1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with power energy problem

  1. Oct 14, 2013 #1
    I was working on the following problem:

    Determine the power output as a function of time from an energy storage device for a four hour discharge period when it is known that for this type of system the power is a linear function of time, (Power) = a0 - bt, where a0 = 1.1(105) W, b is a constant and t is the time. The energy stored in the device at the star of this process is 8.1(108) J. Using your result calculate the power delivered at 2.4 hours into the discharge process.

    Solution:

    What i did was integrate the power equation to come up with an energy equation since

    energy is the integral of power w.r.t time. i got the following expression

    Energy = Energy (at t=0) – (bt2)/2

    Now we are given that At t=0; Energy = 8.1 × 108 J

    t=1; Energy = 8.1 × 108 – b/2
    t=2; Energy = 8.1 × 108 – 2b
    t=3; Energy = 8.1 × 108 – 9b/2
    t=4; Energy = 8.1 × 108 – 8b


    At t=2.4; Energy =8.1 × 108 – 5.76b/2


    I don't know how to calculate 'b' here. Am i doing it right or should i change my approach?


    << Unnecessary link deleted by Moderator >>
     
    Last edited by a moderator: Oct 14, 2013
  2. jcsd
  3. Oct 14, 2013 #2

    BruceW

    User Avatar
    Homework Helper

    Yes, energy is the integral of power w.r.t. time, but you have missed out one of the terms. What happened to the a0 term?

    yes, your approach is mostly right. (apart from the missed out term). They tell you how much energy there is to begin with, and imply how much energy there is at the end. So you can use this to calculate the unknown parameters.

    Oh, and welcome to physicsforums :)
     
  4. Oct 14, 2013 #3
    Thanks man. Yes, i corrected that and i got

    Energy = a0t –bt^2/2

    So here is the thing..when i put t=0 ...energy comes out to zero whereas we are given that energy at the start is 8.1 × 108 J. i am not understanding this.
     
  5. Oct 14, 2013 #4
    Also, if i consider the ao term , how can i write 8.1 × 10^8 in its place?

    at t=1; Energy = 8.1 × 108 – b/2
     
  6. Oct 14, 2013 #5

    BruceW

    User Avatar
    Homework Helper

    It's almost right. There will be another term, the constant of integration. And this will let you make sure that the initial energy is correct.
     
  7. Oct 14, 2013 #6
    Energy = a0t – bt2/2 + c

    Will this mean that c is 8.1 × 10^8? I still don't know how will i calculate b?

    This problem is frustrating!
     
  8. Oct 14, 2013 #7

    BruceW

    User Avatar
    Homework Helper

    keep with it! you've done most of the work. And yes, c is 8.1 × 10^8. To calculate b is not difficult, but they did not ask the question very well. They say a 'four hour discharge period'. I'm pretty sure this is meant to imply there is zero energy left in the device when the 4 hours is done. You are explicitly told how much energy there is at t=0, so you can use these bits of information to get b.
     
  9. Oct 14, 2013 #8
    Can i equate Energies?

    t=0; Energy = 8.1 × 108 J
    t=1; Energy = a0 – b/2 + c

    8.1 × 108 J = a0 – b/2 + c
    1.1 × 105 = b/2
    b= 5500
     
  10. Oct 14, 2013 #9
    that was b = 220000
     
  11. Oct 14, 2013 #10

    BruceW

    User Avatar
    Homework Helper

    No wait, sorry, I think they are implying that when the power output gets to zero, the 'four hour discharge period' has finished. i.e. power=0 when t=4 hours. This makes more sense, otherwise the power would go negative if we assume zero energy at t=4 hours.

    So, use power=0 at t=4 hours. p.s. be careful with the units. They give you hours, but that's not the SI unit of time.
     
  12. Oct 14, 2013 #11
    okay, so i am totally confused now. I thought power and energy the same thing basically with the exception that power gives the amount of energy per time. So, how can energy be negative if power reaches zero? I am sorry i am annoying you repeatedly.
     
  13. Oct 14, 2013 #12

    BruceW

    User Avatar
    Homework Helper

    no, it's important questions. uh, so yeah, power is energy per time. And in this problem, the device initially has some energy stored in it. Then it does this 'four hour discharge period', where energy goes out of the device. And they have used the convention that positive power means energy leaving the device. So, during this 'four hour discharge period', the power is always positive (since energy is always leaving the device). And during this 'four hour discharge period', the power output is slowly decreasing until it reaches zero. In other words, to begin with, the energy is flowing out of the device quickly, then less quickly, then even less quickly, until eventually the energy stops flowing at 4 hours.
     
  14. Oct 14, 2013 #13

    BruceW

    User Avatar
    Homework Helper

    This is what it seems the question is implying. They don't really say explicitly. You should think it through too, and think of what is most likely that they want you to assume. So, anyway, if we are talking about power output, then the rate of change in energy of the device is going to be the negative of this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Help with power energy problem
Loading...