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Technical:
 
Wattage Calculation Formulas

Problem 1:  Basic Heating and Heat Loss


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 70F.

From Table 1Properties 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 @ 400F

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.