# Using plastic pipes as heat exchanger...? - Page 2

Posted by Slow Joe on June 5, 2004, 9:52 pm

wrote: While this is normally correct, about 15 years ago I heard of some
research done because of some anomalous residential fires. Apparently,
wood that is subjected to long term temperatures, above 250f degrees,
can slowly come to have a reduced ignition temperature. Apparently
what was happening is, inadequate insulation allowed the wood
adjacent to fireplaces and chimneys to be subjected to over 250f
degrees. Eventually, the ignition point was reduced until it only took
a few minutes of sub 350f degrees to ignite the wood.

I believe this was manufactured wood, such as plywood, paneling and
chip board, but I can't recall any more details.

regards,
Joe

Posted by John Canivan on June 5, 2004, 6:06 pm

Hi
I was wondering if anyone knows how to calculate the amount of heat that
will be transfered across a given surface are of copper tubing with a given
thickness and a given delta T.

For example copper has a thermal conductivity of about 400 W/mK. How can
this information be used to calculate how long it will take 30 feet of 3/4
inch copper tubing to transfer heat to an 80 gallon heat storage tank and
raise the temperature of the tank 10 degrees F.
The starting temp of the tank is 70 F
The temperature of the water circulating through the copper tube is 120 F so
the delta T is 50 F.
The thickness of the copper tube is .2cm
The surface area of the copper exposed to the inside of the tank is 8 sq ft

Question: How long will it take collected heat to raise the temp of the tank
10 degrees F?
If a coil of heat exschange copper tubing were 60 feet long
how long would it take to raise the tank 10 degrees F?

Posted by daestrom on June 6, 2004, 5:37 pm There are many texts that can help you with this sort of question.  But here
is some info to give you an idea.  First, let's convert everything to one
set of units.  I chose 'English' here but you could go metric if you want.

The thermal conductivity of copper at 400W/mK would be about 230
BTU/hr-ft-F.  And if the tubing is 3/4 inch outside and 0.2 cm wall, then
the outside radius is 0.03125 ft and the inside radius is 0.0306 ft.

For cylinders, the heat transfer through the side wall is
Q = (Touter - Tinner) / Rwall
and
Rwall = ln(router/rinner) / (2* pi* lambda * length)

(I'm using lower case 'r' for radius of the inner/outer, and upper case 'R'
for the R-value (resistance to heat transfer))

So, for just the *copper* portion of the problem we have....

Rwall = ln(.03125/.0306)/(6.28*230*30) = 4.895e-7 hr-F/BTU

And one might be tempted to then calculate Q = (70F - 120F) / 2.43e-7 =
1.021e8 BTU/hr.  That would heat your tank of ~640 lbm of water up by ten
degrees in less than 1 second.  But we know in our 'gut' that this is wrong.

That's because the 'R' we're using is only for the copper.  The inside wall
of the tubing has a small layer of water next to it that isn't moving as
quickly as the bulk water being circulated.  The heat energy has to transfer
through this thin layer of water.  Similarly, the outside surface of the
tubing has another layer of water against it.  And since the tank may not
have any circulation system, this layer can be quite 'thick'.  For forced
convection water flow, the conductance of this layer can be ~289
BTU/hr-ft^2-F and for free convection water (tank side) it can be as low as
~29 BTU/hr-ft^2-F.  (notice how these layers have slightly different units
with the area involved).

So we calculate the 'R' for the inner film and outer films (just invert the
product of conductance and area)
Rinner = 1 / (2*pi*rinner * length * coefficient) =  1/ (6.28*0.03060 * 30 *
289) = 6.00e-4 hr-F/BTU
Router = 1 / (2*pi*router * length * coefficient) = 1/ (6.28*0.03125 * 30 *
29) = 5.875e-3 hr-F/BTU

Total Q = (Touter - Tinner) / (Rinner + Rwall + Router) = (70F - 120F)/
(6.00e-4 + 2.43e-7 + 5.875e-3) = ~7721. BTU/hr.  With *that* for a heat
addition rate, 640 lbm of water would take about 5 minutes to rise just 1
degree.

Of course, as the tank temperature rises, the Q will change since (Touter -
Tinner) will change with tank temperature.  And in reality, the inner
temperature of the water in the tubing isn't going to be a constant 120F
since it loses heat as it flows through the tubing.  Going in one end it
will be 120, but leaving the other it will be less.  If the water is flowing
at 10 GPM, that is 600 GPH or about 4800 lbm/hr.  If we're removing heat
from this flow at the rate of 7721 BTU/hr, that would mean we're removing
7721/4800 = 1.61 BTU/lbm of water. And that would cool the water by about
1.61 F.  So the outlet water is only 118.39 F.  Of course, if the water flow
rate is a lot lower, then the outlet temperature would be lower as well.
With a lower flow rate, the value for the inner film coefficient will be
lower since it is affected by the velocity of the water (resulting in a
higher value for Rinner).

As you can see, the film against the tubing is a *major* part of the heat
transfer calculation.  The numbers I used here for the film coefficient are
representative, but free convection films can vary by a couple of orders of
magnitude depending on the exact setup.  For example, if the coil is in the
bottom of the tank and arranged to promote internal circulation in the tank
by natural convections, better film coefficients can be obtained.  But if
the coil is arranged so that warm tank water from the bottom turn rises and
flows around the next turn, then the temperature 'seen' by this upper turn
of the coil is not 70F but some higher value.  And that would yield poorer
heat transfer.

Bottom line, the calcs I've shown you are only the theoretical.  There is a
lot of things that will reduce performance.

daestrom

Posted by nicksanspam on June 7, 2004, 12:08 pm With a 1/16" wall thickness, 3/4" ID pipe would have a 0.087" OD, so
A = 30Pi0.875/12 = 6.9 ft^2... I measured 30 for two still films in series, timing the temperature fall
of a paper cup full of hot water sitting in a big pot full of cold water.

Say G = 6.9x29 = 200 Btu/h-F, as a first approximation. We might say it's the same, as a first approximation.

C = 80x8.33 = 666 Btu/F makes RC = C/G = 3.33 hours. If 80 = 120+(70-120)e^(-t/3.33), t = -3.33ln((80-120)/(70-120))
= 0.743 hours, ie 44.6 minutes. ...22.3 minutes.

Nick

Posted by John Canivan on June 8, 2004, 10:40 pm

•
• Subject
• Author
• Date   Re: Using plastic pipes as heat exchanger...? Anthony Matonak 06-04-2004    Re: Using plastic pipes as heat exchanger...? NickW 06-04-2004     Re: Using plastic pipes as heat exchanger...? Anthony Matonak 06-04-2004      Re: Using plastic pipes as heat exchanger...? Slow Joe 06-05-2004   Re: heat conductance question John Canivan 06-05-2004   Re: heat conductance question daestrom 06-06-2004    Re: heat conductance question nicksanspam 06-07-2004     Re: heat conductance question John Canivan 06-08-2004     Re: heat conductance question John Canivan 06-08-2004   Re: heat conductance question daestrom 06-08-2004   Re: heat conductance question nicksanspam 06-09-2004  Re: Using plastic pipes as heat exchanger...? N. Thornton 06-11-2004