300 kg/min of steam (2 bar, 0.98 dry) flows through a given stage of a reaction turbine. The exit angle of fixed blades as well as moving blades is 20° and 3.68 kW of power is developed. If the rotor speed is 360 r.p.m. and tip leakage is 5 per cent, calculate the mean drum diameter and the blade

Question

300 kg/min of steam (2 bar, 0.98 dry) flows through a given stage of a reaction turbine. The exit angle of fixed blades as well as moving blades is 20° and 3.68 kW of power is developed. If the rotor speed is 360 r.p.m. and tip leakage is 5 per cent, calculate the mean drum diameter and the blade height. The axial flow velocity is 0.8 times the blade velocity. (Roorkee University)

Accepted Answer

Rate of flow of steam through the turbine, [latex] \dot{m}_{s} = \frac{300}{60} [/latex] = 5 kg/s
Pressure and condition of steam, p = 2 bar, x = 0.98.
The exit angles of fixed blades as well as moving blades, α = φ = 20°
Power developed, P = 3.68 kW
Speed of the rotor, N = 360 r.p.m.
Tip leakage = 5 per cent
Axial flow velocity, [latex] C_{f} = 0.8 C_{bl} [/latex] (blade velocity)
Refer Fig. 49.

Mean drum diameter, D :

Mean blade velocity, [latex] C_{bl} = \frac{πDN}{60} = \frac{πD × 360}{60}[/latex] = 18.85 D m/s

Power developed, P = [latex] \frac{\dot{m}_{s} C_{bl} C_{w}}{1000} [/latex]

or 3.67 = [latex] \frac{(5 × 0.95) × 18.85 D × C_{w} }{1000} [/latex]

∴ [latex] C_{w} = \frac{3.67 × 1000}{(5 × 0.95) × 18.85 D} = \frac{40.988}{D}[/latex]

Assuming Parson’s reaction turbine, we have

[latex]C_{f_1} = C_{1} sin α or C_{1} = \frac{C_{f_1} }{sin α} = \frac{0.8 C_{bl} }{sin α} = \frac{0.8 × 18.85 D}{sin 20°}[/latex] = 44.091 D
[latex](C_{f_1} = C_{f_0} = C_{f})[/latex]

Also, [latex] C_{w} = 2 C_{1} cos α – C_{bl} or \frac{40.988}{D} = 2 × 44.091 D \cos 20° – 18.85 D = 64.01 D[/latex]

or 40.988 = 64.01 D² or D = 0.8 m or 800 mm.

Blade height, h :

Mean steam flow rate, [latex] \dot{m}_{s} = \frac{πDhC_{f}}{x v_{g}} [/latex]

or (5 × 0.95) = [latex] \frac{π × 0.8 × h × C_{1} sin α}{0.98 × 0.885} = \frac{π × 0.8 × h × (44.091 D × sin 20°)}{0.98 × 0.885}[/latex]

(At 2 bar : [latex] v_{g} [/latex] = 0.885 kg/m³)

or h = [latex] \frac{(5 × 0.95) × 0.98 × 0.885}{π × 0.8 × (44.091 × 0.80 × 0.3420)} [/latex] = 0.1359 m or 135.9 mm.

Question 6.35: 300 kg/min of steam (2 bar, 0.98 dry) flows through a given ...

Related Answered Questions

Verified Answer:

In a three-stage steam turbine steam enters at 35 bar and 400°C and exhausts at 0.05 bar, 0.9 dry. If the work developed per stage is equal, determine : (i) Condition of steam at entry to each stage. (ii) The stage efficiencies. (iii) The reheat factor.( iv) Internal turbine efficiency. Assume ...

Verified Answer:

Verified Answer:

The following data relate to a stage of reaction turbine : Mean rotor diameter = 1.5 m ; speed ratio = 0.72 ; blade outlet angle = 20° ; rotor speed = 3000 r.p.m. (i) Determine the diagram efficiency. (ii) Determine the percentage increase in diagram efficiency and rotor speed if the rotor is ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer: