Turbomachinery MCQ Quiz - Objective Question with Answer for Turbomachinery - Download Free PDF
Last updated on Jun 11, 2025
Latest Turbomachinery MCQ Objective Questions
Turbomachinery Question 1:
Mean diameter or the pitch diameter (D) of the Pelton Wheel which rotates at N r.p.m. is given by
(where, Ku is the speed of rotation and H is the net head)
Answer (Detailed Solution Below)
Turbomachinery Question 1 Detailed Solution
Concept:
The pitch diameter of a Pelton wheel is calculated by equating the tangential velocity derived from geometry and speed ratio. The tangential velocity is:
and also
Equating both:
Final Answer:
Turbomachinery Question 2:
Theoretical power required to drive a Single-acting Reciprocating Pump is
(where, ω is the specific weight of the liquid, A is the cross-sectional area of piston or the plunger, L is the length of the stroke, N is delivery stroke per minute, Hs is the static head, Hd is the delivery head)
Answer (Detailed Solution Below)
Turbomachinery Question 2 Detailed Solution
Explanation:
Theoretical Power Required to Drive a Single-Acting Reciprocating Pump:
- The theoretical power required to drive a single-acting reciprocating pump is derived based on the fundamental principles of fluid mechanics and pump operation. The power required is a function of the work done by the pump in lifting the liquid through the given head, considering the weight of the liquid, the stroke volume, and the number of strokes per minute.
The theoretical power for a single-acting reciprocating pump is calculated by multiplying the volume of fluid delivered per second by the specific weight and total head (suction + delivery).
\( P = \frac{w \times A \times L \times N}{60} (H_s + H_d) \)
Where:
- \( w \): Specific weight of the fluid
- \( A \): Cross-sectional area of the piston
- \( L \): Length of stroke
- \( N \): Number of delivery strokes per minute
- \( H_s \): Suction head
- \( H_d \): Delivery head
Turbomachinery Question 3:
The overall efficiency of Francis Turbine is given by
(where, W is the weight of water per second which strikes the runner, P is the power available at the runner shaft and H is the net head at the runner shaft)
Answer (Detailed Solution Below)
Turbomachinery Question 3 Detailed Solution
Concept:
The overall efficiency of a Francis turbine is defined as the ratio of useful power output at the runner shaft to the hydraulic power input from water striking the runner.
Formula:
Hydraulic Power Input = Weight of water per second × Net Head
\( \lambda_0 = \frac{\text{Power Output}}{\text{Hydraulic Power}} = \frac{P}{W \cdot H} \)
Turbomachinery Question 4:
The specific speed, (NS) of a Centrifugal Pump is given by
(where, Hm is the available head from the pump and Q corresponding to the maximum efficiency of the pump at its normal working speed N)
Answer (Detailed Solution Below)
Turbomachinery Question 4 Detailed Solution
Concept:
- The specific speed of a centrifugal pump is defined as the speed of a geometrically similar pump which would deliver one cubic metre of liquid per second against a head of one metre.
- \({N_s} = \frac{{N\sqrt Q }}{{H_m^{3/4}}}\)
Important Points
- The specific speed (Ns) of a turbine is defined as, the speed of a geometrically similar Turbine that would develop unit Power (P) when working under a unit Head (H) (1 m head).
- Specific speed for the turbine is given by:
- \({N_s} = \frac{{N\sqrt P }}{{{H^{\frac{5}{4}}}}}\)
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Turbomachinery Question 5:
Which of the following statement is true about hydroelectric power plant?
Answer (Detailed Solution Below)
Turbomachinery Question 5 Detailed Solution
Explanation:
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Hydroelectric power plants serve multiple purposes beyond electricity generation, such as irrigation, flood control, water supply, and recreation.
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They store water in large reservoirs, which can be regulated to meet seasonal and agricultural demands.
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During heavy rains, controlled release of water helps prevent downstream flooding.
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Stored water can be supplied to nearby towns or industries as a dependable water source.
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Reservoirs created by dams are often used for tourism, boating, and fishery development.
Additional Information
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Hydroelectric power uses the potential energy of stored water at a height, converting it to mechanical energy through turbines and then to electrical energy via generators.
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These plants are built on rivers with dams, which create large reservoirs to control water flow and ensure consistent power generation throughout the year.
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They are considered renewable energy sources, as water is naturally replenished by the hydrological cycle without emitting pollutants.
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Hydroelectric plants have low operating and maintenance costs once installed and can respond quickly to demand fluctuations due to their flexible operations.
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These plants also contribute to grid stability and peak load supply, especially when paired with other sources of energy.
Top Turbomachinery MCQ Objective Questions
Which of the following is a positive displacement pump?
Answer (Detailed Solution Below)
Turbomachinery Question 6 Detailed Solution
Download Solution PDFExplanation:
Positive displacement pump:
- Positive displacement pumps are those pumps in which the liquid is sucked and then it is pushed or displaced to the thrust exerted on it by a moving member, which results in lifting the liquid to the required height.
- Reciprocating pump, Vane pump, Lobe pump are the examples of positive displacement pump whereas the centrifugal pump is the non-positive displacement pump.
A Pelton wheel is to be designed for a pitch diameter of 1 m and jet diameter of 0.1 m. The number of buckets on the runner computed by Taygun's formula is:
Answer (Detailed Solution Below)
Turbomachinery Question 7 Detailed Solution
Download Solution PDFConcept:
Pelton wheel:
It is a tangential flow impulse turbine in which the pressure energy of water is converted into kinetic energy to form a high-speed water jet and this jet strikes the wheel tangentially to make it rotate.
Taygun's formula:
It is used to determine the number of buckets on the runner in the Pelton wheel turbine. It is given by the below formula:
\(Z = 15+{ D\over 2d}\)
Where D = Pitch or mean diameter, d = Nozzle or Jet diameter
Calculation:
Given,
D = 1 m, d = 0.1 m
The number of buckets on the runner by Taygun's formula
\(Z = 15+{ D\over 2d} = 15+{ 1\over 2\times 0.1}\)= 15 + 5 = 20
The change in head across a small turbine is 10 m, the flow rate of water is 1 m3/s and the efficiency are 80%. The power developed by the turbine is approximately:
Answer (Detailed Solution Below)
Turbomachinery Question 8 Detailed Solution
Download Solution PDFConcept:
The overall efficiency ηo of turbine = volumetric efficiency (ηv)× hydraulic efficiency (ηh)× mechanical efficiency (ηm)
\({\eta _o} = {\eta _v} \times {\eta _h} \times {\eta _m}\)
\({{\rm{\eta }}_{\rm{v}}} = \frac{{{\rm{volume\;of\;water\;actually\;striking\;the\;runner}}}}{{{\rm{volume\;of\;water\;actually\;supplied\;to\;the\;turbine}}}}\)
\({{\rm{\eta }}_{\rm{h}}} = \frac{{{\rm{Power\;deliverd\;to\;runner}}}}{{{\rm{Power\;supplied\;at\;inlet\;}}}} = \frac{{{\rm{R}}.{\rm{P}}}}{{{\rm{W}}.{\rm{P}}}}\)
\({{\rm{\eta }}_{\rm{m}}} = \frac{{{\rm{Power\;at\;the\;shaft\;of\;the\;turbine}}}}{{{\rm{Power\;delivered\;by\;water\;to\;the\;runner}}}} = \frac{{{\rm{S}}.{\rm{P}}}}{{{\rm{R}}.{\rm{P}}}}\)
Overall efficiency: \({\eta _o} = \frac{{S.P}}{{W.P}}\)
Water Power = ρ × Q × g × h
Calculation:
Given:
ηo = 0.8, Head h = 10 m, and Q = 1 m3/s.
\({\eta _o} = \frac{{S.P}}{{W.P}} = \frac{{S.P}}{{\rho \times Q \times g \times h}}\)
\(0.8 = \frac{{S.P}}{{1000 \times 1 \times 9.81 \times 10}} \Rightarrow S.P = 78480\;W \approx 78\;kW\)
The flow ratio of a Francis turbine, if it is working under a head of 62 m and velocity at inlet 7 m/s (g = 10 m/s2) is
Answer (Detailed Solution Below)
Turbomachinery Question 9 Detailed Solution
Download Solution PDFExplanation:
Flow ratio
The flow ratio of Francis turbine is defined as the ratio of the velocity of flow at the inlet to the theoretical jet velocity.
\(Flow\;Ratio = \frac{{{V_{f1}}}}{{\sqrt {2gH} }}\)
In the case of Francis turbine,
Flow ratio varies from 0.15 to 0.3
Speed ratio varies from 0.6 to 0.9
Calculation:
\(Flow\;Ratio = \frac{{7}}{{\sqrt {2\times10\times62} }}=0.2\)
Draft tube at the exit of a reaction turbine used for the hydroelectric project is __________.
Answer (Detailed Solution Below)
Turbomachinery Question 10 Detailed Solution
Download Solution PDFExplanation:
Draft tube
It is a conduit which connects the runner exit to the tailrace where the water is being finally discharged from the turbine.
Hence, Draft tube at the exit of a reaction turbine used for the hydroelectric project is always immersed in water.
Function
The primary function of the draft tube is to reduce the velocity of the discharged water to minimize the loss of kinetic energy at the outlet.The machines which transform a power input (e.g. from an electric motor) into a hydraulic power output are:
Answer (Detailed Solution Below)
Turbomachinery Question 11 Detailed Solution
Download Solution PDFPump:
The hydraulic machine that converts mechanical energy into hydraulic energy is called a pump
Nozzle:
It is a pipe or tube of varying cross-sections. It is generally used to control the pressure or rate of flow.
Turbine:
The main function of prime movers or hydro turbines is to convert the kinetic energy of the water into mechanical energy to produce electric power.
Boiler:
It is a closed vessel in which steam is produced from water by the combustion of fuel.
The specific speed (Ns) of the pump is given by the expression
Answer (Detailed Solution Below)
Turbomachinery Question 12 Detailed Solution
Download Solution PDFExplanation:
Specific speed:
- It is defined as the speed of a geometrically similar pump that would deliver one cubic meter of liquid per second against the head of one meter.
- It is used to compare the performances of 2 different pumps.
- Its dimension is M0L3/4T-3/2 and given by the formula and is given by
\(N_{s}=\frac{N\sqrt{Q}}{H_{m}^{3/4}}\)
Where NS = Specific speed, Q = Discharge, H = Head under which the pump is working, N = Speed at the pump is working.
Additional Information
(specific speed for turbines) = \({{\rm{N}}_{\rm{s}}} = \frac{{{\rm{N}}{\sqrt{P}}}}{{{{\rm{H_m}}^{5/4}}}}\)
For a non-dimensional specific speed value of 1, for maximum efficiency, which of the following turbines is preferred?
Answer (Detailed Solution Below)
Turbomachinery Question 13 Detailed Solution
Download Solution PDFNon-dimensional specific speed is given by
\({N_S} = \frac{{N\sqrt P }}{{{{\left( {gH} \right)}^{\frac{5}{4}}} \cdot \sqrt \rho }}\)
The Francis turbine is a type of reaction turbine, and it can operate over a wide range of water flows and height differences, which makes it suitable for a specific speed value of 1. It is more flexible in terms of operation conditions compared to the Pelton wheel.
The overall efficiency of a centrifugal pump when head is 25 m, discharge = 0.04 m3/s and output power p = 16 kW (take g = 10 m/s2? and ρ = 1000) is
Answer (Detailed Solution Below)
Turbomachinery Question 14 Detailed Solution
Download Solution PDFExplanation:
Overall Efficiency (η): It is defined as a ratio of the power output of the pump to the power input to the pump.
The overall efficiency of the pump will be given as,
\({{\rm{\eta }}_{\rm{}}} = \frac{{{\rm{water\;power}}}}{{{\rm{shaft\;power\;}}}} = \frac{{{\rm{\omega QH}}}}{{\rm{P}}}\)
\(P = \frac{{{\bf{\omega QH}}\;}}{{{\eta }}}\)
Calculation:
\(\eta = \frac{{{\bf{\rho g QH}}\;}}{{{P }}} = \frac{{{\bf{1000\times10\times0.04\times25}}\;}}{{{16000}}}=0.625\)
Additional Information
Manometric Efficiency (ηman): It is the ratio of the manometric head to head imparted by the impeller to the water.
\({\eta _{man}} = \frac{{{H_m}}}{{\frac{{{V_{w2}}{u_2}}}{g}}} = \frac{{g{H_m}}}{{{V_{w2}}{u_2}}}\)
Mechanical Efficiency (ηm): It is the ratio of the power available at the impeller to the power at the shaft of the centrifugal pump.
\({\eta _m} = \frac{{{\rm{Power\;at\;the\;impeller}}}}{{{\rm{Power\;at\;the\;shaft}}}} = \frac{{\frac{W}{g}\left( {\frac{{{V_{w2}}{u_2}}}{{1000}}} \right)}}{{{\rm{SP}}}}\)
The hydropower plants which utilise the minimum flow in a river having no appreciable pondage on its upstream are called as:
Answer (Detailed Solution Below)
Turbomachinery Question 15 Detailed Solution
Download Solution PDFExplanation:
Runoff river plants:
- A runoff river plant is defined as a plant that utilized minimum flow in a river and has no appreciable pondage on its upstream side. The excess water is temporarily stored in the pond on the upstream side when the discharge at the site is more than the demand.
Additional Information
Tidal plant:
- A tidal power plant is a plant that produces electricity using a source of renewable energy tides.
Pumped storage plant:
- A pump storage plant is a hydroelectric system in which electricity is generated during periods of high demand by the use of water that has been pumped into a reservoir.
Storage plant:
- A storage plant is defined as a plant used for the storage of energy in the form of water.