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Water temperature

  1. Nov 21, 2004 #1
    Hi! I have a problem that I need to solve. I was really hoping someone could help me get started. It should be simple...

    I have a pipe made out of copper with an inner diameter of 50mm and a thickness of 5mm (i.e. outer diameter is 55mm). Around this pipe is an isolating layer with a thermal conductivity of 0.04W/(mK) followed by a plastic layer with a thermal conductivity of 0.15W/(mK). The total diameter of the pipe is 160mm.

    Now we want to send water with an initial temperature of 80C (about 353K) through the pipe. The water travels at a speed of 0.5m/s and the temperature of the surrounding is 7C (about 280K).

    I need to approximate the temperature of the water after 1km.

    I of course do not expect anyone to solve this for me; I just need help getting started.

    Thanks in advance!
  2. jcsd
  3. Nov 21, 2004 #2
    First, a question to clarify the diameters:

    If the bore diameter of the copper pipe is 50mm, and the wall thickness is 5mm, then the outside diameter would be 60mm.

    On the other hand, if the outside diameter is 55mm, then the wall thickness would only be 2.5mm.

    Now to the problem. Start by calculating the mass flow of water, in kg/sec, and look up the specific heat capacity of water.

    Now if you assume that in the first metre of pipe, the temperature won't change that much, you can work out the total heat loss for the first metre of pipe, where the temperature difference between the water and the air is 73K (80 - 7).

    Knowing the heat loss, the mass flow, and the specific heat capacity, you can calculate the temperature change after the first metre length.

    In principal, you can then repeat this for the next metre, with the new water temperature and so on. A computer program could add together all the temperature drops across each small length of pipe to get the total temperature loss.

    Once you have a feel for the problem, shorten the lengths of pipe you consider down to a tiny increment, instead of one metre. This will make the result more accurate. You'll end up with a tricky integral, if you want an exact answer, and don't wish to use numerical methods.
  4. Nov 21, 2004 #3
    Hi! Thanks for responding! And I'm sorry about the typo... it should be 60mm.

    What about the time factor? If you check at any given point in the pipe, the temperatur is always the same, right? I would like to know the temperatur after 1km, and I know the speed of the water so I know the time the water has traveled when it reaches that point. So, couldn't I just look at that point, compute the time the water has been cooled (the time it has traveled) and calculate the loss from that?

    Thanks in advance!
  5. Nov 21, 2004 #4
    Yes, the temperature at any given point will remain constant. The temperature loss won't be at a constant rate though. It will decrease faster near the start of the pipe, where the pipe is hottest. A graph plotting the temperature against position will show an exponential decrease.

    I've not looked at the figures yet, so I don't know how the numbers will come out, but you might lose say 5 degrees in the first 100 metres (20 seconds) and then you'd lose about 5 * (73 - 5) / 73 degrees in the next 100 metres, or 20 seconds. The heat loss would only be 68/73 of the loss in the first part, as the temperature difference from inside to outside the pipe is now only 68 degrees, rather than 73.

    The example above assumes the temperature against distance graph would be a staircase, but actually, of course, it will be a smooth curve. That's why you'll get a more accurate result by considering small increments of pipe length (or time, which amounts to the same thing).
    Last edited: Nov 21, 2004
  6. Nov 21, 2004 #5
    Thanks again for your help!

    Lets say I split the pipe into section of 5 metres. In a section I would get something like [tex]T_i = f(t) * T_{i-1}[/tex] where [tex]f(t)[/tex] is the decrease in temperatur as a function of time. Given the number of sections, the time can easily be found. But how to I found the decrease in temperatur as a function of this time?

    Thanks again!
    Last edited: Nov 21, 2004
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