Correct Answer - Option 2 : 180 RPM and 365 kW
Concept:
Unit Quantity:
Unit Quantity are the parameters of the turbine which is defined by the turbine operator under the unit head and give maximum efficiency. They are used to find out performance parameters (N, P, Q) i.e. speed, power, discharge for the same turbine under the different-different heads.
Unit speed i.e. speed of the turbine under the unit head:
\({{\rm{N}}_{\rm{u}}} = \frac{{\rm{N}}}{{\sqrt {\rm{H}} }}\)
Unit Power i.e. the power of the turbine under the unit head:
\(P_u = \frac {P}{H^{\frac {3}{2}}}\)
Given:
Calculation:
P = 500 kW, H1 = 100 m, N1 = 200 rpm, and H2 = 81 m, N2 = ?, P2 = ?
\(\frac{{{{\rm{N}}_1}}}{{\sqrt {{{\rm{H}}_1}} }} = \frac{{{{\rm{N}}_2}}}{{\sqrt {{{\rm{H}}_2}} }} \Rightarrow \frac{{200}}{{\sqrt {100} }} = \frac{{{{\rm{N}}_2}}}{{\sqrt {81} }}\)
\( {{\rm{N}}_2} = \frac{{20 \times 9}}{{10}} = 180{\rm{\;rpm}}\)
\( \frac {P_1}{H_1^{\frac {3}{2}}} = \frac {P_2}{H_2^{\frac {3}{2}}} \Rightarrow \frac {500}{100^{\frac {3}{2}}} = \frac {P_2}{81^{\frac {3}{2}}} \)
P2 = \((81)^\frac {3}{2} \times\frac{500}{(100)^\frac{3}{2}}\)
P2 = 364.5 \(\approx\) 365 kW
