Power Plant Engineering

212 POWER PLANT ENGINEERING

Vb

Vt 1

A E

CD

Vr 0

Vr 1

V 1

α φ= 31.5° B θ= 29.5

V = V 0 a 0 Va 1

m = A ×^1

V

V

a

s

= πDh ×^1

V

V

a

s

= 196.82 kg/s

power = m × Vb × ()^10

1000

VVtt− kW

from velocity triangle

(Vt 1 – Vt 0 ) = 348.65 m/s

power = 196.82 × 169.65 ×

348.65

1000

= 11637.6 kW

Blade efficiency = 2 × vb ×^102

1

(V V )

V

tt−

ηb = 2 × 169.64 × 2

348.65

(360.95)

= 90.79 %

ηs = ηb × ηn = 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