# Optimal maintenance temperature and allowable drop for energy efficient hot water storage

I bought an under-sink electric boiler/hot water storage unit, for instant almost-boiling water. This device has an 'eco' mode, which broadens its hysteresis to 10 degrees C: it will allow the stored water temperature to drop to $$T_\text{target} - 10$$ before turning on the element.

This made me curious, since naively I would think that it will turn on less often of course, but for longer (or hotter) when it does; consuming just as much energy overall.

If we assume:

1. the element has a fixed resistance (e.g. it does not deliberately vary with $$T_\text{target} - T_\text{now}$$), i.e. a fixed kW consumption $$P$$ s.t. $$E=P\cdot{t_\text{heated}}$$ and we need only worry about $$t_\text{heated}$$;
2. the unit has a fixed efficiency across all $$T_\text{now}$$;
3. standby power is negligible;

(so as to avoid specifics about the particular device)

I know:

1. the stored unheated water will cool exponentially over time;
2. the heated water will rise in temperature as an inverse exponential, asymptotic to $$T_\text{element}$$;

let:

1. $$t_{\text{fall},x}$$ be the time taken for the unheated water to fall from $$T_\text{target}$$ to $$T_\text{target} - x$$;
2. $$t_{\text{rise},x}$$ be the time taken for the heated water to rise from $$T_\text{target}-x$$ to $$T_\text{target}$$;

then:

1. what is the relationship between these exponents, what else are they a function of if it's really the case that $$\dfrac{t_{\text{rise},x}}{t_{\text{rise},y}} \ne \dfrac{t_{\text{fall},x}}{t_{\text{fall},y}}$$?
2. if they are only a function of time, $$T_\text{target}$$, and $$T_\text{ambient}$$ (room/insulation, or the element, for cooling or heating respectively), then surely 'eco' would be setting an appropriate $$T_\text{target}$$, not the max allowable $$T_\text{target}-T_\text{now}$$?
• Heat transfer is driven by the delta T. Heating or losses... So add insulation. Jan 19, 2023 at 20:53
• @SolarMike I'm not asking how to make it more efficient per se, I'm asking if the manufacturer's assertion that allowing a wider temperature range is more 'eco' is true. And as a tangent to that, whether the choice of $T_\text{target}$ doesn't have at least as much bearing. Jan 25, 2023 at 11:13
• So set up a calculation based on real use - you can measure your water use precisely for a week or two. Then run that data against the ordinary mode and eco mode. Which uses the most power? It’s no good using other people’s data as their use pattern is not identical to yours. Also, is electricity more expensive at certain times of the day? Or cheaper at others? Why not control water heating to the cheaper slots? Jan 25, 2023 at 12:20
• I'm more interested in whether it theoretically makes sense. Jan 25, 2023 at 16:33

the heating is not asymptotic. Its roughly linear with time. The reason is that when you are heating you are supplying constant heat flux (ie. heat energy per unit of time). The underlying equation is that:

$$Q = m \cdot c_p \delta T$$

where:

• Q is the heat energy provided
• m is the mass of the liquid
• $$C_p$$ is the heat capacity of the liquid
• $$\delta T$$ is the temperature difference.

So if you differentiate with time (keeping constant m and constant $$c_p$$ then)

$$\frac{dQ}{dt} = m\cdot c_p \cdot \frac{d T}{dt}$$

The reason that it takes longer is because the losses to the environment increase however, they are significantly less in the heat energy balance.

• Huh, but cooling is exponential? I find that really counter-intuitive. That surely suggests then that allowing the temperature to drop further before heating up again is worse - the 'eco' mode should be 'set $T_\text{target}$ as high as possible and keep it there' - doing linear heating little and often? Jan 25, 2023 at 11:18
• The cooling is exponential because you are starting from a set temperature and you are dropping to a constant environment. If the water was cooler than the environment (ie. heating) you also get asymptotically exponential growth of temperature. The diffference with the boiler is that the heating plate does not have a constant temperature, but the electrical resistance generates a set amount of heat that goes into the liquid.
– NMech
Jan 25, 2023 at 11:58
• So when you drop the temperature you need a lot more time for cooling, but the time for heating is relatively proportional to the temperature drop.
– NMech
Jan 25, 2023 at 12:00
• Right, sorry I got confused - the cooling happens for free (!) so I think it isn't any better (nor worse) to heat 'little and often' (nor a lot infrequently, per the 'eco' mode) - it doesn't make a difference if $Q\propto\delta{T}$. And unaffected by current/target temp. I understand your answer now, thanks! Jan 25, 2023 at 16:40