Equation for Heat Transfer of Water

[freedomtech]

I need to calculate how long it will take to heat water with two concentric heated pipes. Specifically, one pipe 6" in diameter, 6" tall and a second cylinder 5" diameter and 6" tall, so there is a 1" gap between them full of water.

Question: If we heat the outside of the larger cylinder to a specified temp and the inside of the smaller cylinder to a some temp, how long will it take for the water between them to get to 200 deg F? The water starting temp is 45 def F.

Ideally an equation for this would be great so I can run different scenarios of cylinder sizes and temperatures.

I started to figure it out but it's been a while! Thanks for any help!

 

[Yorkman]

I'm not sure I follow this or not, is the 5" dia. pipe inside the 6" dia pipe? I not sure of your diameters either, are these out side diameters? If the 5" goes inside the 6", what is the wall thickness of the 6", what ever the thickness you will have far less that a 1" water gap. You didn't mention what temperatures you were going to heat these pipes to, or the type of material these pipes were made of. That might be helpful in developing your equation.

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int.

 

[desertfox]

Hi freedomtech

You could estimate the heat input required the heat the water to 200F from 45F using :-


Q= m*Cp*temp change


where m = mass of water between cylinders

Cp = specific heat capacity of water

temp change = 200-45

Alternatively you could calculate the heat transfer by conduction through the cylinders into the water but you would need to know the thermal conductivities of the cylinder materials as well as water and in addition you would need cylinder starting temperatures as well.
If you refer to any heat transfer books the formula for
heat through conduction will look like this:-


Q= -k*A*dT/dx

where -k= thermal conductivity

A= area of conducting material

dT/dx = temp gradient across thickness of conducting
material.

Once you know the transfer of heat required you can work out how long it will take from the power rating of the heat source.

regards desertfox

 

[ko99]

okimion,

Solar heating is tough to nail down -- solar intensity is very sensitive to a number of factors and of course it's also a transient problem due to the constantly changing angle of the sun's rays.

Standards have been developed by ASHRAE. You may also want to check out Bellcore Spec GR-487-CORE.

Also, here's an article summarizing how one product was simulated, designed and tested for solar heating:

http://www.flomerics.com/flotherm/technical_papers/t258.pdf

ko (

http://www.ecooling.biz)

 

[ko99]

oops! my reply was intended for a different thread. please excuse...

ko (

http://www.ecooling.biz)

 

[freedomtech]

desert fox,

Thanks for the answer. That is what I was looking for. Can you take a look at the following. It is using the first method your recommended.

Q = m*Cp*TempChange

Assuming:
Temp Change = 85.8 C = 359 K
Water Volume = 2 Liters

m = p*V where p = density of water (1kg/L)
=1kg/L * 2 L
= 2kg

Cp = specific heat capacity of water
= 4.183 Joules/(kg deg K)


Therefore,

Q = m*Cp*TempChange
Q = 2kg * 4.183 Joules/(kg Deg K)*359 deg K

= 3003 Joules = 0.83 Watt Hours

Taking this info to calculate time using
Energy = Power * Time

E = Pt
where

E = Energy in Watt-Hours
P = Power in watts
t= time in hours

I need to heat the 2Liters of water in 3 minutes. So

t = 3 minutes = 0.05 hours

E = Pt
0.83 Watt-Hours = P * 0.05 hours
P = 0.83/0.05 = 16.6 Watts

This means with 16.6 Watts I can heat up 2 Liters of water in 3 minutes?

Is this true? 16.6 Watts does not seem like much power.

Am I doing this correctly?

 

[freedomtech]

desertfox,

I also want to use your other method Q=-k*A*dT/dx

The problem is for 2 concentric aluminum cylinders. Larger has radius of 3 inches, smaller one has radius of 1.6 inches. Both are 6 inches tall. The water is in the area between the 2 cylinders, touching both cylinders. Heat is applied to the outside of the larger cylinder and to the inside smaller cylinder. This way the water gets heated from both cylinders.

Using the equation Q=-k*A*dT/dx

Smaller Cylinder - C1
Q1 = -k*A1*dT/dx where A1 is the area of smaller cylinder

A1 = 2*PI*r1*h
dT/dx = temp difference between inside wall of small cylinder and outside wall of small cylinder ????

-k = thermal conductivity of aluminum ???

Next I calculate Q for outside wall of small cylinder to some point in the water where dT/dx is temp difference between outside wall of small cylinder and the water -k is thermal conductivity of water?

Do same for large cylinder going going the other direction into the water?

I am sure I have this wrong. Where does thickness of cylinder walls come into play? (assume 0.25")


I really want to use this method because it will give me some insight into the wall temperatures.

ANY help will be greatly appreciated.

 

[desertfox]

hi freedomtech

Yes for the first method I agree with you at 16.68 watts.
However you need a good margin to allow for losses etc but the methods correct.
I will espond later to the 2nd method.

regards

desertfox

 

[timbones]

Hate to point out math errors, but 85.8°C = 85.8 K not 359 K if you are dealing with temperature differences.

I think your energy transfer calculation looks fine, but the power formula used is probably not quite right as you will probably need to consider energy losses to the surroundings. I would go to the next step and do some heat transfer calculations.