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X-ray tube used for cancer therapy

  1. Aug 15, 2003 #1
    Heres the problem:
    An x-ray tube used for cancer therapy operates at 4.0 MV, with a beam current of 25 mA striking the metal target.Nearly all of this power is transferred to a stream of water flowing through the holes drilled in the target. What rate of flow, in Kg/sec, is needed if the temperature rise of the water is not to exceed 50 degrees celsius.

    I/m thinking power = 4000 V * .025 A = 100 watts

    then the specific heat of water = 4186 j but this problem doesent give an initial temperature of the water.. Am I right so far?
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Aug 15, 2003 #2

    enigma

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    If it doesn't give you a starting condition, assume a worst case scenario and design from that.

    Also, is it MegaVolts or milliVolts? You've got the equation set up for kiloVolts.
     
  4. Aug 15, 2003 #3
    im confused about worst case scenario. I dont understand what to do...........
     
  5. Aug 15, 2003 #4

    Bystander

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    Re: Confused



    How do you define "temperature rise?"
     
  6. Aug 15, 2003 #5
    I guess I define tempwerature rise as a positive change in temperature. .....right?
     
  7. Aug 15, 2003 #6

    FZ+

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    Precisely - it's the change in temperature, and hence energy of the water that is significant. So you don't need to know the initial temperature of the water.
     
  8. Aug 15, 2003 #7
    Now that that is established, I am stil confused on exactly what I need to do..... How to find the flow of water in Kg/sec that is..Where would I start. The electricity part is not a problem. it is just implementing it.
     
  9. Aug 17, 2003 #8
    50*4186*rate (kg/s)=100KW, hence
    rate=0.477783
     
  10. Aug 17, 2003 #9

    HallsofIvy

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    The "electricity part" IS important- it tells you how much energy goes into the system each second. Now, how much energy can REMAIN in the system (enough to raise the temperature of the water 50 degrees) and how has to leave the system? How fast does the water have to go out to in order to remove that energy.

    To Sonty: Thanks for helping but just giving the answer doesn't help as much as (1) hints, (2) sometimes the solution given in detail with explanation.
     
  11. Aug 19, 2003 #10
    I thought he got the energy almost right and I guess he should know the

    Q=m*c* &Delta * &theta

    formula. What was making him insecure is that "rate" thing which you get from the realtion between energy and power.
     
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