I have an unvented electric immersion heater cylinder which heats the hot water for my household. The capacity is 200L and the temperature is set to 55°C.

The water heater is hard-wired to run overnight during cheaper off-peak hours only. This works reasonably well for most of the time, but occasionally during particularly heavy usage we find we run out of hot water by the evening so we're unable to take hot showers etc. at night.

If I increase the cylinder temperature to (e.g.) 65°C, will this effectively provide us with more "concentrated" hot water which will take longer to run out, assuming we continue to shower at the same comfortable temperature (~43°C)?

If yes, by how much longer might the hot water last?

  • To be honest, it is going to matter more on how long you use the hot water. Also, note that (typically) the hotter you run a storage system, the more you reduce it's effective life. Also, clearly, the cost of heating it goes up DRASTICALLY. If your goal is to reduce cost, maintaining hotter water isn't going to help that. Perhaps the best approach is to shorten water use time, and use low flow valves/heads.
    – noybman
    Oct 31 '18 at 4:47
  • I would doubt it would last longer. It depends on how the hot water is used during the day. Laundry, dishwashers usually demand full hot water, the temperature coming out will not matter. Your only savings would be when you're mixing with cold to get warm water. For me, that is mostly for showers. I'd say, try it for a week and see...
    – Gary Bak
    Oct 31 '18 at 11:46
  • @noybman The heater is only switched on for a finite period (overnight) so electricity is not being used to maintain the temperature; it's mostly just the initial heat-up.
    – WackGet
    Oct 31 '18 at 18:58
  • 1
    Insulating the hot water pipe from heater to shower can help a lot. They make slit foam tubes for just that purpose. Nov 1 '18 at 5:54

Absolutely. It will cost you more money to run the hot water heater, but you will have usable hot water for longer. You can calculate the additional time simply as well, it's an algebra mixture problem. I hated algebra, but let's walk through the math anyway:

Let's assume your cold water is at 10°C. You want the output to be 43°C Your tank holds 200L.

Given the above, without replenishing the tank, how many L of 43°C water can you get?

 temp = (hl*ht + cl*ct) / (hl + cl)

substitute figures to solve for cl: (200*55)+(cl*10) / (200 + cl) cl= 72L of cold.

So 200L+72L = 272L of 43°C before it runs out at 55°C.

Now let's solve for 65°C: (200*65)+(cl*10) / (200 + cl) cl = 133L of cold.

So 200L+133L = 333L of 43°C before it runs out at 65°C.

Comparing 333L of hot to 272L of hot, you will get about 20-25% more hot water before it starts cooling. Of course, there's other losses; hot water cool more proportionally before the faucet, the heater looses more waste heat with hotter water, it takes longer to heat water hotter, affecting the recharge rate, etc.

Using an online cost estimator, it looks like you'll spend 33% more money to get 25% more hot water.

  • This is assuming the heat is not lost at all but it is, and as you mention, its lost fast, so at what (lograthimic) curve is it? The tank mfr will probably have this published. Then looking at the extra cost to get it to this temperature at night will end up costing the same, but the amount of hot water wont be ~133L. You might make out to heat the water to the demand temp but keep that temp lower than 55c (i.e., 49c) nearer to the time of use and not at all when not used.
    – noybman
    Oct 31 '18 at 22:31

You'll have hotter water, which will last longer in the shower because you'll be mixing it with more cold water to get the right temperature when you use it.

The down-side is that:

  • It will cost you more to run. Hotter water "leaks" more energy into the surrounding area.

  • It's easier to get really serious burns.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.