A Francis turbine has the following data: Flow rate 0.4 m^3/s, head = 92 m, runner speed 1260 rev/min. The guide vane angle 20°, the radius at inlet = 600 mm, width of vanes 30 mm. Determine the shaft power, hydraulic efficiency and specific speeds (wT, NT). Assume radial discharge.

Question

A Francis turbine has the following data: Flow rate [latex] 0.4 {m^{3}}/{s}[/latex], head = 92 m, runner speed [latex]1260 {rev}/{min}[/latex]. The guide vane angle [latex]20^{\circ}[/latex], the radius at inlet = 600 mm, width of vanes 30 mm. Determine the shaft power, hydraulic efficiency and specific speeds [latex] \left(w_{T}, N_{T}\right)[/latex]. Assume radial discharge.

Accepted Answer

given [latex] N = 1260 rpm, Q = 0.4 {m^{3}}/{s}, H = 92 m ,\alpha_{1} = 20^{\circ}; R_{1} = 600 mm, B_{1} =30 mm[/latex]

Torque produced by the runner

[latex] T = \dot{m} V_{1w} R_{1} - \dot{m} V_{2w} R_{2}[/latex]

[latex] T = Q ho V_{1w} R_{1} as V_{2w} = 0[/latex]

also[latex] Q = \pi D_{1} B_{1} V_{1f}[/latex]

[latex] herefore 0.4 = \pi imes \frac {2 imes 600}{1000} imes \frac {30}{1000} imes V_{1f}[/latex]

[latex] herefore V_{1f} = 3.5 {m}/{s}[/latex]

From inlet diagram Fig. 6.18

[latex] an \alpha_{1} = \frac{V_{1f}}{V_{1w}}; an 20 = \frac {3.5}{V_{1w}}[/latex]

[latex] herefore V_{1w} = 9.61{m}/{s}[/latex]

[latex] T = Q ho V_{1w} R_{1} = 0.4 imes 1000 imes 9.61 imes \frac{600}{1000}[/latex]

[latex] T = 2304 N\cdot m[/latex]

Shaft power [latex] = T imes w[/latex]

where[latex] w =\frac{2N\pi}{60}; w = \frac{2 imes 1260 imes \pi}{60} = 131.1 {rad}/{s}[/latex]

Shaft power [latex] =\frac {2304 imes 2 imes 1260 imes \pi}{60} = 304 kw[/latex]

Overall efficiency [latex] = \frac {shaft power}{ hydraulic power } = \frac{304}{360} = 0.84[/latex]

[latex] = 84\%[/latex]

Specific speed [latex] w_{T} = \frac{w\sqrt {P/ ho}}{\left(gH ight)^{5/4}}[/latex]

[latex] w=\frac {2 imes 1260 imes \pi}{60} = 131.8 {rad}/{s}[/latex], substituting proper values

[latex] w_{T} = \frac {131.8 \sqrt {\frac{304 imes 10^{3}}{1000}}}{\left(9.8 imes 92 ight)^{5/4}}[/latex]

[latex] w_{T} = 0.46[/latex]

[latex] N_{s} = \frac{N \sqrt P}{H^{5/4}} = \frac{1260 \sqrt{304}}{92^{5/4}} =76.9[/latex]

Question 6.5: A Francis turbine has the following data: Flow rate 0.4 m^3/...

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