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Heat loss through PEX pipe

  1. May 28, 2010 #1
    I want to build a small geothermal cooling unit to cool one small room in my house. If i lay 0.5" pex pipe which is 0.475" ID.

    The big question is how long does the water flowing in the pex tubing have to stay under ground. Constant ground temperature is about 55 degrees F around my house. I have red clay around my house. I know that the temperature as a function of time is a exponential equation.

    Can somebody give me an equation temperature Vs. time spent under ground. This way I can calculate the time needed to get the right temperature water coming out of the pex tubing. Here is as many specs as I think can effect the equation:

    Pex ID = 0.475"
    Red clay soil
    constant ground temp = 55oF
    average water temp flowing into loop = 70oF
  2. jcsd
  3. May 29, 2010 #2
    You need to know the velocity of the flowing water so you can calculate the Reynolds number because it will be necessary for whatever equation you use. I dont know the necessary equation off the top of my head but I am sure if you pick up any heat transfer text book you will be able to find it pretty quickly for your situation. The equation will allow you to solve for the necessary length to achieve a certain temperature change.
  4. May 29, 2010 #3
    Ok cool. Can you give me reynolds number equation? And the heat transfer equation. I Dont have a heat transfer book.
  5. May 29, 2010 #4
    The Reynolds number is Re=[tex]\rho[/tex]*V*D/[tex]mu[/tex]

    [tex]\rho[/tex] is the density of the water,
    [tex]\mu[/tex] is the viscosity of the water,
    V is the velocity,
    D is the inner diameter of the pipe.

    In order to do this you need to determine the convection heat transfer coefficient and there are a couple of ways to do that and the validity of those methods depends on the Reynolds number and whether the flow can be considered laminar or turbulent. So once you calculate the Reynolds number let us know and we can go from there.
  6. May 29, 2010 #5
    Check My Math:

    re = (998.21 KG/M3)(0.254 M/S)(0.012573 M) / (1.00 mPa * S)

    re = 3.18782

    Do the Units and the Math look good?

    v = 30 gallons per hour through .495 ID pipe = .254 M/S
  7. May 30, 2010 #6
    Your Reynolds number is off by a factor of 1000, your viscosity is 1*10^(-3).

    Keep in mind that I am not a professional engineer yet. I have only taken a course in heat transfer but I will try to help as best I can and if anyone has better suggestions hopefully they will let you know.

    So the RE is actually about 3000.

    I am going to be using SI units.

    The next step is to determine the Nusselt number which is the ratio of convective heat transfer to conductive heat transfer. To do this I used the Gnielinski correlation which is an empirical correlation that you can find on this page about half way down.


    f is the friction factor, and Pr is the Prandlt number. Pr is about 7.56 for water and f is about 0.04 (I used the correlation directly below the Gnielinski correlation on that web page to calculate f).

    So Nu=21.5

    The definition of Nu is, Nu = h*D/k

    Where h is the convection coefficient, D is the diameter and k is the conduction coefficient.

    So we can solve for h and I get h=1071 W/(m^2*K)

    Now I am going to assume that this situation is a pipe with a constant surface temperature of 55 degrees F. Hopefully the ground is a decent enough heat sink to justify this. If anything is wrong with what I am doing this assumption is most likely it. I am also ignoring the entry length of the pipe where the flow is not fully developed. But this will allow us to solve for the necessary length to get the desired outlet temp of the water. The equation for a constant surface temperature pipe that relates the inlet, outlet, surface temperatures to the convection coefficient, length and a few other properties can be solved for the length and that is shown below.

    L=-ln((Ts-To)/(Ts-Ti))*mdot*Cp / (pi*D*h)

    L is length, Ts is surface temp, To is outlet temp, Ti is inlet temp, mdot is mass flow rate (0.029 kg/s for this case), Cp is the specific heat (4190 J/(kg*K)).

    So you should just have to pick the desired outlet temperature and plug it into this equation using the h from above.

    Make sure your units are consistent!!! I say this only because the temperatures you listed were in F but the values I have given here are in SI units.

    Hopefully this helps. Let us know how it goes.
  8. Jun 4, 2010 #7
    Hey, I'm not getting the numbers to com out right. It says I need less than 1 meter of pex.

    run the numbers and see what you get:

    ts = 285.777 K
    to = 286.333 K
    ti = 294.111 k
  9. Jun 11, 2010 #8
    Can anybody help me out? The numbers are not working out for me.
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