New pex question

[eeprom]

Hello,
I have a (another) question about heat transfer properties of Pex. I have looked high and low for a heat transfer coefficient for standard 5/8" pex tubing. I was hoping this value would be in terms of BTU/(hr x ft^2 x F). But instead I found a coefficient of thermal conductivity in terms of (3.2 BTU in) / (hr x ft2 x F). Can someone explain how I would apply this coefficient? Where does the "in" come into play? Does this mean a 1" pex would have a transfer coefficient of 3.2 BTU / (hr x ft^2 x F)? Please advise...

thanks,
EEPROM

 

[IRstuff]

Yes

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[ione]

It’s not a heat transfer coefficient unit. It is a thermal conductivity unit. Now the pipe thickness comes into play:

Thermal conductance [Btu/(hr ft^2 F)] = Thermal conductivity/ thickness

 

[moltenmetal]

See my reply to your post in another thread on the same topic, which surmised that you were confused between the thermal conductivity of the PEX material itself and the overall heat tranfer coefficient for a SYSTEM in which the PEX tubing is only a part. The thermal conductivity of the PEX itself may, or may not, be a controlling resistance.

The electrical analogy works here, which might be helpful to you in particular. Think of DC voltage as temperature and thermal resistance as DC resistance, and heat flow as DC current. You have a bunch of resistances in series for which the total resistance determines the flow. If one resistance dominates, the others matter much less. Thermal conductivity (k) and film coefficients (h) are conductances (i.e. inverse resistances).

 

[eeprom]

Ione,
So if I have a known thermal conductivity of pex tubing (thermoplastic) as 3.2 (BTU in)/(hr x ft^2 x F), this is a unit of measure per inch of wall thickness of tubing. Is this correct? As a sample calculation: suppose a 5/8" pex tubing has a wall thickness of 0.0625". Then Uo is equal to 3.2/0.0625 = 51.2 BTU/(hr x ft^2 x F). Can that be right? That's a higher heat transfer rate than steel pipe.

 

[ione]

Nope. This is not the overall heat transfer coefficient, that is what really matters, and that you have been already addressed to (ref your thread

http://www.eng-tips.com/viewthread.cfm?qid=293673&page=1).

I suggest you give another read to Moltenmetal’s post. You have a series of thermal resistances (reciprocal of thermal conductance) as at least three mechanism of heat transfer are involved here:
1. heat transfer from the fluid film to the inner surface (convection)
2. heat transfer from the inner pipe outer surface through the pipe wall thickness (conduction)
3. heat transfer from the pipe outer surface to the (convection and radiation)

In order to apply the formula: Q = U*A*deltaT

You need to know the overall heat transfer coefficient U (Btu/hr/ft^2/F), that you’ll hardly find in any standard and that has to be calculated for your specific field conditions.

 

[eeprom]

Thanks

 

[IRstuff]

The units for the thermal conductivity are correct.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize