- Initial heat for startup of the system, usually from an ambient temperature to a desired processing temperature.
- Losses due to materials changing phase, either during initial heatup or while processing (melting a solid to a liquid or boiling a liquid to a gas).
- Heating of material being processed during operation.
- Heating of material flowing through the process such as a liquid that will be heated and pumped to be used elsewhere.
- Heat losses to the environment due to conduction, convection and radiation.
Before beginning the calculations, it is important to realize the distinction between energy and power and their relationship to wattage requirements. In metric units, power is measured in watts (W) and energy is measured in watts multiplied by hours (W x hr). In English units, energy is measured in British thermal units (BTUs) and power is measured in BTU/hr. Energy is measured in BTU/hr in English units. Power is the rate at which energy is used.
It also should be noted that the difference between the starting temperature and the final temperature during the application of power commonly is referred to as delta T (ΔT). If a process is started at room temperature, say 80oF, and the application process temperature is 800oF, then ?T is 800oF minus 80oF, or 720oF.
Initial Heat for StartupThe first calculation involves determining the heat required to raise the system temperature to the desired temperature. All system components must be considered. In the simplest case, a block of material is heated; in more complex situations, calculations for multiple materials must be performed.
The energy (E) required to raise the temperature of a material by ΔT is calculated using the following formula:
EInitial = Weight of Material x Specific Heat of Material x ΔT
An example will help demonstrate, so I will use examples throughout. In this case, suppose you wanted to calculate the energy required to raise a 23 lb block of tin from an ambient temperature of 70oF to 300oF, given that the specific heat of tin is 0.056 BTU/lb oF. Knowing that 1 BTU equals 0.2930711 W-hr, the specific heat of tin also can be written as 0.0164 W-hr/lb oF. Therefore:
ΔT = 300oF - 70oF = 230oF
EInitial = 23 lb x 0.0164 W-hr/lb oF x 230oF
EInitial = 86.8 W-hr
This means that it will require 86.8 W of power to raise the temperature of a 23 lb block of tin from 70 to 300oF in one hour. Because wattage requirements are calculated based on one-hour increments, the requirements for faster or slower heatup times are calculated proportionally. If the temperature needs to be reached faster, say in a half-hour, then the wattage requirement would be doubled.
Losses Due to Material Phase ChangesIf a system being heated contains material that will pass through a phase change (melt or boil), the calculations are slightly more complex and have to be broken into parts, but they follow the same rules. Imagine a system that is heated from 70 to 300oF. Contained in the vessel being heated is a mass of Alloy ABC, which melts at 200oF. The wattage requirement of heating the container is calculated as shown above, but the heating of Alloy ABC must be done in two parts.
In this case, the wattage requirement is calculated from ambient (70oF) to the melting point (200oF) for Alloy ABC using the specific heat for Alloy ABC as a solid. A second calculation is performed from 200 to 300oF using the specific heat for Alloy ABC in a liquid state. Moreover, there is additional energy required to “push” the material from one state to another and is calculated as follows:
EPhaseChange = Latent Heat of Fusion x Weight
Or, in the case of boiling a liquid, the equation reads:
EPhaseChange = Latent Heat of Vaporization x Weight
Continuing the example, assume the latent heat of fusion of Alloy ABC is 100 BTU/lb and it weighs 6 lb. You need to calculate the energy required to melt it. To solve it, you know 1 BTU is 0.2930711 W-hr. So, the latent heat of fusion of Alloy ABC also can be written as 29.30711W-hr/lb.
EPhaseChange = 29.30711 W-hr/lb x 6 lb EPhaseChange = 175.8 W-hr
Remember that this value does not include the energy required for Alloy ABC to reach its melting point or the desired temperature after melting. That must be calculated separately as stated above.
Material Heating During ProcessingOnce the process temperature is reached, the process can begin running. Energy losses caused by materials entering the system during processing now must be considered, if applicable.
If the process involves heating pieces of plastic to a desired temperature at a rate of 10 pieces/hr, then the wattage required to heat that material must be calculated. The method is the same as for initial heatup. Remember that if each piece of plastic weighs 1.764 oz (50 g), then 17.64 oz (500 g) of material are processed in one hour, which is the time basis for all the calculations. Call this number EProcess.
Similarly, if a fluid is flowing through the process and requires heating to a desired temperature, its wattage requirements must be calculated. The equation is very similar to earlier equations:
EProcess = Weight of Material Flowing per Hour x Specific Heat of Material x ΔT
Suppose a circulating pump passes 150 lb of seawater through a system every 10 min. The water enters at 110oF and must be at 160oF when it exists. Assuming the specific heat of seawater is 0.95 BTU/lb oF, how much wattage is required?
To solve it, you know that 1 BTU is 0.2930711 W-hr, so the specific heat of seawater also can be written as 0.278 watt•hr/lboF. Therefore:
EProcess = 150 lb x 0.278 W-hr/lb oF x (160oF - 110oF)
EProcess = 2088.1 W-hr
If it takes one hour to pass through the system, approximately 2,090 W are required to raise the temperature of the seawater. Because the seawater needs to be heated in 10 min, then the wattage requirement is approximately 2,090 W multiplied by (60 min/10 min), or 6, which is approximately 12,540 W.
Heat Losses to the EnvironmentBesides determining the heat required to raise a material to a desired temperature, the losses to the environment also must be considered.
In the earlier tin block example, as the block warms to 300oF, it also loses heat to the environment. This means that the earlier calculation of 86.8 W-hr is not enough energy for the block of tin to reach 300oF in 1 hr. Some of the energy bleeds into the environment and is not used to raise or maintain the temperature of the block. Losses to the environment occur during heatup and after the process temperature is reached.
There are three types of heat transfer that can result in heat losses from a system to the environment:
Conduction Losses. These occur when a hot object is in touching contact with a cooler object and heat energy passes from the hot object to the cool one.
Convection Losses. These occur when heat energy is carried away from a system by surrounding fluids of gases or liquids. The calculations involved with convection can be complex, but they can be estimated to a satisfactory degree of accuracy.
Radiation Losses. These occur in the absence of any touching contact or transfer medium and is a form of electromagnetic radiation. Radiation is the manner in which heat travels from the sun to the earth through the vacuum of space.
Considering conduction losses first, the heat losses (L) are calculated using the following formula:
LConduction = Thermal Conductivity of Material Accepting Heat x Area of Contact x ΔT x 1 Hour Per Thickness of Material Accepting Heat
To extend the tin example above, assume the block of tin is in the shape of a cube measuring 4.44" on each side and resting on a plate of 0.5" thick mica. Calculate the conduction loss, given that the thermal conductivity of mica is 3.1 BTU x in/(hr x ft2 x oF).
To solve for this, you know that 1 BTU is 0.2930711 W-hr, the thermal conductivity of mica also can be written as 0.91 W-hr x in/(hr x ft2 x oF).
Conduction losses only occur where the tin is in contact with the mica, so the losses only occur through the bottom face of the block and are calculated as follows:
LConduction = 0.91 W-hr•in/(hr•ft2•oF) x (4.44 in2) x 230oF x 1 hr / 0.5" x (1 ft2 / 144 in2)
LConduction = 57.3 W-hr
Heat losses due to convection are more difficult to calculate because there are many ways for heat to be removed from an object through a surrounding medium. The main challenge, among others, is due to the fact that the convection medium can be motionless or flowing, referred to as natural convection or forced convection. Convection rates also are affected by the orientation of the object in the medium due to the effects of convection boundary layers in the medium. For example, a hot steel plate that is oriented horizontally in still air loses heat to its surroundings at a different rate than if it is oriented vertically.
Convection loss calculations can be extremely complex, involving fluid mechanics calculations and numbers with names like Reynolds, Prandlt, Nusselt, Rayleigh, Graetz and Grashof. Luckily, there are ways to greatly simplify these calculations, and many tables and charts have been developed to help with most convection situations. Any good book on heat transfer will assist with these calculations.
A simplified formula for calculating the heat loss due to natural convection is:
LConvection = Surface Loss (W/in2) x Area x 1 hr
Continuing the tin block example, assume the block resting on the mica plate is surrounded by air that is not moving, except for that which is rising after being warmed by the block. Calculate the convection loss, given that the surface loss at 300oF is 0.7 W/in2.
To find the solution, you know that the exposed surface area of the five sides of the tin cube is 5 multiplied by 4.44 in2. Therefore:
LConvection = 0.7 W/in2 x 98.568 in2 x 1 hr
LConvection = 69.0 W-hr
The calculation becomes more complex when a fluid (like air) is moving over a heated object at a known flow rate (usually expressed in cubic feet per minute, or cfm). Losses due to radiation usually are insignificant at low temperatures, but they can be a large contributor of heat loss at temperatures above 500oF and should be considered when calculating wattage requirements. The material and surface finish of an object determines its emissivity. A perfectly radiating “blackbody,” by definition, has an emissivity of 1. Emissivity is a number used to compare the radiation from an object to a perfectly emitting blackbody. Emissivity usually is difficult to know exactly, but many tables exist for various materials and surface finishes. So, the formula for calculating heat loss due to radiation is:
LRadiation = Radiation Loss (W/in2) x Area x Emissivity x 1 hr
For example, assume the tin block has a radiation loss of 1.1 W/in2 and an emissivity of 0.04. Suppose you wanted to calculate the radiation loss. You know the exposed surface area of the five sides of the cube that can radiate heat is 5 multiplied by 4.44 in2, or 98.568 in2.
LRadiation = 1.1 W/in2 x 98.568 in2 x 0.04 x 1 hr
LRadiation = 4.3 W-hr
Final Wattage Requirement DeterminationNow that all the wattage requirements and losses are known, it is time to calculate the wattage required for the tin block application.
To recollect, a block of tin resting on a 0.5" thick mica plate is being heated from 70 to 300oF and has heat losses via conduction, convection and radiation. Two numbers must be calculated.
- The wattage required to achieve the desired temperature (initial heatup power).
- The wattage required to maintain the desired temperature.
The higher of the two is the wattage requirement. When calculating the initial heatup power, the initial power required for the system must be high enough to heat the system and compensate for any losses while reaching that temperature.
When a system is at ambient temperature, its losses are zero. When a system is at the desired temperature, its losses are 100 percent. During the heatup time, the losses increase and must be included in the initial power requirement. Avoiding the integration calculation, it can be approximated that the losses are about 65 percent of the 100 percent loss amount during the heatup period. Therefore, the wattage requirement during heatup is calculated as follows:
WInitial = EInitial + EPhaseChangeDuringHeatup + 0.65 x (LConduction + LConvection + LRadiation)
Applying that formula to the tin block example:
WInitial = 86.8 + 0 + 0.65 x (57.3 + 69.0 + 4.3)
WInitial = 171.7 W-hr
The operating power is calculated by adding all the losses during processing at temperature. Applying that formula to the tin block example:
WOperating = EProcess + EPhaseChangeDuringProcessing + LConduction + LConvection + LRadiation
WOperating = 0 + 0 + 57.3 + 69.0 + 4.3
WOperating = 130.6 W-hr
It can be seen that the power required to bring the block to temperature (171.7 W) is greater than the power required to maintain it at temperature (130.6 W). Therefore, 171.7 W will bring the tin block to temperature in 1 hr and be more than enough to maintain the desired temperature when controlled properly.
It is necessary to increase that amount by a safety factor to cover for any errors in the simplifications made during the calculations. Depending on the critical nature of the application, a good rule of thumb is to apply a safety factor of 10 percent to 25 percent. If a safety factor of 20 percent is used, simply multiply the requirement by 1.2. Applying a safety factor of 20 percent to 171.7 W-hr results in 206.0 W-hr.
Now that the wattage requirement is known, proceed with selecting the best method of applying heat to the system.