# Heat transfer equation

1. Nov 17, 2016

### rancam

1. The problem statement, all variables and given/known data

An aluminum cylinder with an interior area of 4.1875 sq ft (or 603 sq in) contains 7.445 lbs of CO2 gas under pressure.The volume of the cylinder is .582 cu ft.The temperature of this gas is 110 deg F.I want to know how much heat (in btus) will be transferred from the gas to the cylinder walls ,where the temperature is 80 deg F,in one second.I understand that the equation for heat transfer is q = KA (T1 – T2 ) / L. The K is the coefficient of conductivity.For CO2 it is .0081 btu / hr.and for aluminum it is 123 btu/ hr. How should I calculate the heat transfer?

2. Relevant equations q = KA (T1 - T2)

3. The attempt at a solution
I think I should use the co efficient for CO2 so the equation should read q = .0081 btu /hr 4.1875 sq ft ( 110 - 80 ).I am not sure what I should use for L.Should it be half the radius of the cylinder?

2. Nov 17, 2016

### haruspex

Your heat transfer equation is for transfer through a solid of constant cross-section in steady state.
You specify one second. Are you asking about the first second after the initial conditions are created, or are you wanting the flow rate per second over however long?
For the first, you can treat the gas as effectively a solid, but it will definitely not be steady state.
For the second, convection will become important.

3. Nov 19, 2016

### rancam

Sorry I forgot to mention that the pressure is 855 psig and there is about a 10 % exit of the gas with a simlultaneous 10 % entrance of CO2 so that at the end
of the second the pressure and weight of the gas is the same.Can I use the heat transfer formula with the CO2 coefficient (.0081 btu / hr) and half the radius
of the cylinder (which should give me the average distance that the molecules can travel).Since CO2 gas is a good retainer of heat I did not think that the aluminum coefficient would be relevant.Basically what I want to know is how much heat will reach the walls from the gas in one second.I think it would be a very small percentage of the heat content of the gas.

4. Nov 19, 2016

### haruspex

No. 1 second is too short a time for steady state transfer to be established. I would think the distance from the wall over which much transfer will take place is quite short, so you can effectively treat it as transfer through a thin slab. And I would take the wall to be an infinite heat sink, so its temperature does not change.
The solution takes the form $\theta(x, t)=\Sigma A_ne^{-\lambda_n^2t}\cos(\lambda_nx)$, or somesuch, where the constants An and λn are to be found from the boundary conditions.
Are you familiar with Fourier analysis?

One doubt I have, though: as the gas adjacent to the walls cools it shrinks; the more central gas then expands, doing work. My guess is that you should therefore treat the cooling of the peripheral gas as constant pressure.