Question 1.1: A radial flow hydraulic turbine produces 32 kW under a head ......

Assuming constant fluid density, equating head, flow, and power coefficients, using subscripts 1 for the prototype and 2 for the model, we have from Eq. (1.19),
\frac{P_1} {(ρ_1N^3_1D^5_1)} = \frac{P_2} {(ρ_2N^3_2D^5_2)}, where ρ_1 = ρ_2.
Then,
\frac{D_2} {D_1} = \left(\frac{P_2} {P_1}\right)^{\frac{1}{5}}\left(\frac{N_1} {N_2}\right)^{\frac{3}{5}}  \text{or}  \frac{D_2}{D_1} = \left(\frac{0.032}{42}\right)^{\frac{1}{2}}\left(\frac{N_1} {N_2}\right)^{\frac{3}{5}} = 0.238\left(\frac{N_1} {N_2}\right)^{\frac{3}{5}}
Also, we know from Eq. (1.19) that

L_r = \frac{L_p } {L_m} = \frac{B_p}{B_m} = \frac{D_p} {D_m}     (1.19)
\frac{gH_1}{(N_1D_1)^2} = \frac{gH_2}{(N_2D_2)^2}(gravity remains constant)
Then \frac{D_2} {D_1} = \left(\frac{H_2} {H_1}\right)^{\frac{1}{2}}\left(\frac{N_1} {N_2}\right) = \left(\frac{6}{16}\right)^{\frac{1}{2}} \left(\frac{N_1}{N_2}\right)
Equating the diameter ratios, we get
0.238 \left(\frac{N_1} {N_2}\right)^{\frac{3}{5}} = \left(\frac{6}{16}\right)^{\frac{1}{2}} \left(\frac{N_1}{N_2}\right)
or
\left(\frac{N_1} {N_2}\right)^{\frac{3}{5}} = \frac{0.612}{0.238} = 2.57
Therefore the model speed is
N_2 = 100 × (2.57)^{\frac{5}{2}} = 1059 rpm

Model scale ratio is given by
\frac{D_2} {D_1} = (0.238) \left(\frac{100} {1059}\right)^{\frac{3}{5}} = 0.238(0.094)^{0.6} = 0.058.
Model efficiency is η_m = \frac{\text{Power output}} {\text{Water power input}}
or,
0.92 = \frac{42 × 10^3} {ρgQH},
or,
Q = \frac{42 × 10^3}{0.92 × 10^3 × 9.81 × 6} = 0.776 m³/s