Question 6.DE.7: The steam velocity leaving the nozzle is 590 m/s and the noz......

Blade speed U is given by:
U = \frac{\pi DN} {60} = \frac{(\pi × 1050) × (2800)} {(1000) × (60)} = 154 m/s
The velocity diagram is shown in Fig. 6.24.
Applying the cosine rule to the triangle ABC,
V^2_1 = U^2 + C^2_1 – 2UC_1 \cosα_1
= 154² + 590² – (2) × (154) × (590) cos 20°
i.e. V_1 = 448.4 m/s.

Applying the sine rule to the triangle ABC,
\frac{C_1} {\sin (ACB)} = \frac{V_1} {\sin (α_1)}
But sin (ACB) = sin (180° – β_1) = sin (β_1)
Therefore,
\sin (β_1) = \frac{C_1\sin (α_1)} {V_1} = \frac{590 \sin (20°)}{448.4} = 0.450
and: β_1 = 26.75°
From triangle ABD,
C_{w1} = C_1 \cos (α_1) = 590 cos (20°) = 554.42 m/s
From triangle CEF,
\frac{C_{a2}} {U + C_{w2}} = \tan (β_2) = \tan (β_1) = tan (26.75°) = 0.504
or: U + C_{w2} = \frac{C_{a2}} {0.504} = \frac{155} {0.504} = 307.54
so : C_{w2} = 307.54 – 154 = 153.54 m/s
Therefore,
ΔC_w = C_{w1} + C_{w2} = 554.42 + 153.54 = 707.96 m/s
Relative velocity at the rotor outlet is:
V_2 = \frac{C_{a2}} {\sin (β_2)} = \frac{ 155} {\sin (26.75°)} = 344.4 m/s

Blade velocity coefficient is:
k = \frac{V_2} {V_1} = \frac{344.4} {448.4} = 0.768
Work done on the blades per kg/s:
ΔC_{w2}U = (707.96) × (154) × (10^{-3}) = 109 kW
The diagram efficiency is:
η_d = \frac{2UΔC_w} {C^2_1} = \frac{(2) × (707.96) × (154)}{590^2} = 0.6264
or, η_d = 62.64%