Answer to Question #171032 in Molecular Physics | Thermodynamics for Md Asif Iqbal

Question

A model turbine having diameter 0.4 m develops 15 kW when working under a head of 5m at

its best speed of 1600 rpm. If a geometrical similar turbine having diameter 1.6m operates at

the same efficiency under 18 m head. Find the speed and power developed by the actual turbine.

Answer 1

$\dfrac{H_1}{(N_1D_1)²}= \dfrac{H_2}{(N_2D_2)²}$

$\dfrac{5}{(1600× 0.4)²}= \dfrac{18}{(1.6 N_2)²}$

$N_2 = 758.95$ rpm

Equating the power coefficients;

$\dfrac{P_1}{ρ_1N_1^3 D_1^5}= \dfrac{P_2}{ρ_2N_2^3 D_2^5}$

Considering the fluids to be incompressible, and same for both the prototype and model,

we have,

$\dfrac{P_1}{N_1^3 D_1^5}= \dfrac{P_2}{N_2^3 D_2^5}$

$\dfrac{15}{1600^3 × 0.4^5}= \dfrac{P_2}{758.95^3× 1.6^5}$

$3.58 × 10^{-7} = 2.18 × 10^{-10} × P_2$

$P_2 =$ 1639.35 kW

$P_2 = 1.64\ MW$