Calculating heat loss from open tank

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Discussion Overview

The discussion revolves around calculating heat loss from a large open concrete tank used in a wastewater treatment facility. Participants explore various methods to quantify heat loss due to the temperature difference between the water in the tank and the surrounding environment, considering factors such as the tank's dimensions, material properties, and the incoming effluent temperature.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the need to calculate heat loss from three sections: the top of the tank using Newton's law of cooling, the elevated section of the tank walls, and the interaction between the wall and base with the water.
  • Another participant suggests that a full numerical simulation may be necessary for accurate results, especially considering temperature gradients in the water.
  • Concerns are raised about the complexity of the calculations, indicating that calculus may be required due to the continuous inflow of water.
  • A participant mentions the need to account for the thermal conductivity of water in the calculations.
  • Two participants express frustration with the complexity of the equations involved, indicating that they have struggled to make progress on the problem.
  • One participant references two equations relevant to the heat transfer calculations, specifically mentioning Fourier's law and Newton's law of cooling.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to calculate heat loss, with multiple competing views on the necessity of numerical simulations and the complexity of the calculations involved.

Contextual Notes

Limitations include the lack of specific temperature data for the tank walls and the assumption that the wall temperature is a function of the surrounding temperature and the water temperature. There are also unresolved mathematical steps regarding the integration of temperature changes over time.

majikman
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I have a large circular tank made of concrete that has a volume of 9,300,000 litres. The diameter of the tank is 48.7 meters. The side wall of the tank has a height of 4.57 meters. The tank is set into the ground by 3.57 meters and 1 meter of the tank protrudes from the ground. The tanks' walls and bottom are 8" concrete with no insulation. The water in the tank has a temperature of +3degC. The tank has no top and is open to the environment, and using the weather here lately that would mean the temperature is -25degC. How would I go about calculating heat loss on this tank?

Essentially this tank is a clarifier in a waste water treatment facility. There is 1000 cubic meters of effluent that is pumped into the bottom centerwell of this tank at a constant rate over a 24 hour period each and every day. 1000m3 leaves the tank by means of spillage over weirs at the top of the tank. The incoming effluent is 3degC. I need to be able to provide enough heat to increase the eflluent temp to 5degC and then maintain that temp.

All help is much appreciated.
 
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The losses will be form 3 different sections -

1) The top (calculate it using Newton's law of cooling).

2) The elevated section (from the ground) -
You did not provide the temperature of the walls...whatever will it be, you got to calculate the heat lost/gained from the water to the wall using the same Newtons law of cooling.

Actually, since the wall does not generate it's own heat, you can remove the temperature part of the wall...it will actually be a function of the temperature of the surroundings and the water.

This heat transferred will go thought the wall to the surroundings (-25d)...you got to use Fourier's law to calculate that.

3) This section is the interaction between the wall + base and water...similar to case 2, but there instead of air, there's ground.
 
I'm also interested in the answer and I believe only a full numerical simulation can give the result.
I think dE_logics solution is only a correct solution if someone would constantly stirr the water in the container. With temperature gradients in the water the solution would be completely different.
 
Yeah, I forgot about the k of water itself.

It's not that easy...you need to involve calculus in this since water continuously enters.
 
dE_logics said:
Yeah, I forgot about the k of water itself.

It's not that easy...you need to involve calculus in this since water continuously enters.

Tell me about it. I've been at this equation for the last four hours and am no further ahead than I was when I started.
 
I did all this long ago...

We have to use 2 equations basically -

Q = -kA(dt/dx)

Q = hA(dt)

Here dt and dx and not the calculus ones.
 

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