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X-ray tube used for cancer therapy

  • Thread starter jpnnngtn
  • Start date

jpnnngtn

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?
 
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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.
 

jpnnngtn

im confused about worst case scenario. I dont understand what to do...........
 

Bystander

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Re: Confused

Originally posted by jpnnngtn
Heres the problem:
(snip)temperature rise of the water is not to exceed 50 degrees celsius.

(snip)this problem doesent give an initial temperature of the water..


How do you define "temperature rise?"
 

jpnnngtn

I guess I define tempwerature rise as a positive change in temperature. .....right?
 

FZ+

1,550
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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.
 

jpnnngtn

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.
 
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0
50*4186*rate (kg/s)=100KW, hence
rate=0.477783
 

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.
 
95
0
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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