pi(10 feet)^2; 5 gal/min; 1 inch ? minutes = 39.1679 minutes
Assuming you have a 40 gallon 5000 watt electric water heater, set at
140 degrees, and that water comes into the house at 45 degrees (Fahrenheit)...
Note: shv_water is defined as: 1 cal/cc deltaC
40 gal; shv_water; (140-45)deltaF; 5000 watt ? hour = 1.85881 hourWhat's with the semicolons?
You might also ask what it costs to take that shower?
5000 watt; 1.8 hours; 5 cents/kWh ? $ = 0.45 $
40 gal; shv_water; 95 deltaF; 40000 btu/hr ? min = 47.5686 min
(85 - 60) deltaF * shv_air * 100 cfm ? ton_refrig = 0.24212 ton_refrig 0.24212 ton_refrig ? watt = 851.504 watt 851.504 watt / 60 watt ? % = 1419.17 %That's a pretty good energy efficiency factor!
Measured: It takes 2 min 4 sec to bring 3/4 cup of water from 53 degF to boiling.
(3/4)cup; shv_water; (212 - 52)deltaF; 2 min + 4 sec ? watts = 532.552 watts
40 knot / (2500 rpm * 90 %) ? inch/rev = 21.604 inch/rev
First measure how big the leak is, say it drips out 1/4 cup in 1 min 49 sec and the water heater is set at 130 deg F. (water comes in at 50 deg F)
(130-50)deltaF; shv_water; (1/4)cup; 1 min + 49 sec ? watt = 100.973 watt 100 watt; 6 cents/kWh ? $/year = 52.56 $/year
Assume the pipe connecting a faucet to the water heater is 43 feet long, has an internal diameter of 1/2 inch, and that the flow rate is 3 gallons per minute.
43 feet; pi((1/4)inch)^2; 3 gal/min ? sec = 8.77197 sec
Calchemy Software Inc.
Fort Collins, CO
Last Updated 6/21/2009