Power needed for Induction Heating

Question

How would one calculate the temperature created in a metal (assume stainless steel) that is heated by an induction coil? I am looking to figure out power required to reach a given operating temperature (i.e. sustained temperature in stainless steel).

Any help would be appreciated.

Answer 1

The temperature created in your stainless steel sample by induction heating is given by the equation:

Delta(T) = Q/S

  

where:

Delta(T) = temperature rise, degrees fahrenheit (Final temp minus initial temp)

Q= heat added, BTU's per pound

S = Specific heat of the stainless steel [amount of heat (Q) necessary to raise the temperature of one poind of the substance one degree F]

You'll have to put-up with english-units here, since I'm from the 'old school'.

You know the final temperature you want [Delta(T)].  

You can obtain the specific heat value (S) for the stainless steel either from the manufacturer or from published tables.  The manufacturer's figure would be preferred, since stainless steel is made in a wide variety of alloys with various specific heat values.  However, since ball-park accuracy is assumed acceptable here, the table value should suffice.

So you can calculate Q, the amount of BTU's you need to add per pound of material to obtain your temperture goal.  But time becomes a factor, as how quickly you obtain the temperature rise is a function of the power (heat) input per unit time.

The power required to sustain the temperature is determined by the heat losses to the surrounding environment per unit time (heat loss through conduction, convection and radiation) and would be difficult to quantize. However, it can be assumed to be less than the power required to heat the material to the desired temp in a reasonable amount of time. In a practical system, you would need to provide a feed-back power control system (thermostat) to reduce the initial induction power enough to compensate for heat loss and manitain your desired temp.

The BTU's you need will be created by the induction coil causing current to flow internally through the effective resistance (R) of the material.  The heat generated by that current (i) is given by the equation:

Power (watts) = (i)squared X R

Where the R is calculateded from another table given value, the "resistivity" of the material. "resistivity" is defined as "the resistance of a unit cube" of the material .  It would most accurately be provided by the manufacturer of the stainless steel sample, but the generic table value will suffice for our "ball-park" calculation. The resistivity figure can be converted to various units, but for purposes of this discussion, we need the value in "ohms-inch per unit area" usually given in tables as "ohms-inch per circular mill", where a circular mill is the area of a circle of 1/1000 inch diameter.

Armed with that value, you calculate R by measuring the volume, in cubic inches, of your sample of stainless steel and multiplying that result by the resistivity from the table.  

A side note here: The resistivity of any material is not a constant but varies with temperature, so in this case an average value will have to be used.  Another reminder that these calulations are only "ball-park" accurate!

But you really don't need to deal with this power equation at all. Introducing it makes the point that the answer to your problem involves the physical parameters of the material you are using.  for example, you can use it to check that at the power level you calculate, you won't melt the local material before the whole mass reaches your desired target temp because the (i) squared X R losses are so great.

You can more easily derive the power needed from the induction coil by converting the required BTU's per pound, Q , directly to watts.

We do this using common tables of equivalency giving:

BTU = 778 ft-lbs

Watt = 0.737 ft-lbs/sec

The result is the conversion factor that one BTU = 1056 watt-seconds.

You can verify this by putting one pound of water on your electric stove and heating it with one-kilowatt of power for one second. The water temperature should rise approx. one degree F.

So from the orignal equation:

Q = Delta(T) X S (BTU's/lb)

P (watts) = Q X 1056 secs/lb

So choose how many seconds you can wait for your material to reach the desired temperature and you have your answer!

Good luck with the project!

Answer 2

check the "Heat Treaters Guide"...