The outlet angle of the blade of a Parson’s turbine is 20° and the axial velocity of flow of steam is 0.5 times the mean blade velocity. If the diameter of the ring is 1.25 m and the rotational speed is 3000 r.p.m. determine :
(i) Inlet angles of blades.
(ii) Power developed if dry saturated steam at 5 bar passes through the blade whose height may be assumed as 6 cm. Neglect the effect of blade thickness.

Question

Accepted Answer

Refer Fig. 48.
Angles, α = φ = 20°
Axial velocity of flow of steam,
[latex]C_{f_1} = C_{f_0} = 0.5 C_{bl} [/latex] (blade speed)
Diameter of the ring, D = 1.25 m
Rotational speed, N = 3000 r.p.m.

Blade speed, [latex] C_{bl} = \frac{πDN}{60} = \frac{π × 1.25 × 3000}{60}[/latex] = 196 m/s

∴ [latex]C_{f_1} = C_{f_0}[/latex] = 0.5 × 196 = 98 m/s

Velocity diagram is drawn as follows :
• Takes LM ([latex] C_{bl} [/latex] ) = 196 m/s, and α = φ = 20°.

• Draw line 1–2 parallel to LM at a value of 98 m/s (according to scale). The points S and N are thus located on the line 1–2.

• Complete the rest of the diagram as shown in Fig. 48.

(i) Inlet angles of blades :

The inlet angles (by measurement) are :
β = θ = 55°

(ii) Power developed, P :
Area of flow is given by, A = π × D (mean diameter) × h (height of blade)
Mean flow rate is given by,

[latex]\dot{m}_{s} = \frac{ Area of flow × Velocity of flow }{Specific volume of steam } = \frac{πDh × C_{f}}{v}[/latex]

From steam tables, [latex]v_{g}[/latex] = 0.375 m³/kg at 5 bar

∴ [latex]\dot{m}_{s} = \frac{π × 1.25 × (\frac{6}{100} ) × 98}{0.375} [/latex] = 61.57 kg/s

Power developed, P = [latex] \frac{\dot{m}_{s} × C_{w} × C_{bl}}{1000} = \frac{61.57 × 330 × 196}{1000} [/latex] = 3982.3 kW.

Question 6.33: The outlet angle of the blade of a Parson’s turbine is 20° a...

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