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Texas Waste Water Treatment Ch#15.8 Pump Problems

Take the test..... duh....... Study questions for the class C waste water license. Info source: TEEX infrastructure training & safety institute waste water treatment resource book Module #15.8 Pump Problems 2023.
Conversion Factor: 1 hp = 0.746 kilowatts = 33,000 ft lb/min = the amount of power required to move 33,000 pounds of mass one foot in one minute
(1 hp = 33,000 ft lb/min) / 8.34 lb/gal = 3,956.8 GPM/ft of Head which rounds to 3,960 GPM/ft of Head ||| Pump Efficiency (%PE) is a ratio of whp to bhp
Water Horsepower = Water (hydraulic) horsepower (whp), also called pum output, is the horsepower delivered by the pump impeller. ||| whp = ft of Head X GPM / 3,960 GPM/ft of Head
Brake Horsepower = Also called motor output, Brake Horsepower (bhp), is the horsepower delivered to the pump shaft. ||| bhp = ft of Head X gpm / 3,960 gpm/ft of Head X (%Pump Efficiency = whp/bhp)
Quiz by Mandaris
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Last updated: January 17, 2024
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First submittedJanuary 16, 2024
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1. A centrifugal pump recirculates water to a trickling filter at a rate of 1,200 gpm. The head is 25 ft. The pump has an efficiency of 65%. Find the water horsepower and the brake horsepower.
7.6 whp / 11.7 bhp
6.7 whp / 17.1 bhp
17.1 whp / 6.7 bhp
11.7 whp / 7.6 bhp
TLDR: whp = gpm X Head / 3,960 = 1,200 gpm X 25 ft / 3,960 = 7.6 whp! | bhp = gpm X Head / 3,960 X PE = 1,2000 gpm X 25 ft / 3,960 X 0.65 = 11.7 bhp! ||| EXPLANATION: This question wants to know how much horsepower is output by the motor and how much of that horsepower is delivered to the impeller. The pieces of information we were given were the flow rate the motor is causing (gpm), the height of the equivielent water column the impeller is pushing against (ft of head), and ratio of the whp to bhp in the form of pump efficiency in percent. This is all information about the flow side of the pump assembly and the losses mainly due to friction involved in the process so we will have to begin by calculating the whp and using that value to derive the bhp on the motor end of the pump assembly. Start by multiplying the equivelent water column height and the flow rate that the pump impeller is producing against the weight of that same water column and divide that value by the flow rate a 1hp motor can produce against one foot of water column or head to determine the amount of horsepower that must be put out by the impeller to produce the values given in the question.(whp = gpm X Head / 3,960 gpm/ft of Head = 1,200 gpm X 25 ft / 3960 gpm/ft of Head = 7.6 whp! | Now we need to use that value with the pump efficiency given to determine the amount of horsepower the pump motor must produce to provide the impeller horsepower we have just calculated. Start by multiplying your equivelent water column and flow values given in the question and divide that value by amount of flow a 1 hp pump can produce ahainst one foor of head adjusted for real life losses by multiplying it by your real pumps efficiency. (bhp = gpm X Head / 3,960 X PE = 1,2000 gpm X 25 ft / 3,960 gpm/ft of Head X 0.65 = 11.7 bhp!) for a dummy check we can confirm that BHP > WHP. (BHP > WHP due to mechanical and hydraulic losses incurred in the pump therefore since 11.7 > 7.6 we know our answer can be correct in the same way we know a window is open if we feel a breeze through it.)
2. A 15 hp motor is driving a pump at a speed of 1,750 rpm. How many gpm will the pump deliver against a 50 ft. head if the efficiency of the pump is 80%?
119.0
18,210
14.17
950.4
TLDR: gpm = (HP X 3,960 gpm/ft of Head X %PE) / Head = (15 X 3,960 X 0.80) / 50ft = 950.4gpm! ||| EXPLANATION: For this problem the pump RPM is extra info you dont need and we will be rearranging the BHP equation to solve for GPM. (bhp = (gpm X ft of Head) / (3,960 gpm/ft of Head X PE%) => gpm = (bhp X 3,960 gpm/ft of Head X PE%) / ft of Head = (15 X 3,960 X 0.80) / 50ft = 47,520 / 50 = 950.4gpm!)
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