Figuring Btuh Requirements for Water Heating

How do we calculate the capacity of a given potable hot-water storage system for residential applications? SWAG for sizing? Well, maybe in a spec house, but not if you can talk to the owners and ask them about their hot water usage habits. In almost every circumstance, theyll be drawing hot water faster than the recovery rate, which is why a storage tank is needed in the first place. We can treat

How do we calculate the capacity of a given potable hot-water storage system for residential applications? SWAG for sizing?

Well, maybe in a spec house, but not if you can talk to the owners and ask them about their hot water usage habits. In almost every circumstance, they’ll be drawing hot water faster than the recovery rate, which is why a storage tank is needed in the first place.

We can treat the storage tank capacity as we would the standing water in a low-yield well — as draw down capacity. And just like that well, we can add back in the GPM regeneration rate over the duration of the draw down.

But how can we know how much hot water is being drawn per minute? You could turn on just the hot side and time how long it takes to fill a 2-gal. bucket, but that won’t be possible if you’re dealing with a scald-guard faucet that’s designed to shut down upon loss of cold water flow. Furthermore, turning on only the hot side can produce higher flow rates than you’ll get if both hot and cold water are being drawn at the same time.

There’s a much better method that’s extremely accurate. In my March column, “Storing hot water at 120°F doesn’t save energy” (pg. 24), I used the following formula to determine what percentage of a given flow was hot water: R = (BT - IT) ( (ST - IT) where R = ratio; BT = bathing temperature; ST = storage temperature; and IT = inlet temperature — using the coldest inlet temperatures seen during the year.

After asking the homeowner to adjust the bathing water temperature to the setting she likes, you can determine GPM of hot water with accuracy.

If the storage temperature is 138°F and the inlet temperature is 55°F under design conditions and the adjusted bathing temp is 102°F, the formula looks like this: R = (102 - 55) ÷ (138 - 55). R then equals 47 ÷ 83 or 0.566.

Now it’s time to get that bucket and time total GPM out of their faucet(s). If, for example, the total adds up to 21.2 GPM, the net flow of hot water becomes 21.2 x 0.566 or 12 GPM of hot water. That’s the scenario Mike Williamson faced in April’s column (pg. 36), and his customers wanted assurance they’ll have the hot water capacity needed for their guests.

With a 12-GPM flow rate, we can determine net Btu load by multiplying the weight of the water times the Delta-T between the inlet and storage temperatures: (12 x 8.3) x (138 - 55) = 99.6 x 83 = 8,266.8 Btu per minute. We’ll need to multiply by 60 to determine the hourly load, which is how appliances are rated, and that equals 496,008 Btuh. That would take one bodacious instantaneous water heater!

Once we begin adding storage volume, the picture begins to change.

The one crucial piece of the sizing puzzle we’re still missing is how long they expect to be drawing the tempered water. In Mike’s example, we know the volume of the whirlpool and we’ll assume a 20-minute shower. The 75-gal. whirlpool hot water volume is 75 x 0.566 = 42.45, which will take just over 7 minutes to fill at 6 GPM on the hot side. With a shower running concurrently for 20 minutes at 6 GPM of hot water, we add another 120 gal. for a total of 162.45 gal. of hot water.

All of which brings us back to the low-yield well example.

We know Mike’s Delta-T of 83 (138 - 55) means each gallon will require 688.9 Btu (83 x 8.3). If we begin with a single 80-gal. storage tank, we can then subtract about 74-gal. from the required 162.45, which means we are short of what’s needed by 88.45 gal.

Why just 74 gal.? Adjusted volume due to loss as cold incoming water mixes with hot in the tank; 88.45 gal. x 688.9 Btu = 60,933 Btu and that must be multiplied by 3 to adjust for a one-hour net output of 182,799 Btuh.

Suppose we coupled that 80-gal. storage tank up to a copper fin-tube potable water boiler that has a gross input of 225,000 Btuh and an operating efficiency of at least 81%? That’s a net output of 182,250 Btuh, which is about as close to hitting the bulls-eye as you’ll ever get!

And that’s exactly what Mike e-mailed me to say they were planning on installing. The copper fin-tube boiler will be installed outdoors with the 80-gal. storage tank located in the previously cramped mechanical room.

Dave Yates can be reached by e-mail at behler@blazenet.net.