13.3 Pumping Math: Total Dynamic Head, Horsepower, Efficiency & Power Cost

13.3 Pumping Math: Total Dynamic Head, Horsepower, Efficiency & Power Cost

Pumping is usually the largest electrical cost at a water or wastewater utility. WPI exams include pump and motor calculations in the equipment content areas, and the WPI formula sheet lists the horsepower relationships. This section builds on the head and pressure conversions in Section 13.1 and the pump curves and affinity laws in Section 11.1.

Total Dynamic Head (TDH)

Total dynamic head is the total energy, in feet of water, that the pump must add:

$$\text{TDH} = \text{Total static head} + \text{Friction and minor losses} \; (+ \text{velocity head, usually small})$$

• Static suction lift: the pump sits above the water surface it draws from (the lift adds to the static head).
• Static suction head: the water surface is above the pump centerline (it reduces the head the pump must add).
• Static discharge head: height from the pump centerline to the discharge water surface.
• Total static head is the difference in elevation between the discharge water surface and the source water surface.

Example 1. A pump lifts water from a wet well whose surface is 8 feet below the pump centerline and discharges into a tank whose surface is 92 feet above the pump. Friction and minor losses are 14 feet. Total static head is $8 + 92 = 100$ feet, and TDH is $100 + 14 = 114$ feet.

TDH from gauges. With a suction gauge and a discharge gauge at the same elevation:

$$\text{TDH (ft)} = (\text{Discharge psi} - \text{Suction psi}) \times 2.31$$

If the discharge gauge reads 65 psi and the suction gauge reads 5 psi, TDH is $(65 - 5) \times 2.31 = 138.6$ feet. If the suction gauge shows a vacuum, convert it to a negative pressure (or add the equivalent lift) before subtracting.

Horsepower Chain

$$\text{Water horsepower (WHP)} = \frac{\text{Flow (gpm)} \times \text{TDH (ft)}}{3{,}960}$$ $$\text{Brake horsepower (BHP)} = \frac{\text{WHP}}{\text{Pump efficiency}}$$ $$\text{Motor horsepower (input)} = \frac{\text{BHP}}{\text{Motor efficiency}}$$

• Water horsepower is the useful work done on the water. The constant 3,960 comes from 33,000 ft-lb/min per horsepower divided by 8.34 lb/gal.
• Brake horsepower is the power the motor must deliver to the pump shaft.
• Motor input horsepower is the electrical power drawn.
• Wire-to-water efficiency equals pump efficiency times motor efficiency.

Efficiencies are always used as decimals (78 percent becomes 0.78). Dividing by an efficiency always makes the number larger, which is a quick check that you used the right operation.

Power and Cost

$$\text{kW} = \text{Horsepower} \times 0.746$$ $$\text{Cost} = \text{kW} \times \text{Hours of operation} \times \text{Cost per kWh}$$

Example 2. A high-service pump delivers 1,000 gpm against 120 feet of TDH. Pump efficiency is 78 percent and motor efficiency is 92 percent. The pump runs 16 hours a day and power costs 11 cents per kWh.

1. $\text{WHP} = 1{,}000 \times 120 \div 3{,}960 = 30.3$ hp
2. $\text{BHP} = 30.3 \div 0.78 = 38.8$ hp
3. $\text{Motor input} = 38.8 \div 0.92 = 42.2$ hp
4. $\text{kW} = 42.2 \times 0.746 = 31.5$ kW
5. Daily energy is $31.5 \times 16 = 504$ kWh, which costs about 55 dollars and 44 cents per day at 11 cents per kWh.

Wire-to-water efficiency is $0.78 \times 0.92 = 0.718$, or about 72 percent. Raising pump efficiency by rebuilding worn wear rings (Section 11.1) or running closer to the best efficiency point lowers this cost directly.

Pumping Rate from a Wet-Well Drawdown Test

When no flow meter is available, operators time the drop in wet-well level.

Example 3 (inflow stopped). A circular wet well is 8 feet in diameter. With the influent valve closed, the pump lowers the level 2.0 feet in 3.0 minutes.

$$\text{Volume} = 0.785 \times 8^2 \times 2.0 \times 7.48 = 751.6\text{ gal}$$ $$\text{Pump rate} = \frac{751.6\text{ gal}}{3.0\text{ min}} = 250.5\text{ gpm}$$

Example 4 (inflow continuing). If inflow cannot be stopped, measure the inflow first by timing the rise with the pump off, then time the drawdown with the pump on. The pump rate equals the drawdown rate plus the inflow rate. If the level rises at a rate equal to 60 gpm with the pump off, and the drawdown with the pump on equals 190 gpm, the pump delivers $190 + 60 = 250$ gpm.

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