design cooling tunnel

[punk]

I have to figure out to cool sheetmetal cabinets exiting a cure oven. parts are 350F, need to cool parts to 100F in 5 minutes. parts are hung from an overhead conveyor system moving at 10 FPM with 15 lbs/ft of parts. I plan to construct a sheetmetal tunnel (5 min x 10fpm = 50 ft long)with supply ductwork. We will use supply fan (ambient air) to pump in air at the exit and use an additional fan to exhaust air at the tunnel inlet (cross-flow heat transfer). can anyone tell me theoretically how to calculate the required cfm, keep in mind winter ambient (-15F) versus summer ambient (90F). Also, is it a good idea to mix the winter ambient with indoor air to create a 70F temp. so we don't "shock" the 350F parts?

 

[pmover]

a litle ingenuity may be needed here...

it is required to cool X mass of metal to 100°F in 5 minutes from an initial temperature of 350°F. the heat removed from the metal is q = m Cp dT.
primary mode of heat transfer to air is by convection (q = h A dT), with little/negligible radiation.

Soem thoughts...

is it possible to set up some experiment(s) to determine initial rates of cooling? For example, take an unused sheet metal part, heat it to 350°F, and then subject the part to natural ambient conditions, measure the surface temperature with respect to time. then determine what the heat transfer coefficient would need to be in order to remove the heat from the metal within required time frame. in fact, determining the heat transfer coefficient (h) is needed for all conditions (summer/winter/blend).

my initial thought is that during initial exposure to atmospheric air, there would be substantial heat transfer to the air (but must gently remove the air surrounding the metal), until the temperature dropped below a certain point, then perhaps forced convection would be required to further aid in lowering the temperature.

since the heat transfer coefficient can reach a limit (i.e. ~5 Btu/hr-ft^2-°R), it may be necessary to control the temperature difference between the air and metal (i.e. blending of air).
hope this helps.
good luck
-pmover

 

[punk]

it will be hard to predict how the air will flow over the area of the parts since they are on a conveyor and about 3 ft apart. how to design the ductwork to direct the air through the parts so that we can get a "sweeping" effect over each part is another thing. we currently have the system running and in ambient air, the parts will cool to 100F in 12 minutes from leaving the oven.
Can we use Q= m x Cp x (t2-t1)= 1.08 x cfm x (assume 30 delta T) and solve for cfm? I am not sure how to tie in the Q=hA (t2-t1) formula, can you be more specific?

 

[quark]

pmover put good thoughts into the problem. It is always better to have ground work prepared rather than making blind assumptions (I am also weak at that)

You will come to know howmuch heat is to be removed from the cabinet by mxCpx(t2-t1) and if you equate it to
hxAx(T2-T1) you have (T2-T1) i.e raise in air temperature.

Now you can use the formula for sensible heat rise 1.08 x cfm x (T2-T1) to know the cfm.

 

[quark]

I am sorry for the blunder I did. I found it just after pressing the submit button.The idea is to know whether the amount of heat that is to be removed can be convected away in the air.

Once you are sure of it then you can do some trial and error to know the temperature difference of air and then cfm.

If you want a correlation for heat transfer coefficient see Heat Transfer by Holman or let me know. There are approximations for heat transfer coefficient for air flow across square cylinders.

Regards,

 

[punk]

quark, i don't have info. on heat transfer by holman, could you supply info, thanks.

I have the part area, T=350F, cool down time in 70F ambient
is 12 minutes, cooled to 100F. I could find the actual "h" value, do you know the heat transfer convective coeffficient for steel to air in still air and also in moving air (Btu/hr-ft^2-R)

 

[quark]

Punk!


Here goes the equation.
(hxd)/k_f = c(u_ad/v_f)ⁿxPr^1/3

For a square cylinder (I think that approximates your condition) c = 0.102 and n = 0.675

d = side of the square
k_f = conductivity of air at film temperature (average of metal and air temperature)
u_a = velocity of air
v_f = kinematic viscosity of air at film temperatuer
Pr_f = Prandtl number at film temperature
This equation holds good for 5000<Re_df<100000
Re_df is Reynolds number with respect to duct.

You should tell two things for the solution. 1. Surface area of cabinet and 2. MOC or specific heat.

However it is better if you do the rest of the calculation

If your cabinet is covered from all sides then you have to provide cooling to the face which lies back to the airflow by diverting air to this face. If it is open on one side, keep the open face against air flow.

Best of Luck.