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Energy problem

  1. Jun 26, 2009 #1
    Hi everyone!

    I am having a bit of a problem solving the following question, and would be very grateful for any advice given.

    1. The problem statement, all variables and given/known data

    A car is run for 400 hours per year, with a total mileage of 10,000 a year. The car uses diesel and consumes 2,000 litres per year. A litre of diesel cost £1.10 and holds 38.7MJ of primary energy.

    a) Calculate the mean engine power?

    b) If the engine is 20% efficient overall, how much energy would need to be stored in 95% efficient batteries to give a range of 300 miles on full battery charge?




    3. The attempt at a solution

    a) energy in 2000litres of diesel 38.7MJ * 2000litres = 77400000000MJ
    number of seconds in 400hours 3600*400 = 1440000secs
    power = energy/time = 77400000000/1440000 = 53750Watts or 53.75kW

    b) Really stuck with this part
     
  2. jcsd
  3. Jun 26, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi boyblair! Welcome to PF! :smile:

    Yes, that's ok :smile: … but if you write all those zeros, you're very likely to make a mistake so you really ought to write the whole thing as a fraction, and then do a bit of cancelling, rather than doing it in stages. :wink:
    efficiency is energy output divided by energy input …

    so do that for the batteries (95%) and the engine (20%) separately, then combine them, so as to find how much of the energy put into the batteries wil come out of the engine. :smile:
     
  4. Jun 28, 2009 #3
    Hey, thanks for the quick response.
    I have attempted part b) and would be grateful for any feedback.

    Primary energy consumes in 1 year: 300h*53.75kw=16.1MWh
    Energy in to 20% efficient motor: 21.5MWh/20%=80.5MWh
    Energy supplied to 95% efficient batteries: 80.5MWh\95%=84.7MWh

    Many thanks
     
  5. Jun 28, 2009 #4

    tiny-tim

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    The efficiency calculations are correct

    (though, as before, doing the whole thing together, instead of in stages, would be safer and would look better: 16.1 x 100/20 x 100/95 = 84.7 :wink:).

    However, your energy per 300 miles is wrong …

    you have multiplied the miles by the power, which is energy per time, instead of energy per mile. :redface:
     
  6. Jun 29, 2009 #5
    Thanks for your help tiny-tim.

    I have made another attempt at the problem:

    Primary energy consumed in 1 year: 400h*53.75kw=21.5MWh
    Energy supplied to 95% efficient batteries: 21.5 x 100/20 x 100/95 = 113.2MWh
    For 300 miles: 300/10000*113.2=3.4MWh
     
  7. Jun 29, 2009 #6

    tiny-tim

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    eugh! :yuck:

    why so long-winded? :rolleyes:

    do it the easy way … you want the energy for 300 miles, so read the question carefully, and you'll see it goes 10,000 miles on 2,000 litres, so that's 5 miles a litre … carry on from there :smile:
     
  8. Aug 18, 2010 #7
    Tiny Tim

    Hows it going?

    I've found your guidance on this thread and need help with the exact same question funnily enough...
    I got to where BoyBlair got to (the long-winded response) - is this the correct answer for this problem, just the long way round?
    I've got severe brainblock after 7 days of constant studying and don't even feel like i can count to 10 anymore...

    Your response would be greatly appreciated

    Ta!
    :confused: + :zzz: + :cry: = me!
     
  9. Aug 19, 2010 #8

    tiny-tim

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    welcome to pf!

    hi jacstar! welcome to pf! :wink:

    (just got up :zzz: …)

    as Zryn says in the thread you've started on this problem (https://www.physicsforums.com/showthread.php?t=423021"), you need to calculate the energy per distance

    as i said before, it goes 10,000 miles on 2,000 litres, so that's 5 miles a litre …

    so try it that way, and show us what you get :smile:
     
    Last edited by a moderator: Apr 25, 2017
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