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PHONE: (978) 777-0070
Technical: Wattage Calculation Formulas
Problem 1: Basic Heating and Heat Loss
A steel mold is being used to form polyethylene parts. Each hour, 90 ounces of nylon is
introducted to the mold. The mold itself measures
10" x 8" x 4". The mold is attached between two stainless steel platens,
each measuring 15" x 12" x 1 1/2" thick. The platens are insulated
from the press mechanism with 1/2" thick insulation. Operating temperature of the
mold if 400° F and is required to reach this temperature
in 1 hour with an ambient temperature of 70°F.
From Table 1: Properties of Metals
Specific heat of steel: .12 BTU/lb/°F
Specific heat of stainless steel: .12 BTU/lb/°F
Converting cubic inches into pounds (density lb/cu. in.)
From Table 2: Properties of Non-Metallic
Solids
Specific heat of polyethelyne: .55 BTU/lb/°F
From Graph 1: Guide for Heat Losses
Heat losses curves: A + B @ 400°F
Formula A: Wattage required for heat-up
| To Heat Mold (10" x 8" x 4") = 320 cu.in. x 2 x .284 = 181. 7(lbs) x .12 BTU/lb °F x (400-70) °F = 3.412 x 1 |
2,110 watts |
| To Heat Platens (15" x 12" x 1 1/2") = 270 cu.in. x 2 x .286 = 154.5(lbs) x .12 BTU/lb °F x (400-70) °F = 3.412 x 1 |
1,800 watts |
| To Heat Polyethelyne 90 = 5.6 (lbs) x .55 BTU/lb °F x (400-70) °F = 16 3.412 x 1 |
300 watts |
| Compensation Factor 20% (2,110 + 1,800 + 300) = |
840 watts |
| Total wattage required for Heat-up= | _________ 5,050 watts |
Formula B: Wattage losses at operating temperature (See graphs: Guide for Heat Losses)
| Heat Loss From Mold (vertical surfaces) 10" x 4" x 4" + 8" x 4" x 4" = 2 sq. ft. x 350 w/sq.ft./hr. = 144" |
700 watts |
| Heat Loss From Platen (vertical surfaces) 1 1/2" x 15" x 4" + 1 1/2" x 12" x 4" = 1.1 sq.ft. x 350 w/sq./ft./hr. = 144" |
385 watts |
| Heat Loss From Platen (horizontal surfaces, uninsulated) 15" x 12" x 2"- (10" x 8" x 2") = 1.3 sq.ft. x 250 w/sq./ft./hr. = 144" |
350 watts |
| Heat Loss From Platen (insulated surface) 15" x 12" x 2" = 2.5 sq.ft. x 100 w/sq./ft./hr. = 144" |
250 watts |
| Compensation Factor 20% (700w + 385w + 350w + 250w) = |
340 watts |
| Total wattage losses at operating temperature = Total wattage required for heat-up = |
_________ 2,025 watts 5,050 watts |
| Total wattage required = | _________ 7,075 watts |
The number of holes in the mold would dictate the number of heaters required. Dividing the wattage by the number of heaters will equal the wattage rating of each heater. |
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