EGEE 102
Energy Conservation and Environmental Protection

Energy Required for Water Heating

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The amount of energy required to heat water is proportional to the temperature difference of what?

To calculate the Heat Required, use this equation:

Q=m C p ΔT

Where …

m = mass of water heated

C p = the heat capacity of water (1 BTU / lb ºF)

ΔT = temperature difference.

Important Point IconRemember to make your units of measurements consistent. Since Cp is measured in pounds, your mass of water heated should be measured in pounds as well. Thus, if you only know the number of gallons, you must convert it into pounds. One gallon of water = about 8.3 pounds, so multiply number of gallons by 8.3 to determine the weight in pounds.

Example 1

It is estimated by the United States Department of Energy that a family of four each showering for 10 minutes a day consumes about 700 gal of hot water a week. Water for the showers comes into the home at 55ºF and needs to be heated to 120ºF.

To calculate the heat required, determine the variables:
m = mass of water heated = 700 gallons = 5810 lbs
Cp is the heat capacity of water = 1 BTU/lb ºF (given)
ΔT = temperature difference = 120 ºF – 55 ºF

Heat energy required to heat 700 gal can be calculated as follows:

Heat Required = 5810 lbs x 1 BTU/lb ºF x (120 ºF – 55 ºF)
Heat Required = 5810 lbs x 65 ºF
Heat Required = 377,650 BTU/week

The heat requirement for one year is :

377,650 BTU/Week x 52 Weeks/Year = 19,637,800 BTU/year or 5,755 kWh

Assuming that the natural gas costs $ 10/MMBTU (1 MMBTU = 1000000 BTU) and electricity costs 0.092 per kWh, the gas costs would be $196.37 while electric costs would be $529.46. Clearly, electric heat is more expensive than natural gas.

Example 2

Estimate the % energy savings of an electric water heater that heats 100 gallons of per day when the temperature is set back at 110° instead of 120°F. The basement is heated and is at 65°F. The life of the water heater is expected to be about 10 years. Use an appropriate cost for electricity and compare the operating expenses.

Heat required (BTU) = m x Cp x (Temperature Difference)

Where Cp is the heat capacity of water (1 BTU/lb/F) and m is the mass of the water (Assume 1 gal has 8.3 lb of water and the 3,412 BTU = 1 kWh)

Solution:

Energy required for heating the water to 120°F:

=m× C p ×ΔT

= 100  gal day × 8.3  lb gal m × 1 BTU lb   °F C p × ( 12065 ) °F ΔT

= 100  gal day × 8.3  lb gal × 1 BTU lb   °F × ( 12065 ) °F

=45,650 BTU/day

In a year the energy required is:

45,650 BTU day × 365  days year =16,662,250 BTUs per year

In a 10 year period, the energy required is 166,622,500 BTU which is equal to 48,834 kWh .

166,622,500  BTU  × 1 kWh 3,412  BTU = 48,834 kWh

Operating cost over its life time is:

 

48,834 kWh 1 × $0.09 kWh =$4,395.06

 

Energy required for heating the water to 110°F:

=m× C p ×ΔT

= 100  gal day × 8.3  lb gal m × 1 BTU lb   °F C p × ( 11065 ) °F ΔT

= 100  gal day × 8.3  lb gal × 1 BTU lb   °F × ( 11065 ) °F

=37,350 BTU/day

In a year the energy required is:

37,350 BTU day × 365  days year =13,632,750 BTUs per year

In a 10 year period, the energy required is 136,327,500 BTU which is equal to 39,995 kWh .

136,327,500  BTU  × 1 kWh 3,412  BTU = 39,995 kWh

Operating cost over its lifetime is:

39,955 kWh 1 × $0.09 kWh =$3,595.95

Estimated % Energy Savings:

$4,395.06 - $3,595.95 = $799.11 savings

$799.11 $4,395.06  = 18.2% savings