Turbomachinery MCQ Quiz - Objective Question with Answer for Turbomachinery - Download Free PDF

Last updated on May 7, 2025

Latest Turbomachinery MCQ Objective Questions

Turbomachinery Question 1:

Priming of a pump refers to:

  1. removing air from the pump casing and suction line
  2. lubricating the pump bearings
  3. increasing the rotational speed of the pump
  4. adjusting the impeller clearance

Answer (Detailed Solution Below)

Option 1 : removing air from the pump casing and suction line

Turbomachinery Question 1 Detailed Solution

Explanation:

Priming of a Pump

  • Priming of a pump refers to the process of removing air from the pump casing and suction line to ensure that the pump operates efficiently. This process is crucial for the proper functioning of the pump, especially in cases where the pump is used to lift fluids from a lower level to a higher level.
  • Pumps are designed to move fluids by creating a pressure differential. However, when air is trapped in the pump casing or suction line, it can disrupt the pressure differential and prevent the pump from functioning correctly. Priming involves filling the pump casing and suction line with the fluid to be pumped, ensuring that there is no air trapped in the system. This allows the pump to create the necessary pressure differential and move the fluid efficiently.

Methods of Priming:

  • Manual Priming: This method involves manually filling the pump casing and suction line with the fluid to be pumped. This can be done using a priming pump, a priming chamber, or by pouring the fluid directly into the pump.
  • Automatic Priming: Some pumps are equipped with automatic priming systems that use a small auxiliary pump or a priming chamber to remove air from the pump casing and suction line. These systems are designed to maintain the prime automatically, ensuring continuous operation.

Importance of Priming:

  • Prevents Cavitation: Cavitation occurs when air bubbles form in the pump and collapse, causing damage to the pump components. Priming ensures that there is no air in the system, preventing cavitation and extending the life of the pump.
  • Ensures Efficient Operation: Air in the pump casing or suction line can disrupt the pressure differential and reduce the pump's efficiency. Priming ensures that the pump operates at its optimal efficiency, providing consistent performance.
  • Prevents Damage: Running a pump without priming can cause it to run dry, leading to overheating and damage to the pump components. Priming ensures that the pump is always filled with fluid, preventing damage and ensuring reliable operation.

Turbomachinery Question 2:

How does the volute shape in a single volute pump casing affect performance?

  1. It helps maintain a uniform flow distribution.
  2. It increases turbulence within the pump.
  3. It restricts the pump’s operating range.
  4. It serves only an aesthetic purpose.

Answer (Detailed Solution Below)

Option 1 : It helps maintain a uniform flow distribution.

Turbomachinery Question 2 Detailed Solution

Explanation:

Single Volute Pump Casing:

  • A single volute pump casing is a type of pump casing design in which the volute is a single spiral-shaped chamber that surrounds the impeller. The primary function of the volute is to convert the kinetic energy imparted to the fluid by the impeller into pressure energy.
  • In a centrifugal pump, the fluid enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, flowing radially outward into a diffuser or volute chamber, from where it exits into the downstream piping system. The single volute design ensures that the fluid is smoothly transitioned from the high-velocity region near the impeller to the lower-velocity region in the volute, helping in maintaining the flow characteristics.

Advantages:

  • Uniform Flow Distribution: The single volute design helps maintain a uniform flow distribution around the impeller, reducing the chances of flow separation and energy losses. This uniformity in flow minimizes the radial forces acting on the impeller, leading to smoother operation and reduced wear and tear on the pump components.
  • Efficiency: By maintaining a uniform flow distribution, the single volute design helps in achieving higher pump efficiencies. The smooth transition of fluid reduces turbulence and hydraulic losses, ensuring that a significant portion of the energy imparted by the impeller is converted into pressure energy.
  • Cost-Effectiveness: Single volute pump casings are generally simpler and less expensive to manufacture compared to more complex designs like double volute casings. This cost-effectiveness makes them a preferred choice for many industrial applications where cost considerations are critical.

Disadvantages:

  • Limited Operating Range: Single volute pumps may have a restricted operating range compared to double volute designs. At off-design conditions, single volute pumps can experience higher radial forces, leading to increased vibration and wear.
  • Potential for Cavitation: In certain conditions, single volute designs can be more prone to cavitation, especially at low flow rates. Cavitation can cause significant damage to the pump impeller and casing, leading to reduced pump life and efficiency.

Turbomachinery Question 3:

One challenge associated with double volute casings is:

  1. increased complexity in manufacturing and alignment
  2. higher risk of leakage due to multiple seams
  3. decreased efficiency in converting kinetic to pressure energy
  4. reduced flow rates

Answer (Detailed Solution Below)

Option 1 : increased complexity in manufacturing and alignment

Turbomachinery Question 3 Detailed Solution

Explanation:

Double Volute Casings

  • A double volute casing is a type of pump casing where the volute (the spiral-shaped casing that collects fluid discharged from the impeller) is divided into two separate channels or volutes. These channels are designed to balance the hydraulic forces on the impeller, thereby reducing radial loads and prolonging the life of the pump components.
  • In a double volute casing, the fluid is discharged from the impeller into two separate volute channels. These channels help to balance the pressure around the impeller, reducing radial thrust and minimizing the mechanical stress on the pump bearings and shaft. This design is especially beneficial in high-capacity and high-pressure applications.

Advantages:

  • Reduced radial thrust on the impeller, leading to longer bearing and shaft life.
  • Improved hydraulic balance, which enhances the overall reliability and performance of the pump.
  • Better handling of high-pressure and high-flow applications.

Disadvantages:

  • Increased complexity in design and manufacturing, which can lead to higher production costs.
  • More challenging alignment and assembly processes due to the additional components and tighter tolerances required.

Applications: Double volute casings are commonly used in industrial and municipal applications where high flow rates and pressures are required, such as in water treatment plants, chemical processing, and power generation.

Turbomachinery Question 4:

Manometric head is defined as:

  1. the height of a fluid column corresponding to the pump’s output pressure
  2. the weight of the pump
  3. the temperature difference across the pump
  4. the fluid velocity within the pump

Answer (Detailed Solution Below)

Option 1 : the height of a fluid column corresponding to the pump’s output pressure

Turbomachinery Question 4 Detailed Solution

Concept:

Manometric head is the pressure developed by the pump expressed as an equivalent height of the fluid column:

\( H_m = \frac{P}{ρ g} \),

Where, P = Pressure, ρ = Density, g = Acceleration due to gravity

This represents the energy imparted to the fluid by the pump in the form of pressure head.

Turbomachinery Question 5:

Compared to axial flow pumps, radial flow pumps typically:

  1. generate higher pressures with lower flow rates
  2. are used exclusively for pumping gases
  3. have lower efficiency in all applications
  4. produce higher flow rates at lower pressures

Answer (Detailed Solution Below)

Option 1 : generate higher pressures with lower flow rates

Turbomachinery Question 5 Detailed Solution

Explanation:

Radial Flow Pumps:

  • Radial flow pumps are a type of centrifugal pump where the fluid enters axially into the impeller but exits radially, perpendicular to the pump shaft. These pumps are designed to develop high pressures with relatively low flow rates, making them suitable for applications where a significant pressure head is required.

Working Principle: In a radial flow pump, fluid is drawn into the center of the impeller along its axis (axial direction). The rotating impeller imparts kinetic energy to the fluid, converting it into pressure energy as the fluid moves outward in a radial direction. The fluid exits the pump casing at a 90-degree angle to the shaft.

Advantages:

  • Capable of generating high pressures, making them ideal for applications requiring a large pressure head.
  • Compact design and relatively easy to maintain.
  • Well-suited for handling clean liquids with low viscosity.

Disadvantages:

  • Limited flow rate capabilities compared to axial flow pumps.
  • Not suitable for handling large volumes of fluid or highly viscous liquids.

Applications: Radial flow pumps are commonly used in industries such as water supply, chemical processing, boiler feed applications, and irrigation systems where high pressure and low flow rates are required.

Top Turbomachinery MCQ Objective Questions

Which of the following is a positive displacement pump?

  1. Reciprocating pump
  2. Propeller pump
  3. Centrifugal pump
  4. Jet pump

Answer (Detailed Solution Below)

Option 1 : Reciprocating pump

Turbomachinery Question 6 Detailed Solution

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Explanation:

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.​

SSC JE ME Full  Test 4 set 2 Images-Q14

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:

  1. 25
  2. 15
  3. 20
  4. 10

Answer (Detailed Solution Below)

Option 3 : 20

Turbomachinery Question 7 Detailed Solution

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Concept:

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.

turbine

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:

  1. 100 kW
  2. 78 kW
  3. 1 MW
  4. 50 kW

Answer (Detailed Solution Below)

Option 2 : 78 kW

Turbomachinery Question 8 Detailed Solution

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Concept:

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

  1. 0.3
  2. 0.2
  3. 0.1
  4. 0.4

Answer (Detailed Solution Below)

Option 2 : 0.2

Turbomachinery Question 9 Detailed Solution

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Explanation:

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 __________.

  1. above or below the water surface, depends on unit speed of the turbine
  2. always above the water surface
  3. always immersed in water
  4. may either be below or above the water surface

Answer (Detailed Solution Below)

Option 3 : always immersed in water

Turbomachinery Question 10 Detailed Solution

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Explanation:

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:

  1. turbines
  2. pumps
  3. dams
  4. jets

Answer (Detailed Solution Below)

Option 2 : pumps

Turbomachinery Question 11 Detailed Solution

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Explanation:

Pump: 

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

  1. \(N_{s}=\frac{N\sqrt{Q}}{H_{m}^{5/4}}\)
  2. \(N_{s}=\frac{N\sqrt{P}}{H_{m}^{3/4}} \)
  3. \(N_{s}=\frac{N\sqrt{Q}}{H_{m}^{3/4}}\)
  4. \(N_{s}=\frac{N\sqrt{P}}{H_{m}^{5/4}}\)

Answer (Detailed Solution Below)

Option 3 : \(N_{s}=\frac{N\sqrt{Q}}{H_{m}^{3/4}}\)

Turbomachinery Question 12 Detailed Solution

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Explanation:

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?

  1. Pelton wheel
  2. Francis turbine
  3. Kaplan turbine
  4. Tyson wheel

Answer (Detailed Solution Below)

Option 2 : Francis turbine

Turbomachinery Question 13 Detailed Solution

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Non-dimensional specific speed is given by

\({N_S} = \frac{{N\sqrt P }}{{{{\left( {gH} \right)}^{\frac{5}{4}}} \cdot \sqrt \rho }}\)

F1 Eng Arbaz 11-1-24 D1
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

  1. 65%
  2. 55%
  3. 52.5%
  4. 62.5%

Answer (Detailed Solution Below)

Option 4 : 62.5%

Turbomachinery Question 14 Detailed Solution

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Explanation:

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.

\({\eta _m} = \frac{{{\rm{Power\;at\;the\;impeller}}}}{{{\rm{Power\;at\;the\;shaft}}}} = \frac{{\frac{W}{g}\left( {\frac{{{V_{w2}}{u_2}}}{{1000}}} \right)}}{{{\rm{SP}}}}\)

Identify the false statement from the following:

The specific speed of the turbine increases with _______

  1. increase in shaft speed
  2. increase in discharge
  3. decrease in gravitational acceleration
  4. increase in head

Answer (Detailed Solution Below)

Option 4 : increase in head

Turbomachinery Question 15 Detailed Solution

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Concept:

The specific speed of a turbine is defined as, the speed of a geometrically similar Turbine that would develop unit power when working under a unit head (1m head). Is prescribed by the relation 

 \({N_s} = \frac{{N\sqrt P }}{{{H^{\frac{5}{4}}}}}\) 

For Pump, \({N_s} = \frac{{N\sqrt Q }}{{{H^{\frac{3}{4}}}}}\)

Where,

Ns = specific speed, N = speed in rpm, P = power generated, and H = total head developed by the turbine.

Clearly from the formula, it can be stated that

\({{\rm{N}}_{\rm{S}}} ∝ \frac{{{\rm{1}} {\rm{}} }}{{{{\left( {\rm{H}} \right)}^{\frac{5}{4}}}}}\)

Specific speed will decrease if H increases.

Also,

As gravity acceleration decreases the resistance decreases; discharge increases and hence specific speed increases.

Mistake Points It has been asked to identify the false statement from the given options

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