Fiddling with figures…

To quote from a rather sheepish Richard:

“OK. 15 minutes is not enough. Neither is 16.

The specific heat capacity of water is about 4200 J/(kg K), which means a 3kW thermostat will take about 1 hour and 45 minutes to heat up 150 litres from 25 degrees to 55 degrees.

I’ve now set the timer to one hour, mornings and evenings.”

In case you are interested and mathematically on a sound footing (which I am not – I completely fluffed the last equation…), he used the following equations to get these results:

c = P t / (m ΔT)

c = specific heat of water = 4200 J/(kgK)
P = power [watt – in this case we have a 3000 watt geyser]
t = time [seconds]
m = mass [kg – in our case our geyser has a capacity of 150 litres, therefore 150 kg]
ΔT = change in temperature [usually in Kelvin, but as it is the change in temperature, we can just use degrees Centigrade]

In order to calculate how long it will take to heat up the geyser from 25 to 55° C, i.e. by 30°, the equation has to be rewritten like this:

t = c m ΔT / P = 4200 x 150 x 30 / 3000 = 6300 seconds = 105 mins = 1 hour, 45 mins.

And in order to calculate the change in temperature possible in 15 mins (900 seconds), the equation has to be rewritten like this:

ΔT = t P / c m = 900 x 3000 / (4200 x 150) = 900 x 3 / (42 x 15) = 4 degrees (i.e. from 26° to 30°C)

I guess that could explain why this morning’s shower was a little on the lukewarm side.

I'd love to hear your views

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.