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.


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}