I am working on heat removal utilizing forced air. My goal is to find cfm volume necessary. I will push air via cent fan through a pump and do not know how much heat the coils give off as it is just being designed. We have calulated coil mass and current draw. I need a good formula for heat absorbtion of air and find volume of air necessary.I am used to calculating btu/hr of direct fired burners and sizing fans to burner to arrive at proper temp rise. This heat removal has got me stump and can not see the path to complete this work.
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heat removal using air flow 1
- Thread starter rcmac21
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1
- #2
You can read this:
It's free to download. Wikipedia is also a good source.
Is this for school?
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
It's free to download. Wikipedia is also a good source.
Is this for school?
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
StrykerTECH
Mechanical
When you are talking about removing heat using air, you really are concerned about Mass flow rate, which then based on density and air temp you can figure out what CFM you need. Depending on the complexity of the problem and at what temperatures you are rejectin heat, your CFM needed with actually fluctuate. Can you give some more specifics of what you are trying to do with the burners. Maybe walk through the process, Thanks
StrykerTECH Engineering Staff
StrykerTECH Engineering Staff
- Thread starter
- #4
To answer first question, no this is not for school. I have rescently change jobs and I am in a different field of Engineering.
The details requested for help with heat removal are as follows:
I am pushing ambeint air through a linear pump. The pump has 950lb of coils and 221 amps at 240volts. The coils have silicone pole pieces and create heat in this 17" dia tube. I have .75ft3 of air volume in the tube and want to keep outside wall at 104 degreesF. Coils are nicely centered inside tube. The goal is to find cfm required to keep outside wall at 104F. This unit has not been made yet as it is still being designed. I have nothing to test but need to figure out in advance of testing.
The details requested for help with heat removal are as follows:
I am pushing ambeint air through a linear pump. The pump has 950lb of coils and 221 amps at 240volts. The coils have silicone pole pieces and create heat in this 17" dia tube. I have .75ft3 of air volume in the tube and want to keep outside wall at 104 degreesF. Coils are nicely centered inside tube. The goal is to find cfm required to keep outside wall at 104F. This unit has not been made yet as it is still being designed. I have nothing to test but need to figure out in advance of testing.
"do not know how much heat the coils give off as it is just being designed"
I don't think it is possible that no one in your team has design requirements and yet have already designed sufficiently that they're trying to implement cooling. So if your designers wind up dissipating 5 GW, that's OK? Obviously not, so someone has a written requirement for the amount of heat being rejected by the unit.
How are they designing the pump that they can't tell you how much surface area is available for cooling? The CFM is completely a red herring if you have no idea about the surface area to be cooled. You could have gale force air flow and still not cool something if air flow can't get to access. SOMEONE has to be intimately involved in the design of the heat rejection mechanics, if for no other reason than to ensure that the pump seals and electronics don't melt into a slag.
Are you part of the design team? Your earlier comments suggest otherwise.
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
I don't think it is possible that no one in your team has design requirements and yet have already designed sufficiently that they're trying to implement cooling. So if your designers wind up dissipating 5 GW, that's OK? Obviously not, so someone has a written requirement for the amount of heat being rejected by the unit.
How are they designing the pump that they can't tell you how much surface area is available for cooling? The CFM is completely a red herring if you have no idea about the surface area to be cooled. You could have gale force air flow and still not cool something if air flow can't get to access. SOMEONE has to be intimately involved in the design of the heat rejection mechanics, if for no other reason than to ensure that the pump seals and electronics don't melt into a slag.
Are you part of the design team? Your earlier comments suggest otherwise.
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
- Thread starter
- #6
StrykerTECH
Mechanical
O.K. couple things.
What is an accetable temp variance? 104 is a nice target but I doubt you are going to achieve that along the whole length of tube.
221 amp @ 240 volts ? is this constant or can this be varied? Am I to assume 100% heat creation or is there an efficiency on those?
Right now I am at 53kW.
How long is the 17 dia tube?
What is the tube made of?
what is the surface area of the coils?
Wall thickness of the tube?
As a rough guide line I would approach it this way.
Find out what the steady state heat load the tube can handle so that the outer temp is 104 degrees. (pretend there isn't anything inside, purely theorectical)
Lets say that is 5kW (total guess, purely for disucssion)
Then use the internal air temp calculated and that heat load value determine how much heat in Kilowatts you need to dissipate, 53kW - 5Kw = 48kW @ 120 degsF <-- pure example again.
Then use the thermodynamic equation to come up with how much air flow in CFM needs to go through assuming you have a heat gain of [CFM(120 degsF - Ambient)] = 48kW <-- not a formula just for reference.
StrykerTECH Engineering Staff
What is an accetable temp variance? 104 is a nice target but I doubt you are going to achieve that along the whole length of tube.
221 amp @ 240 volts ? is this constant or can this be varied? Am I to assume 100% heat creation or is there an efficiency on those?
Right now I am at 53kW.
How long is the 17 dia tube?
What is the tube made of?
what is the surface area of the coils?
Wall thickness of the tube?
As a rough guide line I would approach it this way.
Find out what the steady state heat load the tube can handle so that the outer temp is 104 degrees. (pretend there isn't anything inside, purely theorectical)
Lets say that is 5kW (total guess, purely for disucssion)
Then use the internal air temp calculated and that heat load value determine how much heat in Kilowatts you need to dissipate, 53kW - 5Kw = 48kW @ 120 degsF <-- pure example again.
Then use the thermodynamic equation to come up with how much air flow in CFM needs to go through assuming you have a heat gain of [CFM(120 degsF - Ambient)] = 48kW <-- not a formula just for reference.
StrykerTECH Engineering Staff
OK, so are you actually trying to cool roughly 53kW through 15.5 ft^2 of area? That's quite a bit:
53 kW/[(15.5 ft^2)*(80F)] = 828 W/m^2-K
I don't there's a fan strong enough or big enough to cool such a large load through such a tiny area using air. This is something that requires a rather humongous radiator, or requires liquid cooling.
Even if the load is only 1/10, or about 5.3kW, you'd still need 82.8 W/m^2-K, which is potentially achievable, but probably only in a wind tunnel.
I think you need to go and ask someone on the team what load are you actually cooling. Either that, or they're just having fun with the new guy on the block...
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
53 kW/[(15.5 ft^2)*(80F)] = 828 W/m^2-K
I don't there's a fan strong enough or big enough to cool such a large load through such a tiny area using air. This is something that requires a rather humongous radiator, or requires liquid cooling.
Even if the load is only 1/10, or about 5.3kW, you'd still need 82.8 W/m^2-K, which is potentially achievable, but probably only in a wind tunnel.
I think you need to go and ask someone on the team what load are you actually cooling. Either that, or they're just having fun with the new guy on the block...
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
- Thread starter
- #9
After your response I started over and throw out all informationhanded to me and went and found really solid model.
OD tube = 17.7"
OD of coils = 14.764"
surface area = 1451.6 ft2
air volume in container is uniform around outside =3.27ft3
air volume around coil surface area = 2.18ft3
Pump power for coils = 105kw
Tube is SS316L wall is .125" thickness
Measured outlet temp =140 F on prototype at 70 ambient.
OD tube = 17.7"
OD of coils = 14.764"
surface area = 1451.6 ft2
air volume in container is uniform around outside =3.27ft3
air volume around coil surface area = 2.18ft3
Pump power for coils = 105kw
Tube is SS316L wall is .125" thickness
Measured outlet temp =140 F on prototype at 70 ambient.
OK, much more amenable. How much that power needs to dissipated? And how much of the area is actually accessible to air flow? To accurately answer that, one would probably need to run a CFD analysis.
If all the area is usable, then something around 8 m/s air flow across the surface would be adequate.
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
If all the area is usable, then something around 8 m/s air flow across the surface would be adequate.
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
StrykerTECH
Mechanical
140F outlet temp, I thought you can exceed 104?
Is that the design problem, you need that 140 to not exceed 104?
What type of air flow were you using on the prototype?
I am coming at this from an "is it feasible" angle now.
105kW is a lot, hopefully you don't have to dissipate all of that. I am guessing you have 90%+ efficiency through the coil so, we are "only" looking at (100%-90+%)*105kW for heat dissipation. So 10kW ? ? ?
StrykerTECH Engineering Staff
Milwaukee, WI
Is that the design problem, you need that 140 to not exceed 104?
What type of air flow were you using on the prototype?
I am coming at this from an "is it feasible" angle now.
105kW is a lot, hopefully you don't have to dissipate all of that. I am guessing you have 90%+ efficiency through the coil so, we are "only" looking at (100%-90+%)*105kW for heat dissipation. So 10kW ? ? ?
StrykerTECH Engineering Staff
Milwaukee, WI
- Thread starter
- #12
This design is linear pump and it is basically straight tube with inlet force cooling air at one end and coils in the tube and then exists exhaust air opposite end. Exhaust air is 140 degrees F and tube temperature is 104 degrees F.
The pump is elctromagnetic and no moving parts, so efficency is very low. Traditional higest preformance I have seen data on is 20%.
I have a problem with the area in the tube air must pass through. The area between coils and outer shell equals 0.166 ft2
That makes air volume at 1575ft/min about 261 cfm.
This number works out but seems extremely low.
The pump is elctromagnetic and no moving parts, so efficency is very low. Traditional higest preformance I have seen data on is 20%.
I have a problem with the area in the tube air must pass through. The area between coils and outer shell equals 0.166 ft2
That makes air volume at 1575ft/min about 261 cfm.
This number works out but seems extremely low.
StrykerTECH
Mechanical
Sounds like I would have to construct a CFD model. You have a complicated problem on your hands. Could you upload a rough SolidWorks Model, either "save as a part" or IGES.
Then I might be able to help you more.
StrykerTECH Engineering Staff
Milwaukee, WI
Then I might be able to help you more.
StrykerTECH Engineering Staff
Milwaukee, WI
- Thread starter
- #14
I have attached the only drawing that I can in pdf format to view the pump and see internal coponents enough to understand item in question. Thanks for all the help, it is nice to know that I am not struggling with something really simple to do at first glance.
But, based on your image and your comments about the clearance between the shell and the coils, it looks like your available surface area is only about 32 ft^2 (using about a 0.4-m diameter and 2.37-m length from the drawing)
If so, we're back to where we started, particularly if the efficiency is only 20%, which means the cooling load is on the order of 84kW.
You would need to add pin fins, or similar, to the exterior of the coils to gain back some area.
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
If so, we're back to where we started, particularly if the efficiency is only 20%, which means the cooling load is on the order of 84kW.
You would need to add pin fins, or similar, to the exterior of the coils to gain back some area.
TTFN
FAQ731-376
Chinese prisoner wins Nobel Peace Prize
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