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Heat transfer out of hot water storage tank

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blondifre

Mechanical
Feb 10, 2012
1
Hello,
I am trying to calculate heat loss from uninsulated hot water storage tanks using Fourier's law. The tank is made of steel with thermal conductivity (k) of 50W/mK with thickness of 5mm. Inside Radius R1=.75m and outside Radius R2=.775m. The length of the tank L=2m Tank water temp Tw=60C and the temperature of the outside of the tank is To=55C
For cylinders, Fourier's law says that heat transfer Qout=[2*pi*k*L*(Tw-To)]/[ln(R2/R1]

Plug and chug and, Qout=95810W... now for an entire year Qout*24*365, we get 839295kW
Calculating energy the lost, at 15cents/kwh:
839295kW*$0.15=$125894!!!???

This does not make any sense, that number is out of control. Does anyone have an idea how to realistically estimate heat loss from the system? Can anyone help me shed some light on this?

Thanks!!
 
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This way you’ve just evaluated conduction through the wall of the tank. You then have to consider convection plus radiation heat transfer mechanisms from the outer surface of the tank to the ambient (this is what will slash heat loss). What’s your ambient temperature?
 
Go to Spirax-Sarco's website and search their discussions regarding heat loss from tanks and vats. I have found their approach to be very good and it simplifies a lot of things, like providing you with a good starting point for the OHTC.

This is a good start. I recently used this approach and ended up with results that were within 5% of those theoretically predicted by others using HYSYS.


Regards,

SNORGY.
 
What makes you say that the inside wall temperature is 60C and outside wall temperature is 55C? Is this from direct measurement? You are enforcing a high gradient across the wall. Is the temperature of fluid in the tank 60C? Fluid temperature and wall temperature are not necessarily the same.
 
You have convection on the inside and outside. Crank the Biot number to see what the actual impact of the tank wall itself is. And there's no way in the world that free convection can support the 2000 W/m^2-K conduction you have going through the wall.

TTFN
faq731-376
7ofakss
 
Your "problem" statement for calculating that cost extrapolated over a year's time is dead wrong: The inside temperature of the tank liquid may be 60C. Your tank might be heated consistently that hot for a year's time. Both reasonable assumptions. Or reasonable conditions.

The "outside" air temperature of the tank is NOT going to be 55 C all year (not even all night) or all day at any location.

To get a "year's cost" for heat loss you need to call out your professor on his/her assumption. Max outside temperature of 110-115 does occur in limited areas during the day. Your plant may be at one of those sites. But nighttime temperatures are not 55.

You can't multiply a heat-loss/second/sq meter times seconds-per-year to get a yearly total.
 
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