Heat Calculation Examples

EXAMPLE 1: A tubular Bushing Heater is required to heat a 10 gallon tank of water. Initial water temperature is at room temperature of approx. 70� F. Final water temperature required is 100� F. Water must be brought up to final temperature in 1 hour. No additional water is added to the system. Customer has an ideal system with zero heat loss. Calculate the wattage required to heat only the water (ignore the tank for this example.).

SOLUTION:

1. Basic equation for initial material heatup:

Wattage =	weight (lbs) x spec. heat (btu/lb �F) x temp. change (�F)
	3.412 (btu to watt/hr conversion) x heat up time in hours

2. From properties of liquids and engineering conversions charts:

a. 1 gallon of water = 8.3 pounds (lbs)
b. specific heat of water = 1.000 btu/lb

3. Plug numbers into equation and solve:

Wattage =	(10 gal x 8.3 lbs/gal) x 1.000 btu/lb �F x (100 �F - 70 �F)
	(3.412 btu/watt hr) x 1 hr

= 7.29.8 watts

Add 20% allowance = 730 watts x 1.20 = 875 watts

EXAMPLE 2: Customer requires Flanged Tubular Heater to heat a 200 gallon tank of water. Initial water temperature is at room temperature of approx. 70� F. Final water temperature required is 150� F. Water must be brought up to final temperature in 3 hours. No additional water is added to the system. Tank has approximately 25 square feet of 1�4� thick uninsulated steel walls and a total tank weight of 254 lbs. Water surface area is approximately 12 square feet and is open to the environment. Tank is installed in a room at 70� F.

SOLUTION:

1. Remember basic equations:

a. For material and tank heatup

Wattage =	weight (lbs) x spec. heat (btu/lb �F) x temp. change (�F)
	3.412 (btu to watt/hr conversion) x heat up time in hours

b. For heat loss from water surface:

average wattage loss =	liquid surface area (sq. ft.) x watts/sq. ft. loss at final temp
	2

c. For heat loss from walls of uninsulated steel tank

average wattage loss =	tank wall surface area (sq. ft.) x watts/sq. ft. loss at final temp
	2

2. From properties of liquids chart, engineering conversions chart, and heat loss graphs:

1 gallon of water = 8.3 pounds (lbs)
Specific heat of water = 1.000 btu/lb
Specific heat of steel = .12 btu/lb
Heat loss from water surface at final temp = 300 w/sq. ft.
Heat loss from uninsulated steel walls at final temp = approx. 25 w/sq. ft. x .83 (averaging factor) = 20.8 w/sq. ft.

3. Plug numbers into equations and solve:

wattage (to heat water) =	(200 gal x 8.3 lbs/gal) x 1.000 btu/lb x (150� F - 70� F)
	(3.412 btu/watt hr) x 3 hr

wattage (to heat water) =	(254 lbs) x .12 btu/lb x (150� F - 70� F)
	(3.412 btu/watt hr) x 3 hr

wattage (ave. loss from water surface) =	12 sq. ft. x 300 w/sq. ft.
	2

wattage (ave. loss from tank walls) =	25 sq. ft. x 20.8 w/sq. ft.
	2

Total wattage = 12,974 w + 238 w + 1800 w + 260 w = 15,272 w (15.3 kw)

Total wattage with 20% contingency = 18,326 watts (18.3 kw)

EXAMPLE 3: Same conditions as example 1 except tank and water are already up to final operating temperature. No additional water or other material is added to system. How much wattage is required to maintain the tank and water at the 150� F temperature in the 70� F room?

SOLUTION:

1. The tank and water are already up to the final operating temperature, therefore, no heat is required to bring the system up to temperature.

Heatup wattage = 0

2. The only factors affecting heat requirements are heat losses from the water surface and tank walls at operating temperature.

3. Heat equations to counteract heat losses at operating temperature

a. For heat loss from water surface:

wattage loss = Liquid surface area (sq. ft.) x watts/sp. ft. loss at final temp

b. For heat loss from walls of uninsulated steel tank:

wattage loss = tank wall surface area (sq. ft.) x watts/sq. ft. loss at final temp

4. Use same heat loss - watts/sq. ft. values from graphs as in example 2:

a. Heat loss from water surface at final temp = 300 w/sq. ft.

b. Heat loss from uninsulated steel walls at final temp = approx. 25 w/sq. ft. x 8.3 (averaging factor) = 20.8 w/sq. ft.