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.

• https://files.engineering.com/getfile.aspx?folder=1595749d-bc24-4ba0-8830-8

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 https://www.mdpi.com/2073-4433/13/2/149. 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.

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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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Quote (MOjo97)

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

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. https://feeco.com/look-indirect-coolers/ 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.

Quote:

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.

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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Quote (IRstuff)

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

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.

Quote (MOjo)

This cube already had 2 fans on its face.

Do those fans have data sheets?

(MintJulep)
Do those fans have data sheets?

Nope, could not pull the data sheet because the nameplate was missing and no indication of what kind of fan it is.

Wouldn't the motor or equipment manufacturer have a cooling solution?

What type of actual application is this? If there are 500hp waste heat, and we assume 90% efficiency, that would be a 4,500 hp motor. I would think an application that large gets a professional design by the OEM and not a DIY solution. You also need to dissipate the 500hp + compressor power somewhere.

I don't know what you mean by "core" of the motor. Motors have rotors, stators, windings etc. that all have temperature limits. You don't have control over heat transfer within the motor. the manufacturer will have requirements on the conditions at their interface to the outside (outside the motor). i suspect a motor that large may have integrated sensors and that is why you know the temperatures? You might want to use those to control cooling.

mojo97,

Can you please stop drip feeding vital bits of information. Post a proper dimensioned diagram of this motor and supply proper information.

This hollow cylinder you refer to in the OP does not appear to be solid. As solid cylinder 58" ID, 66"OD 12 feet long would weight 16.5 tonnes, not one tonne. So what is it made of? How does heat transfer from the inside to the outside?

Is this actually a 500HP electric motor? in which case your heat output is probably about 50hp or less.

You don't need AC here, you just need some bigger fans, but there is no way this core is outputting 500hp / 375 kW.

Without establishing the actual design data, no one will get anywhere other than wild assumptions and calculations using data which is patently incorrect.

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Quote:

The inlet temperature was 78 F and exhaust was at 135 F. With this system the temperature of core came down to 267 F.

Regardless of my calculations, this actual, useful, information that you've been hiding already answers your presumed question; you'd need inlet air close to 0 degF to get the core down to 200 degF. There is no conventional A/C that would get you air that's below freezing.

If nothing else, you could at least get some measurements of airflow to figure out what your apparent heat load is; you have inlet/outlet temperature and if you had airflow, you'd have a much better idea how much heat you're removing as well as how efficiently.

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The details are drip fed at customers information leaking speed.

Anyway, I found out the core is 30 inches in length. So basically, the cylindrical area of focus is 2*pi*r*l. The core is carbon steel.
Any other cooling system is not an option because of space constraints.
I do not have the data for air flow CFM and been asking for it for a while.

Appreciate the help.
I did attach a drawing to the OP.
I understand how the problem has been thrown every which way.

Regardless, you really don't have many options
> Refrigeration unit to cool inlet temp closer to 0 degF -- that seems absurd, since it'll likely be several times the volume of your box and probably require more maintenance, although, transferring the maintenance risk to something that has to stay online might be worth it.
> MASSIVELY increase the size of the fans -- but reliability/noise/vibration will get worse
> Attach fins to the cylinder to massively increase the exchange surface area -- this seems the simplest thing, assuming it hasn't already been done

you got an ostensible 80 degF drop with your existing setup, so that's doing pretty well and I would focus on that, rather than the supposed power consumption, since the deltaT is a measured fact and does not assume any data not in evidence, other than a presumed 350 degF apparent source temperature. Using that, doubling the efficiency of whatever is doing the heat transfer will hypothetically get you to the desired surface temperature, such as doubling the heat transfer area or doubling the air flow, or dropping the inlet air temp to 0 degF. You just have to pick your poison and live with the consequences.

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Let me know what is wrong with these steps:

Calculate the heat rate using the inlet and outlet air temperature. I will assume any existing CFM AC. Example 4000 CFM
Heat loss rate = Mass flow rate x specific heat capacity of air x Delta T

For this particular heat loss rate calculated we can determine what the outside surface temperature to be when the inside surface of the core is 200 F. Thermal conductivity of carbon steel is used.
Use this to outside temperature and inlet air temperature to determine what the heat transfer coefficient is for convection.

From the heat transfer coefficient find what the velocity of air is using the empirical formula.

Keeping this aside before we determine if that velocity is enough, we calculate with the current CFM how fast does the temperature go down. If its too slow then we can probably think of a baffle design to increase the velocity and improve the heat transfer.
If that doesn't work either we increase the CFM and re do the steps.

Older designs likely used experience of how many fins etc. a motor needs to have and what airflow is required. Modern designs may use CFD and heat transfer analysis in addition.

For a simple TEFC motor they basically had the housing with fins and the fan and at a given ambient temperature "that works". it isn't a simple equation to "assume" a coefficient and calculate heat transfer in a complex system. Velocities and turbulences will be all over the place depending on location.

Even if you cool your "cylinder" to 200°F, is that cold enough for the motor? Does it have a 200°F ambient rating?

I understand this. It's an odd situation I haven't been on the site so pulling information like if it has fins on the casing or not is not easy. I don't have the nameplate information either. The one they sent is blurred.
I am told for some reason the motor started operating at 350 F and it needs to come down to 200 F inside (don't know if it's the ambient temperature).Many a times when a motor starts going haywire manufacturers change its ambient temperature based on application and customers situation. I was also told the cylinder likely has no fins and I can see that because many high voltage motors have no fins but internal fans on the rotor and a vent.

I haven't assumed a coefficient for heat transfer. I only back tracked calculated it, assuming a CFM for a 15-inch radius duct.

That would give you an insight of the how flow might be.

Let me know what is wrong with these steps:

Calculate the heat rate using the inlet and outlet air temperature. I will assume any existing CFM AC. Example 4000 CFM
Heat loss rate = Mass flow rate x specific heat capacity of air x Delta T

Or Q = 1.1 (CFM) (delta T) at standard conditions

For this particular heat loss rate calculated we can determine what the outside surface temperature to be when the inside surface of the core is 200 F. Thermal conductivity of carbon steel is used.
Use this to outside temperature and inlet air temperature to determine what the heat transfer coefficient is for convection.

There are many assumptions here:

Assuming an inside surface temperature and outside surface temperature of steel the heat flow is Q

Q= KA dT/dX For the thinckness of the steel cylinder

K is know for steel but what is actual heat transfer area? Heat is being transferred mostly around the core area to the outside but it is being transferred horizontally from the core area along the length of the cylinder first then to the outside so heat transfer is not just one dimensional.

Also heat transfer across outside surface film is

Q = UA (delta T)

Again most of the heat tranfer area is in the center around the core, and conducted away from the core to the cylinder then transferred by convection outward so A versus delta T is not really known. Also the delta T is not just the metal surface temperature minus the mean of the air temperature but the long mean temperature difference. There are handbooks that give calculations for the LMTD based on different flow configurations such as parallel/counter/cross flow configurations, but for such a large single cylinder I don't know if there is any calculations.

From the heat transfer coefficient find what the velocity of air is using the empirical formula.

Again things are not very symetric here. Some surfaces the air is blowing over very fast but some surfaces the air my not be flowing much at all depending on where the fans are located.

Keeping this aside before we determine if that velocity is enough, we calculate with the current CFM how fast does the temperature go down. If its too slow then we can probably think of a baffle design to increase the velocity and improve the heat transfer.

You lost me here. You already know that with the current CFM the temoerature in the core is still too high.

If that doesn't work either we increase the CFM and re do the steps.

I would also look at rearranging the air flow so that all surfaces are exposed to the blast of the air flow from the fan. You could duct the outlet of the fans to all sides of the cylinder surface. Also check for best location of the outlet so air flow path maximizes LMTD.

Can't you just insulate the cylinder with high temp insulation? Why leave it bare metal and then try to cool it?

The OP is trying to remove the heat in the first place.

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After many back and forth clarification queries, here is the true problem. The rotor of the motor is equipped with a fan and moves air axially through the stator. The inlet air temp is 78f and outlet air is 135 F. The material of the core being cooled down to 200 F from 350 f is carbon steel.