Power Plant Engineering

6.17 STEAM TURBINE SPECIFICATIONS
Steam turbine specifications consist of the following:
(i) Turbine rating. It includes :
(a) Turbine kilowatts
(b) Generator kilovolt amperes
(c) Generator Voltage
(d) Phases
(e) Frequency
(f) Power factor
(g) Excitor characteristics.
(ii) Steam conditions. It includes the following:
(a) Initial steam pressure, and Temperature
(b) Reheat pressure and temperature
(c) Exhaust pressure.
(iii) Steam extraction arrangement such as automatic or non-automatic extraction.
(iv) Accessories such as stop and throttle valve, tachometer etc.
(v) Governing arrangement.
SOLVED EXAMPLES
Example 1. In an impulse steam turbine, steam is accelerated through nozzle from rest. It enters
the nozzle at 9.8 bar dry and saturated. The height of the blade is 10 cm and the nozzle angle is 15°.
Mean blade velocity is 144 m/s. The blade velocity ratio is 0.48 and blade velocity coefficient is 0.97.
Find:
(1) Isentropic heat drop.
(2) Energy lost in the nozzles and in moving blades due to friction.
(3) Energy lost due to finite velocity of steam leaving the stage.
(4) Mass flow rate.
(5) Power developed per stage.
(6) Diagram and stage efficiency. Take: Nozzle efficiency = 92%
Blade angles at inlet = Blade angles at out let Speed = 3000 rev/min
Solution.
V.R. = 
1
V
V
b
= 0.48

---

210
POWER PLANT ENGINEERING
Now, V
b
= 144 m/s ; V
1
= 144
= 300 m/s 0.48
V
2
/2 × 10
3 
= Isentropic heat drop × Nozzle efficiency
(1) Isentropic heat drop = 
2
3
300
(2 10
0.92)
×
×
= 48.9 kJ/kg
(2) Energy lost in nozzles = Isentropic heat drop × (1 – 
η
n
) = 48.9 × (1 – 0.92)
= 3.91 kJ/kg
Energy lost in moving blades due to friction = 
2
1
2
3
V
V
(2 10 )
−
×
r
r
kJ/kg
Now Vr
0
= 0.97 × Vr
1
V
b
V
t 1
E
A
C
D
V
r 2
V
r 1
V
1
B
α
φ
θ
V
0
V
a 1
To draw velocity triangles, AB = V
b
= 144 m/s, 
α
= 15°, V
1
= 300 m/s With this triangle ABD can
be completed.
Measure V
r1
, and 
θ
, V
r1 
= 168 m/s, 
θ
= 29°
V
r0
= 0.97 × 168 = 163 m/s, and 
θ
= 29° .
Velocity triangle ABC can be completed
Energy lost in moving blades due to friction = 
2
2
3
168
163
(2 10 )
−
×
= 0.83 kJ/kg 2 × 10
3
(3) Energy lost due to finite velocity of steam leaving the stage 
2
0
3
V
(2 10 )
×
From velocity triangle, V
0
= 80 m/s
Energy lost = 
2
3
80
(2 10 )
×
(4) V
b
= 
π
Dn/60
144 = 
π 
× D × 
3000
60
D = 0.917 m

---

STEAM TURBINE
211
Area of flow, A = 
π
Dh = 3.14 * 0.917 * 
10
100
= 0.288 m
2
Now from steam table, for steam dry and saturated at 9.8 bar,
V
s
= 0.98 m
3
/kg
V
a1
= 77.65 m/s from velocity triangles
Mass flow rate = V
a1
× 
A
V
s
= 77.65 * 
0.288
0.198
= 112.95 kg/s
(5) Power = m × V
b
× (V
t1
– V
t0
) kW
from velocity triangle V
t1
– V
t0
= 288 m/s
power = 112.95 × 144 × 
288
1000
= 4682 kW
(6) Diagram efficiency = 2 × V
b
× 
1
0
2
1
(V
V )
V
t
t
−
= 92.16
stage efficiency = diagram efficiency × 
η
n
= 0.9216 × 0.92