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Pipe heat loss calculation

  1. Jun 28, 2017 #1
    I am new and have a math and finance background. Because of my work I am starting to get into engineering related questions, specifically in piping. I flyfish a lot and can give pointers to anyone that helps me with my questions.

    Say I have a 5-10' length of pvc, metal or copper pipe that is 4 or 5" diameter with x number of feet in pvc pipe attached to it. The 5-10' length is surrounded by 67°F water and the air traveling through the pipe is 70-100°F The air traveling through the pipe is exchanged at a rate of 45 cfm. The air is filtered, so there might be turbulence and I will have to calculated with and without turbulence. My question is how to calculate what 45 cfm flowing through a 4-5" diameter pipe 5-10' length will cool to? 90°F air going in and x temp coming out of the 5-10' section that is surrounded by 67°F water.
    Any help is appreciated.
     
  2. jcsd
  3. Jun 28, 2017 #2

    haruspex

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    Even if there is little or no turbulence there will still be convection, so I would take the temperature to be uniform across each section of pipe. Can you write the equation on that basis?
     
  4. Jun 28, 2017 #3
    Not enough info to go forward as I don't even know the equation. I am lost trying to figure out how much a 5-10' section of 4-5" pipe surrounded by water would cool the air going through it.
     
  5. Jun 28, 2017 #4

    haruspex

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    I would treat it the same as each short section of air being static in an ambient θext = 67F for a period L/v, where v is the velocity of the air and L the length of pipe.
    If its temperature at some instant is θ, its rate of cooling will depend in a straightforward way on:
    - θ-θext
    - the conductivity of the pipe material
    - the radius of the pipe
    - the thickness of the pipe
    - the specific heat (constant pressure) of the gas
    - the density of the section (which again depends on θ)
    Can you write the differential equation now?

    Edit: I had written "water" in two places where I meant "air". Now corrected.
     
    Last edited: Jun 29, 2017
  6. Jun 29, 2017 #5
    I can find all those numbers, but it will take time. Do you have a more detailed equation for me to use. I am leaving the internet webs for the holidays, so it will be a week before I get back to you. Thanks
     
  7. Jul 5, 2017 #6
    Thickness .125" and 4.875" inner diameter and Thermal conductivity at 25°C is 401 as an example.
     
  8. Jul 5, 2017 #7
    36 feet per second
     
  9. Jul 18, 2017 #8
    Or this question might be simpler and help me as well. Say you have a 10’ long 4-5” diameter copper pipe or pvc pipe surrounded by 65°F water and 45 cfm of air flowing through it. What is the temperature of the air at the end of the pipe if the beginning starts at 65°F? How long would it take 45 cfm of air to flow through. If you know the equation it will help a lot.
     
  10. Jul 21, 2017 #9
    In this system the heat transfer resistance will be dominated by the resistance on the gas side. I have done some calculations to estimate the overall heat transfer coefficient, and it will be about ##U=1 Btu/(hr-ft^2-F).## The equation for heat transfer in this system will be $$WC_p\frac{dT}{dx}=\pi D U(T_0-T)$$ where W is the mass rate of flow of air (~200 ##lb_m##/hr), Cp is the heat capacity of air (0.24 ##Btu/(lb_m-F)##), x is distance along the pipe, D is the inside diameter, and ##T_0## is the outside temperature, 67F. The solution to this equation is $$\frac{T_{out}-T_0}{T_{in}-T_0}=e^{-\frac{UA}{WC_p}}$$
    where A is the heat transfer area ##\pi DL##. When you run this calculation, you will find that there is very little cooling of the air stream between the inlet and the outlet of the pipe.
     
  11. Jul 24, 2017 #10
    What if the air coming into the pipe is 90°F? What would the end temperature be if it is surrounded by 65°F water?
     
  12. Jul 24, 2017 #11
    Above 80 F.
     
  13. Jul 24, 2017 #12
    Thank you again for your help. I am not following the math on this. If you have time can you plug the numbers in? I am curious as to the 45cfm of air flow through the pipe and how much that will affect the 80°F+ number.
     
  14. Jul 24, 2017 #13
    The calculations are based on the following properties of air at 90 F:

    viscosity = ##\mu=1.90\times10^{-4}\ \frac{gm}{cm-sec}##

    heat capacity = ##C_p=0.24\ \frac{Btu}{lb-F}##

    thermal conductivity = ##k=0.0459\ \frac{Btu}{hr-ft-F}##

    density = ##\rho=0.0723\ \frac{lb}{ft^3}=1.16\times 10^{-3}\ \frac{gm}{cm^3}##

    Prantdl number = ##\frac{C_p \mu}{k}=0.712##

    Calculation of pipe cross sectional area: ##A=\frac{\pi}{4}D^2=\frac{\pi}{4}\left(\frac{4.875}{12}\right)=0.1296\ ft^2##

    Calculation of mass flow rate: W=(45)(0.1296)=3.258 lb/min = 195 lb/hr = 0.0543 lb/sec = 24.65 gm/sec

    Calculation of Reynolds number for flow: ##Re=\frac{4W}{\pi \mu D}=\frac{4(24.65)}{\pi (1.9\times 10^{-4})(4.875\times\ 2.54)}=13340##

    From table 14.3 in Transport Phenomena by Bird, Stewart, and Lightfoot, the Colburn j-factor at this Reynolds number is 0.004, where the j-factor is defined as:$$j=\frac{\left(\frac{hD}{k}\right)}{Re\ Pr^{1/3}}$$where h is the heat transfer coefficient. In the figure, the j-factor is very insensitive to Reynolds number in this range of Reynolds number. So, we have:

    $$\frac{hD}{k}=0.004(13340)(0.712)^{1/3}=47.6$$So, calculating the heat transfer coefficient, we obtain: $$h=(47.6)(0.0459)/(4.875/12)=5.4\ \frac{Btu}{hr-ft^2-F}$$

    The heat transfer inside of the pipe dominates the heat transfer resistance, so the overall heat transfer coefficient U is also equal to ##5.4\ \frac{Btu}{hr-ft^2-F}##. So the dimensionless group ##\frac{U\pi D L}{WC_p}## is given by:

    $$\frac{U\pi D L}{WC_p}=\frac{(5.4)(\pi)(4.875/12)(10)}{(195)(0.24)}=1.47$$
    So the outlet temperature is $$T_{out}=T_{water}+(T_{in}-T_{water})e^{-1.47}=65+(90-65)(0.2306)=71 F$$

    I guess my earlier calculations had some errors. Sorry about that.
     
  15. Jul 24, 2017 #14
    Wow so 90F to 71F with only 10' of pipe is impressive.... I appreciate the help.
     
  16. Jul 24, 2017 #15
    The reason for this is that the mass flow rate of the air is very low, and its heat capacity is 4x lower than water.
     
  17. Jul 25, 2017 #16
    I wish I studied more of it in college as it interests me now. I am going to have to figure out the amount of water to surround the pipe with as the water will be surrounded with earth. Then to figure the 45cfm - 80 cfm of air flow through the pipe how much the 71°F will increase.
     
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