Question

A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m³/s at 450 rpm. Take the water density and the acceleration due to gravity as 10³ kg/m³ and 10 m/s², respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full-sized turbine while working under a head of 30 m is _______ (correct to two decimal places).

Answer 1

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 (P_m) = 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: H_m = 2 m, D_m = D_o/5, N_m = 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.