(Impulse reaction turbine). The following data relate to a stage of an impulse reaction turbine : Steam velocity coming out of nozzle = 245 m/s ; nozzle angle = 20° ; blade mean speed = 145 m/s ; speed of the rotor = 300 r.p.m. ; blade height = 10 cm ; specific volume of steam at nozzle outlet and

Question

(Impulse reaction turbine). The following data relate to a stage of an impulse reaction turbine :
Steam velocity coming out of nozzle = 245 m/s ; nozzle angle = 20° ; blade mean speed = 145 m/s ; speed of the rotor = 300 r.p.m. ; blade height = 10 cm ; specific volume of steam at nozzle outlet and blade outlet respectively = 3.45 m³/kg and 3.95 m³/kg ; Power developed by the turbine = 287 kW ; efficiency of nozzle and blades combinedly = 90% ; carry over co-efficient = 0.82.
Find : (i) The heat drop in each stage ; (ii) Degree of reaction ;
(iii) Stage efficiency.

Accepted Answer

Steam velocity coming out of nozzle, [latex] C_{1} [/latex] = 245 m/s
Nozzle angle, α = 20°
Blade mean speed, [latex] C_{bl} [/latex] = 145 m/s
Speed of the rotor, N = 3000 r.p.m.
Blade height, h = 10 cm = 0.1 m
Specific volume of steam at nozzle outlet or blade inlet, [latex] v_{1} [/latex] = 3.45 m³/kg
Specific volume of steam at blade outlet, [latex] v_{0} [/latex] = 3.95 m³/kg
Power developed by the turbine = 287 kW
Efficiency of nozzle and blades combinedly = 90%
Carry over co-efficient, ψ = 0.82
Blade speed, [latex] C_{bl} = \frac{πDN}{60}[/latex]

∴ D = [latex] \frac{ 60 C_{bl} }{πN} = \frac{60 × 145}{π × 3000}[/latex] = 0.923 m

Mass flow rate, [latex] \dot{m}_{s} = \frac{ C_{f_1} × πDh }{v_{1}} = \frac{C_{1} sin α × πDh }{v_{1}}[/latex]

= [latex] \frac{245 × sin 20° × π × 0.923 × 0.1}{3.45} [/latex] = 7.04 kg/s

Also [latex] \dot{m}_{s} = \frac{ C_{f_0} πDh }{v_{0}} [/latex]

∴ [latex] C_{f_0} = \frac{\dot{m}_{s} v_{0}}{πDh} = \frac{7.04 × 3.95}{π × 0.923 × 0.1}[/latex] = 95.9 m/s

The power is given by, P = [latex] \frac{\dot{m}_{s} × C_{bl} × C_{w}}{1000} [/latex]

287 = [latex] \frac{7.04 × 145 × C_{w} }{1000} [/latex]

∴ [latex] C_{w} = \frac{287 × 1000}{7.04 × 145 } [/latex] = 281 m/s

Now draw velocity triangles as follows :
Select a suitable scale (say 1 cm = 25 m/s)
• Draw LM = blade velocity = 145 m/s ; ∠MLS = nozzle angle = 20° Join MS to complete the inlet ∆LMS.
• Draw a perpendicular from S which cuts the line through LM at point P. Mark the point Q such that PQ = [latex] C_{w} [/latex] = 281 m/s.
• Draw a perpendicular through point Q and the point N as QN = 95.9 m/s. Join LN and MN to complete the outlet velocity triangle.

From the velocity triangles ;

[latex] C_{r_1} = 117.5 m/s ; C_{r_0} = 217.5 m/s ; C_{0} = 105 m/s. [/latex]

(i) Heat drop in each stage, [latex] (∆h)_{stage} [/latex]:
Heat drop in fixed blades ( [latex] ∆h_{f} [/latex])

= [latex] \frac{ C_{1}² - ψ C_{0}²}{2 × η_{nozzle}} [/latex] , where ψ is a carry over co-efficient

= [latex] \frac{(245)² - 0.82 × (105)²}{2 × 0.9 × 1000} [/latex] = 28.32 kJ/kg

Heat drop in moving blades ( [latex] ∆h_{m} [/latex])

= [latex] \frac{ C_{r_0}² - C_{r_1}²}{2 × η_{nozzle}} [/latex] = [latex] \frac{(217.5)² - (117.5)²}{2 × 0.9 × 1000} [/latex] = 18.61 kJ/kg

Total heat drop in a stage,

[latex]( ∆h)_{stage} = ∆h_{f} + ∆h_{m} [/latex] = 28.32 + 18.61 = 46.93 kJ/kg.

(ii) Degree of reaction, [latex]R_{d} [/latex]:

[latex]R_{d} = \frac{ ∆h_{m} }{ ∆h_{m} + ∆h_{f} } = \frac{18.61 }{18.61 + 28.32} [/latex] = 0.396.

(iii) Stage efficiency, [latex] η_{stage} [/latex] :

Work done per kg of steam

= [latex] \frac{C_{bl} × C_{w}}{1000} = \frac{145 × 281}{1000} [/latex] = 40.74 kJ/kg of steam

∴ [latex] η_{stage} = \frac{Work done per kg of steam}{Total heat drop in a stage} = \frac{40.74}{46.93} [/latex] = 0.868 or 86.8%.

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