Correct Formula?

[PLCKing]

I have a 24" dia x 22" long cyclindrical aluminum tank 1/4" thick. This tank is filled with diesel fuel. I need to place this tank inside a reactangular (26"x26"x16"high) aluminum tank 1/4" thick that will be filled with water. The water tank (rectangular) will have an inlet and an outlet to be connected to a 3kW electrical heater that will circulate the water and heat it to 50 deg C.

I'm trying to figure out:
1: Assuming there is no heat loss from the rectangular tank and the water pipes, how much heat loss will there be from the top 1/3 of the cylindrical tank that is sticking out of the rectangular tank when the diesel fuel is at 50 deg C and the ambient air is at 20 deg C.

2: Assuming no heat loss, how much time will it take the 3kW heater to heat up the water in the recgtangular tank and diesel in the cyclindrical tank from 20 deg C to 50 deg C>

I haven't started calculating part 2 yet because the results I'm getting for part 1 I'm unsure of so I'd like to figure that out first.

For part 1, I'm assuming the diesel is at 50 deg C and the ambient air is at 20 deg C. I'm using the formula Q = (kA(Thot - Tcold))/d
k for aluminum is 205 W/m.C
A = 0.55 m^2
d = 0.25"

When I calculate the heat transfer (Q) I get approx. 530 kW. This seems too high to me. Am I using the correct formula? or is there another way I should be looking at this problem? I appreciate the help in advance.

PKing

 

[IRstuff]

No you're using essentially a thermal conductivity equation. The fact that you came up with watts is a sign that you used the qrong equation. You presumably were trying to calculate total heat, which should have joules as the resultant unit. You need the specific heat for all the elements involved to determine the energy change.

TTFN

FAQ731-376

 

[JStephen]

You're calculating the heat transfer through the aluminum only. You need to calculate the convective heat loss from the outside of the tank instead. You could probably assume the inside and outside of the tank were the same temperature for this. Get a good heat transfer textbook and see what you can find. Be prepared to approximate.

 

[PLCKing]

Thanks for the replys guys. I see so what I've been calculating is the maximum amount of heat transfer through the exposed aluminum at a 30 deg C temperature difference. I agree that convection accross the exposed top cylindrical tank is a better way to go. I'll look through my book to see what I can figure out.

 

[prex]

Can't exactly see how you calculated the exposed top area: the flat(?) top head is 0.29 m[sup]2[/sup] and 6" of shell is 0.29 m[sup]2[/sup].
Assuming anyway your figure (that's not very different), you can use h= 10 W/m[sup]2[/sup]°C as the heat exchange coefficient to calm air in the formula
Q=hA(t[sub]w[/sub]-t[sub]a[/sub])
where the heat resistance of the aluminum wall can be neglected.
This gives 165 W of heat loss, but you'll have to account for other losses of course (through the bottom, from the outside of the external container and from the upper surface of water in the external container).

prex

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[PLCKing]

The area was based on a 1/3 of the cylindrical tank sticking out of the rectangular tank. I worked it out to 0.55 m^2 including the flat ends of the cyclindrical tank.

Now assuming the cylindrical tank is fully submerged inside the rectangular tank that's full of water at 50 deg C, and the cylindrical tank is full of diesel fuel, how much heat is transfered to the cylindrical tank to get the diesel fuel up to 50 deg C from 20 deg C.

I was tring to solve this using the thermal resistance concept for a holow cylinder (cylindrical tank) + a solid cylinder (diesel fuel), but I can't find the thermal coefficient (h) for diesel fuel. Also would this be the correct approach? I'm getting some high numbers for Q which is probably not right.

 

[prex]

The exchange coefficient for natural convection on fuel side should be calculated from the physical properties of the fluid. I guess the order of magnitude is 100 w/m[sup]2[/sup]°C, supposing the viscosity of fuel is substantially higher than that of water, but don't take this as an estimate, I didn't calculate it.
In the calculation of the heat exchange through tank wall you could neglect the heat resistance water side, as the water will be moving under forced circulation.
If you want to estimate the time for the fuel temperature to go steady, than you will need to accept a very rough estimate. Don't try to calculate the temperature distribution inside the fuel mass, it's a very complex (and almost unsolvable) problem.
Another important point is that the water should be substantially hotter than the fuel temperature you want to reach (say at least 60°C if you want to go to 50).
To estimate the time of warming you must determine the amount of heat required from the fuel mass, specific heat and temperature rise. Then you can calculate the heat flux through tank wall from the heat exchange coefficient fuel side assuming the fuel is at an average temperature (35°C).
Finally you'll have to check wether the supplied power is enough by adding all the losses to the power used to heat up the fuel.
This is not an easy calculation for someone not used to thermal problems.

prex

http://www.xcalcs.com

: Online tools for structural design

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads