Unit 5 of ME 3003 (Mechanical/Automobile Engineering)

Unit 5: Centrifugal and Reciprocating Pumps for your ME 3003 (Mechanical/Automobile Engineering) course. These are short notes for revision purpose. please refer you Reference book & College study materials for complete study.

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5. CENTRIFUGAL PUMPS

A centrifugal pump is a type of pump that uses rotational energy from a motor or engine to impart kinetic energy to a fluid, increasing its velocity and converting it into pressure energy.

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5.1 Principle, Working, and Applications of Centrifugal Pumps

Principle of Operation:

• The centrifugal pump works based on the centrifugal force. When the impeller (a rotating disc with curved blades) spins, it imparts velocity to the fluid (water or other liquids).
• The liquid enters the pump through the inlet (suction) and is accelerated by the rotating impeller. The velocity increases as the fluid moves radially outward from the center of the impeller.
• As the fluid exits the impeller, the kinetic energy is converted into pressure energy in the volute casing (or diffuser), increasing the fluid pressure.

Working:

1. The fluid enters the pump casing through the inlet.
2. The impeller, rotating at high speed, accelerates the fluid.
3. As the fluid moves outward through the impeller, its velocity increases.
4. The increased velocity is converted into pressure by the diffuser or volute casing, where the fluid exits with higher pressure and lower velocity.

Applications:

• Centrifugal pumps are widely used for:
  • Water supply (municipal and industrial)
  • Irrigation systems
  • Heating and cooling systems
  • Chemical and petrochemical industries (for pumping chemicals)
  • Firefighting (for high-pressure water flow)

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5.2 Derivation for Work Done and Efficiency of Centrifugal Pumps

Work Done by Centrifugal Pump:

• The work done by a centrifugal pump is the energy imparted to the fluid, which is equal to the increase in the head (or energy per unit weight) as the fluid moves through the pump.
• The head generated by the pump is the difference in pressure between the inlet and outlet of the pump, typically represented as H.

The work done (W) is given by the formula:

$$W = \rho \cdot Q \cdot g \cdot H$$

Where:

• $\rho$ = Density of fluid (kg/m³)
• $Q$ = Flow rate (m³/s)
• $g$ = Gravitational acceleration (9.81 m/s²)
• $H$ = Head generated by the pump (m)

Efficiency of a Centrifugal Pump:

Efficiency ($\eta$) is the ratio of useful energy output to the total energy input:

$$\eta = \frac{\text{Hydraulic Power Output}}{\text{Input Power}} = \frac{\rho \cdot Q \cdot g \cdot H}{\text{Power supplied to pump}}$$

The total efficiency of the centrifugal pump includes mechanical, volumetric, and overall efficiency, which accounts for the losses in the pump.

Mechanical Efficiency is calculated as:

$$\eta_m = \frac{\text{Mechanical Power Output}}{\text{Input Power to Pump}}$$

Volumetric Efficiency is the ratio of the actual volume of fluid pumped to the theoretical volume.

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5.3 Reciprocating Pumps

A reciprocating pump is a positive displacement pump that uses the back-and-forth motion of a piston or plunger to move fluids. It operates by drawing the fluid in during the suction stroke and displacing it during the discharge stroke.

5.3.1 Working Principle and Applications of Reciprocating Pumps:

• Working Principle:
  • During the suction stroke, the piston or plunger moves backward, creating a vacuum that draws the fluid into the cylinder through the inlet valve.
  • During the discharge stroke, the piston moves forward, closing the inlet valve and opening the discharge valve, forcing the fluid out of the cylinder under pressure.
• Applications:
  • Water supply systems (especially where high pressure is required).
  • Chemical plants (for precise flow of chemicals).
  • Oil industry (for pumping viscous fluids).
  • Fuel injection systems in engines.
  • Medical devices (such as infusion pumps).

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5.4 Concept of Slip in Reciprocating Pumps

Slip in reciprocating pumps refers to the difference between the theoretical and actual volume of fluid pumped during one complete stroke. It occurs due to leakage and imperfect operation of the pump.

• Slip is defined as:

$$\text{Slip} = \frac{\text{Theoretical Displacement} - \text{Actual Displacement}}{\text{Theoretical Displacement}} \times 100$$

• Slip increases when the pump operates at lower speeds or with high viscosity fluids.
• Negative slip (also called "over-lapping") occurs when more fluid is pumped than the theoretical displacement due to higher pressure at the outlet than expected.

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5.5 Cavitation and Separation in Pumps

Cavitation in Pumps:

• Cavitation is a phenomenon where the pressure in the pump falls below the vapor pressure of the fluid, leading to the formation of vapor bubbles.
• These bubbles collapse violently when they move into higher-pressure areas, generating shockwaves that can damage the pump components.

Causes of Cavitation:

1. Low pressure at the inlet of the pump.
2. High pump speeds.
3. Improper pump operation (e.g., operating at too high a flow rate).

Prevention of Cavitation:

• Proper NPSH (Net Positive Suction Head) should be maintained.
• Ensure that the inlet pressure remains above the vapor pressure of the fluid.
• Design pumps with larger inlet pipes and lower flow velocities.

Separation in Pumps:

• Separation refers to the loss of fluid from the pump blades due to the fluid not adhering to the curved surface, which can cause a loss in efficiency and potential damage.
• This typically happens in centrifugal pumps if the flow rate is too high, or if the design does not maintain enough pressure.

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5.6 Numericals on Centrifugal Pumps

Example 1: Calculating Power Required for a Centrifugal Pump

A centrifugal pump has a flow rate of 0.5 m³/s, operates at a head of 25 meters, and the density of water is 1000 kg/m³. Calculate the power required to drive the pump.

Solution:

• Given:
  • $Q = 0.5 \, \text{m}^3/\text{s}$
  • $H = 25 \, \text{m}$
  • $\rho = 1000 \, \text{kg/m}^3$

The power required is given by:

$$P = \rho \cdot Q \cdot g \cdot H$$ $$P = 1000 \times 0.5 \times 9.81 \times 25 = 122,625 \, \text{W} = 122.6 \, \text{kW}$$

Thus, the power required to drive the pump is 122.6 kW.

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Example 2: Calculation of Efficiency of a Centrifugal Pump

A centrifugal pump delivers 0.4 m³/s of water with a head of 30 meters. The efficiency of the pump is 80%, and the power supplied to the pump is 150 kW. Calculate the hydraulic power output.

Solution:

• Given:
  • $Q = 0.4 \, \text{m}^3/\text{s}$
  • $H = 30 \, \text{m}$
  • $\eta = 80\% = 0.80$
  • Power supplied to the pump = 150 kW

The hydraulic power output can be calculated as:

$$P_{\text{hydraulic}} = \eta \times \text{Input Power} = 0.80 \times 150 = 120 \, \text{kW}$$

Thus, the hydraulic power output is 120 kW.

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Summary of Key Concepts

• Centrifugal pumps use rotational energy to impart kinetic energy to a fluid, which is then converted into pressure energy.
• Reciprocating pumps are positive displacement pumps that use piston or plunger motion to move fluid.
• Slip occurs in reciprocating pumps when there is a difference between the theoretical and actual displacement due to leakage.
• Cavitation and separation can damage pumps and reduce efficiency; proper NPSH and design can help prevent these issues.
• Power and efficiency calculations for pumps are essential for selecting the right pump for a given application.