Heat transfer from pipe

  1. I am trying to predict the temperature of water coming out of a pipe.

    I know the temperature of the water going into the pipe.
    I know the material and characterisitics of the pipe.
    I know the ambient air temperature.

    I have explored Fourier's Heat Conduction
    [tex]\mbox {\Huge Q= $\frac {2 \pi k L (T_i-T_o)}{ln (\frac{r_o}{r_i}) }$ } [/tex]

    where,
    Q is the heat transfer (BTU/hr)
    k is teh thermal conductivity (BTU/(hr ft deg F)
    L is the length of the pipe (F)
    T are the temperatures inside and outside of the pipe
    r are the radii inside and outside the pipe wall

    I'm not sure how to relate this to a temperature change.

    Does anybody have a suggestion?
     
  2. jcsd
  3. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    Hi sidkdbl07, welcome to PF. In general, this is a relatively complex problem involving heat conduction along the length of the pipe and the water, along with radial conduction and convection through the water, the pipe, and the air. So you're going to want to see if you can make some assumptions to simplify things.

    If the pipe diameter and wall thickness is small, for example, and the water flow is sufficiently large, then convection from the pipe to the surrounding air ([itex]Q=hA(T-T_{\inf} )[/itex]) may be the limiting factor. I'd probably start by calculating this, then comparing it to the values of conduction heat transfer.
     
  4. Here's what I did and someone can tell me what is wrong with it.

    1. Calculate Q using Fourier's Heat conduction equation (see my post above). That give me the heat transfer along the length of the pipe in unit BTU/hr.

    2. Get degrees celsius heat units per hour (C HU/hr) by dividing BTU/hr by 1.799

    3. Since it takes 1 Celsius heat unit to heat 1 pound of water by 1 deg C, we divide Q (in C HU/hr) by the weight of water to get the temperature change in degrees per hour (deg C/hr).

    4. If we know how long it takes the water to traverse the pipe, simply multiple the temperature change rate by this time, and you have your temperature change.

    Here's an example:

    We have 20,000 ft of PVC pipe (2' radius and 3" thick). The water is 40F at the begining of the pipe, and the ambient air temperature is 50F. The water travels at an average speed of 1.5 ft/s. The thermal conductivity of PVC is 0.11 BTU/(hr ft deg F).

    Q = 1173599.20 BTU/hr
    or
    Q = 652361.98 Celcius HU/hr

    The volume of water is 251,327 cu.ft. (V=2 pi r) or 1,875,574 lbs.

    It takes 1 C HU/hr to heat 1 lb of water by 1 C, so
    temperature change rate is,
    652361.98 / 1,875,574 = 0.35 deg C / hr

    It takes 20000/1.5 = 13333.33 s = 3.7 hr for water to traverse the pipe so the water will gain 0.35*3.7 = 1.30 deg C of temperature.
     
  5. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    1) According to this approach, if the flow rate is decreased by a factor of ten, the temperature increase will be ten times as much, or 13°C, which is impossible given the initial conditions.

    2) Assuming the outside of the pipe is a 50°C and the ambient temperature is the same as assuming infinite convective heat flux. In reality, the outside of the pipe will be less than 50°C.

    However, I would buy a claim that you have determined an upper bound for the temperature increase.
     
  6. Your point is well received Mapes.

    I made an error above:
    251,327 cu.ft of water is 15,689,592 lbs

    This makes a BIG difference, since the temperature gain would be 0.15C.

    It also means that decreasing the flow rate by ten fold, results in 1.54C change, which makes more sense. There is a limit however to how slow the water can move. Heuristic testing will have to determine this.

    Another problem I have with this model is this:
    k_PVC = 0.11 BTU/(hr ft deg F)
    k_Steel = 24.8 BTU/(hr ft deg F)

    so if I simply swap out the PVC pipe with a steel pipe, the temperature change is 34.7C, which is of coarse impossible. Any suggestions here?
     
  7. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    Well, you're applying a steady-state equation to a non-steady-state problem. You're assuming the temperature difference across the pipe wall is 10°F, but this isn't the case for at least two reasons: (1) the outside of the pipe is at <50°F because heat transfer to the air by convection isn't infinite; (2) the water temperature is increasing with time. So you'll always overestimate the amount of water heating, including above what is physically possible.

    So again, I'd recommend investigating the convective heat transfer from the pipe to the air; if this is low, the water will simply cool the pipe to approximately 40°F, and the water will emerge at 40°F. Also, the time-dependent equation is a partial differential equation that incorporates transient heating of the water. It may be necessary to switch to this PDE.
     
  8. Good suggestion re: PDE Mapes.

    I've turned Fourier's heat law into a first order differential equation:
    [tex]\frac{dT(x)}{dx} = \frac{k(T_i - T(x))}{404301.5752 \cdot r_i \cdot ln(\frac{r_o}{r_i}) \cdot v_w }[/tex]

    If we re-arrange things so that it is in the form
    a dT(x)/dx + b T(x) + c = 0
    I get the following
    [tex]0 = \frac{dT(x)}{dx} + \frac{k}{404301.5752 \cdot r_i \cdot ln(\frac{r_o}{r_i}) \cdot v_w} T(x) + \frac{-T_ok}{404301.5752 \cdot r_i \cdot ln(\frac{r_o}{r_i}) \cdot v_w}[/tex]

    For the homogenous solution,
    [tex]T(x) = T_i e^{\frac{-kx}{...}}[/tex]

    For the particular solution,
    [tex]T(x) = T_i e^{\frac{-kx}{...}}+T_o[/tex]

    So I add the two solutions together
    [tex]T(x) = 2T_i e^{\frac{-kx}{...}}+T_o[/tex]

    This doesn't make sense.

    I expected the equation to be something like this
    [tex]T(x) = T_o - (T_o-T_i)e^{\frac{-kx}{...}}[/tex]

    EDIT:
    v_w is the speed of the water in the pipe
    x is the distance from the begining of the pipe
    T_i is the initial Temperature of the water (constant)
    T_o is the temperature of the outside of the pipe
     
    Last edited: Apr 21, 2010
  9. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    With such a thin pipe wall compared to the radius, you can assume the pipe wall to act like a plane wall (the error in replacing [itex]1/\mathrm{ln}(r_o/r_i)[/itex] with [itex](r_o+r_i)/2(r_o-r_i)[/itex] is about 0.1%), which lets you get rid of those logarithms.

    An energy balance for the water gives

    [tex]\pi r_i^2 c \rho \frac{dT}{dt}=\dot{Q}[/tex]

    where [itex]r_i[/itex] is the inside radius, [itex]c[/itex] is the specific heat, [itex]\rho[/itex] is the density, [itex]T[/itex] is the water temperature, [itex]t[/itex] is time, and [itex]\dot{Q}[/itex] is heat rate (in units of power per unit length) from the pipe. From conduction through the pipe and convection to the air, we also have

    [tex]\dot{Q}=2\pi k\left(\frac{r_o+r_i}{2}\right)\frac{T_\mathrm{pipe-air}-T}{r_o-r_i}=2\pi r_oh(T_\mathrm{air}-T_\mathrm{pipe-air})[/tex]

    where [itex]k[/itex] is the pipe thermal conductivity, [itex]r_o[/itex] is the outside radius, [itex]T_\mathrm{pipe-air}[/itex] is the temperature at the outside of the pipe, [itex]h[/itex] is the convection coefficient, and [itex]T_\mathrm{air}[/itex] is the air temperature.

    Are you on board with this? From here you should be able to solve for [itex]T_\mathrm{pipe-air}[/itex] (symbolically) and end up with a differential equation for [itex]T(t)[/itex]. Please let me know if you have any questions.
     
    Last edited: Apr 22, 2010
  10. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    Continuing along these lines, we find that the solution is

    [tex]T_\mathrm{water,final}=T_\mathrm{air}-(T_\mathrm{air}-T_\mathrm{water,initial})\exp (-k^\prime L/r_i^2 c\rho v)[/tex]

    where [itex]T_\mathrm{water,final}[/itex] is the water exiting the pipe, [itex]T_\mathrm{water,initial}[/itex] is the water entering the pipe, [itex]L[/itex] is the pipe length, [itex]v[/itex] is the water velocity, and

    [tex]k^\prime =kr^\prime\left(\frac{2hr_o}{2hr_o+kr^\prime}\right)[/tex]

    [tex]r^\prime=\frac{r_o+r_i}{r_o-r_i}[/tex]

    Looking at a few extreme cases:

    [tex]L/v\rightarrow 0\mathrm{, }\quad T_\mathrm{water,final}\rightarrow T_\mathrm{water,initial}[/tex]

    [tex]L/v\rightarrow \infty\mathrm{, }\quad T_\mathrm{water,final}\rightarrow T_\mathrm{air}[/tex]

    [tex]h \rightarrow 0\mathrm{~or~}k\rightarrow 0\mathrm{, }\quad T_\mathrm{water,final}\rightarrow T_\mathrm{water,initial}[/tex]

    [tex]h \rightarrow \infty\mathrm{, }\quad T_\mathrm{water,final}\rightarrow T_\mathrm{air}-(T_\mathrm{air}-T_\mathrm{water,initial})\exp (-kr^\prime L/r_i^2 c\rho v)[/tex]

    [tex]k \rightarrow \infty\mathrm{, }\quad T_\mathrm{water,final}\rightarrow T_\mathrm{air}-(T_\mathrm{air}-T_\mathrm{water,initial})\exp (-2hr_0 L/r_i^2 c\rho v)[/tex]

    [tex]r_i \rightarrow r_o\mathrm{, }\quad T_\mathrm{water,final}\rightarrow T_\mathrm{air}-(T_\mathrm{air}-T_\mathrm{water,initial})\exp (-2h L/r_o c\rho v)[/tex]

    so everything seems to work out right. Remember that we are assuming the water is well mixed in its cross section and therefore has a uniform temperature at any point in the pipe. Does this all make sense?
     
  11. This is a great contribution Mapes. Thank you.

    So the desity and specific heat capacity are for water. These values are temperature dependant, although as long as the water doesn't lose too much temperature along the pipe these values can be looked up on a table.

    Are the density and specific heat capacity of the surrounding air included in the convection coefficient (h)? This value seems hard to compute.

    If my pipe is buried, (i.e. surrounded by rock) this will affect my results, no? Of coarse it will, I'm simply asking "how" it will affect it.
     
  12. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    The convection coefficient is notoriously hard to compute, and is best determined experimentally. Values of 10-25 W m-2 K-1 are probably a good approximation for natural convection.

    If the pipe is buried, things get a little trickier because heat is being conducted to or from an infinite medium. Depending on the relative thermal conductivity values of the pipe, the water, and the rock, this configuration may be easier or more difficult to analyze. Specifically, if one component has a relatively low thermal conductivity, it will be the limiting component for heat flow, and it may be possible to ignore the other components. However, if all the values are similar, the entire system may need to be analyzed to get accurate results.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook