Heat Loss from uninsulated sch 80 CPVC
Heat Loss from uninsulated sch 80 CPVC
(OP)
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: Heat loss from underground pipe (insulated versus non insulated)) 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?
I found a similar thread (thread391-289057: Heat loss from underground pipe (insulated versus non insulated)) 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?





RE: Heat Loss from uninsulated sch 80 CPVC
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
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RE: Heat Loss from uninsulated sch 80 CPVC
RE: Heat Loss from uninsulated sch 80 CPVC
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
RE: Heat Loss from uninsulated sch 80 CPVC
Should read
Your final overall coefficient of conductance would be the
reciprocal of 1/(hr+hc)+lw/k