.2)
(6.8844 – 6.6732) (3034.8 – 2919.2)
h = 2971.56
at P = 1.5 Mpa
(1.5 – 1.4) = (h – 2939.3)
(1.6 – 1.4) (2971.56 – 2939.3)
h1 = 2955.48
Cai1 = √2 x (ho – h1) x 100 = 718 m/s
ηstage = Power Output / Input Power
= 2 x 173 x 1000 / (718)2
ηstage = 67.1 %
PROBLEM # 11.6
For the turbine of problem 11.5 the mass flow of steam for
each of nozzles is 4.5 kg/s. calculate the length of the arc
occupied by the nozzle if the nozzle height is 25 mm and
the wall thickness between them is negligible. If the blades
of the wheel have a pitch of 25 mm and the blade tip
thickness at exit is 0.5 mm, calculate the blade exit height
for each row.
GIVEN DATA:
Mass flow rate, mo
= 4.5 Kg/s
Length of arc occupied by the nozzle, I = 15 m
p = 25 mm Thickness, t = 0.5 mm
REQUIRED:
l1 = ?, lf = ?
SOLUTION:
Across the Nozzle, volume flow rate will be the same
mo
Ve = (Cai1Sinαi) x nl
at P = 10 bar & S1 = 6.7690, from the previous
question, we know that the steam is super heated.
Therefore we can find v from double interpolation at
P = 1.4 Mpa
6.7690 – 6.7467 = v – 0.1635
6.9534 – 6.7467 0.1822 – 0.1635
v = 0.1655 m/s & V at 1.6 Mpa
V = 0.14365 m/s
Now 1.5 – 1.4 = V – 0.1655
1.6 – 1.4 0.1465 – 0.1655
V = 0.15677 m/s
mo
v = Cai1Sinαi(nl)

8. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
n = (mo
v) / (lCai1Sinαi) = 0.13045m = 130.45 mm
To find the length of the blade height:
mo
ve = n/p(pSinαi2 – t)Cai2 lf
lf = 0.0335 m = 33.50 mm
And similarly the height of the next blade can be found.
PROBLEM # 11.7
In a reaction stage of a steam turbine the nozzle angle is 20o
and the absolute velocity of the steam at inlet to the moving
blades is 240 m/s. the blade velocity is 210 m/s. if the
blading is designed for 50% reaction, determine:
(i) the blade angle at inlet and exit;
(ii) the enthalpy drop per unit mass of steam in the moving
blades and in the complete stage;
(iii) The diagram power for a steam flow of 1 kg/s;
(iv) The diagram efficiency.
GIVEN DATA:
Device: Reaction Stage Turbine
Nozzle angle, αi1 = 20o
Ab. velocity at inlet to moving blade, C1 = 240 m/s
Blade velocity, Cb = 210 m/s
Degree of reaction, A = 0.5
REQUIRED:
1. Blade angle at inlet, βi1 & Exit, βe1
2. Enthalpy drop in moving blade and also in total size
3. P =? for mo
= 1 kg / s
4. Diagram Efficiency, ηd.
SOLUTION:
As Cb / Cai1 = 210 / 240 = 0.845
& Cosαi1 = Cos20o
= 0.939
So Cb / Cai1 > Cosαi
Now For Reaction Turbine:
Cri = Cae, αi = βe
Cre = Cai, αe = βi
Applying Cosine Law on inlet Triangle,
Cri1 = √{Cb
2
+ Cai
2
– 2CbCaiCos20o
Cri1 = √{210o
+ 240o
– 2(210)(240) Cos20o
= 83.5 m/s
To find β
Cai1 = √{Cb
2
+ Cri
2
– 2CbCriCosβi}
2402
= 2102
+ 83.52
– 2(210)(83.5)Cosβi
βi = 79.2o
1. Blade inlet angle, βi = 79.2o
As in the reaction turbine βe = αi = 20o
Enthalpy drop in the Rotor
Applying Steady State Steady Flow equation:
h1 + Cri
2
/2 = h2 + Cre
2
/2
h1 – h2 = (V2
2
– V1
2
) / 2
Enthalpy drop per unit mass of steam in moving
blades, h1 – h2 = 25.313 Kg/Kj
Enthalpies drop in Stage;
A = enthalpy drop in rotor / enthalpy drop in stage
A = (h1 – h2) / (ho – h2)
(ho – h2) = (h1 – h2) / A
Enthalpy drop per unit mass of steam in complete
stage; ho – h2 = 50.6 KJ/Kg
2. Power = mo
Cb ∆Cw
Power = mxCb(CriCosβi + CreCosβi2) = 50.60 KW
3. Diagram Effciency, ηd = Power out / Power in.
ηd = mxCbx∆Cw / m{Cai
2
/ 2 + (h1 – h2)}
ηd = 50600 / (2402
/ 2 + 25313)
ηd = 0.935 = 93.5 %

9. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
PROBLEM # 11.8
The speed of rotation of a blade group of a reaction turbine
is 3000 rpm. The mean bade velocity is 100 m/s. the blade
speed rations is 0.56 and the exit angle of the blades is 20o
.
If the mean specific volume of the steam is 0.65 m3
/kg, and
the mean height of the blade is 25 mm, calculate the mass
flow of steam through the turbine. Neglect the effect of
blade thickness on the annulus area, and assume 50%
reaction blading.
GIVEN DATA:
Device: Reaction Turbine (50%)
Speed of rotation of group of blades;
N = 3000 rpm = 3000 / 60 = 50 rev./sec.
Blade velocity, Cb = 100 m/s
Blade speed ratio, Cb / Cai = 0.56
Exit angle of blades, βe = 20o
Specific volume, v = 0.65 m3
/kg
Height of blade, l = 25 mm
REQUIRED:
Mass Flow Rate, mo
.
SOLUTION:
As Cb / Cai = 0.56, Cai = 178.56 m/s
To find “r”, Cb = 2πrN
r = Cb / 2πN = 100 / 2π 50 = 0.381 m
To find mass flow rate
mv = 2πrlSinαi(Cai)
m = 2πrlCaiSinαi / v = 2π*25*10-3
*178.5Sin20o
/ 0.65
m = 14.75 kg / s =
PROBLEM # 11.9
In the blade group of problem 11.8 there are five pairs of
blades. Calculate the useful enthalpy drop required for the
group and the diagram power.
GIVEN DATA:
The data is the same as in Problem # 11.8
mo
= 4.694 kg/s & Five pairs of blades.
REQUIRED:
Total Useful Enthalpy Drop.
Diagram Power.
SOLUTION:
From the figure
Cri = √(Cb
2
+ Cai
2
– 2CbCaiCos20o
Cri = √(1002
+ 178.572
– 2x100x178.57Cos20o
= 91.25 m/s
To find βi, Cai
2
= Cb
2
+ Cri
2
– 2CbCriCosβi
178.572
= 1002
+ 91.252
– 2x100x91.25 Cosβi
βi = 42o
To find enthalpies drop in the 1st
stage in the rotor.
Applying steady state equation
h1 + Cri
2
/ 2 = h2 + Cre
2
/ 2
h1 – h2 = Cre
2
– Cri
2
/ 2 = 11.78 KJ / Kg
To find enthalpy drop in 1st
stage, we have
A = (h1 – h2) / (ho – h2)
For 50% Reaction Turbine, A = 0.50
ho – h2 = 23.56 KJ / Kg
Total enthalpy drop in five stages = 5x23.56 = 117.8 KJ /
Kg
To find diagram Power;
Power = m2
Cb∆Cw
= 4.694x100(91.25Cos42o
+ 178.57Cos20o
)

10. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
Power = 110.562 KW
Total Power in five stages = 5x110.562 = 552.81 KW
PROBLEM # 11.10
Calculate the optimum diagram efficiency for the reaction
stage of problems 11.8 and 11.9, assuming the nozzle angle
and the axial velocity remain unchanged, the blade speed
being adjusted. Calculate also the blade at optimum
efficiency.
GIVEN DATA:
Assume the Turbine is working at optimum conditions
Nozzle Angle, αi = 20o
, Inlet velocity, Cai = 178.57m/s All
the data of Problem # 11.8.
REQUIRED:
Blade Velocity, Cb., Efficiency, η.
SOLUTION:
At optimum conditions, Cb / Cai = Cosαi
Cb = Cai (Cosαi), Cb = 167.8 m/s
To find Power;
Power = mo
Cb∆Cw
= 4.694x167.8(178.57Cos20o
) = 132.168 KW
To find enthalpies drop in the rotor;
m(h1 – h2)=m(Cre
2
/2 - Cri
2
/2)=4.694x14.5=65.984 KW
m(Cai
2
/2) = 4.694x178.572
/2 = 74.839 KW
Diagram Efficiency, ηd = Power out / Power in
Power output = mo
Cb∆Cw = 132.168 KW
Power input = (mo
Cai
2
/2) + {m2
(h1-h2)}
Power Input = 65.984 + 74.389 = 140.823 KW
ηd = 132.168 / 140.823 = 0.938 = 93.8 %
PROBLEM # 11.11
Ten stages of an ideal reaction turbine develop 3000 kW
when the mass flow of steam is 18000 kg/h. The mean
value of the blade velocity is 0.8 of the steam velocity from
the fixed blades. The exit angle of each blade is constant at
20o
and the axial velocity is constant throughout the
turbine.
Calculate the inlet angle of the blades and the
enthalpy drop in each moving row.
GIVEN DATA:
Device: Reaction Turbine-10 Stages.
Power, P = 3000 KW, Blade velocity, Cb = 0.8Cai
Mass flow rate, mo
= 18000 kg/h = 5 kg/s
Exit angle of the blades, βe = αi = 20o
Axial velocity is constant throughout Turbine.
REQUIRED:
Inlet angle of blades, βi = αe.
Enthalpy drops in single stage.
SOLUTION:
Power = mo
Cb∆Cw
3000000 = 5(0.8Cai)(CaiCos20o
+ CaeCosαe).....(A)
In Reaction Turbine, Cae = Cri & βi = αe
3000000 = 5(0.8Cai)(CaiCos20o
+ CriCosβi)......(B)
To find Cri in terms of Cai
Cri
2
= Cb
2
+ Cai
2
– 2CbCaiCos20o
Cri = √{(0.8Cai)2
+ Cai
2
– 2(0.8Cai)CaiCos20o
} = 0.396Cai
To find inlet angle of blade;
Cai
2
= (0.8Cai)2
+ (0.369Cai)2
+ 2(0.8Cai)(0.369Cai)Cosβi
Taking Cai
2
common and solving for βi;
Βi = Cos-1
[{1-(0.8)2
– (0.369)} / {2(0.8)(0.369)}] = 67.72o

11. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
Putting in equation (B)
3000000 = 5(0.8Cai)(CaiCos20o
+ CaiCos67.72o
) = 4.31Cai
2
Cai = 834.29 m/s,
As Cb = 0.8 Cai = 667.43 m/s.
Putting the values of Cai in (A)
Cri = 307.85 m/s
To find Enthalpy drops in total moving row;
Applying Steady State equation;
h1 + Cri
2
/ 2 = h2 + Cre
2
/ 2
h1 – h2 = (Cre
2
– Cri
2
) / 2 = 300.63 KJ/Kg
Enthalpies drop in single moving row;
As there are 10 stages, so
h1 – h2 = 300.63 / 10 = 30.063 KJ/Kg.
PROBLEM # 11.14
At a particular stage of an axial flow compressor the
required temperature rise is 20 K. The blade velocity at the
mean radius is 200 m/s and the flow velocity is 150 m/s.
assuming that the degree of reaction is to be 50% at the
mean radius and the work done factor is0.9, calculate the
required blade angles at the mean radius.
GIVEN DATA:
Device; Axial Flow Compressor
Temperature rise = 20 k
Blade speed, Cb = 200 m/s
Degree of Reaction, A = 50% = 0.5
Work done Factor, Y = 0.9
REQUIRED:
Blade angle at Inlet, βi = αi.
Blade angle at exit, βe = αe.
SOLUTION:
Since it is 50% reaction Turbine, so we know that;
Cae = Cre, Cai = Cri
As we know that,
Work done = Actual Power / Work input calculated from
the diagram.
Y = (mCp∆T) / (mCp∆Cw). As all values are
known so; ∆Cw = 111.67 m/s
From figure; ∆Cw = Cf(Tanβi – Tanβe)
Tanβi – Tanβe = 0.74-----------------------------------
(1)
Now A = Cf(Tanβi – Tanβe) / 2xCb.
After putting the values, we get;
Tanβi – Tanβe = 1.333---------------------------- (2)
By adding eq.(1) & eq.(2);
Tanβi = 1.0385 → βi = 46.08o
(put in eq.1)
βe = 16.4o
PROBLEM # 11.19
(a) Show that for an axial compressor of 50% reaction
design with blade velocity Cb, axial flow component of air
flow, Cf, and inlet blade angle, i, that
Specific power input per stage = Cb
2
[2(Cf/Cb)tani – 1] x
(work done factor)
(b) A 10-stage axial flow compressor of 50% reaction
design has a mean blade velocity of 250 m/s and the blade
inlet angle for each row is 45o
. The ration of flow velocity
to blade velocity is 0.75, the work done factor for each
stage is 0.87 and the isentropic efficiency of the compressor
is 0.85. Assuming an air inlet temperature of 20 C,
calculate:
(i) the exit angle of the blades;

12. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
(ii) the pressure ratio of the compressor;
(iii) The pressure ratio of the first stage.
GIVEN DATA:
Device: Axial Compressor of 50% Reaction
Blade velocity = Cb. Inlet angle = βi
Axial flow component = Cf
SHOW THAT
Specific Power input / stage = Cb
2
[2(Cf / Cb)Tanβi – 1] x
Work done.
PROOF:
(A)As Sp. Power input / stage = Cbx∆CwxW.d. Factor
= Cb(CriSinβi – CreSinβe)x W.d. factor
= Cb(CfTanβi - Cb + CfSinβe)xW.d. Factor
= Cb(2CfTanβi - Cb) x W.d. Factor
So we can write now
Specific Power input / stage = Cb
2
[(2CfTanβi – 1) / Cb] x
Work done factor.
(B) 10 stage Axial flow compressor of 50% Reaction
A = 50% = 0.5, Cb = 250 m/s, Βi = αe = 45o
Cf / Cb = 0.75 → Cf = 0.75 x 250 = 187.5 m/s
Work done = Y + 0.85
Ti = 20 Co
= 293 k, ηisen. = 0.85
REQUIRED:
αe = βe
Pressure Ratio of Compressor, P2 / P1
Pressure Ratio of the first stage.
SOLUTION:
i) A = Cf(Tanβe – Tanβi) / (2Cb)
→ βe = 18.4o
ii) Now
Sp. Power input / stage = Cbx∆CwxW.d.factor
∆Cw = Cf(Tanβi – Tanβe) = 125.13 m/s
So, Sp. Power input / stage = 20411.8 J
ηisen.=(Power input/stage)/ (Isentropic Sp.Power/stage)
After putting values, we get the value of the change in
temperature, ∆T = 23.9 k
→ T2s – T1 = 23.9 → T2s = 23.9 + 293 = 316.9 K
Now as, P2 / P1 = (T2s / T1)(r-1/r)
After putting the values, we get;
Pressure ratio of the 1st
stage = P2 / P1 = 1.3158
iii) Sp.Power input = 10 Cb ∆Cw x W.d.factor
Sp.Power input = 204118 J/kg
ηisen. = 204118 / (Cp ∆T)
→ ∆T = 238 K→ T2s – T1 = 238 K
→ T2s = 531.9 K
P2 / P1 = (T2s / T1)(r-1/r)
. After putting values we get;
P2 / P1 = 8
PROBLEM # 11.20
A centrifugal compressor running at 16000 rpm takes in air
at 17 C and compresses it through a pressure ratio of 4 with
an isentropic efficiency of 82%. The blades are radially
inclined and the slip factor is 0.85. Guide vanes at inlet
give the air an angle of pre-whirl of 20o
to the axial
direction; take the mean diameter of the impeller eye as 200
mm and the absolute air velocity at inlet as 120 m/s.
Calculate the impeller tip diameter.
GIVEN DATA:
Speed, N = 266.66 rev./sec.
Inlet Temperature, T1 = 290 k

13. Rotodynamic Machinery Designed by Sir Engr. Masood Khan
Pressure Ratio, P2 / P1 = 4
Isentropic Efficiency, ηisen. = 0.8
Angle of whirl = 20o
, Cai = 120 m/s
Diameter of Impeller eye, deye = 0.2m
REQUIRED:
Impeller Tip Diameter = dtip.
SOLUTION:
For Isentropic compression, T2s / T1 = (P2 / P1)(1 / r)
From this we get; T2s = 430.94 k
T2s – T1 = 140.94 k
Power input per unit mass flow rate = Cp(T2 – T1)--(1)
First find (T2 – T1)
ηisen. = (T2s – T1) / (T2 – T1)
(T2 – T1) = 171.888 k
(1) →Power input per unit mass flow rate = 172739.4J
Cwi = CaiSin20o
= 41.04 m/s
Cbi = 2πNR = 2πN (deye / 2) = 168.55 m/s
Slip factor = Cwe / Cwi
Cwe = 0.85 Cwi
As blades are radial so; Cwe = Cbe
Now Cwe
/
= 0.85 Cbe
Power input / in. = (Cbe x Cwe
/
- Cbi x Cwi)
Cbe = 441.74 m/s
As Cbe = 2πNR = 2πN (dtip / 2)
Diamter of Tip, Dtip = 0.5273

14. Rotodynamic Machinery Designed by S