Technical Specification
Technical: Wattage Calculation Formulas
Wattage Calculation Data
Basic Heat Formulas
The following
formulae can be employed in determining wattage capacity required for different
materials.
Formula A: Wattage required for heatup =
Weight
of material (lbs) x Specific Heat x Temperature Rise°F
3.412
x Time (hours of fraction thereof)
For specific heat and
weights of each material being heated, see 
Formula B: Wattage losses at operating temperature = Wattage
loss/sq.ft. x Area in sq.ft.
Formula C: Wattage for melting or
vaporizing =
Weight
of material (lbs) x Heat of fusion or vaporization (BTU/lb)
3.412
x Heatup time (hours of fraction thereof)
When the specific
heat of a material changes at some temperature during the heatup, due to
melting (fusion) or evaporation (vaporation),perform Formula A for heat
absorbed from the initial temperature up to the temperature at the point of
change, add Formula B, then repeat Formula A for heat absorbed from the point
of change to the final operating temperature. For heats of fusion and
vaporization and temperatures at which these changes in state occur see
Tables 1, 2 and 3 as
refered to under Formula A.
Specific Applications
For specific applications, substitute the Basic Heat Formulas (A, B, or C
above) into the following:
To Heat Liquids
Wattage for initial heatup = (a) + (b)
2
Wattage for operating
requirements = (a) for new material added + (b)
To insure adequate capacity, add 20% to final wattage figures. This will
compensate for added heat losses not readily computed.
To Melt Soft Metals
Wattage for initial
heatup = (a) to melting point + (c) to melt (a) to heat above melting
point + (b)
2
Wattage for operating
requirements = [(a) to melting point = (c) to melt + (a) to heat above melting
point] for added material + 11. To insure adequate capacity, add 20% to final
wattage figures. This will compensate for added heat losses not readily
computed.
To Heat Ovens
Wattage = (a) (for air) + (a) (all material introduced into oven) + (b)
Add 25% to cover door heat losses
Forced Air Heating
Wattage = C.F.M.
x temperature rise (°F)
3
Mathematical
Conversions
Areas and Volume
Circles
To find circumference  multiply the diameter by 3.1416; or, divide diameter by 0.3183.
To find diameter  multiply the circumference by
0.3183; or, divide circumference by 3.1416.
To find radius  multiply the circumference by
0.15915; or, divide circumference by 6.28318;
or, divide diameter by 2.
To find the side of a square
to be inscribed in a circle  multiply diameter by 0.7071; or,
multiply the circumference by 0.2251; or, divide the circumference by 4.4428.
To find the side of a square
to equal the area of a circle  multiply the diameter by 0.8862;
or, divide diameter by 1,1284; or, multiply the circumference by 0.2821; or,
divide circumference
by 3.545.
To find the area of a circle  multiply the circumference by
onequarter of the diameter; or,
multiply the square of the diameter by 0.7854; or, multiply the square of the
circumference by
0.7958; or, multiply the square of onehalf the diameter by 3.1416.
Doubling the diameter of a circle increases the area 4 times.
Squares
A side multiplied by 1.412 = the diameter of a circle which will
circumscribe the given square.
A side multiplied by 4.443 = the circumference of its circumscribing circle.
A side multiplied by 1.1284 = the diameter of a circle equal in area to that
given square.
A side multiplied by 3.545 = circumference of an equal circle.
To find diagonal of a square  multiply side by 1.4142.
Measurements from Other Geometrical Forms
To find the area of an ellipse  multiply the product of its axes by 0.7854; or, multiply the
product of its semiaxes by 3.14159.
Contents of a cylinder = area of end X length.
Contents of a wedge = area of triangular base X altitude.
Surface of a cylinder = length X circumference plus area of
both ends.
Surface of a sphere = diameter squared X 3.1416; or,
diameter X circumference.
Contents of a sphere = diameter cubed X 0.5236.
Contents of a pyramid or
cone, right or oblique, regular or irregular = area of base
X onethird of the altitude.
Area of a triangle = base X onehalf the altitude.
Area of parallelogram = base X altitude.
Area of trapezoid = altitude X onehalf the sum of
parallel sides.
To find distance across the
corners of hexagons 
multiply the distance across the
flats by 1.1547.
Conversion Factors
1 gal. water = 8.3 lb.
1 hp = 745.2 watts.
1 BTU = .252 kg calories = 0.2930 watt hours.
1 BTU per lb. = 1.8 cal per gram.
1 kwhr = 3412 BTU per hour.
1 kwhr will evaporate 3.5 lb of water at 212°F.
1 kwhr will raise 22.75 lb of water from 62°F to 212°F.
1 gal. = 231 cu.in. = 3.785 liters = .1337 cu.ft.
1 cu.ft. = 1728 cu.in. = .03704 cu.yd. = 7.481 gal.
To find
the equivalent, in terms of a unit in the customary system, of a given number
of
metric units, multiply or divide their number (as indicated) by the factor
shown.
Thus: 10 millimeters are equivalent to 10 x 0.03937 inches or to 10 ÷ 25.4
inches.
Millimeters
x .03937 = inches; or ÷ 25.4 = inches.
Centimeters x .03937 = inches; or ÷ 25.4 = inches.
Meters x 39.37 = inches.
Meters x 3.28 = feet.
Kilometers x 3280.8 = feet.
Square meters x 10.764 = square feet.
Cubic centimeters ÷ 16.387 = cubic inches.
Cubic centimeters ÷ 3.70 = fluid drams (U.S.P.).
Cubic centimeters ÷ 29.57 = fluid ounces (U.S.P.).
Cubic centimeters ÷ 3.531 x 10 5 = cubic feet.
Cubic meters x 35.314 = cubic feet.
Liters x 61.025 = cubic inches.
Liters x 33.81 = fluid ounces (U.S.P.).
Liters x .2642 = gallons (231 cubic inches).
Liters ÷ 3.785 = gallons (231 cubic inches).
Liters ÷ 28.317 = cubic feet.
Grams x 15.432 = grains.
Grams (water) ÷ 29.57 = fluid ounces.
Grams ÷ 28.35 = ounces avoirdupois.
Grams per cubic centimeter ÷ 27.7 = pounds per cubic inch.
Kilograms x 2.2046 = pounds.
Kilograms x 35.3 = ounces avoirdupois.
Kilograms per square centimeter x 14.223 = pounds per square inch.
Kilo per meter x .672 = pounds per foot.
Kilo per cubic meter x .062 = pounds per cubic foot.
Kilowatts x 1.34 = h.p. (33,000 foot pounds per minute).
Watts ÷ 746 = horse power.
Centigrade x 1.8 + 32 = degrees Fahrenheit.
Technical: Ohms Law and Wiring Diagrams
Ohms Law
E
= Volts, W = Watts, I = Amperes, R = Ohms
To Determine Watts
(W):
W = E I

W = I² R

W = E²
R

To Determine Volts
(E):
E =

E = W
I

E = IR

To Determine Ohms
(R):
R = W
I²

R = E²
W

R = E
I

To Determine
Amperes (I):
I = E
R

I = W
E

I =

Resistance
wire  Current vs. Temperature
Current Carrying Capacity of Straight Nickel
Chromium Wire
Approximate Amperes to heat a straight, oxidized wire in quiet air to given
temperature
Degrees F

400

600

800

1000

1200

1400

Degrees C

205.

315

427

538

649

760

A.W.G.
or
B. & S.

Inches
Dia.

Amperes

15

.057

7.2

10.0

12.8

16.1

20.0

24.5

16

.051

6.4

8.7

10.9

13.7

17.0

20.9

17

.045

5.5

7.5

9.5

11.7

14.5

17.6

18

.040

4.8

6.5

8.2

10.1

12.2

14.8

19

.036

4.3

5.8

7.2

8.7

10.6

12.7

20

.032

3.8

5.1

6.3

7.6

9.1

11.0

21

.0285

3.3

4.3

5.3

6.5

7.8

9.4

22

.0253

2.9

3.7

4.5

5.6

6.8

8.2

23

.0226

2.58

3.3

4.0

4.9

5.9

7.0

24

.0201

2.21

2.9

3.4

4.2

5.1

6.0

25

.0179

1.92

2.52

3.0

3.6

4.3

5.2

26

.0159

1.67

2.14

2.60

3.2

3.8

4.5

27

.0142

1.44

1.84

2.25

2.73

3.3

3.9

28

.0126

1.24

1.61

1.95

2.38

2.85

3.4

29

.0113

1.08

1.41

1.73

2.10

2.51

2.95

30

.0100

.92

1.19

1.47

1.78

2.14

2.52

31

.0089

.77

1.03

1.28

1.54

1.84

2.17

32

.0080

.68

.90

1.13

1.36

1.62

1.89

33

.0071

.59

.79

.97

1.17

1.40

1.62

34

.0063

.50

.68

.83

1.00

1.20

1.41

35

.0056

.43

.57

.72

.87

1.03

1.21

36

.0050

.38

.52

.63

.77

.89

1.04

37

.0045

.35

.46

.57

.68

.78

.90

38

.0040

.30

.41

.50

.59

.68

.78

39

.0035

.27

.36

.42

.49

.58

.66

40

.0031

.24

.31

.36

.43

.50

.57

Current Carrying Capacity of Ribbon Nickel Chromium
Wire
At 1200°F approximate
.

WidthInches

Thickness
Inches

1/64

1/32

1/16

3/32

1/8

3/16

.

Amps

.0063

1.56

2.89

5.5

8.2

10.1

16.6

.0056

1.45

2.69

5.2

7.2

9.5

15.6

.0050

1.35

2.52

4.9

6.8

9.0

14.7

.0045

1.26

2.38

4.6

6.4

8.5

14.0

.0040

1.18

2.23

4.1

6.0

8.0

13.1

.0035

1.09

2.07

3.8

5.6

7.5

12.3

.0031

1.01

1.94

3.6

5.3

7.0

11.5

.0020













.0015

4











The following two formulae will help you to calculate the diameter of wire under certain Capacity and Voltage.
Where:
d — Diameter of heating wire, mm
p — Resistivity of heating wire，Ω·mm^{2}/m
P — Power per phase, KiloWatts
U — Voltage，Volts
W — Surface loading，W/cm^{2}
L — Length of heating wire, m
For Example:
1, Supposing the condition is 800W/240V and resistance per meter you required is 8.7 Ω/m, you are going to use Nichrome 80 as the heating wire because of its high and stable performance. Refer to Nichrome Wire Data Table, you will find that dia. 0.4mm is what you need. Here according to technical data of Nichrome 80, p is about 1.09 Ω·mm2/m. So according to the formula 2, L is 8.3 m. To verify the surface loading, W is 7.68 W/cm^{2}, you can check whether it is ok for you. If your calculated surface loading is too high or too low, please change the grade of alloy.
2, Surface loading is related with heat transfer and element's service life. If you have requirement on surface loading such as W is 5 W/cm^{2}，then you can know d is 0.46mm under the same condition as 1.
If you have any questions, please contact our sales.
We are wiling to provide our esteem customer with any kind of heating related Technical services .
The current values in these are based
on actual teats of single strands of oxidized wire mounted in quiet air and
operated at 1200°F. The tables are calculated for wire having a
resistivity at 1200°F and a total surface wattsdensity of 28 watts per square
inch.
