Power Plant Engineering

214 POWER PLANT ENGINEERING

Steam flow rate through blades = 0.92 ×

16500

3600

= 4.216 kg/s

Power output = 3.31 = m × Vb ×^10

(V V )

1000

tt−

(Vt 1 – Vt 0 ) = 3.31 ×

1000

(4.216 15.7 D)×

= 50/D m/s

Flow velocity

Va 1 = Va 0 = 0.72 Vb = 11.30D m/s

From velocity triangle (Vt 1 – Vt 0 ) = 2 × Va cot 20° – Vb

= 2 × 11.30 D × 2.7475 – 15.7D

= 46.415 D m/s

50

D

= 46.415 D

D = 1.03 m

Flow velocity Va = 11.30 × 1.03 = 11.64 m/s

Now Vs (at 1.765 bar and 0.9 dry, from steam table)

= 0.9975 × 0.9 m^3 /Kg

A = πdh = 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. Vb = 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,

Vtl – Vto = 212 m/s