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Kilowatt hr

  1. Jun 16, 2008 #1
    I am new to the forum,so if i posted the question not according to some specific format, i apologize.

    Power companies typically bill customers based on the number of kilowatt-hours used during a single billing period. A kilowatt is a measure of how much power (energy) a
    customer is using, while a kilowatt-hour is one kilowatt of power being used for one hour.
    For constant power use, the number of kilowatt-hours used is calculated by kilowatt-hours=kilowatts * time (in hours). Thus, if customers use 5 kilowatts for 30 minutes, they'll have used 5 kilowatts * (1/2)hrs =2.5 kilowatt-hours.

    Suppose the power use of a customer over a 30-day period is given by the continuous
    function P(t) where P is kilowatts, t is time in hours, and t =0 corresponds to the
    beginning of the 30 day period.

    A.
    Approximate, with a Riemann sum, the total number of kilowatt-hours used by the customer in the 30 days.
    B.
    Derive an expression representing the total number of kilowatt-hours used by the
    customer in the 30-day period. (This expression should not be an approximation.)



    for A:
    I did R=720 sigma t=0 to t=720 f(t)

    for B: i just did T(t)= integral sign 0 to 720 f(t)dt

    Could someone confirm my answers? Thanks in advance.
     
  2. jcsd
  3. Jun 16, 2008 #2

    Dick

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    Almost. But why does A have an overall factor of 720?
     
  4. Jun 16, 2008 #3
    I thought that the Riemann sum from 0 to 720 would give me the total number of kilowatt used, so I multiplied that by 720 hrs to get kilowatt hours. Is part B correct then? Thanks for your help.
     
  5. Jun 16, 2008 #4

    Dick

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    B is certainly correct. A has two problems. Suppose the time interval were only three hours and you know f(0), f(1), f(2) and f(3). How would you approximate the power used?
     
    Last edited: Jun 16, 2008
  6. Jun 16, 2008 #5
    Hmm...I would add f(0), f(1), f(2) and f(3) up and times that sum by 3 to get kilowatt hours.
     
  7. Jun 16, 2008 #6

    Dick

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    Maybe you would. I wouldn't. There are three one hour intervals. So I would write either f(0)*(1hr)+f(1)*(1hr)+f(2)*(1hr) or f(1)*(1hr)+f(2)*(1hr)+f(3)*(1hr). Those are Riemann sums. Put in numbers for f(t). Suppose e.g. f(t) is a constant 10kw. Then the exact answer is 30kw*hr. Do you see why what you are doing doesn't give the correct answer, even approximately?
     
  8. Jun 16, 2008 #7
    Oh, I think I see what you mean. So for A, it should just be R=sigma t=0 to t=720 f(t).
     
  9. Jun 16, 2008 #8

    Dick

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    Much closer. But now your sum has 721 terms in it and there are only 720 hours. And the units are still in kw. You need to multiply by hours somehow. Look back at the 3hr example.
     
  10. Jun 16, 2008 #9
    So sigma t=0 to 719 f(t)* (t1-t0)? by t1-t0 i mean the increase in the number of hrs from the preceding hr which is 1 hr for each.
     
  11. Jun 16, 2008 #10

    Dick

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    Why do you want to increase the number of hours? Each interval has only one hour in it. Just think about how you would approximate it using a pencil and paper. Multiply an approximation of the power usage over each hour by the interval (1 hr).
     
    Last edited: Jun 16, 2008
  12. Jun 16, 2008 #11
    I think that it is 1 hr times sigma t=0 to 719 f(t)
     
  13. Jun 16, 2008 #12

    Dick

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    Right! Could also be 1 hr times sigma t=1 to 720 f(t), yes? That would be just as good. The '1 hr' in the Riemann sum corresponds to the 'dt' in the integral.
     
  14. Jun 16, 2008 #13
    yes it could also be 1 hr times sigma t=1 to 720f(t). Thanks so much for ur help and being so patient! :smile:
     
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