Calculating Heat Loss in Pipes with Air Flow and Water Surrounding

  • #1
Flyfisherman
15
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.
 
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  • #2
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?
 
  • #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.
 
  • #4
Flyfisherman said:
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.
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.
 
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  • #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
 
  • #6
Thickness .125" and 4.875" inner diameter and Thermal conductivity at 25°C is 401 as an example.
 
  • #7
36 feet per second
 
  • #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.
 
  • #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.
 
  • #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?
 
  • #11
Flyfisherman said:
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?
Above 80 F.
 
  • #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.
 
  • #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.
 
  • #14
Wow so 90F to 71F with only 10' of pipe is impressive... I appreciate the help.
 
  • #15
Flyfisherman said:
Wow so 90F to 71F with only 10' of pipe is impressive... I appreciate the help.
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.
 
  • #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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