Pumps MCQ Quiz in తెలుగు - Objective Question with Answer for Pumps - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Power required to run pump = ρ_w g Q (H + h_f)

Given:

g = 10 m/s², Q = 6 m³/min, H = 10 m, h_f = 10 m

Calculation:

Power required to run pump \(= {10^3} \times 10 \times \frac{6}{{60}}\left( {10 + 10} \right)\)

∴ P = 20 kW

Explanation:

Reciprocating pump​

• A reciprocating pump is a hydraulic machine which converts mechanical energy into hydraulic energy.
• Here a certain volume of liquid is collected in the enclosed volume and is discharged using pressure to the required application.
• Reciprocating pumps are more suitable for low volumes of flow at high pressures (low discharge and high head).
• Example of reciprocating pump:
  • Windmill pump
  • Hand pump
  • Axial piston pump
  • Radial piston pump

Advantages of a reciprocating pump:

• low efficiency.
• No priming is needed.
• Can deliver water at high pressure.
• Can work in a wide pressure range.
• The continuous rate of discharge

Disadvantages of reciprocating pump:

• More parts mean high initial cost
• High maintenance cost
• No uniform torque
• Low discharging capacity
• Pulsating flow
• Difficult to pump viscous fluid
• High wear in parts

Additional Information Submersible pump:

• Pumps that are designed to be submersible are placed within the reservoir of water that needs pumping out. As such, they are often used for drainage in floods, sewerage pumping, emptying ponds, or even as pond filters. Filtration pumps found inside fish tanks are a type of submersible pump.

Rotary pump:

• Rotary pumps are a type of positive displacement pump where for each revolution, a fixed volume of fluid is moved. These pumps are self-priming and provide near-constant delivered capacity no matter the pressure.

Axial flow pump:

• Axial pumps are used for the promotion of incompressible fluids and are employed for large volume flows at relatively low delivery heads. As with all types of centrifugal pumps, the energy transmission in axial flow pumps is carried out exclusively through flow-related processes.

Explanation:

Reciprocating Pump:

• A reciprocating pump is a positive displacement pump as sucks and raises the liquid by actually displacing it with a piston or plunger that executes a reciprocating motion in a closely fitted cylinder.
• The amount of liquid pumped is equal to the volume displaced by the piston.
• The reciprocating pump is best suited for relatively small capacities and high heads.

Single-Acting reciprocating pump:

• A single-acting reciprocating pump has one suction pipe and one delivery pipe.
• Initially, the crank is at the inner dead centre and rotates in the clockwise direction. As the crank rotates the piston moves towards the right and a vacuum is created on the left side of the piston. This vacuum causes the suction valve to open and consequently the liquid is forced from the sump to the left side of the piston.
• When the crank is outer dead centre the suction stroke is completed.
• When the crank turns from ODC to IDC, the piston moves inward to the left, and high pressure builds up in the cylinder.

Double-acting reciprocating pump:

• In a double-acting reciprocating pump, suction and delivery strokes occur simultaneously.
• When the crank rotates from IDC in a clockwise direction, a vacuum is created on the left side of the piston and the liquid is sucked in from the sump through valve S₁. At the same time, the liquid on the right side of the piston is pressed and high pressure causes the delivery valve D₂ to open and the liquid is passed on to the discharge tank. This operation continues till the crank reaches ODC.
• Because of continuous delivery strokes, a double-acting reciprocating pump gives a more uniform discharge.

Air vessels:

• An air vessel is a closed chamber containing compressed air in the upper part and liquid being pumped in the lower part.
• One air vessel is fixed on the suction pipe just near the suction valve and one is fixed on the delivery pipe.

The air vessels are used for the following purposes:

• To get a continuous supply of liquid at a uniform rate.
• To save the power required to drive the pump (By the use of air vessels, the acceleration and friction heads are considerably reduced and the work is also reduced).
• To run the pump at a much higher speed without any danger of separation (By fitting the air vessels as close to the pump as possible, the length of the pipe in which acceleration takes place is reduced due to which acceleration head is reduced, and the pump can run at a high speed without separation).

​

​The percentage of work saved in pipe friction by fitting air vessels in the case of a single-acting reciprocating pump is approximately 84.8% and in the case of double-acting, it is approximately 39.2 %.

Explanation:

If a centrifugal pump consists of two or more impellers, then the pump is called a multistage centrifugal pump.

It performs the following two functions:

i) To produce a high head = To achieve this, impellers are connected in series

Concept:

The following are the important efficiencies of a centrifugal pump:

1) Manometric Efficiency (ηman): It is the ratio of the manometric head to head imparted by the impeller to the water.

\({\eta _{man}} = \frac{{{H_m}}}{{\frac{{{V_{w2}}{u_2}}}{g}}} = \frac{{g{H_m}}}{{{V_{w2}}{u_2}}}\)

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.

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

3) Overall Efficiency (ηo): It is defined as a ratio of the power output of the pump to the power input to the pump.

ηo = ηman × ηm

Calculation:

Given:

ηman = 0.74, ηm = 0.68

ηo = ηman × ηm = 0.74 × 0.68 = 0.5032

The overall efficiency of the pump is 50.32%.

Explanation:

Theoretical discharge from a single-acting reciprocating pump:

\(Q_{th}={LAN\over 60}\)

Where L = Stroke length = 2 × crank radius

A = Area of Piston

N = Pump rotation in rpm

Calculation:

Given data:

Piston diameter (d) =15 cm or 0.15 m

Crank radius (r) = 15 cm or 0.15 m

Diameter of delivery pipe (D) = 10 cm

L = 0.30 m

Pump rotation (N) = 60 rpm

Actual discharge (Q_act) = 310 liters/minute

Lifting height (h) = 15 cm

Coefficient of discharge (C_d) =?

\(Q_{th}={0.30× 60×({\pi \over 4})× 0.15^2\over 60}\)

\(Q_{th}=0.00530\, m^3/s\)

given, \(Q_{act}=310\, liters/minute\)

1000 liters = 1 m³

\(Q_{act}={310\over 1000× 60}\, m^3/s\)

\(Q_{act}=0.00516\, m^3/s\)

\(Coefficient\, of\, discharge\,(C_d)={Actual\, discharge\,(Q_{act})\over Theoretical\, discharge\,(Q_{th})}\)

\(C_d={0.00516\over 0.00530}=0.9735\approx 0.974\)

\(C_{d}=0.974\)

Hence, the most appropriate answer is option 4.

Concept:

The centrifugal pump converts the mechanical energy into fluid energy. Also, it increases manometric head by converting the pressure head into manometric head. Hence, during its operation there is a chance that the pressure at the eye of the impeller may reaches to zero or negative pressure which may lead to the cavitation effect (if it reaches below vapour pressure). Hence to reduce this effect, it must be placed at certain height from the sump level called as suction height.

This minimum manometric head is required to make possible the suction which is represented by NPSH (net positive suction head).

Hence, the centrifugal pump height is such that the negative pressure does not reaches below vapour pressure.

Important points:

NPSH is defined by the Thomas’s cavitation number times the manometric head.

Cavitation is a low-pressure phenomenon which causes wear and tear of the equipment.

Explanation:

Characteristics curve of a centrifugal pump:

• These curves are plotted from the result of a test on the centrifugal pumps.
• It helps to predict the behavior and performance of the pump when the pump is working under different flow rate, head and speed.

There are 2 types of Characteristics curve.

• Main characteristic curve 
• Operating characteristic curve 

Main characteristic curve:

• This curve consist of variation of head, power, and discharge with respect to speed.

Operating characteristic curve:

• This curve is plotted at a constant speed.

Explanation:

Cavitation

• Cavitation is the phenomenon of the formation of vapor bubbles of a flowing liquid in a region where the pressure of the liquid falls below the vapor pressure of the fluid and sudden collapsing of these bubbles in the region of higher pressure.
• In centrifugal pumps, the cavitation may occur at the inlet of the impeller of the pump or at the suction side of the pump where the pressure is considerably low. So to avoid cavitation pressure on the suction side should be high. In order to determine whether cavitation will occur in any portion of the suction side of the pump, the critical value of Thoma’s cavitation factor is calculated.
• The hydraulic machines subjected to cavitation are reaction turbines and centrifugal pumps.

Thoma’s cavitation factor:

• Thoma's cavitation parameter (σ): It is the ratio of the Net Positive Suction Head (NPSH) to the total head.
• NPSH: It is defined as the net head developed at the suction port of the pump, in excess of the head due to the vapor pressure of the liquid at the temperature in the pump.

From the definition of σThoma,

\({\sigma _{Thoma}} = \frac{{{H_a} - {H_{vp}} - H_s}}{H}\)

• So, the correct answer is option 4.

Where Hvp is the vapor pressure head, Ha is the atmospheric pressure head, and Hs is the suction head

• If, Hs = Hs, max then σ = σC

​\({\sigma _{c}} = \frac{{{H_a} - {H_{vp}} - ​​H_{s, max}}}{H}\)

• Condition for no cavitation, Hs ≤ Hs, max

⇒ σ ≥ σC

• Where, σC is the critical cavitation factor
• The turbine manufacturer will specify the critical value of Toma's cavitation factor after performing testing. If the height of the Draft tube is increased then the pressure of the pressure at the exit of the turbine decreases and if it falls below the vapor pressure then the cavitation starts. The critical cavitation factor is the cavitation factor corresponding to the 10% drop in the efficiency of the turbine due to cavitation.

 Additional Information

• if Thoma's cavitation factor is greater than its critical value then there is no cavitation.
• The draft tube is used in the reaction turbine to convert the exit velocity head into the pressure head, the height of the draft tube should be such that pressure at the inlet of the draft tube is greater than vapor pressure.
• if pressure is less than vapor pressure then cavitation takes place.
• Draft tube is not used in impulse. impulse turbines have no cavitation because they are open to the atmosphere. everywhere there is atmospheric pressure. 

Concept:

Specific speed

• The specific speed of a centrifugal pump is defined as the speed of a geometrically similar pump which would deliver one cubic meter of liquid per second against a head of one meter.

\({N_s} = \frac{{N\sqrt Q }}{{H_m^{3/4}}}\)

Calculation:

Given:

Q = 625 ltr = 0.625 m³, H = 16 m, N = 800 rpm

\({N_s} = \frac{{N\sqrt Q }}{{H_m^{3/4}}}=\frac{800 \times \sqrt{0.625}}{16^{3/4}}=79.05=80\)

Specific speed for the turbine is given by:

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