Centrifugal Pump MCQ Quiz in मल्याळम - Objective Question with Answer for Centrifugal Pump - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Closed impellers (Two-sides shrouded):

• In the closed or shrouded impellers, the vanes are covered with shrouds (side plates) on both sides
• The back shroud is mounted into the shaft and the front shroud is coupled by the vanes
• This ensures full capacity operation with high efficiency for a prolonged running period
• This type of impeller is meant to pump only clear water, hot water and acids

Semi-open impeller (One-side shrouded):

• It has a plate (shroud) only on the backside
• The design is adapted to industrial pump problems which require a rugged pump to handle liquids containing fibrous material such as paper pulp, sugar molasses and sewage water etc.

Open impeller:

• In open impeller, no shroud or plate is provided on either side i.e. the vanes are open on both sides
• Such pumps are used where the pump has a very rough duty to perform i.e. to handle abrasive liquids such as a mixture of water, sand, pebbles and clay, wherein the solid contents may be as high as 25%.

Thus, centrifugal pumps dealing with water containing slurry and sewage​ have an open impeller.

Explanation:

The figure shows the impeller and outlet velocity triangle of a pump. Let us assume that the impeller is rotating in a clockwise direction such that the positive sign shows the leading face and the negative sign shows the trailing face. 

When liquid passes through the passage of impeller vanes which is rotating in a clockwise direction, high pressure is created on the leading face and low pressure is created on the trailing face. This difference in pressure is known as vane loading.

On the high-pressure side: 

• The liquid follows the blade contour and it leaves the blade tangentially.

On the low-pressure side: 

• The liquid leaves the vane with a certain circumferential component.

Hence liquid which should leave the impeller at β₂ ideally leaves the impeller at β'₂ in the actual case.

Due to this deviation in the flow path, the tangential component gets reduced from V_w2 to V'_w2 which is known as a slip in the centrifugal pump which results in the reduction of energy transfer.

Slip factor (SF):

It is the ratio of the ratio of actual & ideal values of the whirl velocity components at the exit of the impeller.

\(SF=\frac{V_{w2}^{'}}{V_{w2}}\)

Explanation:

Priming: 

• Priming is the process of removing air from the pump and suction line. 
• In this process, the pump is been filled with the liquid being pumped, and this liquid forces all the air, gas, or vapour contained in the passageways of the pump to escape out.
• Priming may be done manually or automatically. Not all pumps require priming but mostly do.
• Priming reduces the risk of pump damage during start-up as it prevents the pump impeller to becomes gas-bound and thus incapable of pumping the desired liquid.

Surging: Sudden drop in delivery pressure and violent aerodynamic pulsations. Surging usually starts to occur in the diffuser passages.

Concept:

• For fluids in incompressible regime of operation such as backward curved vanes are used for maximum efficiency for radial and forward vanes, the power requirement increases monotonically as discharge increase.
• But backward curved the power requirement peaks at maximum efficiency.

Explanation:

Characteristic curves are necessary to predict the behaviour and performance of the pump when the pump is working under different flow rates, heads and speeds.

If the speed is kept constant, the variation of manometric head, power and efficiency with respect to discharge gives the operating characteristic curves of a pump.

Concept:

Model laws of the pump:

\(\frac{Q}{ND^3}~=~constant\),

\(\frac{H_m}{N^2D^2}~=~constant\),

\(\frac{P}{N^3D^5}~=~constant\)

Calculation:

Given:

N₁ = 1000 rpm, H₁ = 40 m, N₂ = 3000 rpm

∵ \(\frac{H_m}{N^2D^2}~=~constant\)

⇒ \(\frac{H_2}{H_1}~=~(\frac{N_2}{N_1})^2\)

⇒ H₂ = \((\frac{3000}{1000})^2~×~40\)

⇒ H2 = 9 × 40 = 360 m

Concept:

Tangential velocity at inlet \({u_1} = \frac{{\pi {D_1}N}}{{60}}\)

\(\left( {\tan \theta } \right) = \frac{{{V_{f1}}}}{{{u_1}}}\)

Calculation:

Given:

N = 1200 rpm, Inner Diameter = 200 mm = 0.2 m, Outer Diameter = 400 mm

Now,

Tangential velocity at inlet \({u_1} = \frac{{\pi {D_1}N}}{{60}}\)

\(= \frac{{\pi × 0.2 × 1200}}{{60}} = 12.56\;{m}/{s}\)

Now,

∴ Vane angle at inlet \(\left( {\tan \theta } \right) = \frac{{{V_{f1}}}}{{{u_1}}} = \frac{{V_{f1}}}{{10.47}}\)

∴ tan 20° × 12.56 = V_f1

∴ Vf1= 4.57 m/s

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 rates, head and speed.

There are 2 types of Characteristics curves.

• Main characteristic curve 
• Operating characteristic curve 

Main characteristic curve:

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

Operating characteristic curve:

• This curve is plotted at a constant speed.

Concept:
As per the affinity law, the relationship between the power and diameter of the impeller is given by:

\(P \propto D^{3}\)
Where,

P is shaft power, D is the diameter of the impeller,
Given:
P1 = 10 hp, D1 = 125 mm, D2 = 250 mm
In the given question it is said that the only diameter is changed and other parameters are constant

\(\frac{P_{2}}{P_{1}} \) = \(\frac{D_{2}^{3}}{D_{1}^{3}}\) 

\(P_{2} = 10\times \frac{250^{3}}{125^{3}}\) = 80 hp

Important Points The three Affinity Laws describe how changes in pump shaft speed or impeller diameter affect different aspects of pump performance:

Affinity Law 1: Flow Proportionality - Flow rate (Q) changes proportionally to changes in shaft speed (N) or impeller diameter (D). Specifically, if shaft speed or impeller diameter increases by a certain percentage, the flow rate increases by the same percentage.

\({Q_1 \over Q_2} = {N_1 \over N_2} = {D_1 \over D_2} \)

Affinity  Law 2: Pressure Proportionality - Pressure or head (H) changes proportionally to the square of the change in shaft speed or impeller diameter. For example, a 10% increase in shaft speed or diameter results in a 21% increase in pressure or head.

\({H_1 \over H_2} = {({N_1 \over N_2})}^2 OR\ {({D_1 \over D_2})}^2\)

Affinity  Law 3: Power Proportionality - Power (P) changes proportionally to the cube of the change in shaft speed or impeller diameter. Thus, a 10% increase in shaft speed or diameter leads to a 33.1% increase in power required.

\({P_1 \over P_2} = {({N_1 \over N_2})}^3 OR\ {({D_1 \over D_2})}^3\)

Explanation:

A Centrifugal pump works on the principle of force vortex where an external force is applied to the impeller with the help of a prime mover.

In a pump, there are two important parts, the first is the impeller which generates velocity through rotation and the second is the casing which converts this velocity into pressure by the change in the area.

When fluid is inside the impeller, then the speed of the fluid experiences a rotational motion because of the torque provided by the prime mover, and the flow is characterized as a forced vortex flow.

When the fluid comes out from the rotating impeller at that time fluid also has a vortex motion because of the inertia of the fluid but since the external torque is absent outside the casing therefore it becomes free vortex flow.