(Tough Problem) Temperature change of heated car metal due to forced convection

[JC1500]

Here's a tough one.

A 334.69kg car body with a surface area of 113 m^2 exits a oven at 124 deg C. While the car is moving at .0366 m/s, 27 deg C air is forced toward the vehicle from the lower sides at 15.24 m/s from within .30 meters away. What is the new average temperature of the car body after 3 meters or 83 seconds.


My equation: I EITHER BLEW SOMETHING SOMEWHERE OR USED THE WRONG EQUATION.

Heat Transfer Coefficient hc=10.45-V+101/2 hc=34.25 W/(m2K)

Heat Transfer (q)
q=(hc)(area)((surface temp)-(air temp))
q=(34.25)(113)((124)-(27))
q=352,506 W

Tf= Ti + [q/(m*specific heat of steel)]
Tf=124 +[352,506/(334690 grams*.5108 J/(g-deg C)
deltaT=2.06 deg C
Tf=121.94 deg C
Tff = 121.94 - (2.06*83) = -49.04 deg C (WHICH OBVIOUSLY CANNOT BE THE NEW TEMP)

After the first 3M the vehicle continues to a second 3M section but the forced air velocity goes down to 12.7 m/s

 

[SwinnyGG]

The above 'textbook' heat transfer equations are valid only for an instantaneous point in time.

Think about it- your end result is a direct function of the temperature difference between the part and the cooling air. As soon as the cooling air touches the part, the temperature difference changes.

In the real world, a temperature calculation like this is highly nonlinear. The answer to this problem is the solution to a differential equation- and it's a complicated one, since this part is very complicated geometrically and is exposed to cooling air incrementally (at a very slow rate, at that). At the end of 83 seconds, unless this is a little car, the whole body won't be exposed yet. Even if it is, it won't be the same temperature everywhere, because every part of the surface won't see the same actual heat transfer coefficient. So you'll have to deal with conduction in the part as well.

If this is a real problem you're trying to solve, you're going to have to do a CFD analysis on a very powerful computer, or you're going to have to actually measure.

 

[IRstuff]

How about doing a dimensional analysis before you go off the deep end? Is this for school? Student posting is forbidden.

Firstly, are your input values making sense? If the car body is steel, then 334.69kg/8050kg/m^3/(113m^2)/2= 735.9 micrometers, which is about 14 sheets of paper. The good news, if this is correct, is that the fact that you ignored thermal conductivity is OK, because the body is so thin.

Second, Your math in the calculation of the heat flow appears faulty; I get 375.4 kW

Third, dimensional analysis; your second equation is dividing joules by watts/degC, which should result in degC/s, which should tell you that your time resolution is completely honked up, i.e., you need to be running this calculation with a delta time of no more than about 10 seconds.

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[JC1500]

Thanks for the responses. No, it is definitely not a homework question, although I did write it like one. This is a real life scenario at an automotive paint shop. We are trying to cool a car body coming out of an oven. All details are existing conditions except for the air velocity. The plant is asking for projected cool down after our proposed work is completed.

As you guys said, maybe a answer isn’t achievable.

 

[IRstuff]

Where did I say THAT?

TTFN (ta ta for now)
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[JC1500]

IRstuff, I'm giving this one last try before telling the plant I cannot do this.

You're correct on my heat transfer equation. 352.2 kW was for another temp difference.

Here is the curve I'm working off of.
[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1525804862/tips/Oven_Curve_r1yh0u.pdf[/url]


Do you know of another formula that would help in this situation. Trying to figure out the maximum temperature a vehicle will be cooled after going to a modified cooling tunnel with new low level duct.

Another question: How would you factor in the 15.24 m/s of air being supplied on the opposite side at the same time?

 

[IRstuff]

As i mentioned before, the biggest issue was your time scale. You need to be running those equations on the time scale of single-digit seconds, say, 5; if your resultant change is less than 5%, then you can do the next time step.

The air flow will change the heat transfer coefficient, but there are curves for that sort of thing.

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[racookpe1978]

Is this very thin, curved part completely surrounded by the cooling air flow? If so, the thickest part of the thickest piece is going to have the greatest thermal mass. All of the thin metal sections will cool independently at very close to the same rate since heat flow is from a "middle" of the sheet out to the two surfaces.

 

[chicopee]

As stated above dimensional analysis should been done. This is how the last equation should look Q(watts= 0.9992cal/sec) = M(grams)* Cp(cal/gram-dC) * deltaT(dF or dK) / t (sec); You forgot to include time when you moved that equation around; the units are in parenthesis; watt is the rate of heat transfer which in this case =0.9992 cal/sec. the letter "d" stands for the degree symbol.

 

[JC1500]

The thickest part is of the vehicle is the frame work. The hood, doors, fenders tailgate, roof, etc. are thin. The area I am focusing on is the rocker, the area under the door edge the runs front to back. The ductwork we are proposing to install will supply high velocity air toward this section of the vehicle. The specific area and weight of this part of the vehicle is unknown, so I am using the total weight and total surface area of the vehicle to get and estimate of what the cooling will be.

Thanks

 

[chicopee]

Oops, my conversion from watt to cal/sec is incorrect; it should be 1 watt= 0.2389 cal/sec and not 0.9992cal/sec.