Calculate the daily energy saving (in units of kWh)

Question

You make a decision to reduce your daily shower time from 15 minutes to 5 minutes. For each shower the water is heated from 20oC to 50oC and an energy-saving shower head is used which limits the water flow rate to 7 L/min. Calculate the daily energy saving (in units of kWh) resulting from the decision to take shorter showers.

Note: To raise the temperature of one litre (L) of water by 1oC requires 4200 J of heat energy

Answer 1

First, determine the total volume of water used by each shower duration.

Longer shower: Volume = 15 min * 7 L/min = 105 L

Shorter shower: Volume - 5 min * 7 L/min = 35 L

*

Energy consumed = Q = Vol * energy/vol of water * Temp Change  (units must cancel)

 

Longer shower: (105 L) * (4200 J / L - deg C) * (50 - 20 deg C) = 1.32 x 10^7 J = 13,200 kJ

Shorter shower:  (35 L) * (4200 J / L - deg C) * (50 - 20 deg C) = 4.4 x 10^6 J = 4,400 kJ

 

Daily savings = 8,800 kJ