KW to Temperature (Degree F)

Heat transfer, and therefore the final temperature, depends upon a number of factors including how big the <... 25KW system...> is, whatever it is, how well it is insulated, what the ambient temperature is and what is flowing past the <... 25KW system...>, and at what velocity, that might cool it.

In withholding such vital information, at the moment the forum can produce only WAGs.

3.41 btu per kw

25,000 * 3.41 = 85,250

85,250 * 5% = 4262.5 btu's

https://www.mcmaster.com/high-temperature-insulation

"3.41 btu per kw"

The btu is an energy unit and the kW is a power unit, so the above can't be correct.

I didn't check the conversion, but I suspect you intended to say something like "3.41 btu per hour = 1 Watt" (NOT kW)

Yes that would be watt, not kilowatts...good catch

Understanding BTU to watts conversion is important when comparing cooling capacity with electrical power usage. For example, air conditioners are usually rated in BTU, but power consumption is often shown in watts. By converting BTU to watts, you can better estimate energy usage, electricity costs, and system efficiency.

In engineering and science, one can ONLY expect to obtain correct answers to calculations if the units on one side of the equation are exactly equivalent to the units on the other side of the "=".

I was unable to post an image of the site of your link, but the headline indicates a conversion from BTU to Watts. There can be no conversion from one side to the other of BTU to Watts, because the BTU is a unit of energy, while the Watt is a unit of power.

If you look at the smaller print a bit farther down the page, you'll see that it is converting BTU/hr to Watts. These are both values of power, so can be valid.

Another valid conversion would be BTU to Watt-hours; both of these are values of energy.

To measure the quantity of heat,you must have a time factor.KW/H,or BTU/Hr.

Degrees is only a measurement of the temperature,not of the quantity.

You could have a small pinhole hole with 1400F air or gas escaping,but this will not tell you anything about total losses.

If you only want to know the temperature of the heat escaping,use a digital non contact thermometer(infrared scanner). The losses are going to vary from place to place on the machine,and expose weak or insufficient insulation.

Some IR scanners are equipped with cameras or computer interface,thumb drives,etc. so you can easily document the areas of highest loss.

You can then calculate the areas of temperature differences and derive a close approximation of the losses per time unit.

Please redefine your request to get a better answer.

I know its metric RedNek but you are usually brighter than this

kW = kJ/s Energy (kilojoules) per time

I have no idea what you are measuring in energy per time squared especially with the two different time units wobble

I guess like acceleration is must be rate of change of work!

Otherwise lots of good stuff as expected

I presumed he wanted to know the amount of energy lost,which requires a time factor.

Maybe it was a bad presumption.

Energy is a unit of power times a unit of time.

I presumed he wanted to know energy lost.

Electrical energy is sold in the USA by the KWH or MWH, plus demand factors,power factor,etc.

There terms are easily confused and sometimes are incorrectly used interchangeably.

Great Feedback. Thank you.

I'm limited on the amount of information i can share due to IP restrictions, but what was provided thus far puts me in the right direction. i will try yo add some more detail shortly.

First off, what in hell is a "system"?

..."a set of things working together as parts of a mechanism or an interconnecting network."..

You will be losing the heat from the surface of the insulation.

So you need to check your ambient conditions and calculate the heat transfer to the surroundings.

You will need to know the outside area of your system and make assumptions on convective heat transfer.

Work your way backwards from there.

But all things said you will at some point put 25 kW in the system you will lose 25kW only at a higher temperature.

With other words just put 5 kW in the system and it will lose only 5 kW.

<...amount of heat lost in Degree F...>

A rate of heat loss cannot be measured using units of temperature.

<...only allowed...> by what authority?

thanks for all the feedback. I think i now have what i need.

Again, my goal here was to spec out the right insulation coverage to not allow more than 5KW of heat lost.

Example sketch.

Whatever.

Unfortunately, your sketch only adds to the confusion! You originally said nothing about a fuel; if you are burning a fuel inside a closed furnace, there must be an exhaust to remove the combustion products. That exhaust will carry a great deal of heat, over and above the heat losses through the insulation.

You show the 1400° zone near the bottom of the enclosure. Since hot gasses are lighter than cooler ones, they tend to rise to the top of the enclosure. What is the temperature there?

You show the electrical line well up the side of the furnace, but don't show the placement of the heating elements. The placement of those heating elements would have a significant effect on the temperature distribution inside the furnace. I'm assuming the current 25kW and desired 5kW both refer to the electrical power only.

You show "Pumps" at the top of the furnace. What product are you able to pump, and what kind of pump do you use that can cause flow in that 1400° material?

Consider a nuclear reactor system.

It could fit the model, not that I am any kind of expert.

If you are looking to only lose 5kw heat through the walls and the temperature inside and outside the walls will be held constant then you just need the surface area of the tank to know the insulation value needed.

Using an ambient temp of 70°F, insulation is needed to an r-value equal to:

(Surface area in sq. ft. × ΔT of 1330°F × any safety factor)÷ (3412 × 5 kW)

....just ignore the awkward units for now and put in your square footage and safety factor and the result will be the required r-value.

A BTU to Watts converter helps you quickly convert British Thermal Units (BTU) into watts for heating and cooling calculations. BTU measures thermal energy, while watts measure power. This conversion is commonly used for air conditioners, heaters, HVAC systems, and other electrical appliances.