Rise in Temperature due to Heat Losses 2

[DBeas3]

I have a very unusual problem. I was asked to compute the temperature rise in an enclosed cylinder that has a motor running inside. And the cylinder is underwater (seawater) with no ventilation.

I am a CE and have no experience in this area. I know that the heat losses due to the motor inefficiency is 1HP and have the cylinder dimensions and material props.

I need to figure out when the system reaches steady-state and what the temperature inside will be at that time.

 

[IRstuff]

You'd need to consult a heat transfer text, of which, one is available for download:

http://web.mit.edu/lienhard/

http://www/ahtt.html

You have, more or less, a liquid cooling problem, which does require knowing the temperature of the water, and the flow thereof, around the cylinder.

TTFN

FAQ731-376

 

[DBeas3]

Thanks IRstuff. I'll take a look at the text.

 

[DBeas3]

I took a look at the Lienhard text but didn't find what I was looking for in terms of an equation. A colleague had mentioned I might be able to use the lumped parameter method but it didn't seem applicable since the air inside the cylinder will be increasing in temperature as the motor runs, meanwhile the seawater (40 F) will be cooling the inside. This is part of a feasibility study, so we are just bounding the problem, as a starting point I'm assuming still water -- it's not acting to specifically cool the system.

I was thinking that if I could relate the power losses to the change in temperature over time, and compute the heat transfer rate between the cylinder (we are trying both aluminum and stainless steel) and the water I could figure out when the system would reach steady state. Am I totally off base in this thinking?

 

[Zesti]

As a rough, but educated guess of the temperature inide the cylinder just treat the whole thing as heat transfer through a wall with the same area as the motor casing/cylinder. Hot air on one side, cool water on the other side.

You should be able to find the equations needed in the above mentioned book.

Then all you have to do is calculate at which temperature difference 1 HP is transfered.

I am sure this will be close enough for your purposes.

 

[FredRosse]

A relevant issue here is where the motor energy is going. Within the enclosure, the motor will throw off heat to the inside of the cylinder due to its inefficiency. For an 80% efficient motor at 1 HP (746 Watts) shaft output, this is 186 Watts. The total electric energy to the motor is 186 + 746 = 932 watts.

Depending on what the motor is driving, some of the motor's 1 HP output may be deposited inside the enclosure cylinder. For example, if the motor is driving a ventilation fan within the enclosure, then all of the fan's energy will be converted to heat, and the enclosure must reject 932 Watts. If the motor is driving a sea water circulating pump, then virtually all of the 1 HP energy will leave the enclosure with the water, and only 186 Watts must be removed from the enclosure via heat transfer.

 

[prex]

You have natural convection on the outside of a horizontal tube for water and natural convection on the inner surface for air. The cylinder material and thickness is unrelevant as it is not insulated.
For calm air you can take a heat exchange coefficient of 10 W/m²°C. For water I guess that the coefficient is beyond 100 W/m²°C (can be determined by using a textbook, possibly the one referenced is also OK).
So the air side exchange dominates, and you can conservatively use an overall coefficient of H=10 W/m²°C.

prex

http://www.xcalcs.com

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[IRstuff]

A typical natural convection coefficient ranges from about 10 W/m^2-K for a small object, but drops to below 4 W/m^2-K for a larger object >= 300mm in extent (ASHRAE SE-99-15-05).


TTFN

FAQ731-376