The recovery time of a hot water heater is the time required to heat all the water in the unit to the desired temperature. Suppose that a 62-gal (1.00 gal = 3.79 x 10-3 m3) unit starts with cold water at 11 °C and delivers hot water at 53 °C. The unit is electric and utilizes a resistance heater (120 V ac, 3.4 Ù) to heat the water. Assuming that no heat is lost to the environment, determine the recovery time (in hours) of the unit.
To determine the recovery time of the unit, we need to calculate the amount of heat energy required to heat the 62-gallon water from 11°C to 53°C, and then divide it by the power output of the heater.
The equation for heat energy is given as:
Q = mcΔT
Where:
Q = heat energy required (in joules)
m = mass of water (in kg)
c = specific heat capacity of water (in J/kg°C)
ΔT = temperature change (in °C)
Converting the volume of water to mass:
Mass = Volume × Density
Density of water = 1000 kg/m³
Mass = 62 gal × (3.79 x 10^-3 m³/gal) × (1000 kg/m³) = 235.78 kg
The specific heat capacity of water is approximately 4.18 J/g°C.
Next, we calculate the temperature change:
ΔT = 53°C - 11°C = 42°C
Now, let's calculate the heat energy required:
Q = (235.78 kg) × (4.18 J/g°C) × (42°C) = 43161.33 J
The power output of the heater is 120 V ac, with a resistance of 3.4 Ω. Using Ohm's Law, we can find the power (P) using the formula:
P = (V²) / R
P = (120 V)² / 3.4 Ω = 4235.29 W
Now, we divide the heat energy by the power output to get the recovery time:
Recovery time = Q / P = 43161.33 J / 4235.29 W
Let's convert the units for consistency:
Recovery time = (43161.33 J) / (4235.29 W) = 10.197 seconds
To convert this to hours, we divide by 3600 (the number of seconds in an hour):
Recovery time (in hours) = 10.197 seconds / 3600 = 0.0028 hours
Therefore, the recovery time of the unit is approximately 0.0028 hours, or about 10 seconds.