windgat wrote:I don't watch rugby...
Instead
I did something useful...
Pity . . you missed a really good game. But . . . some horses are "working" horses, while others are "luxury" horses.
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?
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.
Back to the formula:
Geyser heat loss= (T[n] - T[ambient])/R, In which R represents the thermal resistance of the geyser. MIne (100L) is about 0.4°/Watt. 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)
Geyser heat gain= 3.6x0.24xHxL°/hr, In which H= heater element in kW, and L = liters of water