Suppose I have a heat generator with a connected heat storage tank. If the heat generator remains the same and the heat storage tank is dimensioned larger, the running time of the heat generator increases due to the higher storage losses. Why is this so?

Normally, the surface with exponent 2 and the volume increases with exponent 3 - should the heat storage tank not have fewer losses?

  • 3
    $\begingroup$ More material to heat.. $\endgroup$
    – Solar Mike
    Commented Nov 8, 2019 at 8:36

1 Answer 1


Heat loss due to conductivity of the walls of your container assuming the ambient outside temperature, $$ \frac{\Delta Q}{\Delta t*A}=-k \frac{\Delta T}{\Delta x} =C_{constant}$$

In the above equation, the left part is heat loss per unit of area per second and the right side is the thermal conductivity, K multiplied by the temperature gradient which is constant. x is the coordinate perpendicular to the temperature gradient of the tank wall.

So every unit of the surface of the tanks loses heat at the same rate.

Meaning the larger tank loses more heat because it has more surface area.

  • $\begingroup$ Thank you very much. Would you mind giving me an exemplary calculation for two cylindrical heat storages - one with A = 6$m^2$ (~1000 liters) and the other one with A = 11$m^2$ (~2000 liters); both start at 90 °C and have an ambient temperature of 20 °C. $\endgroup$
    – MerklT
    Commented Nov 8, 2019 at 23:11

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