Pity . . you missed a really good game. But . . . some horses are "working" horses, while others are "luxury" horses.windgat wrote:I don't watch rugby...
Instead I did something useful...
That would make sense. The geyser insulation relates to the area coverered (m²), and the content to m³. So, if you double the linear sizes of the tank, you need 4X the amount of insulation to cover 8X the amount of water.windgat wrote:One surprising result so far in the model is that the bigger the geyser capacity, the less the electricity used! Do you think that's accurate?
Temperature decrease is not constant. It is ruled by the balance of heat gain vs heat loss. You need to work out the heat loss based on the current temperature (ie: (Th-Ta)/R) and then convert that loss to temperature drop via the formula we just corrected (ie: ?T=3.6X0.24XLoss/L)windgat wrote:You have given me heat loss in Watts - what I really need is temperature decrease per time unit.
Greystoke wrote:You can measure that value by doing a "cooling " exercise. It's not too difficult, but you do need a thermocouple thermometer. (some electronic multimeters come with themocouple probes)
4.8 kW looks OK to me.windgat wrote:The total daily power use comes in at 4.8kW - seems too low. I would expect 10 - 15 kWh.
Greystoke wrote:you do need a thermocouple thermometer.
I have faith in your ingeniouty. So, kindly tell me how you did it.windgat wrote:I have one, but how can I get my probe into the how water geyser? (sounds like a line from a Pam Anderson movie...)
Did the same thing trying to get an estimate for the thermal resistance of my geyser. I'm sure its the right order of magnitude, but I truly haven't a clue how accurate it is.windgat wrote: . . To get a smooth temperature at midnight I manually fiddle the starting temp to about match the ending temp. It may need to be over multiple days for more accuracy.
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