This set of Hydraulic Machines Multiple Choice Questions & Answers (MCQs) focuses on “Workdone by Francis Turbine”.
1. 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.
2. 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.
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.
3. Work done per second by a Francis turbine can be given by ρAVf (Vw1u1 + Vw2u2).
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.
4. Which of the following terms is considered to be zero while deriving the equation for work done per second for Francis Turbine?
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.
5. 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.
6. 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.u1te is directly given.
7. 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.
8. 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.
9. 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.
10. The available head of a Francis Turbine is 100 m. Velocity of the flow at the runner inlet is 15 m/s. Find the flow ratio.
Explanation: Flow ratio is given by ψ = Vf1 / sqrt(2gH). Substituting the given values and taking the value of g = 10 m/s2, we get ψ = 0.33.
11. How does the flow ratio (ψ) of a Francis turbine vary with available head (H)?
a) ψ α H
b) ψ α 1/H
c) ψ α sqrt (H)
d) ψ α 1/(sqrt (H))
Explanation: Flow ratio is given by ψ = Vf1 / sqrt(2gH). Hence, the flow ratio is inversely proportional to the square root of available head.
12. What is the typical value for flow ratio in a Francis turbine?
a) 0.05 – 0.1
b) 0.15 – 0.30
c) 0.35 – 0.45
d) 0.50 – 0.60
Explanation: Flow ratio denoted by ψ is given by Vf1 / sqrt(2gH). Sqrt (2gH) is called the spouting velocity. The practical values of the flow ratio for Francis turbine lie in the range of 0.15 – 0.3.
13. The available head of a Francis Turbine is 120 m. The blade velocity is given 35 m/s. Find the speed ratio of the turbine.
Explanation: The speed ratio φ = U/ sqrt(2gH). Hence, substituting the given values into this equation, we get φ = 0.71.
14. The speed ratio (φ) varies directly with which of the following parameters?
c) N (RPM)
d) H (Available head)
Explanation: The speed ratio is given by φ = U/ sqrt(2gH). Speed ratio directly depends upon U which in turn depends directly upon RPM of the turbine (N).
15. The typical value range of speed ratio for a Francis turbine is:
a) 0.3 – 0.6
b) 0.5 – 0.6
c) 0.1 – 0.4
d) 0.6 – 0.9
Explanation: Speed ratio denoted by φ is given by U / sqrt(2gH). Sqrt (2gH) is called the spouting velocity. The practical values of the speed ratio for Francis turbine lie in the range of 0.6 – 0.9.
Sanfoundry Global Education & Learning Series – Hydraulic Machines.
To practice all areas of Hydraulic Machines, here is complete set of 1000+ Multiple Choice Questions and Answers.