# Heat time of steel in Watts

1. Apr 13, 2015

### JustLearning2

Hello. New here and trying to figure out if my formula is right. I'm a controls guy so this other stuff keeps me on the ropes!

Trying to heat 179lb's of steel 1440F in 30 minutes and find wattage needed. It appears the formula is:

#Lb's X specific heat X differential temp
3.415 x .5 (minutes)

179 x .12 x 1440
1.7075

18,130.8 Watts

Does this seem correct? This is all from trying to determine if a heat treat furnace can bring these parts up to temp in the required amount of time. Thanks for any replies.

2. Apr 13, 2015

### JustLearning2

Oops-sorry about the thread title. "heating time of steel in Watts" doesn't make any sense... I've been thinking too hard again ;-)

3. Apr 13, 2015

### Baluncore

mass = 179 lbs = 81.193 kg
d_temp = 1440°F = 800°C
time = 30 min = 1800 sec
At room temperature, the specific heat of Iron/Steel = 452. J/kg/°C

Energy needed is 452.(J/kg/°C) * 81.193(kg) * 800.(°C) = 29359388.8 joule.
This energy must be delivered over 1800 seconds. The rate is therefore;
29359388.8(J) / 1800(s) = 16310.771 watt = 16.31 kW.

There will be changes of the specific heat with temperature. They must be considered.
Any phase transitions in the steel will require additional energy.
The specific heat is also dependent on alloy.

See figures 1 to 5 in "On the Specific Heat of Carbon Steels", Saburo Umino.
http://publikationen.ub.uni-frankfu...deliver/index/docId/14044/file/E001892563.pdf

Last edited: Apr 13, 2015
4. Apr 13, 2015

### Randy Beikmann

JustLearning2, I'd just comment that the way Baluncore did the problem makes it much easier to do it correctly, and to follow. He took all the input data and converted it into consistent units. Then he split it up into calculating the total amount of energy Q needed, and finally divided by the time allowed. This way also, you can do a sense-check on each portion of the problem. This will make your life easy. ;-)

5. Apr 13, 2015

### Baluncore

JustLearning2, welcome to PF.

Randy Beikmann, thanks for the commentary.

If the fundamental principles are not understood, memorising formulae will lead to misapplication.
I kept my units in the numerical solution. Following the unit cancellations through the computation makes it obvious that the dimensional analysis is correct.

I was not completely happy with my post because;
I failed to make it clear that one joule per second is one watt.
I carried too many digits through the computation, I did it to avoid scientific notation in Windows Calculator.