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

wrote:

*>You can find data sheets on the web and they will indicate the*

*>melting points. I believe wood and paper self ignite around 451F.*

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

*> 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?*

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

*>> 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... The surface area of the copper exposed to the inside*

*>> of the tank is 8 sq ft*

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...

*>...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*

*>outside surface of the tubing has another layer of water against it...*

*>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...*

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.

*>...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.*

We might say it's the same, as a first approximation.

C = 80x8.33 = 666 Btu/F makes RC = C/G = 3.33 hours.

*>> How long will it take collected heat to raise the temp of the*

*>> tank 10 degrees F?*

If 80 = 120+(70-120)e^(-t/3.33), t = -3.33ln((80-120)/(70-120))

= 0.743 hours, ie 44.6 minutes.

*>> If a coil of heat exschange copper tubing were 60 feet*

*>> long how long would it take to raise the tank 10 degrees F?*

...22.3 minutes.

Nick

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

*> >> 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... The surface area of the copper exposed to the inside*

*> >> of the tank is 8 sq ft*

*> 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...*

*> >...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*

*> >outside surface of the tubing has another layer of water against it...*

*> >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...*

*> 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.*

*> >...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.*

*> We might say it's the same, as a first approximation.*

*> C = 80x8.33 = 666 Btu/F makes RC = C/G = 3.33 hours.*

*> >> How long will it take collected heat to raise the temp of the*

*> >> tank 10 degrees F?*

*> If 80 = 120+(70-120)e^(-t/3.33), t = -3.33ln((80-120)/(70-120))*

*> = 0.743 hours, ie 44.6 minutes.*

*> >> If a coil of heat exschange copper tubing were 60 feet*

*> >> long how long would it take to raise the tank 10 degrees F?*

*> ...22.3 minutes.*

*> Nick*

>You can find data sheets on the web and they will indicate the>melting points. I believe wood and paper self ignite around 451F.