A Pelton wheel is to be designed for the following specifications. Shaft Power= 735.75 kW, Head= 200 m, Speed= 800 r.p.m., Overall efficiency = 0.86 and jet diameter is not to exceed one-tenth the wheel diameter. Determine : (i) Wheel diameter, (ii) The number of jets required, and (iii) Diameter of the jet. Take C_v = 0.98 and speed ratio = 0.45.

$$ \text{Jet Velocity, } V_1 = C_v \sqrt{2gH} = 0.98 \sqrt{2 \times 9.81 \times 200} \approx 61.38 \, \text{m/s} $$ $$ \text{Bucket Velocity, } u = \text{Speed Ratio} \times \sqrt{2gH} = 0.45 \times \sqrt{2 \times 9.81 \times 200} \approx 28.19 \, \text{m/s} $$

$$ u = \frac{\pi D N}{60} \implies D = \frac{60 u}{\pi N} $$ $$ D = \frac{60 \times 28.19}{\pi \times 800} \approx 0.673 \, \text{m} $$

$$ \text{Water Power, } P_w = \frac{\text{Shaft Power}}{\eta_o} = \frac{735750}{0.86} \approx 855523 \, \text{W} $$ $$ P_w = \rho g Q H \implies Q = \frac{P_w}{\rho g H} $$ $$ Q = \frac{855523}{1000 \times 9.81 \times 200} \approx 0.436 \, \text{m}^3/\text{s} $$

$$ \text{Maximum jet diameter, } d_{max} = \frac{D}{10} = \frac{0.673}{10} = 0.0673 \, \text{m} $$ $$ \text{Area of one jet (max), } a_{max} = \frac{\pi}{4} d_{max}^2 = \frac{\pi}{4} (0.0673)^2 \approx 0.003558 \, \text{m}^2 $$ $$ \text{Discharge of one jet (max), } q_{max} = a_{max} \times V_1 = 0.003558 \times 61.38 \approx 0.2184 \, \text{m}^3/\text{s} $$ $$ \text{Number of jets, } n = \frac{\text{Total Discharge}}{\text{Discharge per jet}} = \frac{Q}{q_{max}} = \frac{0.436}{0.2184} \approx 1.996 $$ $$ \text{Since the number of jets must be an integer, we take } n = 2 \text{ jets}. $$

$$ \text{Actual discharge per jet, } q_{actual} = \frac{Q}{n} = \frac{0.436}{2} = 0.218 \, \text{m}^3/\text{s} $$ $$ \text{Actual area per jet, } a_{actual} = \frac{q_{actual}}{V_1} = \frac{0.218}{61.38} \approx 0.003551 \, \text{m}^2 $$ $$ a_{actual} = \frac{\pi}{4} d^2 \implies d = \sqrt{\frac{4 a_{actual}}{\pi}} = \sqrt{\frac{4 \times 0.003551}{\pi}} \approx 0.0672 \, \text{m} $$