# Calculating heat loss from open tank

• majikman
In summary, the tank will lose heat to the surroundings (ground and air) and to the water in the tank. The heat transferred will go thought the wall to the surroundings (-25d).
majikman
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.

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.

## 1. How do you calculate heat loss from an open tank?

To calculate heat loss from an open tank, you need to know the surface area of the tank, the ambient temperature, the temperature of the liquid inside the tank, and the overall heat transfer coefficient. Then, you can use the following formula: Q = U x A x (Ts - Ta), where Q is the heat loss, U is the overall heat transfer coefficient, A is the surface area of the tank, Ts is the temperature of the liquid, and Ta is the ambient temperature.

## 2. What is the overall heat transfer coefficient?

The overall heat transfer coefficient is a measure of how easily heat can transfer between two materials. It takes into account factors such as the thermal conductivity of the materials, the thickness of the materials, and the contact resistance between them. In the context of calculating heat loss from an open tank, it is used to describe the rate at which heat can transfer from the liquid inside the tank to the surrounding environment.

## 3. How does the surface area of the tank affect heat loss?

The surface area of the tank is directly proportional to the heat loss. This means that the larger the surface area, the more heat will be lost. This is because a larger surface area allows for more contact between the liquid inside the tank and the surrounding environment, resulting in a higher rate of heat transfer.

## 4. Can the ambient temperature affect heat loss from an open tank?

Yes, the ambient temperature plays a significant role in the calculation of heat loss from an open tank. The larger the difference between the temperature of the liquid inside the tank and the ambient temperature, the higher the rate of heat loss. This is because a larger temperature difference creates a greater driving force for heat transfer.

## 5. What are some ways to reduce heat loss from an open tank?

There are several ways to reduce heat loss from an open tank. One method is to insulate the tank, which can significantly decrease the overall heat transfer coefficient. Another way is to cover the tank with a lid or other material to reduce the surface area and minimize contact between the liquid and the surrounding environment. Additionally, maintaining a consistent ambient temperature and keeping the liquid at a higher temperature can also help reduce heat loss.

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