13.2 Pumps & Hydraulic Machinery

13.2 Pumps & Hydraulic Machinery

Pumps are fluid machinery that transfer energy to liquids to increase their static pressure, elevation, and velocity head. For mechanical engineering licensure candidates, pump system hydraulics, head calculations, affinity laws, and cavitation prevention are paramount.

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1. Classification of Pumps

Pumps are broadly divided into two main fluid dynamic categories:

A. Dynamic (Kinetic) Pumps

Energy is continuously added to the liquid by rotating impellers to increase fluid velocity, which is subsequently converted into static pressure head in a volute casing or diffuser.

• Centrifugal Pumps (Radial Flow): Fluid enters axially at the impeller eye and discharges radially outward. Used for high head and moderate flow.
• Mixed Flow Pumps: Fluid flow contains both radial and axial velocity components. Used for medium heads and high flows.
• Axial Flow (Propeller) Pumps: Fluid moves parallel to the shaft axis. Developed for low heads ($H < 10\text{ m}$) and very large flow rates.

B. Positive Displacement (PD) Pumps

Fluid is trapped in fixed volume chambers and physically pushed from suction to discharge. Volume flow is virtually independent of system backpressure.

• Reciprocating Pumps: Piston, plunger, or diaphragm designs.
• Rotary Pumps: Gear, lobe, sliding vane, or screw pumps. Preferred for high-viscosity oils and extreme pressures.

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2. Centrifugal Pump Performance & Energy Equations

Total Dynamic Head ($TDH$)

Applying Bernoulli's energy equation between the suction water surface ($s$) and discharge reservoir surface ($d$): $TDH = h_s + h_d + h_{fs} + h_{fd} + \frac{V_d^2 - V_s^2}{2g}$ Where:

• $h_s$: Static suction lift (if water level is below pump centerline) or static suction head (negative lift, if water level is above pump).
• $h_d$: Static discharge height above pump centerline.
• $h_{fs}, h_{fd}$: Friction head losses in suction and discharge piping systems (including valves and fittings).
• $\frac{V_d^2 - V_s^2}{2g}$: Velocity head difference between discharge and suction nozzles.

Hydraulic Power & Efficiency

• Water Horsepower ($WHP$): The theoretical power imparted to the fluid: $WHP = \frac{\gamma \cdot Q \cdot TDH}{1000} \text{ (kW)} = \frac{Q \text{ (gpm)} \times TDH \text{ (ft)} \times SG}{3960} \text{ (HP)}$ Where $\gamma = 9.81\text{ kN/m}^3$ for water, $Q$ is flow rate in $\text{m}^3/\text{s}$, and $TDH$ is in meters.
• Brake Horsepower ($BHP$): The shaft mechanical power required to drive the pump shaft: $BHP = \frac{WHP}{\eta_p}$ Where $\eta_p$ is pump efficiency ($0 < \eta_p < 1$).
• Electrical Motor Input Power ($P_{in}$): $P_{in} = \frac{BHP}{\eta_m}$

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3. Pump Affinity Laws

The affinity laws govern centrifugal pump performance variations when changing rotational speed ($N$) or impeller diameter ($D$).

Case 1: Constant Impeller Diameter ($D = \text{const}$), Variable Speed ($N_1 \to N_2$)

$\frac{Q_2}{Q_1} = \frac{N_2}{N_1}, \quad \frac{H_2}{H_1} = \left(\frac{N_2}{N_1}\right)^2, \quad \frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^3$

Case 2: Constant Speed ($N = \text{const}$), Variable Impeller Diameter ($D_1 \to D_2$)

$\frac{Q_2}{Q_1} = \frac{D_2}{D_1}, \quad \frac{H_2}{H_1} = \left(\frac{D_2}{D_1}\right)^2, \quad \frac{P_2}{P_1} = \left(\frac{D_2}{D_1}\right)^3$

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4. Specific Speed & Impeller Geometry

Pump specific speed ($N_{sp}$) defines the geometric shape of the impeller regardless of pump size: $N_{sp} = \frac{N \sqrt{Q}}{H^{3/4}}$ (Using $Q$ in gpm, $H$ in ft, $N$ in rpm):

• Radial Impellers: $N_{sp} = 500 - 4000$ (High head, narrow impeller).
• Mixed Flow Impellers: $N_{sp} = 4000 - 8000$ (Medium head, medium width).
• Axial / Propeller Impellers: $N_{sp} > 8000$ (Low head, wide open propeller).

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5. Net Positive Suction Head (NPSH) & Cavitation

Cavitation Mechanism

Cavitation occurs when the absolute local pressure at the suction inlet (impeller eye) drops below the saturated vapor pressure ($P_v$) of the liquid at that operating temperature. Liquid flashes into vapor bubbles. As bubbles move into higher pressure regions of the impeller, they violently collapse, causing severe acoustic pitting, noise, vibration, and loss of pump discharge head.

NPSH Formulations

• Available NPSH ($NPSH_{avail}$): Set by suction piping system hydraulics: $NPSH_{avail} = \frac{P_{atm}}{\gamma} - \frac{P_v}{\gamma} - h_s - h_{fs}$ (If suction level is above pump centerline, $+ h_s$ is used instead of $- h_s$).
• Required NPSH ($NPSH_{req}$): Specified by manufacturer test curves to prevent cavitation.
• Cavitation Prevention Criterion: $NPSH_{avail} \ge NPSH_{req} + 0.5\text{ m}$

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6. Pumps Operating in Series vs. Parallel

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Worked Step-by-Step Pump Design Calculation

Problem: A centrifugal pump delivers $Q = 0.08\text{ m}^3/\text{s}$ of water ($\rho = 1000\text{ kg/m}^3, \gamma = 9.81\text{ kN/m}^3, P_v = 3.17\text{ kPa}$) from an open sump ($P_{atm} = 101.3\text{ kPa}$). The suction static lift is $h_s = 3.5\text{ m}$ and suction friction loss is $h_{fs} = 1.2\text{ m}$. The discharge static elevation is $h_d = 28.5\text{ m}$, discharge friction loss is $h_{fd} = 4.8\text{ m}$, and discharge velocity is $V_d = 2.5\text{ m/s}$. The pump efficiency is $\eta_p = 78%$ and motor efficiency is $\eta_m = 90%$. Calculate (a) Total Dynamic Head, (b) Water Horsepower, (c) Motor Input Power, and (d) $NPSH_{avail}$.

Step-by-Step Solution:

1. Calculate Total Dynamic Head ($TDH$): $TDH = h_s + h_d + h_{fs} + h_{fd} + \frac{V_d^2}{2g} = 3.5 + 28.5 + 1.2 + 4.8 + \frac{2.5^2}{2(9.81)}$ $TDH = 38.0 + 0.3185 = 38.3185\text{ m}$

2. Calculate Water Horsepower ($WHP$): $WHP = \frac{\gamma \cdot Q \cdot TDH}{1000} = \frac{9810\text{ N/m}^3 \times 0.08\text{ m}^3/\text{s} \times 38.3185\text{ m}}{1000} = 30.072\text{ kW} = 40.33\text{ HP}$

3. Calculate Brake Horsepower ($BHP$): $BHP = \frac{WHP}{\eta_p} = \frac{30.072\text{ kW}}{0.78} = 38.554\text{ kW} = 51.70\text{ HP}$

4. Calculate Motor Electrical Power Input ($P_{in}$): $P_{in} = \frac{BHP}{\eta_m} = \frac{38.554\text{ kW}}{0.90} = 42.838\text{ kW}$

5. Calculate Available Net Positive Suction Head ($NPSH_{avail}$): $\frac{P_{atm}}{\gamma} = \frac{101.3 \times 10^3\text{ Pa}}{9810\text{ N/m}^3} = 10.326\text{ m}$ $\frac{P_v}{\gamma} = \frac{3.17 \times 10^3\text{ Pa}}{9810\text{ N/m}^3} = 0.323\text{ m}$ $NPSH_{avail} = 10.326\text{ m} - 0.323\text{ m} - 3.5\text{ m} - 1.2\text{ m} = 5.303\text{ m}$ Check: If manufacturer $NPSH_{req} = 3.8\text{ m}$, then $NPSH_{avail} = 5.303\text{ m} > 3.8\text{ m}$, ensuring cavitation-free operation.