Power Plant Engineering

STEAM TURBINE 211

Area of flow, A = πDh = 3.14 * 0.917 *

10

100

= 0.288 m^2

Now from steam table, for steam dry and saturated at 9.8 bar,

Vs = 0.98 m^3 /kg

Va 1 = 77.65 m/s from velocity triangles

Mass flow rate = Va 1 ×

A

Vs

= 77.65 *

0.288

0.198 = 112.95 kg/s

(5) Power = m × Vb × (Vt 1 – Vt 0 ) kW

from velocity triangle Vt 1 – Vt 0 = 288 m/s

power = 112.95 × 144 ×

288

1000

= 4682 kW

(6) Diagram efficiency = 2 × Vb ×^102

1

(V V )

V

tt− = 92.16

stage efficiency = diagram efficiency × ηn = 0.9216 × 0.92
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, Vb =

DN

60

π

= 3.14 × 0.9 ×

3600

60 = 169.65 m/s

α = 15° ; Va 1 = 93.42 m/s

Now Va 1 = 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° Vr 1 = 202.5 m/s

φ = 29.5 + 2 = 31.5 ; Vr 0 = 0.966 × Vr 1 = 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; Vs = 0.1392 m^3 /kg