Heat Loss from uninsulated sch 80 CPVC

[BronYrAur]

I am trying to determine the heat loss from a 2-1/2" schedule 80 CPVC pipe that is carrying a 27 deg F solution of 30% propylene glycol. The pipe is inside a 70 deg F plant.

I found a similar thread (thread391-289057) but it's focus was on buried pipe and no mention of glycol. I have seen the formulas where the heat loss is calculated based on pipe surface temperature, ambient temperature, wind speed, etc

In my case I don't know the surface temp, just the fluid temp inside. It's probably fair to assume zero wind, since we are inside the plant.

Can anyone point me to a reference or provide a formula?

 

[IRstuff]

It's actually heat gain, right, since your fluid temperature is less than the ambient temperature?

This is still essentially a convection and possibly radiation HT problem. The surface temperature is not a priori required. In the most expanded case, there are at least 3 simultaneous equations using T.fluid, T.insidewall, T.outsidewall, T.ambient. Since energy is conserved, there is only one set of temperatures that satisfies the three equations:

Convective/conductive transfer from innerwall to fluid
Conduction from outerwall to innerwall
Convection/radiation from ambient to outerwall

You may need to throw in something to deal with the temperature increase as the fluid flows along the pipe, i.e., some sort of temperature differencing. Length of piping and flowrate will directly affect the solution. You could assume a worst-case heat gain by keeping the fluid temperature constant.


TTFN
faq731-376

 

[ElSid1]

It is an "insulation" style calculation ([link Pipe Insulation]

http://www.raeng.org.uk/education/diploma/maths/pdf/exemplars_engineering/2_SteamPipe.pdf

[/url]). Once you know the layer temperatures, you need to apply the glycol mixture specific heat to get a temperature rise. Basic thermodynamics

 

[zekeman]

If the glycol is flowing , then the inside temperature is close to 27 degrees. Now the problem is to first find the temperature at the surface which can be done iteratively. First assume the temperature is 27, get the radiation and natural convection coefficients like
hr=.173[(T0/100)^4-(Ts/100)^4]/(T0-Ts) for radiation
A good estimate of convective coefficient is
hc=0.27[(T-Ts)/D]^.25
D=2.87/12 for 2.5 schedule 80
Now the conductance across the cpvc wall is
k/l=0.11/lw
lw=wall thickness in feet
I got
k/lw=4
and
hr =0.66
and
hc =.988

hr+hc=1.65
k/l=4=4
From this first iteration get the surface temperature
(T0-Ts)/(T0-Ti)=(70-Ts)/(70-27)=(1/(hr+hc)/[1/(hr+hc)+l/k]
This gives the new Ts from which you get the new hc,hr .
You probably need one more iteration to get an accurate Ts


Your final overall coefficient of conductance would be the
reciprocal of 1/(hr+hc)+k/lw

Note: strictly speaking l/k is only an estimate of the resistance through the wall but is accurate for lw/D small, the case here

 

[zekeman]

Correction

Should read

Your final overall coefficient of conductance would be the
reciprocal of 1/(hr+hc)+lw/k