# How to calculate necessary heat power

• incalculavel
In summary, the problem with the factory is that the company contracted to solve the problem was unable to do so, and Jose seeks help from the forum in solving the problem. The problem is that the water mix in the tank needs to remain at a specific temperature, and 24000 watts of electrical heating power is required to do so. After two hours and thirty minutes of using the power, the temperature had dropped to 30 degrees Celsius, with all of the heaters still working. The calculation for the amount of power needed to maintain the temperature at 80 degrees Celsius was not possible to do using just the information given. Jose would need a drawing of the pipes to help with the calculation.
incalculavel
Hi

I have a problem in my factory, and contracted a company to solve it, but unfortunately they were unable to do it.

I have a tank with 1.000 liters of water mixed with sand, that is pumped with a high capacity pump, through a compressed air system, to clean some small metal parts.

My problem is that I need the water mix to remain at 80ºC, and for that I have four 6.000 W electrical heaters, submerged in the water, for a total of 24.000 W of heating capacity. The temperature was my requirement, and 24.000 W was the solution the contracted company installed.

After 2 hours and 30 minutes working, the temperature drops to 30ºC, with all the electrical heaters working.

What I would like to know is the way to calculate the amount of heating power needed to maintain the water mix at 80ºC.
Knowing that the 1.000 liters water mix, starting at 80ºC, with 24.000 W of heating power turned on, after 2 hours and 30 minutes drops to 30ºC, is it possible to calculate it?

I really don't want to guess (like the contracted company did), and keep trying until I get enough power... I think that the correct way is to do the math.

Can anyone help me out?

Thanks

Jose Marques

In theory the calculation is easy - that is, when you neglect practical factors like the heat capacity of your tank, heat loss to the environment, dependence on the mixture of water/sand, etc.
Probably assuming that the dissipated power is constant in time, you can estimate an effective heat capacity of the "non-water" in the system and try to do the calculation. However, I think that there would still be dependences on the type of water (sweet / salt, clear / polluted), the amount of sand.

However, I am a theoretician, not an engineer, so maybe someone can offer a practical solution

Yes, it is possible to calculate this.

If you have a drawing of the pipes, it would help greatly in figuring this one out.

I need to know this:
*How much water per hour you use.
*The correct mix of water and sand(does sand go through your piping as well?)
*Celsius or fahrenheit
*Is this a closed system?

Add the following questions to the list:

How often do you replenish the water content and what temperature is that new water at?

Ok, let's see:

*How much water per hour you use.​

The pump has a power of 28 cubic meters / hours of circulation.

*The correct mix of water and sand(does sand go through your piping as well?)​

The sand goes to the pipe also, and we have 100kg of sand in the water.

*Celsius or fahrenheit​

Celsius.

*Is this a closed system?
*How often do you replenish the water content and what temperature is that new water at?
This is a closed system, so no water comes into the system. We start with 1000 liters of water, at 80ºC, and the temperature keeps dropping.

I missed one. I stir the tank, to keep the sand mixed.

So, what you are asking is, how much more kW you need to have this stay at 80 C?

Sorry, but one more bit of confusion on my part...

I am envisioning a wet type of sandblasting operation. Is this correct? The slurry mix is sprayed at parts at atmospheric pressure and then collected through a drain to be sent back to the reservoir. Is this what is really happening?

So, what you are asking is, how much more kW you need to have this stay at 80 C?

That is exactly what I need to know... I am trying not to guess, or to test trial.

FredGarvin said:
Sorry, but one more bit of confusion on my part...

I am envisioning a wet type of sandblasting operation. Is this correct? The slurry mix is sprayed at parts at atmospheric pressure and then collected through a drain to be sent back to the reservoir. Is this what is really happening?

That's it, wet sandblasting. I was having trouble finding the correct English word.

You correctly described the operation.

Did the temperature level out at 30C
What is the environmental temperature?

I will try my hand at the deduction.

Assuming 10C for the environmental temperature.
And assuming that the temperature did level out at 30C

Power loss per deg C. = 24000/(30-10) = 1250W

Power required = (80-10)*1250 = 87500W

I of course could be wrong, and it would be easy to prove. You can use the math above to predict what would happen to the temperature when one of the heaters are turned off.

I have designed and built quite a few specialty furnaces, but clearly do not have the experience that other members of this forum have so I would still seek a concurring opinion before using this advice.

I might add to the previous post that the prediction assumes specific things, and one of them is total power input. One significant source of heat is the pump, and can change the prediction. So one would need to include actual power consumed by the pump, multiplied by the electrical efficiency of the pump (perhaps 75%).

RayRoc said:
Did the temperature level out at 30C
What is the environmental temperature?

I will try my hand at the deduction.

Assuming 10C for the environmental temperature.
And assuming that the temperature did level out at 30C

Power loss per deg C. = 24000/(30-10) = 1250W

Power required = (80-10)*1250 = 87500W

I of course could be wrong, and it would be easy to prove. You can use the math above to predict what would happen to the temperature when one of the heaters are turned off.

I have designed and built quite a few specialty furnaces, but clearly do not have the experience that other members of this forum have so I would still seek a concurring opinion before using this advice.

Ok, this is a very good idea. If I try with only 2 heaters turned on, with a total of 12000W, using the same math the temperature should stabilize at 19.6ºC.

If I have to use around 90.000W, I must prove it first, because I cannot use electric heaters for that kind of power... and the investment in a furnace is much higher.

That's all well and good, but I see a major issue in this whole scenario; it seems that there is no real effort on anyone's part to really understand your heat losses. Here are some of the areas that would need to be pinned down to do a proper heat transfer analysis:

- Size, material, insulation of the main tank
- Piping run distances, pipe insulation to end use
- Temperature of returning liquids and what rate (you are shooting this stuff through a high pressure nozzle, it will expand and drop in temperature among other losses).
- There has to be make up water entering the system. You have to be experiencing some evaporative losses in this system.

To me it looks like you will run into one of two results...
1) The total power input is theoretically correct, but does not account for all of the losses in your system, so the heaters are undersized.
2) You will grossly oversize the heaters to get it to work and will not be as efficient as you could be.

If your contracting company was worth the money you spent, they should have spent some time characterizing your system and not only putting big immersion heaters in your tank, but also looking at ways to minimize heat loss throughout the system. It looks like they did half the job. I hope you paid them only half of the money.

I think the first thing you need to do is to understand the energy balance of your system, and to estimate the different parameters of it. You can draw imaginary control volume around your system and ask you self what cross the borders of this control volume. For example: X Watts are lost from the pipes by natural convection Y Watts enter your C.V from the heater Z Watts enter from the pumps etc. After you understand the energy balance of your system you can you solve the problem.

incalculavel said:
Hi

(snip), that is pumped with a high capacity pump, through a compressed air system, to clean some small metal parts.

(snip)

What's the exhaust rate (volume per unit time) from the system?

Other thing --- 2 1/2 hrs X 24 kW ~ 200 MJ ~ 50 C X 4MJ/ton water K --- del Hvap ~ 2MJ/kg which implies you could be losing 100 liters water over the 2 1/2 hrs; you say 28 m3/hr, and what? 10 times that volume of air? If you're using a "dry" compressed air system you can easily lose that much water --- saturating it at 80 C before running it into the blasting cabinet is one approach you can take, you'll still need another 24 kW to do it.

Whoa! I missed that one! Bystander you have quite the point. The compressed air must saturate after expansion, and that could take allot of water away. On top of that it also indicated that the system is not closed, and can carry allot of heat out through evaporation alone without the aid of the compressed air flow.

If all of this is correct. That will take allot of heat to maintain temperature!

## 1. What is heat power and why is it important to calculate?

Heat power, also known as thermal power, is the rate at which thermal energy is transferred or converted from one form to another. It is important to calculate because it helps determine the amount of heat needed for a specific process or system, such as heating a room or powering an engine.

## 2. What are the units of heat power and how do you convert between them?

The most commonly used unit for heat power is the watt (W), which measures the rate of energy transfer in joules per second. To convert between different units of heat power, you can use conversion factors such as 1 W = 3.412 BTU/hr or 1 W = 0.001341 horsepower.

## 3. What factors are involved in calculating necessary heat power?

The necessary heat power is dependent on several factors including the desired temperature difference, the material properties of the object being heated, the duration of heating, and any external factors such as wind or insulation. Other variables, such as the specific heat capacity and density of the material, also play a role in determining the necessary heat power.

## 4. How do you calculate the necessary heat power for a specific application?

To calculate the necessary heat power, you can use the formula Q = mcΔT, where Q is the heat energy in joules, m is the mass of the object being heated, c is the specific heat capacity of the material, and ΔT is the desired temperature difference. You can also use online calculators or consult with a thermal engineer for more complex calculations.

## 5. Are there any safety precautions to consider when calculating necessary heat power?

Yes, it is important to take into account any safety precautions when calculating necessary heat power. This includes ensuring that the heating system or equipment can handle the required amount of heat power without overheating or causing damage. It is also important to follow proper safety protocols when handling high temperatures and heat sources to prevent accidents or injuries.

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