Posted by *Simon D* on February 24, 2006, 6:54 pm

*> I appreciate the help but am still having a problem. I don't understand *

*> how you arrive at an area of 4.8 cm in this calculation. Would you please *

*> explain.*

It's the cross sectional area of the plate.

Don't let it worry you though, the calculations are based on linear

temperature gradients, which will not exist in this scenario. As a result,

the answer is not to be trusted.

Besides, I'd defy you to hold your hand over a pot of rapidly boiling water

for any significant time, with or without a copper plate.

You need to look up things like partial differential equations and Laplace

to get a grip on this kind of problem. See

http://faculty.gg.uwyo.edu/dueker/tensor%20curvilinear%20relativity/pde%20tutorials.pdf

section 2.

Even in 1 dimension it's a bit tricky, with your problem more so. You also

need to allow for the fact that heat is not just being applied to the part

of copper that is submerged, but the section above water that is being

heated by the steam and condensation.

How long?? I wouldn't like to say.

Simon.

Posted by *Alan Combellack* on February 25, 2006, 2:04 pm

Simon,

Thanks again but I'm still confused. Comments are inserted below.

Alan C

*>> I appreciate the help but am still having a problem. I don't understand *

*>> how you arrive at an area of 4.8 cm in this calculation. Would you *

*>> please explain.*

*> It's the cross sectional area of the plate.*

The plate was 1 ft square or 929 cm^2. Thickness was 1/16 inch, or .15875

cm. Cross section area is thus147.48 cm^2. Am I still screwed up?

*> Don't let it worry you though, the calculations are based on linear *

*> temperature gradients, which will not exist in this scenario. As a result, *

*> the answer is not to be trusted.*

I realise that. The actual temperature change will be very complex and I am

actually trying to find a reasonably accurate but reasonably simple

expression to determine how fast I can get heat to flow from hot water into

gas in a cylinder. I am fairly sure that it will turn out to be an

expression similar to the exponential rise (or fall) of the voltage across a

capacitor fed through a resistor such as Vc=Vs(1-e^(-t/Time constant)), but

maybe not!. The time constant here would be R*C and I need to find the

thermal equivalent of the time constant figure. I am no mathematician so am

having problems with this. The gas bit is easy enough to deal with but I

need to evaluate the very significant effects of the mass of metal in the

heat exchanger, which also has to be heated (and cooled) and the speed with

which heat flows through the metal. This was the reason for my original

simplified question

*> Besides, I'd defy you to hold your hand over a pot of rapidly boiling *

*> water for any significant time, with or without a copper plate.*

OK. Please replace my fingers with some form of thermometer.

*> You need to look up things like partial differential equations and *

*> Laplace to get a grip on this kind of problem. See*

*>*

*
http://faculty.gg.uwyo.edu/dueker/tensor%20curvilinear%20relativity/pde%20tutorials.pdf *
*> section 2.*

*> Thanks. I will look this up. See comment above also*

*> Even in 1 dimension it's a bit tricky, with your problem more so. You also *

*> need to allow for the fact that heat is not just being applied to the part *

*> of copper that is submerged, but the section above water that is being *

*> heated by the steam and condensation.*

I think I can handle the bits about heat going into and out of the parts of

the plate OK but am still unable to figure out how the heat will actually

move with time from the hot end to the cold end, particularly the time

intervals involved

*> How long?? I wouldn't like to say*

Thats the problem. I suspect you are better able to guesstimate it than I

am since you took the trouble to reply.

I expect you are getting fed up with this by now but may I impose a little

more on you obvious expertise snd re-define my example a little?

I will use cgs units since you seem to prefer this.

I have two 100 cm squre plates with a thickness of, say 0.2 cm. These are

connected to each other through a strip of the same gauge copper which is 20

cm long and 2 cm wide. One plate is in air at 20 deg C.and the air volume

is large so generally, except in the immediate vicinity of the plate, Has a

very slow rise in temperature ( air to surface R value, in the British

system is about 0.67 due to the surface film. I don't know the value of R

in cgs offhand). The other plate is also at 20 deg C initially but is now

plaunged into my boiling water. The plate will heat up at a rate depending

on its mas and specific heat and heat will start to flow along the

interconnecting strip at a rate depending on its thermal conductivity and

cross sectional area. The incoming heat will start warming the plate in air

at a rate depending on a lot of factors including its mass; specific heats,

thermal conductivity; surface area and temperature... To ask for a complete

equation relating all this stuff is, I think, a bit much but all I really

need might be what you have already given me which is how to relate the

surface areas, specific heat, temperature gradients; and actual temperatures

and themal conductivity to time, particularly through the connecting strip.

If you can give me this I will be most grateful.

Maybe I will do some actual testing to see what happens.

Alan C

Posted by *Alan Combellack* on February 25, 2006, 3:21 pm

My apologies about the first comment. I seem to be unable to see the

difference between area and volume. Your 4.8 cm^2 for the cross sectional

area is, of course, quite correct and I am wrong. Mea culpa. I still have a

problem with the rest of it though.

Alan C

*> Simon,*

*> Thanks again but I'm still confused. Comments are inserted below.*

*> Alan C*

*>>> I appreciate the help but am still having a problem. I don't *

*>>> understand how you arrive at an area of 4.8 cm in this calculation. *

*>>> Would you please explain.*

*>>*

*>> It's the cross sectional area of the plate.*

*> The plate was 1 ft square or 929 cm^2. Thickness was 1/16 inch, or .15875 *

*> cm. Cross section area is thus147.48 cm^2. Am I still screwed up?*

*>>*

*>> Don't let it worry you though, the calculations are based on linear *

*>> temperature gradients, which will not exist in this scenario. As a *

*>> result, the answer is not to be trusted.*

*> I realise that. The actual temperature change will be very complex and I *

*> am actually trying to find a reasonably accurate but reasonably simple *

*> expression to determine how fast I can get heat to flow from hot water *

*> into gas in a cylinder. I am fairly sure that it will turn out to be an *

*> expression similar to the exponential rise (or fall) of the voltage across *

*> a capacitor fed through a resistor such as Vc=Vs(1-e^(-t/Time constant)), *

*> but maybe not!. The time constant here would be R*C and I need to find *

*> the thermal equivalent of the time constant figure. I am no mathematician *

*> so am having problems with this. The gas bit is easy enough to deal with *

*> but I need to evaluate the very significant effects of the mass of metal *

*> in the heat exchanger, which also has to be heated (and cooled) and the *

*> speed with which heat flows through the metal. This was the reason for my *

*> original simplified question*

*>>*

*>> Besides, I'd defy you to hold your hand over a pot of rapidly boiling *

*>> water for any significant time, with or without a copper plate.*

*> OK. Please replace my fingers with some form of thermometer.*

*>>*

*>> You need to look up things like partial differential equations and *

*>> Laplace to get a grip on this kind of problem. See*

*>>*

*
http://faculty.gg.uwyo.edu/dueker/tensor%20curvilinear%20relativity/pde%20tutorials.pdf *
*>> section 2.*

*>> Thanks. I will look this up. See comment above also*

*>> Even in 1 dimension it's a bit tricky, with your problem more so. You *

*>> also need to allow for the fact that heat is not just being applied to *

*>> the part of copper that is submerged, but the section above water that is *

*>> being heated by the steam and condensation.*

*> I think I can handle the bits about heat going into and out of the parts *

*> of the plate OK but am still unable to figure out how the heat will *

*> actually move with time from the hot end to the cold end, particularly the *

*> time intervals involved*

*>>*

*>> How long?? I wouldn't like to say*

*> Thats the problem. I suspect you are better able to guesstimate it than I *

*> am since you took the trouble to reply.*

*> I expect you are getting fed up with this by now but may I impose a little *

*> more on you obvious expertise snd re-define my example a little?*

*> I will use cgs units since you seem to prefer this.*

*> I have two 100 cm squre plates with a thickness of, say 0.2 cm. These are *

*> connected to each other through a strip of the same gauge copper which is *

*> 20 cm long and 2 cm wide. One plate is in air at 20 deg C.and the air *

*> volume is large so generally, except in the immediate vicinity of the *

*> plate, Has a very slow rise in temperature ( air to surface R value, in *

*> the British system is about 0.67 due to the surface film. I don't know *

*> the value of R in cgs offhand). The other plate is also at 20 deg C *

*> initially but is now plaunged into my boiling water. The plate will heat *

*> up at a rate depending on its mas and specific heat and heat will start to *

*> flow along the interconnecting strip at a rate depending on its thermal *

*> conductivity and cross sectional area. The incoming heat will start *

*> warming the plate in air at a rate depending on a lot of factors including *

*> its mass; specific heats, thermal conductivity; surface area and *

*> temperature... To ask for a complete equation relating all this stuff is, *

*> I think, a bit much but all I really need might be what you have already *

*> given me which is how to relate the surface areas, specific heat, *

*> temperature gradients; and actual temperatures and themal conductivity to *

*> time, particularly through the connecting strip. If you can give me this I *

*> will be most grateful.*

*> Maybe I will do some actual testing to see what happens.*

*> Alan C*

*> *

Posted by *Simon D* on February 25, 2006, 7:35 pm

news:ePZLf.31844$%

*>> How long?? I wouldn't like to say*

*> Thats the problem. I suspect you are better able to guesstimate it than I *

*> am since you took the trouble to reply.*

The reason I'm aware of the maths involved is because I did this kind of

thing at University some 15+ years ago. To say I'm rusty on the details

would be putting it kindly <g>. I would recommend you ask your question in a

physics group, where you will get a better answer/answers, and they'll be

checked by others, so you can be more confident of the values you're given.

Good luck.

Simon.

Posted by *andre_54005* on February 24, 2006, 7:25 pm

Alan Combellack wrote:

*> I appreciate the help but am still having a problem. I don't understand*

*> how you arrive at an area of 4.8 cm in this calculation. Would you please*

*> explain. The expression "= 251 * 60 * 79 = 191" should read*

*> "= 251 * 60 / 79 = 191" Silly little thing but I am a bit out of my depth*

*> here and easily confused. Thanks again*

*> Alan C*

*> You should just try it and see.*

*> Here's my approx answer: about 3 minutes.*

*> heat flow rate = thermal conductivity x area temperature difference*

*> / distance*

*> therm cond = 3.86 W/cm-K*

*> area = 4.8 cm^2*

*> min temp diff = 100 C - 50 C = 50 C*

*> max temp diff = 100 - 30 = 70*

*> avg = 60 C*

*> dist = 14 cm*

*> max heat flow rate = 92.4 Watts*

*> avg heat flow rate = 79 Watts*

*> min heat flow rate = 66 Watts*

*> spec heat = 0.385 J/deg.C/g*

*> density = 8.9 g/cm^3*

*> mass = Volume * density = 73 cm^3 * 8.9 = 651 g*

*> heat capacity = mass * spec heat = 251 J/deg.C*

*> time ~= heat cap (J/deg.C) * avg temp diff (C) / avg heat flow rate*

*> (J/s)*

*> = 251 * 60 * 79 = 191 seconds (3 min 11 sec)*

The copper plate is very likely irrelevant. Try just holding your hand

6 inches above the surface of a "rapidly" boiling 13 inch diameter pot

of water.

______________

Andre' B.

> I appreciate the help but am still having a problem. I don't understand> how you arrive at an area of 4.8 cm in this calculation. Would you please> explain.