The water in a large lake is to be used to generate electricity by the installat
ID: 2993658 • Letter: T
Question
The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a location where the depth of the water is 50 m. Water is to be supplied at a rate of 5000 kg/s. Neglecting the kinetic and flow energy, if the
shaft power generated is measured to be 2000 kW and the overall efficiency is 75 percent, please determine:
(a) the mechanical efficiency of the turbine;
(b) the electrical efficiency of the generator;
(c) the electrical power generated;
(d) the electrical current if the output voltage of the generator is 1200V; (e) the shaft rotation rate (in rpm) if the torque is 20N.m .
Explanation / Answer
The speed of the water at the bottom of the 50 m fall is
v = sqrt(2*g*h) = sqrt(2*9.81*50) =31.32 m/s
The mechanical power generated by this falling water is
P = Ec/t = (m/t)*v^2/2 =5000*31.32^2/2 =2452356 W =2452.356 kW
The electric power generated is
Pe =eta*P =0.75*2452.356 =1839.267 kW
The mechanical efficency is
eta(m) = 2000/2452.356 =0.8155 =81.55%
The electrical efficiency is
eta(e) = 1839.267/2000 =0.9196 =91.96%
CHECK: 81.55%*91.96% =75 % overall efieciency. CORRECT.
The electrical current is
Pe =U*I
I =Pe/U =1839267/1200 =1532.72 A
The shaft rotation rate is
P =T*omega (analogous P =F*v for linear motion)
omega = P/T = 2000*10^3/20 =10^5 rad/sec
F = omega/(2*pi) =10^5/(2*pi)=15915.5 rps =15915.5/60 rpm =265.26 rpm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.