Reciprocating Pump MCQ Quiz - Objective Question with Answer for Reciprocating Pump - Download Free PDF

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

Explanation:

To determine the manometric efficiency of the pump, we need to use the formula for manometric efficiency (ηm" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">ηmηm $\eta_m$ ). The manometric efficiency is defined as the ratio of the measured manometric head to the theoretical manometric head.

The formula for manometric efficiency is given by:

ηm=Measured Manometric HeadTheoretical Manometric Head" id="MathJax-Element-2-Frame" role="presentation" style="position: relative;" tabindex="0">ηm=Measured Manometric HeadTheoretical Manometric Headηm=Measured Manometric HeadTheoretical Manometric Head $\eta_m = \frac{\text{Measured Manometric Head}}{\text{Theoretical Manometric Head}}$

In this case, the theoretical manometric head is 30 meters, and the measured head is 27 meters.

Substituting these values into the formula, we get:

ηm=27meters30meters" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">ηm=27meters30metersηm=27meters30meters $\eta_m = \frac{27 \, \text{meters}}{30 \, \text{meters}}$

ηm=2730" id="MathJax-Element-4-Frame" role="presentation" style="position: relative;" tabindex="0">ηm=2730ηm=2730 $\eta_m = \frac{27}{30}$

ηm=0.9" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0">ηm=0.9ηm=0.9 $\eta_m = 0.9$

Therefore, the manometric efficiency of the pump is 0.9 or 90%, which corresponds to option 3.

Now, let's analyze the other options to understand why they are incorrect:

• Option 1: 0.75 - This value would be obtained if the measured head was 22.5 meters (0.75 × 30 meters), which is not the case here.
• Option 2: 0.85 - This value would be obtained if the measured head was 25.5 meters (0.85 × 30 meters), which is not the case here.
• Option 4: 0.8 - This value would be obtained if the measured head was 24 meters (0.8 × 30 meters), which is not the case here.

All these options do not match the given data, which indicates that the correct answer is indeed option 3.

Important Information:

Understanding manometric efficiency is crucial for evaluating the performance of pumps. Manometric efficiency helps in determining how effectively a pump converts the energy supplied to it into useful work. High manometric efficiency indicates that the pump is effectively converting the input energy into the desired output in terms of head, while low manometric efficiency suggests energy losses within the pump system. The efficiency can be impacted by various factors such as mechanical losses, hydraulic losses, and leakage, which need to be minimized to ensure optimal pump performance. Regular maintenance and proper design can help in achieving and maintaining high manometric efficiency.

Explanation:

Connecting rod

1. In a reciprocating pump, the connecting rod connects the piston to the crankshaft.
2. It transmits the rotational motion of the crankshaft to the linear motion of the piston, enabling the pump to move fluid through the system.

 Additional Information

Reciprocating Pump:

1. A reciprocating pump is a positive displacement pump that uses a piston to move fluid.

2. It operates by the back-and-forth (reciprocating) motion of the piston inside a cylinder, which creates a vacuum to draw in fluid during the suction stroke and forces it out during the discharge stroke.

Connecting Rod:

1. The connecting rod plays a crucial role in converting the rotary motion of the crankshaft into the linear reciprocating motion of the piston.

2. It connects the piston to the crankshaft and is typically made of strong materials like steel.

3. During the operation of the pump, the crankshaft rotates, and the connecting rod transmits the rotational force to the piston, causing it to move back and forth within the cylinder.

Crankshaft and Bearings:

1. The crankshaft is responsible for converting rotational motion into the reciprocating motion of the piston.

2. Bearings support the crankshaft and reduce friction during its rotation.

Suction and Delivery Pipes:

1. The suction pipe brings fluid into the pump, and the delivery pipe carries the fluid out of the pump once it’s displaced by the piston.

2. These pipes are connected to the inlet and outlet of the pump, but they don’t connect the piston to the crankshaft.

Applications:

1. Reciprocating pumps are commonly used in applications requiring precise fluid transfer, such as in hydraulic systems, high-pressure water pumps, and in industries where accurate flow control is important.

Concept:

Reciprocating pumps are positive displacement pumps that use a piston or plunger to move fluid. Understanding the dynamics and losses associated with reciprocating pumps is essential for efficient design and operation.

∴ Frictional losses, inertia effects, and slip are critical factors to consider in reciprocating pumps.

Statement (1): In a reciprocating pump, the velocity of the fluid in the suction or delivery pipe varies with the position of the piston during its stroke. The fluid velocity is maximum at the middle of the piston stroke, as this is when the piston reaches its highest speed. Frictional losses are proportional to the square of the fluid velocity: h_f ∝ v². Since the fluid velocity is highest at the middle of the stroke, the frictional losses in the pipe are also highest at this point.

Statement (2): Maximum inertia effect occurs in place with zero frictional losses. This is correct because the inertia effect is related to the acceleration and deceleration of the piston, which is maximum where there is no frictional loss to counteract it.

Statement (3): Negative slip may occur when the delivery head is high. This is incorrect because negative slip usually occurs when the pump is running at a higher speed, causing the actual discharge to be higher than the theoretical discharge due to the fluid's inertia.

Concept:

Slip:

• It is the difference between the theoretical discharge and actual discharge.

Slip = Qthe - Qact

Positive slip:

• When the theoretical discharge is more than actual discharge is known as positive slip. i.e.  Qthe > Qact

Negative slip:

• When the theoretical discharge is less than actual discharge is known as negative slip. i.e.  Qthe < Qact

Conditions of "Negative slip":

• Due to High speed
• Due to short delivery pipe
• Due to long suction pipe

This happen because of inertia pressure in suction pipe will be large as compared to pressure of the delivery valve. This causes delivery valve open before suction stroke is completed.  

Coefficient of discharge

Cd = \(\frac{Q_{act}}{Q_{the}} \) and % slip = \(\frac{{Q_{the}}-~Q_{act}}{Q_{the}}\)

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.​

Concept:

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.

Discharge through a pump per second:

\(Q = \frac{{ALN}}{{60}}\)

Discharge for a double-acting pump:

\(Q = 2\frac{{ALN}}{{60}}\)

where, A = Area of piston; L = Stroke length

Hnet = hd + hs = Total height through which water is lifted

Hence, the power required is given by

P = Q ρ g Hnet

P = Q ρ g (hd + hs) (in W) = \(\frac{{wQ\space ({H_s}\space +\space {H_d})}}{{75}}\)(in h.p.)

The hydraulic machines which convert the mechanical energy into hydraulic energy are called pumps.

Classification of pump.

Explanation:

• The air vessel, in a reciprocating pump, is a cast iron closed chamber having an opening at its base.
• These are fitted to the suction pipe and delivery pipe close to the cylinder of the pump.

The vessels are used for the following purposes,

• To get continuous supply of liquid at a uniform rate.
• To save the power required to drive the pump. This is due to the fact that by using air vessels, the acceleration and friction heads are reduced. Thus, the work is also reduced.

Note:
It may be noted that by fitting an air vessel to the reciprocating pump, the saving of work and subsequently the power is about 84.8 % in case of a single acting reciprocating pump and 39.2 % in case of double acting reciprocating pump.

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:

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:

Slip: It is the difference between the theoretical discharge and actual discharge.

Slip = Q_the - Q_act

Positive slip: When the theoretical discharge is more than actual discharge is known as positive slip. i.e.  Qthe > Qact

Negative slip: When the theoretical discharge is less than actual discharge is known as negative slip. i.e.  Qthe < Qact

Conditions of "Negative slip":

• Due to High speed
• Due to short delivery pipe
• Due to long suction pipe

This happen because of inertia pressure in suction pipe will be large as compared to pressure of the delivery valve. This causes delivery valve open before suction stroke is completed.  

Coefficient of discharge

C_d = \(\frac{Q_{act}}{Q_{the}} \) and % slip = \(\frac{{Q_{the}}-~Q_{act}}{Q_{the}}\)

Explanation:

Air vessel:

• It is a closed chamber fitted on the suction as well as delivery side, near the pump cylinder to reduce the accelerating head.
• To reduce the acceleration head we need to reduce the length of the suction pipe and the length of the delivery pipe in which fluctuation of velocity occurs.
• This is done by fitting air vessels as close as possible.
• Thus between air vessel and cylinder only fluctuating velocity will occur. Below air vessel on the suction side and above air vessel on delivery side velocity will be constant.

Functions of air vessel:

1) On the suction side:

• Reduction of the possibility of separation of flow
• The pump can run at a higher speed.
• The length of the suction pipe below the air vessel can be increased.

2) On the delivery side:

• A constant rate of discharge can be ensured.
• Save power required to drive the pump (For single-acting pump, percentage work saved is 84.8% and for double-acting pumps the percent work save is 39.2%)

Explanation:

In the case of a centrifugal pump, the power is transmitted from the shaft of the pump to the impeller and then from the impeller to the water. 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.

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

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.

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

\(\rm n_m=\frac{p-p_{mech\ loss}}{p}\)

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:

ηo = 88%, ηvol = 92%

88 = 92 × η_m

η_m = 95.65%

Concept:

Discharge through a reciprocating pump:

• The discharge in m3/min through a single acting reciprocating pump is given as
• Q = Volume of water delivered in one revolution × number of revolutions per minute
• Q = V × N
• Q = (Length of stroke × Area) × N
• Q = LAN

For a single-acting reciprocating pump

• Q = LAN

For a double-acting reciprocating pump

• Q = 2LAN

Calculation:

Given:

L = 250 mm = 0.25 m, D = 150 mm = 0.15 m, N = 60 rpm

\(A=\frac{\pi}{4} \times D^2\)

\(A=\frac{\pi}{4} \times 0.15^2=0.0176~m^2\)

Q = 2 × L × A × N

Q = 2 × 0.25 × 0.0176 × 60 = 0.53 m3/min = 530 liter/min = 8.8 liter/sec

Hence the required discharge will be 8.8 liter/sec