7.3 Energy Required for Water Heating

$$\text{Q}=\text{m}\times {\text{C}}_{\text{p}}\times \text{ΔT}$$

Where …

$\text{m}$= mass of water heated

${\text{C}}_{\text{p}}$= the heat capacity of water (1 BTU / lb ºF)

$\text{ΔT}$= temperature difference

Remember to make your units of measurements consistent. Since C_p 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
C_p is the heat capacity of water = 1 BTU/lb ºF (given)
ΔT = temperature difference = 120 ºF – 55 °F

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

Heat Required = 5,810 lbs × 1 BTU/lb ºF × (120 ºF – 55 ºF)
Heat Required = 5,810 lbs × 65 ºF
Heat Required = 377,650 BTU/week

The heat requirement for one year is :

377,650 BTU/Week × 52 Weeks/Year = 19,637,800 BTU/year 
or 5,755 kWh/year.  (Remember 1kWh = 3412 BTUs) 

Assuming that the natural gas costs $3.5 /MMBTU (1 MMBTU = 1,000,000 BTU) and electricity costs 0.145 per kWh, the annual natural gas costs would be $68.73 while annual electric costs would be $834.54. Clearly, electric hot water is much 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 × C_p × (Temperature Difference)

Where C_p is the heat capacity of water (1 BTU/lb ºF) 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:

$$=\text{m}\times {\text{C}}_{\text{p}}\times \text{ΔT}$$

$$=\underset{\text{m}}{\underbrace{\frac{\text{100 } \text{gal} }{\text{day}}\times \frac{\text{8}\text{.3 } \text{lb} }{\text{gal} }}}\times \underset{{\text{C}}_{\text{p}}}{\underbrace{\frac{\text{1 BTU}}{\text{lb} \text{°F} }}}\times \underset{\text{ΔT}}{\underbrace{\left( 120-65 \right) \text{°F} }}$$

$$= \frac{\text{100 }\cancel{ \text{gal} }}{\text{day}}\times \frac{\text{8}\text{.3 }\cancel{ \text{lb} }}{\cancel{\text{gal} }}\times \frac{\text{1 BTU}}{\cancel{\text{lb} }\cancel{ \text{°F} }}\times \left( 120-65 \right)\cancel{ \text{°F} }$$

$$=\text{45,650 BTU/day}$$

In a year the energy required is:

$$\frac{\text{45,650 BTU}}{\cancel{\text{day}}}\times \frac{\text{365 } \cancel{\text{days}}}{\text{year}}=\text{16,662,250 BTUs per year}$$

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

$$\text{166,622,500 } \cancel{\text{BTU}} \times \frac{\text{1 kWh}}{\text{3,412 } \cancel{\text{BTU}}}\text{= 48,834 kWh}$$

Operating cost over its lifetime is:

$$\frac{\text{48,834}\cancel{\text{kWh}}}{\text{1}}\times \frac{\text{\$0}\text{.09}}{\cancel{\text{kWh}}}=\text{\$4,395.06}$$

Energy required for heating the water to 110°F:

$$=\text{m}\times {\text{C}}_{\text{p}}\times \text{ΔT}$$

$$=\underset{\text{m}}{\underbrace{\frac{\text{100 } \text{gal} }{\text{day}}\times \frac{\text{8}\text{.3 } \text{lb} }{\text{gal} }}}\times \underset{{\text{C}}_{\text{p}}}{\underbrace{\frac{\text{1 BTU}}{\text{lb} \text{°F} }}}\times \underset{\text{ΔT}}{\underbrace{\left( 110-65 \right) \text{°F} }}$$

$$= \frac{\text{100 }\cancel{ \text{gal} }}{\text{day}}\times \frac{\text{8}\text{.3 }\cancel{ \text{lb} }}{\cancel{\text{gal} }}\times \frac{\text{1 BTU}}{\cancel{\text{lb} }\cancel{ \text{°F} }}\times \left( 110-65 \right)\cancel{ \text{°F} }$$

$$=\text{37,350 BTU/day}$$

In a year, the energy required is:

$$\frac{\text{37,350 BTU}}{\cancel{\text{day}}}\times \frac{\text{365 } \cancel{\text{days}}}{\text{year}}=\text{13,632,750 BTUs per year}$$

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

$$\text{136,327,500 } \cancel{\text{BTU}} \times \frac{\text{1 kWh}}{\text{3,412 } \cancel{\text{BTU}}}\text{= 39,995 kWh}$$

Operating cost over its lifetime is:

$$\frac{\text{39,955}\cancel{\text{kWh}}}{\text{1}}\times \frac{\text{\$0}\text{.09}}{\cancel{\text{kWh}}}=\text{\$3,595.95}$$

Estimated % Energy Savings:

$$\text{\$4,395.06 - \$3,595.95 = \$799.11 savings}$$

$$\frac{\text{\$799.11}}{\text{\$4,395.06}}=\text{ 18.2\% savings}$$