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).

  • 1
    I think we'd also need to know the temperature of the hot water, cold water temperature and recovery rate of tank.
    – Steven
    Commented Nov 10, 2014 at 14:58
  • Does the tank have a thermostatic mixing valve at its output, and what is that set to (ie, what percentage of the hot water actually comes from the tank)? What's the mixture of hot and cold water being used at the shower head? How much heat are you losing to the pipes between tank and shower? I think the better question would be to discard the shower and the house, and look only at a 2.5GPM draw upon the tank itself (with its mixing valve, if any). I'm not sure whether stating the recovery rate is sufficient to let us ignore insulation and burner BTU....
    – keshlam
    Commented Nov 10, 2014 at 16:37
  • Knowing how much turbulence there is in the tank -- how much new water gets mixed with the old water, reducing temperature until the burner catches up -- might also be important if you're looking for a serious model. It's starting to sound like "try it and see what happens" is easier.
    – keshlam
    Commented Nov 10, 2014 at 16:42
  • @keshlam I have infinitely insulated pipes, which lose/gain no heat. Also the formula Hot Water % = (Mixed Water Temp. - Cold Water Temp.) / (Hot Water Temp. - Cold Water Temp.) might be helpful. You can also assume the showerer, is constantly adjusting the knobs to maintain a 105°F water temperature for as long as possible. Once the cold knob is all the way off, and the hot is all the way on. The showerer will stay in the shower until the water temperature drops below 80°F, at which point they'll get out of the shower.
    – Tester101
    Commented Nov 10, 2014 at 16:52
  • OK, you've got an integral. Which may be a straight line, but I'm not sure I'd bet on it.
    – keshlam
    Commented Nov 10, 2014 at 16:54

2 Answers 2


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.

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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.

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Note that, because enter image description here does not appear in the expression for enter image description here, enter image description here is now a constant.

  • Note that the comment establishes a second stopping criterion of 80°F, which introduces a second phase that is constant-flow-rate instead of constant-power. This becomes a first-order ODE which is easy to solve but not as easy as the constant-heat version. Commented Nov 14, 2014 at 15:32
  • @Tester101, yes! The tank-temperature dependence appears in both the hot water flow rate from the water heater and the cold water flow rate to the water heater. They cancel out. Commented Nov 14, 2014 at 17:33

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.

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