Chapter 12

Q. 12.20

A 1:10 scale model was made for testing a turbine which develops 1 MW while running at 120 rpm under a net head of 12 m. The overall efficiency at the best operating point is 90%. The model is tested under a head of 4 m. Find out the discharge, speed and power output of the model, if it runs under the condition similar to the prototype. Also, find the specific speed of the turbine.

Step-by-Step

Verified Solution

Given: \text { Scale ratio }=1: 10 ; P_p=1 \times 10^3  kW ; N_p=120  rpm ; H_p=12   m \text {; }\eta_o=0.9 ; H_m=4  m

From the equation for overall efficiency,

\eta_o=\frac{S P}{W P}=\frac{S P}{W Q H}=\frac{1 \times 10^6}{9810 \times Q \times 12}

Discharge of prototype  Q_p=\frac{1 \times 10^6}{9810 \times 12 \times 0.9}=9.44   m ^3 / s

Specific speed  N_s=\frac{N \sqrt{P}}{H^{5 / 4}}=\frac{120 \times \sqrt{10^3}}{12^{5 / 4}}=169.9

While testing geometrically similar turbines, the dimensionless parameters, head coefficient, flow coefficient and power coefficient must be the same for both prototype and model.

\left(\frac{H}{N^2 D^2}\right)_m=\left(\frac{H}{N^2 D^2}\right)_p

Speed of model  N_m=N_p\left(\frac{D_p}{D_m}\right)\left(\frac{H_m}{H_p}\right)^{1 / 2}=120 \times 10 \times\left(\frac{4}{12}\right)^{1 / 2}=692.82  rpm

From the equation for discharge coefficient,

\left(\frac{Q}{N D^3}\right)_m=\left(\frac{Q}{N D^3}\right)_p

Discharge of model Q_m=Q_p\left(\frac{N_m}{N_p}\right)\left(\frac{D_m}{D_p}\right)^2=9.44 \times\left(\frac{692.82}{120}\right) \times\left(\frac{1}{10}\right)^3 = 0.0545   m^3/s

From the equation for power coefficient,

\left(\frac{P}{N^3 D^5}\right)_m=\left(\frac{P}{N^3 D^5}\right)_p

Power developed by model

\begin{aligned}P_m &=P_p \times\left(\frac{N_m}{N_p}\right)^3 \times\left(\frac{D_m}{D_p}\right)^5 \\&=10^3 \times\left(\frac{692.82}{120}\right)^3 \times\left(\frac{1}{10}\right)^5=1.9245   kW\end{aligned}