Given a 40 gallon natural gas water heater, and a shower head with a flow rate of 2.5 gallons per minute. How long should the shower be able to maintain 105°F water temperature?
Assume the average cold water supply temperature is 58.7°F. The tank will be set at 140°F, to avoid Legionella. The tank recovery is typical of a natural gas heater, so it can recover its volume in an hour (40 gph in this case).
Model the water heater as a continuously stirred tank reactor (CSTR), so it is always at a uniform temperature. Assume the recovery time is not dependent on temperature and completely accounts for insulation losses and the like. Neglect losses in pipes and assume the operator controls the shower temperature to 105°F perfectly. Taking the stopping criterion from the question, the shower is over when the water in the tank becomes 105°F.
The recovery rate raises 140 gallons of water by 81.3°F in one hour. We'll say this is a constant heat input of 9118 W.
The key is that a constant-temperature shower removes a constant rate of heat from the tank, that which is associated with raising 2.5 gpm of water from 58.7°F to 105°F. This is 16986.5 W.
The difference is 7868.5 W. With constant heat, you don't need any complicated integrals.
The time for the tank to drop from 140°F to 105°F is 1568 s or 26 minutes.
The (maybe counterintuitive) fact that the variable flow rate from the water heater does not influence the rate of heat removal from the water heater comes from the fact that the incoming cold water is the same temperature at the shower and at the water heater.
Note that, because 
does not appear in the expression for 
, is now a constant.
To determine the heat transfer rate (Q) of the open system in BTUs per hour, you can use the following formula:
Q (BTU/hr) = Flow Rate (GPM) x (Temperature Leaving Process (F) - Temperature Entering Process (F)) x 500.4
If the flow rate of the open system is 2.5 GPM, the temperature leaving the process is 105°F, and the temperature entering the process is 58.7°F, then the heat transfer rate can be calculated as follows:
Q = 2.5 GPM x (105°F - 58.7°F) x 500.4 = 2.5 GPM x 46.3°F x 500.4 = 59,526.85 BTU/hr
So the heat transfer rate in this case would be approximately 59,526.85 BTUs per hour.
A typical gas-fired 40 gallon water heater produces 34,000 BTUs per hour. The difference in the heat rates reduces the temperature in the 40 gallon tank. The difference equals 25,526.85 BTUs per hour, or 7.0907 BTU/sec.
To determine the value of dt (the change in time in seconds) using the formula:
dt = mcpdT/Q
Where m is the mass of the water in pounds, cp is the specific heat capacity of water (1 BTU/pound-F), dT is the change in temperature in degrees Fahrenheit, and Q is the heat transfer rate in BTUs per second, you will need to know the values of m, dT, and Q.
If m equals the mass of 40 gallons of water, the temperature change is 35 degrees Fahrenheit, and Q equals 7.0907 BTU/sec, then the value of dt can be calculated as follows:
dt = 40 gallons * 8.3454 pounds/gallon * 1 BTU/pound-F * 35°F / 7.0907 BTU/sec = 334.18 pounds * 1 BTU/pound-F * 35°F / 7.0907 BTU/sec = 4,347 seconds
To convert this value to minutes, you can divide it by the number of seconds per minute:
Minutes = dt / seconds/minute = 4,347 seconds / 60 seconds/minute = 72.45 minutes
So in this case, dt would equal approximately 72.45 minutes.
