I'm trying to figure out the efficiency of energy usage between heating up 2 tons of water from 5°C to 70°C everyday (during winter) within a certain given time (let's say one hour) and stabilizing 2 tons of water at 70°C all the time, given that there is always an amount of new water being supplied into the tank.

I calculated the first part and it would cost me 151.7 kWh everyday to heat up 2 tons. But I couldn't find how to calculate the energy it would take to stabilize water at a certain temperature.

  • 6
    The answer to your question (and also to your calculation) depends on how much heat you lose by storing alone, and how much cold water is being added. Thus you should really supply more information.
    – Earthliŋ
    Commented Jan 17, 2017 at 16:51
  • 3
    Can you give the context? How does this relate to sustainability?
    – LShaver
    Commented Jan 17, 2017 at 19:03

1 Answer 1


The stabilisation energy rate will be exactly exactly equal to the rate at which energy is lost.

Assuming the tank is closed, so there are no evaporative losses, there are two other ways energy is lost.

  1. Firstly, from conductive losses through the surface of the tank. That will depend on the thermal conductivity of the tank, and on difference between the water temperature and the ambient temperature of the air around the tank. So if your tank loses 2 Watts per Kelvin per square metre of surface area (i.e. has a U-Value of 2.0), and has ten square metres of surface area, and the ambient air is 15 degrees Celsius, then the heat loss rate from conduction is 2 x 10 x (70 - 15) = 1100 Watts.

  2. Secondly, from the hot water being taken away and replaced by cold. If you were to use that full two tonners each day, that would be your 151.7 kWh/day. If you want to know the daily energy consumption based on a lower usage, multiply that 151.7 kWh by the amount of water used on average in a day, expressed as a proportion of the total 2 tonnes. So if 0.5 tonnes got used in a day, that's a quarter of the total water. So the energy lost in that day, i.e. the extra energy needed to heat the incoming cold water, would be a quarter of 151.7 kWh.

Add these two together to get the rate at which energy is lost.

  • While the question needs more detail, my intuition is that OP wants to know how much energy to raise the water from 5C to 70C once every 24 hours, vs maintaining the water at 70C, against an ambient temperature of 5C, over the 24hr period. The U value would still need to be known.
    – LShaver
    Commented Jan 18, 2017 at 17:16
  • 1
    @LShaver ah, good spot, I think you're right. I've amended my answer, and I think this answers it now.
    – 410 gone
    Commented Jan 19, 2017 at 9:00

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