John Rumm wrote in message ...
Charles Middleton wrote:
measurement that I can take of it to determine its maximum rating. Also a
Got a good thermometer and a bucket?
Measure the temperature of the cold water from a tap on the same mains
supply as the shower (kitchen will do) - let it run for a few mins to
get a stable reading of the actual incoming water temp.
Now calibrate your bucket with a marker so you know where 5L is.
Turn on your shower at maximum temperature and let it run for a bit to
get a stable temperature. Measure this temperature.
Now time how long it takes to fill your bucket to the 5L mark.
Now do some sums:
Find out how much energy the shower has added to the water. To do that
multiply the temperature rise, by the mass of water (assume 1L = 1kg),
and then by the specific heat capacity of water (4200).
Say the temperature difference between incoming and heated water was 40
degrees C you will get:-
40 * 5 * 4200 = 840,000 J
Now divide this by the time it took to fill in seconds. Say it took 80
seconds, you get:-
840000 / 80 = 10500 J/second (or Watts)
Thus you would have a 10.5kW shower.
(The above procedure will tell you the heat put into the water by the
shower - it will not tell you the energy lost in the supply cable to it.
So for example if you had a 10kW shower, you may only find 9.8 kW
getting as far as the water).
--
Cheers,
John.
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Perhaps a measurement of the DC resistance across the electrical input
terminal to the shower with a meter might solve the mystery. You would
need to isolate the shower from the mains cable in order to do this.
So the power is calculated by multiplying the 230v mains by itself and
then dividing it by the resistance of the shower heating element.
Jason
shower