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How many kilowatts needed to heat water
Mr Kilowatt does water heater math
|Formula for heating water inside tank:|
Gallon Per Hour (Electric) = (KW x 3413) divided by (temp. rise x 8.25) or (KW x 414) divided by (temp rise.)
Let's say you have 5500 watt elements.
let's say incoming water temperature is 65°F and you want to heat tank to 105°, or 40° rise in temperature.
5500 watt or 5.5Kw elements x 3413 = 18771.5
40 degree temperature rise x 8.25 = 330
So in this specific situation, the element would heat 56.88 gallons per hour
If you have 4500 watt elements and incoming water temperature is 55°, and you want to heat to 120°, then the temperature rise is 65°
4.5 KW x 3413 = 15358.5
65 degree temperature rise x 8.25 = 536
So one element would heat 28 gallons per hour
This number is for EACH element, but does NOT consider heat loss from tank, room temperature, or thickness of insulation etc.
Remember, with residential simultaneous tank, only ONE element is ON at one time.
Upper element is ON, or lower element is ON, or BOTH elements are off. At no time are both elements on.
Kilowatt hours of electricity needed to heat water ... the math formula:1.) Basic formula > it takes .0002931 Kwh to raise 1 pound of water 1°F
2.) Basic number >A gallon of water weighs 8.34 lbs (pounds)
3.) "A BTU is amount of heat needed to raise 1 pound of liquid water by 1° from 60° to 61° F at constant pressure of one atmosphere. Other definitions of BTU exist which are based on different water temperatures and cause the results to vary by .5%." (For analysis of water heaters, .5% is ignored. Variation in atmospheric pressure is also ignored. Water heater 'science' is an approximation since calcium carbonate in tank will vary electrical efficiency and tank capacity.)
4.) BTU formula > 1 BTU = 2.931 x 10–4 Kwh (kilowatt hours).
BTU = 2.931 x .0001 = .0002931 Kwh
Therefore it takes .000239 Kwh to raise 1 pound of water 1° F
5.) 30 Gallons of water x 8.34 = 250.2 lbs
40 Gallons of water x 8.34 = 333.6 lbs
Calculation using 40 Gallon water heater:
> Ordinary 40 gallon water heater in attic. Temperature in attic = 50° F. Temperature of water in tank = 50° F.
> Water in tank weighs 333.6 lbs.
> How much electricity is needed to raise temperature of full tank to 120° F.
> 120° minus 50° = 70° > so the temperature needs to go up 70°
> Multiply 333.6 x 70 x .0002391 = 5.58 Kwh
Calculation using 30 Gallon water heater:
> Same problem as above
> Water in tank weighs 220.2 lbs
> Multiply 250.2 x 70 x .0002391 = 4.18 Kwh
|Temperature of water
||Cost if you use 40 Gallon per day||Cost if you use 30 Gallon per day|
|Raise from 32° to 120°||7.01 Kwh x 14¢ = $1.20||5.26 Kwh x 14¢ = 73¢|
|Raise from 50° to 120°||5.58 Kwh x 14¢ = 78¢||4.18 Kwh x 14¢ = 58¢|
|Raise from 70° to 120°||3.99 Kwh x 14¢ = 56¢||2.99 Kwh x 14¢ = 42¢|
|Raise from 85° to 120°||2.79 Kwh x 14¢ = 39¢||2.09 Kwh x 14¢ = 29¢|
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