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 designed to run at the best theoretical speed, the exit angle being 20°.

Question

Accepted Answer

Mean rotor diameter, D = 1.5 m
Speed ratio, ρ = [latex]\frac{C_{bl}}{C_{1}}[/latex] = 0.72
Blade outlet angle = 20°
Rotor speed, N = 3000 r.p.m.
(This example solved purely by calculations (Fig. 52) is not drawn to scale.)
(i) Diagram efficiency :
Blade velocity, [latex] C_{bl} = \frac{πDN}{60} = \frac{π × 1.5 × 3000}{60}[/latex] = 235.6 m/s
Speed ratio, ρ = [latex]\frac{C_{bl}}{C_{1}}[/latex] = 0.72

∴ [latex] C_{1} = \frac{C_{bl}}{0.72} = \frac{235.6}{0.72} [/latex] = 327.2 m/s
Assuming that velocity triangles are symmetrical
α = φ = 20°
From the velocity ∆LMS

[latex] C_{r_1}² = C_{1}² + C_{bl}² - 2 C_{1} C_{bl} \cos α [/latex]

[latex] C_{r_1} = \sqrt{(327.2)² + (235.6)² - 2 × 327.2 × 235.6 \cos 20°} [/latex]

= 100 [latex] \sqrt{ 10.7 + 5.55 - 14.48 }[/latex] = 133 m/s

i.e., [latex] C_{r_1}[/latex] = 133 m/s

Work done per kg of steam = [latex]C_{bl} C_{w} = C_{bl} (2 C_{1} \ cos α - C_{bl})[/latex]

= 235.6 (2 × 327.2 [latex]\cos 20° [/latex]– 235.6) = 89371.3 Nm.

Energy supplied per kg of steam

= [latex] \frac{C_{1}² + C_{r_0}² - C_{r_1}²}{2} [/latex]

= [latex]\frac{2 C_{1}² - C_{r_1}²}{2} ( ∵ C_{1} = C_{r_0}) [/latex]

= [latex] \frac{2 × (327.2)² - (133)² }{2} [/latex] = 98215.3 Nm

∴ Diagram efficiency = [latex] \frac{89371.3}{ 98215.3 } [/latex] = 0.91 = 91%.

(ii) Percentage increase in diagram efficiency :

For the best diagram efficiency (maximum), the required condition is

ρ = [latex]\frac{C_{bl}}{C_{1}} = \cos α[/latex]

∴ [latex]C_{bl} = C_{1} \cos α = 372.2 \cos 20° = 307.46 m/s [/latex]

For this blade speed, the value of [latex] C_{r_1}[/latex] is again calculated by using eqn. (i),

[latex] C_{r_1} = \sqrt{(327.2)² + (307.46)² - 2 × 327.2 × 307.46 × \cos 20°}[/latex]

= 100 [latex] \sqrt{10.7 + 9.45 - 18.906} [/latex] = 111.5 m/s

Diagram efficiency = [latex] \frac{2 C_{bl} (2 C_{1} \cos α - C_{bl} )}{(C_{1}² + C_{r_0}² - C_{r_1}² } [/latex]

= [latex] \frac{2 × 307.46 (2 × 327.2 \cos 20° - 307.46)}{(327.2)² + (327.2)² - (111.5)² } [/latex] = 0.937 or 93.7%

Percentage increase in diagram efficiency

= [latex] \frac{0.937 - 0.91}{0.91} [/latex] = 0.0296 or 2.96%.

The diagram efficiency for the speed can also becalculated by using relation

[latex]η _{blade} = \frac{2 \cos² α }{ 1 + \cos² α } = \frac{2 × \cos² 20°}{1 + \cos² 20°} = \frac{1.766}{1 + 0.883} = 0.937 or 93.7%.[/latex]

The best theoretical speed of the rotor is given by,

[latex] C_{bl} = \frac{πDN}{60}[/latex] or N = [latex] \frac{ 60 C_{bl} }{πD} = \frac{60 × 307.46}{π × 1.5}[/latex] = 3914.7 r.p.m.

Question 6.38: The following data relate to a stage of reaction turbine : M...

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