14.5 Pump Hydraulics & Horsepower Calculations

10.5 Pump Hydraulics & Horsepower Calculations

Pumping systems represent the largest single consumer of electrical energy in water and wastewater facilities. Operators must understand hydraulic head principles, pump performance curves, and the horsepower energy chain to evaluate pump health, diagnose mechanical wear, and calculate energy operating costs.

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1. Total Dynamic Head (TDH)

Total Dynamic Head (TDH) is the total equivalent height in feet of liquid that a pump must overcome to move fluid from a suction source to a discharge point at a specified flow rate. TDH consists of three distinct hydraulic components:

$\text{TDH (ft)} = \text{Total Static Head} + \text{Total Friction Loss} + \text{Velocity Head}$

Components of TDH:

1. Total Static Head: The net vertical height difference (in feet) between the suction water level and the discharge point.
   • Static Suction Lift: Vertical distance when the liquid level is below the pump centerline.
   • Static Suction Head: Vertical distance when the liquid level is above the pump centerline.
   • $\text{Total Static Head} = \text{Static Discharge Elevation} - \text{Static Suction Elevation}$.
2. Friction Loss ($h_f$): Energy lost due to liquid turbulence against internal pipe walls, valves, elbows, tees, and check valves.
3. Velocity Head ($h_v$): Energy required to accelerate liquid from rest to operating pipe velocity ($h_v = \frac{V^2}{2g}$, where $g = 32.2\text{ ft/sec}^2$).

Measuring TDH in the Field

Operators determine actual field TDH by reading pressure gauges on the pump suction and discharge manifolds:

$\text{TDH (ft)} = \text{Discharge Head (ft)} - \text{Suction Head (ft)}$ $\text{Discharge Head (ft)} = \text{Discharge Pressure (psi)} \times 2.31\text{ ft/psi}$ $\text{Suction Head (ft)} = \text{Suction Pressure (psi)} \times 2.31\text{ ft/psi}$ Note: If the suction gauge reads a vacuum (negative pressure in inches of Mercury, in. Hg), multiply in. Hg by 1.13 ft/in. Hg and add this suction lift value to the discharge head.

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2. The Horsepower Energy Chain

Energy conversion through a pumping system follows a strict descending hierarchy due to mechanical and electrical losses. Energy moves from the electrical grid to the motor, from the motor shaft to the pump impeller, and finally from the impeller into the liquid.

$\text{Motor Horsepower (MHP)} > \text{Brake Horsepower (BHP)} > \text{Water Horsepower (WHP)}$

Electrical Power (kW) ➔ Motor (Motor Eff) ➔ Pump Shaft (BHP, Pump Eff) ➔ Liquid (WHP)

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A. Water Horsepower (WHP)

Water Horsepower is the theoretical work output imparted directly into the liquid to lift a given flow rate against a specified head.

$\text{WHP} = \frac{\text{Flow (gpm)} \times \text{TDH (feet)}}{3,960}$

Derivation of the Constant 3,960:

By definition, 1 Horsepower = 33,000 ft-lbs/minute. Since 1 gallon of water = 8.34 lbs: $\text{Work (ft-lbs/min)} = \text{Flow (gpm)} \times 8.34\text{ lbs/gal} \times \text{TDH (ft)}$ $\text{WHP} = \frac{\text{gpm} \times 8.34 \times \text{TDH}}{33,000} = \frac{\text{gpm} \times \text{TDH}}{\frac{33,000}{8.34}} = \frac{\text{gpm} \times \text{TDH}}{3,956.8} \approx \frac{\text{gpm} \times \text{TDH}}{3,960}$

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B. Brake Horsepower (BHP)

Brake Horsepower is the actual mechanical horsepower required at the pump shaft to overcome mechanical friction, impeller drag, and hydraulic turbulence inside the pump volute.

$\text{BHP} = \frac{\text{Water Horsepower (WHP)}}{\text{Pump Efficiency (decimal)}}$ $\text{BHP} = \frac{\text{Flow (gpm)} \times \text{TDH (feet)}}{3,960 \times \text{Pump Efficiency (decimal)}}$

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C. Motor Horsepower (MHP)

Motor Horsepower is the electrical power input required at the motor electrical terminals to account for electrical resistance, magnetic losses, and cooling fan drag inside the electric motor.

$\text{MHP} = \frac{\text{Brake Horsepower (BHP)}}{\text{Motor Efficiency (decimal)}}$ $\text{MHP} = \frac{\text{Flow (gpm)} \times \text{TDH (feet)}}{3,960 \times \text{Pump Eff} \times \text{Motor Eff}}$ $\text{Wire-to-Water Efficiency} = \text{Pump Efficiency} \times \text{Motor Efficiency}$

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3. Worked Math Examples: Horsepower Chain

Worked Example 1: Water Horsepower

Problem: A high-service centrifugal pump delivers 1,200 gpm against a Total Dynamic Head of 165 feet. Calculate the Water Horsepower (WHP).

Solution Step-by-Step: $\text{WHP} = \frac{1,200\text{ gpm} \times 165\text{ ft}}{3,960} = \frac{198,000}{3,960} = 50.0\text{ WHP}$

Worked Example 2: Complete Horsepower Chain & Motor Selection

Problem: A lift station pump must deliver 800 gpm against a TDH of 150 feet. Pump manufacturer curves indicate a pump efficiency of 80% (0.80) at this operating point. The drive motor has a rated efficiency of 90% (0.90). Calculate:

1. Water Horsepower (WHP)
2. Brake Horsepower (BHP)
3. Motor Horsepower (MHP)

Solution Step-by-Step:

1. Calculate WHP: $\text{WHP} = \frac{800\text{ gpm} \times 150\text{ ft}}{3,960} = \frac{120,000}{3,960} = 30.303\text{ WHP}$
2. Calculate BHP: $\text{BHP} = \frac{30.303\text{ WHP}}{0.80\text{ pump eff}} = 37.879\text{ BHP}$
3. Calculate MHP: $\text{MHP} = \frac{37.879\text{ BHP}}{0.90\text{ motor eff}} = 42.088\text{ MHP}$ Operator Decision: Motor sizes are standardized (30, 40, 50, 75 hp). An operator would select a 50 HP nameplate motor to supply the required 42.1 MHP without overloading the motor.

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4. Electrical Power Consumption and Operating Costs

To calculate electric utility power consumption and operating expenses:

$\text{Kilowatts (kW)} = \text{Motor Horsepower (MHP)} \times 0.746\text{ kW/hp}$ $\text{Daily Energy Consumption (kWh/day)} = \text{kW} \times \text{Operating Hours per Day}$ $\text{Daily Operating Cost (\$/day)} = \text{kWh/day} \times \text{Electricity Rate (\$/kWh)}$

Worked Example 3: Monthly Electrical Operating Cost

Problem: A booster pump operates continuously (24 hours/day) powered by a motor drawing 40.0 MHP. The electrical utility charges $0.10 per kWh. What is the daily electrical operating cost of this pump?

Solution Step-by-Step:

1. Convert MHP to Kilowatts: $\text{Power (kW)} = 40.0\text{ MHP} \times 0.746\text{ kW/hp} = 29.84\text{ kW}$
2. Calculate daily energy consumption in kilowatt-hours (kWh): $\text{Daily Energy} = 29.84\text{ kW} \times 24\text{ hours/day} = 716.16\text{ kWh/day}$
3. Calculate daily electrical cost: $\text{Daily Cost} = 716.16\text{ kWh/day} \times \$0.10\text{ /kWh} = \$71.616\text{ per day}$ Result: The pump operating cost is $71.62 per day (or approximately $2,148.48 per 30-day month).