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Can anyone give me a few pointers to calculating the time take for an electric shower to reach it's set temperature?

The set temperature is actually a power setting and a flow rate setting.

I know that with 3500 watts of heating power and a flow rate of 3.5 litres per min, I will get a 14.35 degree rise.

3.5 l/min = 58.3 g/sec
58.3 * specific heat of water 4.18 = 243.8
3500w / 243.8 = 14.35 degrees rise

So my question is, is it possible to work out the time taken to reach this temp? I assume I need to know something about the efficiency of the heat exchanger to calculate this as well as the specific heat of the materials between the heater and my measurement point?

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    $\begingroup$ I'm not familiar with the unit gms, it's neither metric nor SI. I'm assuming it could be grams, for which the unit is g (for singular & plural) $\endgroup$ – Fred Aug 17 '17 at 14:29
  • $\begingroup$ Just for you fred, yes it's grams. $\endgroup$ – Richard Aug 17 '17 at 17:16
  • $\begingroup$ The difficulty is that you only have one data point. It is not true that half the power, for a constant flow rate, will produce half the temperature rise (nor will double the power double the temperature rise). And as you suggest, the time constants involved to bring the heat exchanger to a steady-state thermal profile depend critically on the exact design of the hardware. $\endgroup$ – Carl Witthoft Aug 18 '17 at 12:55
  • $\begingroup$ I am just about to set up some temp recording kit, so I can actually measure the rate of temp rise. Just by using mk1 finger and stopwatch, the water takes approx 10 seconds to being warming, 10 seconds to close to target temp and maybe 5 seconds to get to the last 1-2 degrees. $\endgroup$ – Richard Aug 19 '17 at 13:46
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With SI units you can derive the formula for yourself.

  • More power -> lower time. $ t \; \alpha \; \frac {1}{P} $.
  • More water (mass) -> longer time. $ t \; \alpha \; {m} $.
  • Higher specific heat capacity -> longer time. $ t \; \alpha \; {SHC} $.
  • Higher temperature -> longer time. $ t \; \alpha \; \Delta T $.

Putting them all together you get

$$ t = \frac {\Delta T \cdot m \cdot SHC}{P} $$

For your numbers:

$$ t = \frac {14.35 \cdot \frac {3.5}{60} \cdot 4180}{3500} = 1 \; s $$

This makes sense as there at steady state we must heat the inflow per unit time to the steady state temperature otherwise it won't be steady state!

The information we're missing is the capacity of the heating tank. If it held 1 L of water then:

$$ t = \frac {14.35 \cdot 1 \cdot 4180}{3500} = 17 \; s $$

Then add in the copper of the cylinder, losses to ambient, ...


My sanity check on this is to calculate time to boil 1 L of water from 20°C using a 2.1 kW electric kettle. Since we're in kW mode we'll convert the SHC to k as well.

$$ t = \frac {\Delta T \cdot m \cdot SHC}{P} = \frac {80 \cdot 1 \cdot 4.2}{2.1} = 160 \; s $$

A little under three minutes sounds about right.

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  • $\begingroup$ I don't think he's talking about a tank system but rather heating an active flow. $\endgroup$ – Carl Witthoft Aug 18 '17 at 12:53
  • $\begingroup$ They usually have a small in-line cylinder with the element inside. That's why I gave the 1 L example. 0.5 L might be more realistic. $\endgroup$ – Transistor Aug 18 '17 at 16:27

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