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# How Much Heat do You Need? Calculations for Heat Energy

## Defining the heating problem

- Minimum start and finish temperatures expected
- Maximum flow rate of material(s) being heated
- Required time for start-up heating and process cycle times
- Weights and dimensions of both heated material(s) and containing vessel(s)
- Effects of insulation and its thermal properties
- Electrical requirements — voltage
- Temperature sensing methods and location(s)
- Temperature controller type
- Power controller type
- Electrical limitations

## Calculating power requirements

## Short method

Start-up Watts = A + C + 2/3L + Safety Factor

Operating Watts = B + D + L + Safety Factor

Safety Factor is normally 10 percent to 35 percent based on application.

A = Watts required to raise the temperature of material and equipment to the operating point, within the time desired

B = Watts required to raise the temperature of the material during the working cycle

** **

**The equation for A and B (Absorbed Watts for raising temperature)**

* **Weight of material (lbs) x Specific heat of material (°F) x temperature rise (°F)*

* ––––––––––––––––––––––––––––––––––––––––––––––*

* Start-up or cycle time (hrs) x 3.412*

C = Watts required to melt or vaporize material during the start-up period

D = Watts required to melt or vaporize material during the working cycle

** **

**The equation for C and D (Absorbed watts for melting or vaporizing) **

* **Weight of material (lbs) x heat of fusion or vaporization (Btu/lb)*

* ––––––––––––––––––––––––––––––––––––––––––*

* Start-up or cycle time (hrs) x 3.412 *

L = Watts lost from surfaces by conduction use, radiation use heat loss curves or convection use heat loss curves

** **

**The equation for L (Lost conducted Watts)**

* **Thermal conductivity of material or insulation (Btu x in./ft ^{2} x °F x hr) x Surface area (ft^{2}) x Temp. differential to ambient (°F)*

* –––––––––––––––––––––––––––––––––––––––––––––*

* Thickness of material or insulation (in.) x 3.412*

## Power Calculations

**Absorbed Energy, Heat Required to Raise the Temperature of a Material**

Because substances all heat differently, different amounts of heat are required in making a temperature change. The specific heat capacity of a substance is the quantity of heat needed to raise the temperature of a unit quantity of the substance by one degree. Calling the amount of heat added Q, which will cause a change in temperature ∆T to a weight of substance W, at a specific heat of material Cp, then Q = w x Cp x ∆T.

Since all calculations are in Watts, an additional conversion of 3.412 Btu = 1 W-hr is introduced.

*Q _{A} or Q_{B} = w x Cp x ∆T *

* –––––––––*

* 3.412 *

QA = Heat Required to Raise Temperature of Materials During Heat-Up (Wh)

QB = Heat Required to Raise Temperature of Materials Processed in Working Cycle (Wh)

w = Weight of Material (lb)

Cp = Specific Heat of Material (Btu/Ib x °F)

∆T = Temperature Rise of Material (T_{Final} - T_{Initial})(°F)

**Heat Required to Melt or Vaporize a Material**

The heat needed to melt material is known as the latent heat of fusion and represented by H_{f}. Another state change is involved in vaporization and condensation. The latent heat of vaporization H_{v} of the substance is the energy required to change a substance from a liquid to a vapor. This same amount of energy is released as the vapor condenses back to a liquid.

*Q _{C} or Q_{D} = w x H_{f or v }*

* –––––*

*3.412 *

Q_{C} = Heat Required to Melt/Vaporize Materials During Heat-Up (Wh)

Q_{D} = Heat Required to Melt/Vaporize Materials Processed in Working Cycle (Wh)

w = Weight of Material (lb)

H_{f} = Latent Heat of Fusion (Btu/Ib)

H_{v} = Latent Heat of Vaporization (Btu/lb)

**Conduction Heat Losses**

Heat transfer by conduction is the contact exchange of heat from one body at a higher temperature to another body at a lower temperature, or between portions of the same body at different temperatures.

*Q _{L1} = k x A x ∆T x te[1] *

* ––––––––––*

* 3.412 x L *

Q_{L1}= Conduction Heat Losses (Wh)

k = Thermal Conductivity (Btu x in./ft^{2} x °F x hour)

A = Heat Transfer Surface Area (ft^{2})

L = Thickness of Material (in.)

∆T = Temperature Difference Across Material (T_{2}-T_{1})°F

**Convection Heat Losses**

Convection is a special case of conduction. Convection is defined as the transfer of heat from a high-temperature region in a gas or liquidas a result of the movement of the masses of the fluid.

*Q _{L2} = A • F_{SL} • C_{F} *

Q_{L2}= Convection Heat Losses (Wh)

A= Surface Area (in2)

F_{SL} = Vertical Surface Convection Loss Factor (W/in2) Evaluated at Surface Temperature

C_{F}= Surface Orientation Factor: Heated surface faces up horizontally (1.29), Vertical (1.00), Heated surface faces down horizontally (0.63)

**Radiation Heat Losses **

Radiation losses are not dependent on the orientation of the surface. Emissivity is used to adjust for a material’s ability to radiate heat energy.

*Q _{L3} = A x F_{SL} x e *

Q_{L3} = Radiation Heat Losses (Wh)

A = Surface Area (in2)

F_{SL} = Blackbody Radiation Loss Factor at Surface Temperature (W/in2)

e = Emissivity Correction Factor of Material Surface

**Combined Convection and Radiation Heat Losses**

If only the convection component is required, then the radiation component must be determined separately and subtracted from the combined curve.

*Q _{L4} = A x F_{SL} *

Q_{L4} = Surface Heat Losses Combined Convection and Radiation (Wh)

A = Surface Area (in^{2})

F_{SL} = Combined Surface Loss Factor at Surface Temperature (W/in^{2})

**Total Heat Losses **

The total conduction, convection and radiation heat losses are summed together to allow for all losses in the power equations.

*Q _{L} = Q_{L1}+ Q_{L2} + Q_{L3 }*If convection and radiation losses are calculated separately. (Surfaces are not uniformly insulated and losses must be calculated separately.)

OR

*Q _{L} = Q_{L1}+ Q_{L4} *If combined radiation and convection curves are used. (Pipes, ducts, uniformly insulated bodies.)

## Power evaluation

After calculating the start-up and operating power requirements, a comparison must be made and various options evaluated.

Shown in Reference 1 are the start-up and operating Watts, in a graphic format, to help you see how power requirements add up. With this graphic aid in mind, the following evaluations are possible:

Compare start-up watts to operating watts.

Evaluate the effects of lengthening start-up time such that start-up watts equals operating watts (use timer to start system before the shift).

Recognize that more heating capacity exists than is being utilized. (A short start-up time requirement needs more wattage than the process in wattage.)

Identify where most energy is going and redesign or add insulation to reduce wattage requirements.

Having considered the entire system, a review of start-up time, production capacity, and insulating methods should be made. Once you have your required heat, you should consider the application factors of your heater.