Power Plant Engineering

Example 2. An impulsive stage of a steam turbine is supplied with dry and saturated steam at
14.7 bar. The stage has a single row of moving blades running at 3600 rev/min. The mean diameter of
the blade disc is 0.9 m. The nozzle angle is 15° and the axial component of the absolute velocity leaving
the nozzle is 93.42 m/s. The height of the nozzles at their exit is 100 mm. The nozzle efficiency is 0.9 and
the blade velocity co-efficiency is 0.966. The exit angle of the moving blades is 2° greater than at the
inlet. Determine:
(1) The blade inlet and outlet angles.
(2) The isentropic heat drop in the stage.
(3) The stage efficiency.
(4) The power developed by the stage.
Solution. Mean blade velocity, V
b
= 
DN
60
π
= 3.14 × 0.9 × 
3600
60
= 169.65 m/s
α
= 15° ; V
a1 
= 93.42 m/s
Now
V
a1
= V
1
sin 
α
; V
1
= 
1
V
sin
a
α
= 360.95 m/s
With this, inlet velocity triangle can be completed
From there: 
θ
= 29.5° V
r1
= 202.5 m/s
ϕ
= 29.5 + 2 = 31.5 ; V
r0
= 0.966 × V
r1
= 195 m/s
With this, the outlet velocity triangle can be completed
θ
= 29.5°, 
ϕ
= 31.5°
Now for dry and saturated steam at 14.7 bar; V
s
= 0.1392 m
3
/kg

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212
POWER PLANT ENGINEERING
V
b
V
t 1
E
A
C
D
V
r 0
V
r 1
V
1
B
α
φ
θ
= 31.5°
= 29.5
V = V
0
a 0
V
a 1
m = A × 
1
V
V
a
s
= 
π
Dh × 
1
V
V
a
s
= 196.82 kg/s
power = m × V
b
× 
1
0
(
)
1000
t
t
V
V
−
kW
from velocity triangle
(V
t1 
– V
t0
) = 348.65 m/s
power = 196.82 × 169.65 × 
348.65
1000
= 11637.6 kW
Blade efficiency = 2 × v
b
× 
1
0
2
1
(V
V )
V
t
t
−
η
b
= 2 × 169.64 × 
2
348.65
(360.95)
= 90.79 %
η
s
= 
η
b 
× 
η
n
= 0.9079 × 0.9 = 0.817
heat drop for the stage = 
15822.7
0.817
= 14244.3 kJ/s
= 72.37 kJ/kg
Example 3. In a simple steam Impulse turbine, steam leaves the nozzle with a velocity of 1000 m/s at
an angle of 20° to the plane of rotation. The mean blade velocity is 60% of velocity of maximum effi-
ciency. If diagram efficiency is 70% and axial thrust is 39.24 N/kg of steam/sec, estimate:
(1) Blade angles.
(2) Blade velocity co-efficient.
(3) Heat lost in kJ in friction per kg.
Solution. for max. efficiency,
V.R = 
1
V
V
b
= cos 
2
α

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STEAM TURBINE
213
1
V
V
b
= cos 
20
2
°
= 0.47
For the present problem = V
b
= 0.6 × 0.47 × V
1
= 282 m/s
α
= 20°, V
1
=1000 m/s
with this, the inlet velocity triangle can be completed. From here
V
r1 
= 740 m/s, V
a1
= 350 m/s, V
t1
= 940 m/s
V
b
E
A
C
D
V
r 0
V
r 1
V
1
B
β
φ
θ
V
a 0
V
a
1
α
= 20°
V
0
= 1000 m/s
V
t 1
V
r 0
now axial thrust = 39.24 = V
a1
– V
a0
V
a0
= 350 – 39.24 = 310.76 m/s
Now diagram efficiency = 0.70 = 2 × V
b
× 
1
0
2
1
(V
V )
V
t
t
−
From here, V
t0
= 301 m/s
Now the outlet velocity can be drawn
From here
θ
= 280, 
ϕ
= 28°, V
r0 
= 660 m/s
Blade velocity co-efficient 
0
1
V
V
r
r
= 660/740 = 0.89
Heat lost in kJ in friction per kg per s.
= 
2
2
1
0
V
V
2 1000
−
×
r
r
= 
2
2
(740)
(660)
2000
−
= 55.94 kJ/kg/s
Example 4. A reaction steam turbine runs at 300 rev/min and its steam consumption is 16500
kg/hr. The pressure of steam at a certain pair is 1.765 bar (abs.) and its dryness fraction is 0.9. and the
power developed by the pair is 3.31 kW. The discharge blade tip angel both for fixed and moving blade
is 20° and the axial velocity of flow is 0.72 of the mean moving blade velocity. Find the drum diameter
and blade height. Take the tip leakage as 8%, but neglect area blocked by blade thickness.
Solution.
V
b
= 
DN
60
π
= 15.7 D m/s

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214
POWER PLANT ENGINEERING
Steam flow rate through blades = 0.92 × 
16500
3600
= 4.216 kg/s
Power output = 3.31 = m × V
b
× 
1
0
(V
V )
1000
t
t
−
(V
t1
– V
t0
) = 3.31 × 
1000
(4.216 15.7 D)
×
= 50/D m/s
Flow velocity
V
a1
= V
a0
= 0.72 V
b
= 11.30D m/s
From velocity triangle (V
t1
– V
t0
) = 2 × V
a
cot 20° – V
b
= 2 × 11.30 D × 2.7475 – 15.7D
= 46.415 D m/s
50
D
= 46.415 D
D = 1.03 m
Flow velocity V
a
= 11.30 × 1.03 = 11.64 m/s
Now V
s
(at 1.765 bar and 0.9 dry, from steam table)
= 0.9975 × 0.9 m
3
/Kg
A = 
π
dh = m × 
V
V
s
a
h = 4.216 × 0.9957 × 
0.9
11.64
× 
π
× 1.03
= 0.1 m
Example 5. At a particular ring of a reaction turbine the blade speed is 67 m/s and the flow of
steam is 4.54 kg/s, dry saturated, at 1.373 bar. Both fixed and moving blades have inlet and exit angles
of 35° and 20° respectively.
Determine:
(a) Power developed by the pair of rings.
(b) The required blade height which is to be one tenth of the mean blade ring diameter.
(c) The heat drop required by the pair if the steam expands with an efficiency of 80%.
Solution.
V
b
= 67 m/s, m = 4.54 kg/s 
ϕ
= 
α
= 20°, 
θ
= 
β
= 35°
With this given data, the velocity triangles can be drawn From the velocity triangles,
V
tl
– V
to
= 212 m/s

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STEAM TURBINE
215
V
b
E
A
C
D
V
r
0
V
r
1
V
1
V
a
0
V
a
1
V
0
V
t
1
V
r
0
B
35°
35°
20°
20°
F
Power = m × V
b
× 
1
0
(V
V )
1000
t
t
−
kW
= 4.54 × 67 × 
212
1000
= 64.47 kW
at 1.373 bar, dry saturated
V
s
= 1.259 m
3
/Kg (from steam table)
Now
m = 
π
D × D × 
V
V
a
s
× 10
From velocity triangle,
Va = 50 m/s
D
2
= 4.54 × 1.259 × 
10
50
×
p
= 0.3
D = 0.5477 m
h = 5.477 cm
Heat drop required = 
74.47
0.8
= 80.58 kJ/kg.
EXERCISE
1. Steam is supplied to a turbine at a pressure of 58.42 bar abs and tetnperature of 440°C. It is
expanded in a H.P. turbine to 6.865 bar abs., the internal efficiency of the turbine being 0.85.
The steam is then reheated at constant pressure upto 300°C. Its is then expanded to 0.049 bar
abs. In L.P. turbine having internal efficiency of 0.80. If the mechanical efficiency of the
turbine is 98% and alternator efficiency is 96%, calculate the amount of steam generated by
the boiler per kWh output.
[Ans. 3.68 kg/hr]
2. A steam power plant working on regenerative heating cycle utilizes steam at 41.2 bar and
400°C and the condenser pressure is 0.944 bar vacuum. After expansion in the turbine to 4.90
bar, a part of steam is extracted from the turbine for heating feed water from condenser in the
open heater. Draw the cycle on T-0 diagram and find thermal efficiency of the plant. Assume
the heat drop to be isentropic and the atmospheric pressure may be taken as 1.013 bar.
[Ans. 38.89%]

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216
POWER PLANT ENGINEERING
3. Steam at pressure of 30.23 bar and 400°C temperature is supplied to a steam turbine and is
exhausted at a pressure of 0.06865 bar. A single bleed is taken between the H.P. cylinder and
L.P. cylinder of the turbine at 2.45 bar for regenerative feed heating. The isentropic efficiency
for both the cylinders of the turbine is 85%. The temperature of the bleed condensate coming
out of the heat exchanger is 10°C lower than the temperature of the bled steam. Determine:
(a) Amount of bled steam per kg of steam supplied to the steam turbine.
(b) The thermal efficiency of the plant.
Consider no losses and pump work as negligible. Let the condensate coming out from the
heat exchanger and condenser be led to the hot well.
[Ans. 0.158 kg/kg of steam, 32%]
4. In a condenser test, the following observations were made:
Vacuum = 69 cm of Hg
Barometer = 75 cm of Hg
Mean temperature of condenser = 35°C
Hot well temperature = 28°C
Amount of cooling water = – 50,000 kg/hr
Inlet temperature = 17°C
Outlet temperature = 30°C
Amount of condensate per hour = 1250 kg
Find
(a) the amount of air present per m
3
of condenser volume.
(b) the state of steam entering the condenser.
(c) the vacuum efficiency .
R for air = 287 J/kgK.
[Ans. 0.0267 kg, 0.883, 97.48%]
5. In a power plant, steam at a pressure of 27.46 bar abs. and temperature 370°C is supplied by
the main boilers. After expansion in the H.P. turbine to 5.$84 bar abs., the steam is removed
and reheated to 370°C. Upon completing expansion in L.P. turbine, the steam is exhausted at
a pressure of 0.0392 bar abs. Find the efficiency of the cycle with and without reheating.
[Ans. 37.19%, 36.38%]
6. A fuel contains the following percentage of combustibles by mass C: 84%, H
2
: 4.1 %. If the
air used from burning of coal in a boiler is 16.2 kg per kg of fuel, find the total heat carried
away by dry flue gases if they escape at 300°C. The specific heats of CO
2
, O
Z
, N
Z
are 0.892,
0.917 and 1.047 respectively. Find the minimum amount of air required from the complete
combustion of 1 kg of this fuel and the excess O
2
supplied. [Ans. 4806.68 kJ, 11.16, 1.158]
7. A sample of coal has the percentage analysis by mass, C-85%, H-5%, incombustible-10%. In
a combustion chamber, the coal is burnt with a quantity of air 50% in excess of that theoreti-
cally required for complete combustion. Obtain the volumetric composition of dry flue gases.
When final temperature of flue gases is 307°C and boiler house temperature is 27°C, estimate
the maximum quantity of heat available for steam raising per kg of coal. Take C
P
for dry flue
gases as 1.005 and assume that the total heat of vapour in flue gases as 2683.87 kJ/kg. The
C.V. of coal is 32784.2 kJ/kg.

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