Power Plant Engineering

210 POWER PLANT ENGINEERING

Now, Vb = 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 – ηn) = 48.9 × (1 – 0.92)

= 3.91 kJ/kg

Energy lost in moving blades due to friction =

2

12

3

VV

(2 10 )

−

×

rr

kJ/kg

Now Vr 0 = 0.97 × Vr 1

Vb

Vt 1

A E

CD

Vr 2

Vr 1

V 1

α φ B θ

V (^0) Va
1
To draw velocity triangles, AB = Vb = 144 m/s, α = 15°, V 1 = 300 m/s With this triangle ABD can
be completed.
Measure Vr 1 , and θ, Vr 1 = 168 m/s, θ = 29°
Vr 0 = 0.97 × 168 = 163 m/s, and θ = 29°.
Velocity triangle ABC can be completed
Energy lost in moving blades due to friction =
22
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) Vb = πDn/60
144 = π × D ×
3000
60
D = 0.917 m