1. In the velocity diagrams for Francis turbine, which of the following velocity directions is along the blade curvature?
a) Vr1
b) Vw1
c) V1
d) u1
Explanation: Vr1 is the relative velocity of the water flow as seen from the blade. Thus, relative velocity is along the direction of the curvature of the blade.
2. Francis turbine is typically used for which of the following values of available heads?
a) 300 m
b) 100 m
c) 30 m
d) 5 m
Explanation: Francis Turbine is a medium head turbine, typically used for heads in the range 60 m to 240 m. Hence, only 100 m from the above options fit in that range.
3. Water flow velocity is given 10 m/s. The runner diameter is 3 m and the width of the wheel is 25 cm. Find the mass of water (kg) flowing across the runner per second.
a) 7500π
b) 50π
c) 300π
d) RPM of the turbine needs to be given
Explanation: Area of the flow (A) = πDB = 0.75π m2. Mass flow rate = ρ.A.Vf = 1000*0.75π*10 = 7500π kg/s.
4. Work done per second by a Francis turbine can be given by ρAVf (Vw1u1 + Vw2u2).
a) True
b) False
Explanation: The work done per second is given by ρAVf (Vw1u1 – Vw2u2). Hence, the outlet term is subtracted from the inlet term and not added to it.
5. Which of the following terms is considered to be zero while deriving the equation for work done per second for Francis Turbine?
a) Vr1
b) Vw2
c) Vf2
d) Vr2
Explanation: Since the flow out of the runner of the Francis turbine is axial in nature, the whirl velocity at outlet is zero. Hence, Vw2 is ignored in the derivation of work done for Francis Turbine.
6. Power developed by Francis turbine are calculated for a certain set of conditions. Now, the inlet whirl velocity is doubled, the blade velocity at inlet is doubled and the flow velocity is quartered. The power developed:
a) Is 4 times the original value
b) Is 2 times the original value
c) Is ½ times the original value
d) Is same as the original value
Explanation: The power developed by a Francis Turbine is given by P = ρAV (Vw1.u1). Hence, if inlet whirl velocity is doubled, the blade velocity at inlet is doubled and the flow velocity is quartered, then the power developed will remain the same as its original value.
7. Volume flow rate of water in a Francis turbine runner is 25 m3/s. The flow velocity, whirl velocity and blade velocity are 11 m/s, 10 m/s and 5 m/s respectively, all values given at runner inlet. Find the power developed by the turbine.
a) 25 kW
b) 1.25 MW
c) 1.25 kW
d) 25 MW
Explanation: P = ρQ (Vw1.u1) = 1000*25*10*5 = 1.25 MW.
8. The flow rate of the water flow in a Francis turbine is increased by 50% keeping all the other parameters same. The work done by the turbine changes by?
a) 50% increase
b) 25% increase
c) 100% increase
d) 150% increase
Explanation: The Power developed in a Francis turbine directly depends on the flow rate of water. If flow rate is increased by 50%, i.e. made 1.5 times the original value, then the power developed becomes 1.5 times its original value too. Hence, a 50% increase.
9. A student performs an experiment with a Francis turbine. He accidently set the RPM of Francis turbine to 1400 rpm instead of 700 rpm. He reported the power to be 1 MW. His teacher asks him to perform the same experiment using the correct RPM. The student performs the same experiment again, but this time the erroneously doubled the flow velocity. What does the student report the power to be?
a) 0.5 MW
b) 0.25 MW
c) 2 MW
d) 1 MW
Explanation: The Power developed by the turbine varies directly with both flow velocity as well as the blade velocity (which in turn varies directly with RPM). So, if all parameters were correct, the reported value should be 0.5 MW. But, flow velocity is again doubled, so the student again reports 1 MW
10. Velocity of whirl at the runner inlet is given to be 10 m/s and blade velocity to be 5 m/s. The volume flow rate of water in Francis turbine is given to be 25 m3/s. Find the power generated by the turbine?
a) 1700 HP
b) 800 HP
c) 3400 HP
d) 1000 HP
Explanation: P = ρQ (Vw1.u1) = 1.25 MW. It is important to know the 1 HP = 736 W. Hence, the answer is 1.25 MW/ 736 = 1700 HP.