Question 16.SP.8: A small Francis runner with a diameter of 2.0 ft is tested a......
A small Francis runner with a diameter of 2.0 ft is tested and found to have an efficiency of 0.893 when operating under most efficient conditions. Approximately what would be the efficiency of a homologous runner having a diameter of 6.0ft ?
Step-by-Step
Eq. (16.15): \quad \frac{1-\eta_{m}}{1-\eta_{p}}=\left(\frac{D_{p}}{D_{m}}\right)^{1 / 5}=\left(\frac{6}{2}\right)^{1 / 5}=3^{0.2}=1.246
1-\eta_{p}=(1-0.893) / 1.246
From which: \eta_{p}=0.914=91.4 \%
\frac{1-\eta_m}{1-\eta_p} \approx\left(\frac{D_p}{D_m}\right)^{1 / 5} (16.15)
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