P = V x I (W)
I = V / R (A)

P = V²/ R = I²x R
R = V²/ P = P / I²

Ex) What is the resistance when voltage is 220V, and power is 500W ?
      R = (220x220) / 500 = 96.8 Ω
Ex) If resistance is 48.4Ω and voltage is 220V, what is the capacity ?
      P = V²/ R = (220x220) / 48.4 = 1,000W
* When only the resistance is available, voltages of 110V, 127V, 220V, 254V, 380V, etc are substituted to gain capacity. Then, power should be compared.

WD = P / (π x D x HL) (W/Cm²)

Line voltage = Vs , Phase voltage = Vw, Line current = As, Phase current = Aw

※ Vw = Vs / √3 (√3 = 1.732050808)
※ As = Aw

Ex) When AC is 380V 3Φ and total capacity is 1200W, the individual capacity for heater is 1200W / 3 = 400W and voltage (Vw) is 380 / √3 = 219.393V
Therefore, three heaters of 220V 400W should be prepared, and Star wired.

Line voltage = Vs , phase voltage = Vw, Line current = As, phase current = Aw

※ Vw = Vs
※ Aw = As / √3 (√3 = 1.732050808)

Ex) When AC is 380V 3Φ and total capacity is 1200W, individual capacity of heater is 1200W / 3 = 400W and voltage (Vw) is Vw=Vs, 380V
Therefore, three heaters of 380V 400W should be prepared and Delta wired.

• 
  AC : Effective value = maximum value / √2
  (General alternating voltage is effective value)
• 
  DC
• 
  DC(rectifier)

Q = I² x R x T (J) = 0.239 x I² x R x T (cal)

Q = Calorie(J) / I = Current(A) / R = Resistance(Ω) / T = Time(sec)

*If 0 ℃ of ice is to be 0 ℃ of water, 80 cal/g of heat of fusion is required.
  If 100 ℃ of saturated water is to be 100 ℃ of vapor, 539 cal/g of latent heat for evaporation is required.

※ 1lb=453g
※ 1BTU=0.252Kcal
※ 1CHU=1.8BTU=0.454Kcal
※ 1Kcal=3.968BTU=427kg.m=4.2KJ

* Cal : calorie for heating 1 g of pure water from 14.5 ℃ to 15.5 ℃
* BTU : calorie for heating 1 lb of pure water from 60 ℉ to 61 ℉
* CHU : calorie for heating 1 lb of pure water from 14.5 ℃ to 15.5 ℃