Size of Air Conditioner 1

I would not use AC but just 100% ventilation air with blower on one side and exhaust air on other side. The theoretical air flow can be calculated using the following heat capacity equation of air:

Q = 1.1 (CFM) (delta T)

Q = heat to be absorbed/removed by the ventilation air = 500 HP = 1,273,349 BTU/H
CFM = required air flow in cubic feet per minute
delta T = raise in temperature of air in deg. F

So if you have say 90F outside air (hottest air during summer) and you want to limit the increase in temperature to 200F the delta T = 110F

Therefore solving for air flow = 10,524 CFM.

Note an AC to cool the inside would require 1,273,349/12,000 = 106 tons which would be an enourmous AC unit that would not be much better than ventilation air. The minimum supply temperature of AC will only be about 50F which does not do you any much better than ventilation air and ventilation air is free except for the blower cost and power.

I understand your approach and reasons why I did not take that route is for the following reasons.

The inside of surface of the cylinder is 350 F. The outside of the cylinder is the surface that is in contact with the air.

The air would have to take in enough heat energy from the outside surface of cylinder so that the heat travels from the inside of the cylinder to the outside.
I would need to calculate how much heat must be transferred from the outside surface to get 200 F on the inside surface.

Wouldn't this be a problem of conductivity and convection? (Ignoring radiation)

A drawing or sketch would help as would an explanation of this huge heat generation. Where does that come from?

This makes zero sense to me at the moment.

And a 4" thick vessel weighing nearly a ton isn't something you come across every day.

Time of this cool down is a key factor.

One day js s whole heap different to one hour...

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The heat is generated from electricity. We just know that the temperature is as a result of, loss of power (500 hp) and assuming that this loss in power is heat. The temperature is also measured on the inside surface.
Here is a sketch.

I don't know the whole picture. As Little Inch indicated it would be best to show some kine of diagram of what the system looks like and where is the heat source.

So the inside of the cylinder is where the heat is being produced like some kind of resistance heating effect? And you want to maintain the inside of the cylinder at 200F? Then put most or all of the ventilation air on the inside of the cylinder.

Snickster, the first part is correct. The air will be blown on the outside surface of the cylinder and blown out from the ventilation outside. So I believe the heat needs to transfer via conductivity (passing the thickness of cylinder to the outside surface) and then through convection.

So you are saying that the inside of the cylinder in not accessible to add ventilation and the only heat removal mode possible is through conduction and convection through the cylinder to the outside surface where ventilation or colder air is input?

Well I think it just becomes a HVAC problem with the cylinder inside being like a space/room that has an internal heat source of 106 tons. The temperature will rise in the cylinder until the heat transfer out equal 106 tons at that temperature. The heat transfer will be based on the equation:

Q = UA (delta T)

U is the overall heat transfer coefficient based on conduction and covection across the cylinder wall. You can only increase this value a little by blowing air at a high velocity across the surface of the cylinder to increase the outer surface convection coefficient, but this would only increase U by a few percentage points. I would think that what will happen is that equillibrium would occur at a very high internal temperature of hundreds or thousand plus degrees on inside of cylinder (delta T) before heat transfer out equals 106 tons.

The bottom line is that 373 kW needs to be removed from cylinder. Using 11 C air temp, you'd need 102 W/m^2-K convection coefficient to get the required surface temperature, which requires something on the order of ~17 m/s air velocity, which is 38 mph air speed, according to the chart below, from I can't speak to the veracity of the chart from the paper, since the Engineering Toolbox chart would seem to imply needing something closer to supersonic airflow.




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As with many problems, a complete and unambiguous problem statement is oh so helpful in getting to a solution.

Is this a steady-state or transient problem? If transient, what are the time limits? What are the initial conditions?
Does the ID surface temp need to get up to 350°F, then be brought down to 200°F, or is it never allowed to go above 200°F?

Electrically generated heat source, yet you're using HP as units?

Radiation is likely not negligible.

Trusting Snickster's calc as a decent starting ball park, this is what a 105 ton chiller looks like:


However, from your "On one face of the cube is attached an AC" I have the impression that your expectation is something like this:


@IRStuff: The air temperature of that chart is a long way from 350°F.

The chart is showing air temperature, not surface temperature. However, while the desired surface temperature is 200F, the exit temperature isn't anywhere near that high. Assuming a 6-inch layer air interacting with the surface, I get something like 43C temperature rise, so earlier calculation works with approximately -30C supply temp.

If we take the temperature rise into account and use 25C air, then we'd assume something like 50C average air temp and need something like 25 to 30 m/s air flow, since the convective coefficient would need to increase to about 130 W/m^2-K. At 25 m/s flow rate, the outlet air rise temperature should be around 30C.

It's hard extrapolating from the chart, so there's lots of opportunities for error factors of 2x; regardless, the calculation approach would be the same, just with different convection coefficients and different air velocities.


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Also depends on which way the air is blowing inside your strange cube?

Blowing perpendicular to the large tube won't get an even air flow so there is a loss of efficiency compared to blowing air axially along the tube from one end.

But this seems like a lot of wasted heat here??

Air is probably not dense enough for your purpose and you end up with very high air velocities so you might need to consider water cooling.

Air conditioning for this use seems not to be very beneficial.

The conduction through the metal will be an order of magnitude more than the convective heat loss so is not a major factor here. The mass of the steel though is if you're looking to cool this strange item down whilst still generating heat.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

Thank you Snickster and IRsruff.
MintJulep and LittleInch

Let me put it in more perspective. 500Hp is 373KW. I am not sure why that is odd to note. It's an electromechanical system. Some amount of voltage is applied to convert it into mechanical power. Some power applied is lost in heat which is 500 hp or 373 KW. This loss of heat or heat generation is continuous. The temperature inside the cylinder is 350 F initially because of the heat.

The cylinder itself is placed in the "Strange Cube" I don't why that is. The AC is attached to the face and blowing in the cube that has the cylinder and vents out on the opposite face. The air will over the cylinder and yes efficiency is lost, hence you can consider turbulent flow. I need to bring the temperature down to 200 F inside the cylinder.

You are more than welcome to attempt the problem or help like Snickster and IRstuff have so far. Thanks

OK, then what I calculated is basically the approach.

> you use 500 hp to determine the heat transfer coefficient required of the cooling configuration, and you might need to adjust the surface area downward to account for the flow inefficiency
> you then use the thermal mass capacity of the airflow to determine the linear velocity or volumetric flow of the air and iterate the assumptions as needed, including the boundary layer depth.

So, note that my last calculation wound up with a 56 mph linear velocity, which is pretty crazy; note also the volumetric flow of about 30,000 cfm, which is likewise crazy. These suggest that you need a MASSIVE cooler with an outlet temperature below 0 degC, like say -30 degC. However, since that's below freezing, there will be a crap-ton of condensation to deal with, which does suck cooling capacity from the air stream.

Note also, it's typical that if you want higher transfer efficiency, you'd consider massively increasing the surface area for the heat transfer or consider liquid cooling, which would be 100 times easier vis-a-vis heat transfer, although, obviously, plumbing and pumps are another matter altogether. Note that this latter case simply prolongs the agony, to some degree, since the heat must then be removed from the liquid, but you would have fewer limitations on surface area, and you could use some evaporative transfer at that point.

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MOjo97 said:

You are more than welcome to attempt the problem or help like Snickster and IRstuff have so far. Thanks

Click to expand...

Both Snickster and IRstuff are assuming that all of the air flowing through the enclosure will have the same effectiveness at cooling the cylinder, and that isn't the case.

In "ideal" conditions, air flowing over a cylinder looks like one of these cases:


Your case is far from ideal.

Much of the air will not contact the cylinder at all. Much of the cylinder's surface is likely to be exposed to relatively stagnant eddy's, and thus have a far lower convective coefficient that IRstuff shows that you'll need.

I'll join LittleInch in saying that air cooling probably won't do the job.

Your configuration has some similarities to an indirect rotary drum cooler with a deluge system, other than your drum isn't rotating. You might consider a similar deluge water system.

Assuming this isn't homework, what is the actual project? Sounds like OP is trying to build a heat exchanger. Normally you use something with large surface areas and turbulent flow.

Is water an option? What is the gas in question? Would a cooling coil be an option? Is use of the waste heat an option and desired?

Much more information required to suggest a good method.

IRstuff are assuming that all of the air flowing through the enclosure will have the same effectiveness at cooling the cylinder, and that isn't the case.

Click to expand...

I knew that at the get-go. The point was to show that even a perfect case would result in near-absurd requirements; you either wind up with absurd velocities or absurd temperatures.

The only possible alternative to liquid cooling might be to increase the effective surface area by a factor of 10 or so, such as possibly using pin fins everywhere.

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IRstuff said:

The point was to show that even a perfect case would result in near-absurd requirements; you either wind up with absurd velocities or absurd temperatures

Click to expand...

I got that.

The OP, maybe not so much.

Additional information.

To answer every question so far, yes, I am trying to build a heat exchanger for an AC motor. The motor has a core that gets to 350 F on the inside surface. This core is enclosed by a cylindrical frame.
It is assumed that the core and cylindrical frame are in contact with each other throughout the surface area. This frame has another frame outside which is the 144*144*144 Cube. Ambient temperature inside the cube was measured as 85 F.

This cube already had 2 fans on its face. The inlet temperature was 78 F and exhaust was at 135 F. With this system the temperature of core came down to 267 F.
I am looking to replace these fans with AC so that the temperature of the inlet air does not change with the ambient temperature.

There is no option for water system because of supply and space constraints.

Missing information from my end: Don't know if there is a baffle inside that is redirecting the air to cover the cylindrical surfaces. I have so far assumed a turbulent flow and that somehow it does flow over cylinder out to the vent.

The 500 hp loss of heat is another assumption based on the rated power of motor and its efficiency. I thought that is helpful information. Originally, I assumed that the temperature inside the core (350 F) is being maintained somehow. It probably is a better option than to assume 500 HP is the power loss, generating the heat to get it to that temperature.