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questions about electric heaters | Hotwatt

 

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Technical:  Wattage Calculation Formulas

Basic Heating and Heat loss

 

 

 

 

 

 

 

 

 

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 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