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Homework Help: 2 Physics word problems

  1. Nov 19, 2012 #1
    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 motor-pump set lifts 25m3 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.

    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2


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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 :smile:
    Last edited: Nov 19, 2012
  4. Nov 19, 2012 #3
    I hope this suffices what was required to ask a question on this subforum.
  5. Nov 19, 2012 #4


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    Well, no, that's the power required to raise the water at the specified rate. Since the pump is only 75% efficient the output from the motor must be 34.0625/0.75 W. But you got the right answer at the end.
  6. Nov 19, 2012 #5
    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
  7. Nov 19, 2012 #6


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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.
  8. Nov 20, 2012 #7
    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
  9. Nov 20, 2012 #8


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    You have:
    power input to motor, Pmi
    power output from motor, Pmo = Pmi * motor efficiency = Pmi * 85%
    power input to pump, Ppi = Pmo
    effective power of pump in pumping water = Ppo = Ppi * pump efficiency = Ppi * 75% = Pmi * 85% * 75%
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