The given problem can be solved by model laws.
We (know) power (P) = ρQgH
‘ρ’ is the density of fluid, ‘Q’ is the discharge volumetric, ‘H’ is the head, ‘P0’ be the power of full-sized turbine.
Power of the model (Pm) = ρQgH = 103 × 1 × 10 × 2 = 0.02 MW
\({\left. {\frac{{{P_0}}}{{D_0^5N_0^3}}} \right|_{prototype}} = {\left. {\frac{{{P_m}}}{{D_m^5N_m^3}}} \right|_{model}}\)
Also given: Hm = 2 m, Dm = Do/5, Nm = 450 rpm, H0 = 30 m
\(\frac{{{H_0}}}{{D_0^2N_0^2}} = \frac{{{H_m}}}{{D_m^2N_m^2}}\)
\(\Rightarrow \frac{{30}}{{D_0^2N_0^2}} = \frac{2}{{{{\left( {\frac{1}{5}} \right)}^2}D_0^2}} \times {\left( {450} \right)^2}\)
⇒ N0 = 348.56 πpm
Using this in equation:
\({\left. {\frac{{{P_0}}}{{D_0^5N_0^3}}} \right|_{prototype}} = {\left. {\frac{{{P_m}}}{{D_m^5N_m^3}}} \right|_{model}}\)
\(\frac{{{P_0}}}{{D_0^5 \times {{\left( {348.56} \right)}^3}}} = \frac{{0.02}}{{{{\left( {\frac{1}{5}} \right)}^5}D_0^5 \times {{\left( {450} \right)}^3}}}\)
⇒ P0 = 29.045 MW
Points to remember: To remember and apply all the formulas correctly. Any mistakes in the formula will result into incorrect answer.