- #1

- 43

- 0

So does that mean if I input 4000 watts for 30 seconds that will output 12000 Joules?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter math111
- Start date

- #1

- 43

- 0

So does that mean if I input 4000 watts for 30 seconds that will output 12000 Joules?

- #2

- 51

- 0

Almost. 120,000 J.

- #3

- 43

- 0

Almost. 120,000 J.

Yeah I missed the extra 0! so you're value is right along with my concept

Thanks...

- #4

- 51

- 0

So in general power (P) is similar to energy (E) the same way velocity (V) is similar to displacement (x). The derivative of energy with respect to time is equal to power.

dE/dt = P vs. dx/dt = V

- #5

- 43

- 0

So in general power (P) is similar to energy (E) the same way velocity (V) is similar to displacement (x). The derivative of energy with respect to time is equal to power.

dE/dt = P vs. dx/dt = V_{x}

I want to take this a little further and see if you see what I do....

(1)http://www.nexthermal.com/page/resource-center/wattage-formulas.aspx

Where it has the formula for KW to heat up a material on top(1). I want to heat up S/S 304.

(2)http://www.aksteel.com/pdf/markets_products/stainless/austenitic/304_304L_Data_Sheet.pdf

I use their constants...Density=.29, Specific Heat=.12

My point:

I have an Electrical Engineering who used a 4.0 KW/208V input for S/S material for an entire weekend and it did not melt. The melting temperature of S/S is 2550F - 2650F and with the formula if we used the 4KW input for only 24 hours it would of reached almost 10,000 F and that would mean the S/S would of 100% melted, right?

Conclusion that I am questioned about:

1 W = 1 J/S from above so if I input this into a material for 1 second it will be 1 J and if I input it for 10 Seconds it will be 10 J so the more time I input Watts the more hot it will get(I am assuming!!) but at what point does it level off and why because if it does not won't the material in the formula go right to infinity with time unless joules & watts has no relation to heating the material?????

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 967

- #7

- 43

- 0

Inputting the energy: a bolt is connected to the material and a wire is connected to the bolt from a power source of 208V. When plugged in does energy keep being added over time or is this constant? This might help me understand it more because all I know is...

1. More time = more energy = higher temperature to infinifnty

or

2. This energy input is constant and will increase the temperature of the S/S up to a certain temp and hold it. If so how do I find that temp where 4.0 KW = ????

- #8

uart

Science Advisor

- 2,797

- 21

Inputting the energy: a bolt is connected to the material and a wire is connected to the bolt from a power source of 208V. When plugged in does energy keep being added over time or is this constant? This might help me understand it more because all I know is...

1. More time = more energy = higher temperature to infinifnty

or

2. This energy input is constant and will increase the temperature of the S/S up to a certain temp and hold it. If so how do I find that temp where 4.0 KW = ????

If something is unable to cool itself by radiating or conducting away energy then yes it will continue to heat up as long as you supply power. Your stainless steel sheet however can obviously radiate away excess energy once it's sufficiently hot.

If your steel didn't melt then you probably underestimated the amount of heat that it was able to radiate or conduct away, something which is a function of how you're physically holding the steel, what it's in contact with and even it's spatial orientation.

You also may have not accounted for the increase in electrical resistance of the material as it's heated.

- #9

- 43

- 0

If something is unable to cool itself by radiating or conducting away energy then yes it will continue to heat up as long as you supply power. Your stainless steel sheet however can obviously radiate away excess energy once it's sufficiently hot.

So when the material's temperature is rising the mode of radiation is ocurring which is a reason why the temperature is leveling off??

Can anyone relate the amount of KW input to the temperature rise? Like if I input 4Kw what will the temp of the material be at or am I missing information to help answer this like a size needed or how long am I inputting the power...

- #10

RonL

Gold Member

- 1,104

- 216

Inputting the energy: a bolt is connected to the material and a wire is connected to the bolt from a power source of 208V. When plugged in does energy keep being added over time or is this constant? This might help me understand it more because all I know is...

1. More time = more energy = higher temperature to infinifnty

or

2. This energy input is constant and will increase the temperature of the S/S up to a certain temp and hold it. If so how do I find that temp where 4.0 KW = ????

uart beat me to the answer, and he is correct.

Let me give another example for you to consider.....an incandescent light bulb has a filament inside that will reach a temperature of close to 5,000 F degrees, it does not melt, nor does the glass around it.

It is a matter of thermal discharge rate.

Ron

- #11

- 43

- 0

uart beat me to the answer, and he is correct.

Let me give another example for you to consider.....an incandescent light bulb has a filament inside that will reach a temperature of close to 5,000 F degrees, it does not melt, nor does the glass around it.

It is a matter of thermal discharge rate.

Ron

OK so as the temperature is rising due to the Kw input over time I have to figure out the radition rate of temperature given off to find this level temperature...

- #12

- 43

- 0

- #13

QuantumPion

Science Advisor

Gold Member

- 902

- 42

If the steel is at its maximum temperature with the power on, then it is at thermodynamic equilibrium which means that power in = power out. If the beam is supported in such a way to minimize conduction and convection than power in ~= radiation out.

If you want to calculate what the maximum temperature of the steel would be given your input power, you need to know its surface area and emissivity. Some helpful info here: http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

Share: