Combustion expert Dick Bennett explains how to know how much heat is needed.



Over the years, I've done a lot of columns on how heat is generated in ovens and furnaces, how it's transferred from one point to another and what losses are incurred in the process of transfer. But on looking back, it seems I've never discussed how the heating requirements of the load itself are determined. Time to fix that.

The design of process heating equipment and the energy input systems have to begin with the requirements of the product load. Figuring how much energy is needed to heat a certain product flow to a certain temperature requires knowing:

• The load's weight.
  
• The time available to do the heating.
  
• The temperature the load has to reach.
  
• The thermal properties of the load (specific heat, heat required for changes of state or other transformations).

Weight and heating time combine to give us a heating rate. If the process is continuous, that's pretty straightforward -- pounds or kilograms per hour production rate specified for the line. For batch processes, you have to divide the weight of the load by the time allowed to heat it. Again, this will give you pounds or kilograms per hour.

Next, look at the thermal properties of the load. For openers, consider a simple situation where the product doesn’t melt, vaporize or go through any phase changes. The key property is the material’s specific heat, the amount of energy required to raise the temperature of one unit weight of that material through one degree. In the English system, it will be expressed as BTU per pound, ^oF or BTU/lb-^oF. Where the metric system prevails, it may be expressed as calories, kilocalories or Joules per gram (or kilogram), ^oC, or cal/g-^oC, kcal/g-^oC or J/g-^oC. (By the way, if you’re using BTU/lb-^oF, cal/g-^oC or kcal/kg-^oC as your system of units, the numerical value of the specific heat is the same. If you’ve ever wondered why a handbook listed specific heats without units, this is why.) Finally, you need the temperature rise of the load.

Say you wanted to heat 1,000 lb/hr of aluminum from room temperature (70^oF) to 250^oF (21 to 121^oC). From a handbook, you find aluminum’s specific heat is 0.22 BTU/lb-^oF. The temperature rise is 250^oF minus 70^oF, or 180^oF (100^oC). Multiply everything together

1,000 lb/hr x 0.22 BTU/lb-^oF x 180^oF

and you get 39,600 BTU/hr.

The specific heats of most materials are not constant; they usually increase with temperature. That’s why many handbooks show several different average specific heats for different temperature ranges. In many cases, the change isn’t that great, so using a specific heat for a temperature range different from your process doesn’t introduce much of an error. There are exceptions, however, so it’s wise to use data for a temperature range as close as possible to your process.

The state (solid, liquid, gas or vapor) of a material adds another wrinkle. Take water as an example, because it can be in all three states over a short range of temperatures. As ice, water has a specific heat of about 0.44 BTU/lb-^oF at -25^oF (-31^oC). As a liquid, it stays pretty close to 1.0 BTU/lb-^oF between 32 and 212^oF (0 and 100^oC). As a vapor (steam), it’s around 0.48 BTU/lb-^oF at temperatures just above boiling. To figure how much energy it takes to heat a material through two or more different states, you have to figure the heat absorbed in each of the states and add them together.

There’s more -- you also have to add the energy the material absorbed in making the transformation from one state to the other. These energy inputs are known as the latent heats of fusion (melting) and vaporization (evaporation or boiling). The term “latent” means hidden, and it is called that because there is no temperature change to tip you off to the fact that the material is absorbing heat. To use water as an example again, its latent heat of fusion is 144 BTU/lb, and its latent heat of vaporization is 970.2 BTU/lb.

Say you start with 100 lb of ice at 0^oF and heat it until it melts and then boils (figure 1). You have 2 hr to do the job. The water will absorb heat in four separate stages (figure 2). If you continue to superheat the steam above 212^oF, 100 lb will absorb an additional 48 BTU for every degree of temperature rise.

Is that all there is to it? Usually, but with some materials, there’s even more to consider. I’ll save that for next time (see link at end of article).

Links

• Janus Technology Group
• Energy Notes: The Special Situations