
#1
Nov1912, 03:25 PM

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Hello gentlemen good afternoon, I'm a newbie on this forum I have been scratching my head all weekend on how to do these problems. I am wondering if someone could show me the right way.
Q1) A motorpump set lifts 25m^{3} of water in 6 minutes through a vertical height of 5 cm. The efficiency of the pump is 75% and the efficiency of the motor is 85%. 1.) Calculate the kW input to the motor. 2.) Motor HP (Calc. value) 3.) Cost of running the system for 8 hrs (The cost per kWh is .10 cents) Q2) Will be revealed after the first one to avoid confusion :) 1. The problem statement, all variables and given/known data Well I've given all known data I was given so that's met. 2. Relevant equations Not sure, perhaps the word problem already encompasses the relevant equations already. 3. The attempt at a solution Well my attempt was, for "1) Calculate kW input to the motor." 25,000kg * 9.81 = 245,250 N 245,250 * 0.05m / 360sec = 34.0625 Watts output to the motor Multiplying both eff. to come up with a combined .75 * .85 = 0.6375 % 34.0625 / 0.6375 = 54.4314 Watts or 0.053 kW input to the motor I'm not sure if this is correct, I haven't moved onto 2) and 3) because of said reason. Regards, Waffles 



#2
Nov1912, 03:34 PM

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You need to post this in the proper format for a homework problem. Please read the forum rules.
EDIT: NOW you're looking good 



#3
Nov1912, 03:46 PM

P: 4





#4
Nov1912, 09:51 PM

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2 Physics word problems 



#5
Nov1912, 11:34 PM

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haruspex, thank you for taking interest in my thread.
Could you help me on the next one "2) Motor HP (Calc. value)" My attempt was, Combined the efficiencies 1.25 * 1.15 = 1.4375% 34.0625W Output * 1.4375 = 48.875W Input 48.875 / 746 = 0.065 HP 



#6
Nov1912, 11:47 PM

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As it happens, I calculated the output power from the motor for you in my previous post.
Where do you get 1.25 and 1.15 from? You can't do that with efficiencies. An efficiency is a ratio. If A = B*75% then B = A*1/.75 = A*133%. Also, the two efficiency factors, 75% and 85%, are for going all the way from input to the motor to useful work done by the pump. The output from the motor (= input to the pump) sits between them. 



#7
Nov2012, 01:58 AM

P: 4

Hello haruspex,
I'll be honest I can't say I follow what you meant, it's the price I pay for studying long period into the A.M. haha.., could you show me how you got "Motor HP" that means you'd just use the motor eff. right when it just simply asks for Motor Horsepower. I just really need this problem solved before I get my much needed sleep. I like to work with the solutions in reverse is how I learn them better. Thank you 



#8
Nov2012, 02:21 AM

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You have:
power input to motor, P_{mi} power output from motor, P_{mo} = P_{mi} * motor efficiency = P_{mi} * 85% power input to pump, P_{pi} = P_{mo} effective power of pump in pumping water = P_{po} = P_{pi} * pump efficiency = P_{pi} * 75% = P_{mi} * 85% * 75% 


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