Calculating Heat Loss in Pipes with Air Flow and Water Surrounding

[Flyfisherman]

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.

 

[haruspex]

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?

 

[Flyfisherman]

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.

 

[haruspex]

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 θ_ext = 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-θ_ext
- 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.

 

[Flyfisherman]

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

 

[Flyfisherman]

Thickness .125" and 4.875" inner diameter and Thermal conductivity at 25°C is 401 as an example.

 

[Flyfisherman]

36 feet per second

 

[Flyfisherman]

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.

 

[Chestermiller]

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.

 

[Flyfisherman]

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?

 

[Chestermiller]

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.

 

[Flyfisherman]

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.

 

[Chestermiller]

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.

 

[Flyfisherman]

Wow so 90F to 71F with only 10' of pipe is impressive... I appreciate the help.

 

[Chestermiller]

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.

 

[Flyfisherman]

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.