Problem of the Day Dec.09

Transcription

1 Problem of the Day Problem of the Day 2015.Dec.09 The reclaimed water pump at the wastewater treatment plant pumps 3,600 gpm to a storage reservoir. Suction lift at the plant is 10 feet and the reservoir is 225 feet above the pump. Friction head is 8 feet. The motor and pump are 92 and 81 percent efficient, respectively. Calculate the pump input power in horsepower Brookhurst Street, PMB 573 Fountain Valley, California WWTT

2 Introduction It occurred to me that I needed to work the solution bridge, step by step, for one of these pumping problems. To reiterate, whether you re faced with an activated sludge problem, a trickling filter problem, a chemical dosing problem, a pumping problem, whatever, it is a really good idea to draw the same picture each time for each kind of problem the exact same way. Doing so will help you understand what you need to know to solve a problem. WasteWater Technology Trainers puts every pumping problem in terms of the following graphic: Q Q Hfriction PW = Q x TDH PW = Q x (Hsuction + Hdischarge + Hfriction) Hdischarge Meter PM PB PW $/kwh EM EP Hsuction Generic graphic for all pumping problems. From left to right: telephone (utility) pole, electric meter, motor (complete with air-cooling fins), pump, pump suction (negative suction shown), and pump discharge. Abbreviations: kwh = kilowatt hours (kilowatts times hours NEVER kilowatts per hour), PM = input power to motor, PB = brake power (output power from motor same as input power to pump), PW = output power from pump = power delivered to the water), Q = pumping flow rate, Hsuction = suction head, Hdischarge = discharge head, Hfriction = friction head, and TDH = total dynamic head = Hsuction + Hdischarge + Hfriction. The equation given in the graphic for calculating the water power (PW), which is the power actually delivered to the water, is very straightforward and is used over and over again (for example, in today s Problem of the Day): PW = Q TDH While ftłlb/min are units of power, in water and wastewater problems power is expressed either as HP or kw; the two are interchangeable. The two most important conversion factors for doing these problems are as follows (WWTT suggests you memorize these). The first is used to convert ftłlb/min to HP and vice versa, and the second is used to convert HP to kw and vice versa. or and kw HP or HP kw

3 Power is the rate of doing work: ftłlb are a unit of work, ftłlb/min are a unit of power. When I think about doing work, I think about going to the gym to work out. When I go to the gym, I don t lift gallons, I lift pounds. Although pumps pump gallons, the work they are doing is lifting pounds of water, and, at 8.34 pounds per gallon, water is pretty heavy stuff! So I use the equation just given (PW = Q TDH) to start the solution bridge. While flow (Q) can be in different units, TDH is always in feet. You ll see why that s important below. Solution Does anybody spot the trick in today s problem? You have to be very careful about what an exam question is asking for. While the problem statement gives the efficiencies of the motor and the pump, because the question is asking to calculate the brake power, there is no need for the motor efficiency. Very, very tricky and very, very typical. Here is the list of information given in the problem following the graphic above starting on the left and moving right: 1. Motor efficiency = 92% = 0.92 (not needed; total trickery) 2. Pump efficiency = 81% = Suction head, Hsuction = 10 ft 4. Discharge head, Hdischarge = 225 ft 5. Friction head, Hfriction = 8 ft 6 TDH (must be calculated) = Hsuction + Hdischarge + Hfriction = ( )ft = 243 ft 7. Pumping rate = 3,600 gal/min 8. Density of reclaimed water = 8.34 lb/gal (assumed since not given) Because PW = Q TDH and PB = PW/.EP, we know that PB = Q TDH/.EP. This is the equation we will use to populate the solution bridge. Once it is populated with the equation, we convert units until we have the units we want, HP, which are put here between heavy vertical lines, as always, followed by an equals sign and the blank solution bridge. Problem of the Day. The reclaimed water pump at the wastewater treatment plant pumps 3,600 gpm to a storage reservoir. Suction lift at the plant is 10 feet and the reservoir is 225 feet above the pump. Friction head is 8 feet. The motor and pump are 92 and 81 percent efficient, respectively. Calculate the pump input power in horsepower. I ve decided I m not going to use the equation for PB above to populate the solution bridge. Instead, I m going to pretend that I m having a severe case of test anxiety and I can t remember the equation. It happens all the time. So, new rule: If you have to calculate PW, PB or PM (either in HP or kw) and it is going to entail using a flow rate and a head, you will always start the solution bridge with the conversion factor used to convert ftłlb/min to HP. In today s problem, this is consistent with the way we ve been doing problems, getting the units on the solution bridge that you need in the answer (shown in bold) before canceling units. Here s how it looks. You have to remember here that we re calculating brake power (PB) not water power (PW), so we have to divide by the pump efficiency expressed as a decimal (No. 2). Percents expressed as decimals never have units.

4 0.81 The units needed in the answer are now on the solution bridge so we proceed by canceling unwanted units until they re all gone. TDH will almost always be in ft, so it (No. 6) is entered in the numerator to cancel the ft in the denominator of the conversion factor we used to start the solution bridge. 243 ft 0.81 The pumping rate given in today s problem (No. 7) has min as the unit of time. This isn t always the case, but we ll use it today to cancel the min in the numerator. 243 ft 3,600 gal 0.81 min Now we have lb and gal to cancel and we do it by entering the density of the reclaimed water being pumped (No. 8) so the units cancel numerator and denominator. 243 ft 3,600 gal 8.34 lb 0.81 min gal Since all the units have now canceled except those needed in the answer, HP, we know the solution bridge is complete. The arithmetic gives the answer. Remember: This is input power to the pump (i.e., brake power, PB). Problem of the Day. The reclaimed water pump at the wastewater treatment plant pumps 3,600 gpm to a storage reservoir. Suction lift at the plant is 10 feet and the reservoir is 225 feet above the pump. Friction head is 8 feet. The motor and pump are 92 and 81 percent efficient, respectively. Calculate the pump input power in horsepower. 243 ft 3,600 gal 8.34 lb 0.81 min gal 243 3, , = 273 HP. Discussion When doing pumping problems, some of you use what I will call the 3,960 conversion factor. Notice that if you move 8.34 in the solution bridge into the denominator it goes in the denominator of the denominator. Notice that 33,000/8.34 = 3,957. This is where the 3,960 comes from (3,957 3,960). But there s a catch that you absolutely need to be aware of: In order to use the 3,960, the flow rate must be in gal/min. The reason WWTT doesn t teach these problems using the 3,960 is because we don t want for our students to have to memorize that in order to use 3,960, they must convert all flows to gal/min. We want you to have to memorize as little as possible. For that reason, you should know that the / will always work no matter what units of flow are given in the problem (although conversions will be necessary to cancel all the unwanted units). Happy calculating! Let us know, by leaving a comment, if you want us to do a specific problem, if you see a mistake, or if you have a question on any of the Problems of the Day you are looking at.

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