Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Increasing supply and return temperatures

  1. Sep 12, 2006 #1
    I'm trying to find the size of pipes that are to carry a certain amount of heat in them, 6000 W. This formula should give me the amount of water needed:

    m=W/(4200*delta T)

    Delta T is 20 degrees (the diff between supply and return), but what I don't get is that if I increase the supply and return temperatures to the extreme, shouldn't the radiator or whatever get increased effect? According to this formula it won't, since only the diff between the two temperatures is used. Can someone explain this to me??
  2. jcsd
  3. Sep 12, 2006 #2
    The question is trying to get you to change the mass of transfer fluid rather than the temperature difference. I think you are missing the velocity limitation to determine the pipe size after finding the mass of transfer fluid.

    Theoretically you could move any amount of mass through your pipe, to give that wattage and hold a 20 deg Delta T. Practically, you will be limited by the velocity and resistance in the pipe.
  4. Sep 12, 2006 #3


    User Avatar

    Staff: Mentor

    You didn't provide any information about a radiator...

    The equation does indeed tell you what your heat flow rate is, but I think the question you really are asking is how do you find the delta-T? The delta-T does depend on the functioning of the heat exchanger. There are several factors:

    -Approach temperature is the difference between the radiator (or whatever the heat exchanger is) temp and the air temp. Bigger approach temp=more heat transfer.
    -Flow rate: Over a short range and large approach temperature, flow rate and delta-T are directly proportional (so heat flow is constant), but if flow rate goes way down, delta-T widens a lot and the heat transfer goes down.

    So... there are a lot of interdependent variables there. There are software packages available to calculate how that will all work out. Perhaps we could help more if we knew what exactly you are trying to do.

    Oh.... and if that's for a homework question, they may not want you to think about any of that stuff...
  5. Sep 12, 2006 #4
    Where did you get the 20 deg F(or C?) delta T? This number (20 deg F) is a convention used (and still used) to simplify pipe sizing. 20 deg F inserted in the calculation below simplifies down to a denominator of 9996 (which is rounded to 10000):

    GPM = ((KWh*1000*3.412) / (DT * SpHtW * 60 * 8.33)

    which will be one tenth of the numerator (the 3.412 converts the KW into MBH). Then the equation works out that the GPM is 1/10 that number MBH and from there it is an easy step to the pipe size, if you know the velocity you want to use. So designers would use the 20 deg DT so they could see the MBH and say the GPM is 1/10 of that.

    In this case:

    (6.0 * 1000 * 3.412)/ (20 *1 * 60 * 8.33) =
    20472 mbh/10000 =
    2.05 GPM

    Is your delta T in F or C? What are the units for m that you are solving for? As you can see, I don't work in Metric very often.
    Last edited: Sep 12, 2006
  6. Sep 20, 2006 #5
    It should all be metric, and the Delta T is in Celsius.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Increasing supply and return temperatures
  1. Liquid Nitrogen Supply (Replies: 4)