Calculating Soak Time in Oven to Reach Ambient Temperature 4

[KimWonGun]

Is the following logic correct to estimate the time required for a metal object in an oven (convection) to be heated up to ambient temperature?

Using the equation (in steady state) P = k* A * T^4, where k is the Stefan-Boltzmann constant, A is the surface area, and T is the ambient temperature, I solve for P to get power.

Using the equation W = (T2-T1)*C * m, where T2 is the oven temperature, T1 is room temperature, C is the object's specific heat, and m is the object's mass, I solve for W to get energy.

Dividing power P into energy W gives me the time required to raise the object's temperature (uniform) to the oven temperature.

I just need a reasonable estimate to ensure that the entire object is heated to the ambient temperature; overshooting is acceptable in this case.

 

[zekeman]

OK, I finally got around to reading your original post and some of the comments, and now realize that all you want is a conservative time for the surface to reach near the oven temperature ( you would have to wait an infinite time to get there).
Typically, we talk about getting to 90% or 95% of the oven temperature.

Now , your Biot parameter is in the range that makes a "lumped"solution valid,so you must solve the heat transfer differential equation

rho*A*c*w*dT/dt=sigma*e*A*(a^4-T^4)
For this problem, the assumption of a constant equivalent h will lead to a less conservative answer since , as the object temperature approaches the oven temperature, the equivalent h drops significantly.and, accordingly,you would get too short a time. So, a closed form solution is more appropriate as follows

rho*c*w dT=sigma*e*(a4-T^4)

t time
T temp of mass at time=t
a oven temperature
rho density
c specific heat
w thickness of mass
sigma 0.173*10^-8

let Q=rho*c*w/sigma*e for convenience
The equation becomes
QdT/(a^4-T^4)=dt

Integrating both sides from T0, initial temp to Tf final temp

t=Q/4a^3*[ln{(a+T)/(a-T)}]+Q/2a^3*arctan(T/a) limits T0 to Tf

In imperial units, I got

Q=50*.22/0.5/0.173*10^-8=127*10^8
where e assumed =.5

Now, as an example
Tf/a=0.9; i.e. Tf= 90% of the oven temperature
T0=530 R

Oven temperature a=1200deg R

t=12700/1728/4*(1.79)+12700/1728/2*(0.733-.38)=1.87*1.79+3.67*.353=4.64 hours.
I checked this against the h equivalent solution (using graphs bt Schneider) and found their answer to be
t= 6.45 hrs
which looks like I made an error someplace.
Will look at the math later when I get some time,or if you haven't gone to sleep, maybe you can.

 

[ione]

Zekeman,

What I have got is

t = Q/(4a) *[2arctg(T/a) + ln(a+T) – ln(T-a)] which is a bit different from what you’ve got


Moreover, taking

rho = 8000 kg/m^3
c = 500 J/(kg*K)
w = 0,075 m
sigma = 5.67*10^(-8) J*m^(-2)*K^(-4)*s^(-1)

Q= 105820.1058*10^8 K^3*s = 617142.857 R^3*s

 

[KimWonGun]

zekeman and ione:

Many thanks for your help.

Using partial fraction decomposition and the Pythagorean identity 1+tan2(theta)=sec2(theta), I was able to replicate zekeman's solution.

However, when using zekeman's imperial values, I calculate 4.84 hours. Where is my error?

 

[ione]

I've forgotten an exponent 3 in my previous post

t = Q/(4a^3) *[2arctg(T/a) + ln(a+T) – ln(T-a)]

This time I'm pretty sure it's ok and it's still sligthly different from zekeman's one

 

[KimWonGun]

ione:

Do you mean ln(a-T) rather than ln(T-a)?

Otherwise, since ln((a+T)/(a-T)) = ln(a+T)-ln(a-T) the solutions would be identical.

 

[ione]

Kim,

Yeah, that's what I meant.

I'm still puzzled on Q value reported by zekeman (without any unit I find it hard to sort it out how he got this)

Q=50*.22/0.5/0.173*10^-8=127*10^8

 

[KimWonGun]

ione:

I think zekeman reported the following:

50 is the product of the density and specific heat (BTU/lb R)
0.22 is the thickness (ft)
0.5 is emissivity
0.173*10^-8 is the Stefan-Boltzmann constant (BTU/hr-ft^2-R^4)

 

[zekeman]

Ione,

"t = Q/(4a) *[2arctg(T/a) + ln(a+T) – ln(T-a)] which is a bit different from what you've got"

I got Q/4a^3*[ln{(a+T)/(a-T)}]+Q/2a^3*arctan(T/a) limits T0 to Tf

The only difference is your term ln(a+T) – ln(T-a) vs my ln{(a+T)/(a-T)}

which from laws of logs are equal.
As far as the Q's they are almost the same except my time is hrs, not seconds.

 

[zekeman]

"However, when using zekeman's imperial values, I calculate 4.84 hours. Where is my error?"

What error; I posted 4.64 hrs which is close enough.

Ione,
I also used 0.5 for the emissivity included in my Q computation, not in yours.


BTW, I found the "error" in my comparison with the graphical one. My solution assumes that the surface temp reaches 90% of the oven temperature, so the parameter

(Tf-T0)/(a-T0)=(.9a-T0)/(a-T0)=(1080-540)/(1200-540)=0.818

I incorrectly used 0.9 in previous post. The corrected value yields a time of 5.60 hrs. Still a problem Should be less than 4.64 hrs or the 4.64 is too low.

 

[ione]

I reassure you I’ve entered emissivity e=0.5 in the formula and now I can see that I worked you’ve worked with hours and I worked with seconds. I finally got 6.52 hours (using 0.075 m as characteristic length) which falls to 5.83 hours (using 0.067 m as characteristic length).

 

[zekeman]

"which falls to 5.83 hours (using 0.067 m as characteristic length). "

I had an inversion typo(1.97 instead of 1.79) which changes my result to 5.00 hrs and since there is a discrepancy of between the rho*c values of 20%, that would explain the difference.

 

[zekeman]

Mystery solved.

Finally got to the actual graphical case for radiation from other Schneider curves and found

t= 4.84 hour

which remarkably is identical to Kim's.

 

[ione]

I have taken:

rho = 8000 kg/m^3= 499 lb/ft^3

c = 500 J/(kg*K) = 0.12 Btu/(lb*R)

I’m not that sure the specific heat of stainless steel remains constant in the temperature range of interest. Indeed I think it should increase with temperatures above 670 R, and this would imply an increase in soaking time, but this is splitting air. What matters is that results seems to match quite well.