There is a heat storage tank that is directly loaded from the top and the heat is also taken from the top. The colder water from the heating circuit return flow enters the heat storage tank at the bottom. This creates a layered water temperature in the heat storage tank.

There are three temperature sensors inside the heat storage tank.

Is it possible to calculate the stored energy via these three temperature sensors?

Edit - Calculation Attempt according to Solar Mike:

$V_{storage} = 1000\ l = 1\ m^3$

$t_{top} = 82\ ^\circ C$, $t_{middle} = 70\ ^\circ C$, $t_{bottom} = 55\ ^\circ C$

$t_{average} = \frac{82+70+55}{3} = 69\ ^\circ C$

$Q = M \cdot c_{H_2O} \cdot \Delta T = V_{H_2O} \cdot p_{H_2O} \cdot c_{H_2O} \cdot \Delta T $

$\ \ \ = 1\ m^3 \cdot 1.163\ \frac{kWh}{m^3K} \cdot (69\ ^\circ C - 20\ ^\circ C) = 56.987\ kWh$

Does this calculation give me a fairly correct value for the stored energy?

  • 2
    $\begingroup$ Yes, use the difference between what the cylinder temperature will be when the hot water is removed and either the 3 regions average temperature or the overall average hot temperature. $\endgroup$
    – Solar Mike
    Jan 20, 2020 at 8:17
  • 1
    $\begingroup$ Thank you, I was hoping it would be that simple. Do you have a reference I can refer to for this? $\endgroup$
    – Emma
    Jan 20, 2020 at 10:32
  • 2
    $\begingroup$ If you’re truly looking for the amount of energy being stored and not just what to use for the temperature in the calculation, then you need to incorporate the fluid’s heat capacity which means identifying the fluid. Is it actually water or were you just using “water” in your description? $\endgroup$
    – Secundus
    Jan 20, 2020 at 12:27
  • 4
    $\begingroup$ So Q = M * Cp * (T1 - T2) where Q is energy, M is mass, Cp is specific heat capacity and T are the temperatures. Cp is available for various temperatures - 4.18 KJ /Kg / K at 20 deg C. Any textbook covering heat teansfer will cover this. $\endgroup$
    – Solar Mike
    Jan 21, 2020 at 5:04
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    $\begingroup$ This is how 'on demand coffee machines' work - stratifying the water and keeping the bit at the top hot. There can be a quite sudden change between hot and cold layers of water, and one temperature sensor in the middle would not be sufficient to determine the height of this transition zone. You could calculate a range from this information here, but not an absolute answer. $\endgroup$ Jan 22, 2020 at 8:45

2 Answers 2


It is difficult to calculate the heat capacity because we have two regimens contributing to the temperature gradient inside the tank.

Heat conductivity of the water establishes a temperature gradient descending from the core of the tank to the tank wall which would cause slow convection up, and advection by the agitation of the circulating pump which causes a fast and likely turbulent flow of hot water down and cold water up.

The sensors' readings are not very helpful because of the spiral flow mixing the hot and cold water and disturbing a uniform predictable pattern of water circulation in the tank one would never know if the reading is of a random plume of water or a steady read.

A very rough estimate would be modeling the core temperature an average of the difference between the heat generator's output and the three sensors'.

Depending on the power of the circulation pump things will vary by a large amount.

I would recommend using simulation software.

  • $\begingroup$ Thanks for your answer. Do you think I'm way off with my calculations in the edit? Even with additional temperature sensors? $\endgroup$
    – Emma
    Jan 21, 2020 at 14:04
  • $\begingroup$ because the sensors are on the wall and not the inside of the tank they measure the colder temps of each level. consider the topography of the temp distribution inside the tank, like un unsymmetrical union with layers starting from the core with temps of heat generator to colder skin temps. So I would pick T1 much higher than 69, more in between 69 and the heat generator temp. because the core is hotter than the skin of the tank. so if the furnace temp is 100, I would choose T1=69 +100-69/2=85 $\endgroup$
    – kamran
    Jan 21, 2020 at 17:26

I feel like this may be an X-Y problem. You are asking for the energy stored in a vessel, but what will you do with the information once you know?

Energy in water

A glass of room temperature tap water has an energy content, but few would find this information useful. Engineers tend to look at the change in energy either put in to water or that can be taken from it. Like how much home heating we could produce from a quantity of hot water, or how many gallons of 65 degree water would it take to give a 45 degree C shower. From a pure physics standpoint, the total energy in liquid water at atmospheric pressure is the energy required to heat it from absolute zero to its melting point as ice, the energy to melt the ice, and the energy required to heat it to its current temperature. But this isn't very useful because we are unlikely to encounter extremely low temperatures and so it would be difficult to extract all that energy. Remember that heat always flows from hot to cold.

Useful energy in water

Your calculation shows that a temperature change of 49 degrees C will be an energy change of 57 KW-hr. But what is that exactly? You wont be able to run a 1000 watt blow drier for 57 hours using that energy. You could use it to heat up some other volume of water, or possibly to heat up a room using a heat exchanger, but it becomes hard to extract all of that energy when the temperatures converge. You use 20 degrees in the calculation, but don't describe what that represents.

What we need

So what we need is to know what you are trying to do with this heated water. Is the circle in the diagram taking out heat and then pumping cold water back into the system? Once we know what you want to do with all the hot water we can better answer how much capacity the water has to do it. Also, let us know how accurate you need the answers to be. Moon shots and swimming pool heaters have different accuracy requirements ;^)


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