How long to cool Steel down to -50C 1

[TheBeardedOne]

Hi
I have an application where i need to have a steel bar at -50C.
The steel is 420 Stainless Steel
The Freezer is set at -50
The dimensions of the bar are 134.8mm diameter x 470mm long
It will start at ambient temp 22C

Can anyone tell me how long i should leave it in the freezer for to ensure that the core of the bar is at the required temperature.
Thanks

 

[IRstuff]

To what uncertainty? Since the target temperature is the same as the driving temperature, the answer could be infinity.

Presumably, your freezer's cooling system actually goes below -50°C in order to maintain -50°C. So what happens if the bar gets below -50°C?

TTFN

FAQ731-376

 

[TheBeardedOne]

The Freezer has the capability to go down to -70C.

I am not concerned if the steel is below -50C

Cheers

 

[chicopee]

I normally dont like to give a direct answer since I believe that you are not learning how solve such problem. This type of unsteady state problem is highly emanable to spreadsheet application. So the basic premise of such problem is that the rate of change of the bar's internal energy equals the rate of heat transfer to its environemnt. Also to simplify this problem assume that there is no temperature gradient in the bar as long as the Boit's number is less than .1. The basic equation would be:
dU/dt=Qr+Qc >m*Cp*dT/dt= e*s*As*F*(T1^4-TA^4)+Hc*As(T1-TA)
Get a book on HT for an explanation of each term. If there is a temperature gradient than the internal energy side will become a lot more complicated.

 

[corus]

In this case the Biot modulus appears to be small such that the transient heat flow equation reduces to one of p.Cp.V.dT/dt = -h.A.(T-Ta) as chicopee suggests. V is the volume of the body (Pi.r^2.L for a cylinder) and A is the cooled surface area (2.Pi.r.L). h is the 'nett' heat transfer coefficient, probably from radiation, and Ta is -50C. The solution is an exponential function.

corus

 

[chicopee]

Corus, you may have to consider radiation loss at -50C or at -70C as IRStuff explains the reason well.

 

[corus]

yep, I mentioned that, and radiation is probably the means of heat flow, but solving for radiation analytically is a bit tricky. On second thoughts though, if you seperate the radiation into a cubic term x (T-Ta), and sucessivelly estimate the cubic term as the constant h, then you could do it by iteration in a spreadsheet using goal seek. You can improve convergence by using one step of Newton's method, if I remember.

corus

 

[corus]

This link might help

http://www.cs-pi.org/uces/archive/classes/CNA/dir3.7/uces3.7.html#3.7.2a

corus

 

[IRstuff]

I think that even a distillation into Excel would allow the Solver to crank out a numerical solution. You would set up the heat balance equations and whichever program you use to solve for the temperature as required. With enough brute force, you could set up a time series that you could run in Excel, or Mathcad, or Matlab.

TTFN

FAQ731-376

 

[corus]

I'm not sure what IRstuff is referring to as it's not a heat balance type problem as it's a transient thermal problem. One way I found to solve the non-linear differential equation you get, was to use a Runge-Kutta finite difference scheme in excel. There is an analytical solution for radiation transient heat flow but it involves finding the roots of an equation involving tan and logs etc., and not worth the effort. The answer you get depends on the emissivity of the material, but using 0.35 as a typical value for stainless, I get the answer to be about 58 hours (2 1/2 days) for the core temperature to get down to -50C from an ambient of -70C if heat flow is by radiation from the outer surfaces alone.

corus

 

[TheBeardedOne]

Thanks Corus,
Can you post the equations etc that you used, as i am going to have to do this calc with various different steels and different dimensioned parts so a step through would be a great help.
Thanks

 

[corus]

I used a macro with these functions (simplified)

Function f(x)
'x is the temperature at time i
f = sig * eps * ((x - Tzero) ^ 4 - (Ta - Tzero) ^ 4)
f = -2. * f / (density * Cp * r)
End

then in the main section, the 4th order Runge Kutta scheme :

k(1) = delt * f(T)
k(2) = delt * f(T + k(1) * 0.5)
k(3) = delt * f(T + 0.5 * k(2))
k(4) = delt * f(T + k(3))
T(i+1) = T(i) + (k(1) + 2# * k(2) + 2# * k(3) + k(4)) / 6.

where delt is the time increment and T(i+1) is the temperature at time T(i) + delt

sig is the stefan boltzman number, eps is emissivity, Cp is specific heat, Ta is the ambient, Tzero is absolute zero.

The equation for f(x) is for a long bar. For any other section profile, or shape, the equation would have to be adapted based on Surface area/Volume

corus

 

[IRstuff]

For transient problems, heat flow balance usually referred to as "continuity" which simply means that conservation of energy is maintained at each point in the system.

TTFN

FAQ731-376

 

[moltenmetal]

If the freezer is at identically -50 and the desired temperature is identically -50, the correct answer is never...

 

[corus]

Correct. I assumed the ambient was -70C.

corus