Power Plant Engineering

STEAM TURBINE 213

1

V

V

b = cos^20

2

°

= 0.47

For the present problem = Vb = 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

Vr 1 = 740 m/s, Va 1 = 350 m/s, Vt 1 = 940 m/s

Vb

A E

CD

Vr 0

Vr 1

V 1

B

β φ θ

Va 0 Va^1

α = 20°

V 0

= 1000 m/s

Vt 1

Vr 0

now axial thrust = 39.24 = Va 1 – Va 0

Va 0 = 350 – 39.24 = 310.76 m/s

Now diagram efficiency = 0.70 = 2 × Vb ×^102

1

(V V )

V

tt−

From here, Vt 0 = 301 m/s

Now the outlet velocity can be drawn

From here

θ = 280, φ = 28°, Vr 0 = 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.

=

22

VV 10

2 1000

−

×

rr =

(740)^22 (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. Vb =

DN

60

π

= 15.7 D m/s