# How many Watts do I need to heat this pipe?

1. Nov 6, 2013

### steves1080

Reference attached image.

In this simple system, I have a large reservoir of water that feeds a small section of 4" dia. pipe. There is about 1 ft of length until there is a gate valve to isolate the flow. When not flowing water, this 1-ft x 4-in section is at risk of freezing when temperatures are low enough. What I am looking to is basically add an immersion heater to prevent freezing, but I need to know how much power I need. Are there any quick heat transfer calc I can do to estimate this?

Thanks a ton-

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2. Nov 6, 2013

### steves1080

It's carbon steel, so thermal conductivity is like 16 or so

3. Nov 8, 2013

### acc0untnam3

Hi,

The following equation might help.

q = 2 π k (ti - to) / ln(ro / ri) (1)

where

q = heat transferred per unit time per unit length of cylinder or pipe (W/m, Btu/hr ft)

k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft))

to = temperature outside pipe or cylinder (K or oC, oF)

ti = temperature inside pipe or cylinder (K or oC, oF)

ln = the natural logarithm

ro = cylinder or pipe outside radius (m, ft)

ri = cylinder or pipe inside radius(m, ft)

If you know what the coldest temperature that the water will be the temperature that you wish to keep the pipe at, you can find the heat transfered per length then multiply that by the length of your pipe (4ft). This will give you a watt value to heat the pipe.

4. Nov 8, 2013

5. Nov 8, 2013

### steves1080

Yea, I know I shouldn't be reinventing the wheel. But I'm very old fashioned and like to make sure I run my own numbers. The problem, acc0untnam3, with that approach (albeit the seemingly correct one), is that I am getting a massively high wattage of heat loss. Like WAY too high. But I am not seeing an issue with my numbers, so it must be the approach I am taking. See attached for my calculation. Any help would be extremely appreciated.

#### Attached Files:

• ###### HeatTrace.xlsx
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6. Nov 15, 2013

### Missurunha

You forgot the natural logarithm..
ln(ro/ri) = 0.110311, you're using the thickness 0.005969 m...
The area is also wrong, you should calculate 2*PI()*L. Still Q is around 22kW.

I think Q is high because you have a 20ºC change in 5mm, and your thermal conductivity is quite high,,