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Heat transfer- determine heat of chip using convection

  1. Aug 16, 2012 #1
    1. The problem statement, all variables and given/known data

    A silicon chip measuring 5mm by 6mm and of thickness 1mm is embedded in a ceramic substrate. At a steady state, the chip has an electrical power input of 0.3W. The top surface of the chip is exposed to a coolant whose temperature is 20°C. The heat transfer coefficient for convection between the chip and the coolant is 150W/(m^2.k). If the heat transfer by conduction between the chip and the substrate is negligible, determine the surface temperature of the chip



    2. Relevant equations



    3. The attempt at a solution

    my attempt at this question is using the formula:

    Q= h(c)A (ΔT)

    where h(c) is the heat transfer coefficient

    I am unsure what to do with the electrical power that has been put in the question?
     
  2. jcsd
  3. Aug 16, 2012 #2
    The unit of power is the watt. A watt has the unit of work/second. What are the units of work and how do these units compare with Q in your equation above?

    Since there is negligible heat being conducted to the substrate, all must leave by convection.
     
  4. Aug 16, 2012 #3
    work = joules?
     
  5. Aug 16, 2012 #4
    Yes, work is joules which is also a unit of energy. Power is work/second or energy/second. So, electrically, you are inputting energy per second into the wafer. Since the problem indicates the situation is steady state, all that comes in MUST leave. So what does this tell you to do?
     
  6. Aug 16, 2012 #5
    would you divide the heat transfer coefficient by the power input?
     
  7. Aug 16, 2012 #6
    No you would not do that.

    What are the units of Q? What does Q represent?
     
  8. Aug 16, 2012 #7
    Q represents the amount of heat loss?

    and units are watts?
     
  9. Aug 16, 2012 #8
    Yes, that is correct. So you have a steady state situation here. All the power that comes into the chip via electricity must leave via heat transfer. Equate them.

    Q= h(c)A (ΔT)

    You have Q, you have h, you have A, and you have the coolant temperature. The only part of the above equation you don't have is the surface temperature. That is what you solve for.
     
  10. Aug 16, 2012 #9
    so would it be

    0.3= 150*(5mm*6mm)(T-293)
     
  11. Aug 16, 2012 #10
    and also what is the best/easiest way to re arrange for T?
     
  12. Aug 16, 2012 #11
    Whenever you use an equation you must insure that the units match. Look at the film coefficient of 150 watts/m^2-K. But you are plugging millimeters into the equation.

    You can use the 20 C directly because 1 degree change in C is 1 degree change in K.
     
  13. Aug 16, 2012 #12
    What class are you taking?
     
  14. Aug 16, 2012 #13
    Thank you, and also what is the best/easiest way to re arrange for T? I always struggle trying to figure this out!
     
  15. Aug 16, 2012 #14
    You have

    A = B*(C*D)*(X - E) where A,B,C,D, and E are constants.

    X = A/(B*C*D) + E

    You should know this. If you plan to continue in the sciences, you will certainly see much more of this and other equations that are considerably more complicated. This is trivial.
     
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