# Water heater and mass flowrate

#### tom89

Hello everybody!

I was thinking: if I change the mass flow rate (by opening the tap when I'm going to take a shower) the water will take more or less time to heat?

I've proceeded as follows:
by making the heater my control volume:
$$\displaystyle \dot{Q}=\frac{d(nU)}{dt} + \dot{n}(H^{out}-H^{in})$$
since the mass, and consequently the moles, inside the heater does not change with time,
$$\displaystyle \dot{Q}=n\frac{dU}{dt}+\dot{n}(H^{out}-H^{in})$$
stating the hypothesis of incompressible fluid,
$$\displaystyle \dot{Q}=nC_P\frac{dT}{dt}+\dot{n}\int\limits _{T_0}^{T} C_Pd\tau$$
where $$\displaystyle T_0$$ is the initial temperature, let's say 293.15 K;
it is common to express, for a liquid,
$$\displaystyle \frac{C_P}{R}=A+BT+CT^2$$
in this case I'll consider only the first term (since my goal here is not to get an A+ in differential equations):
$$\displaystyle \dot{Q}=nRA\frac{dT}{dt}+\dot{n}RA(T-T_0)$$
just making $$\displaystyle \theta (t)=T(t)-T_0$$, and my initial condition being $$\displaystyle \theta (0)=0$$,
$$\displaystyle \frac{\dot{Q}}{nRA}=\dot{\theta}+\frac{\dot{n}}{n}\theta$$
then,
$$\displaystyle T(t)=T_0+\frac{\dot{Q}}{\dot{n}RA}\left[1-\exp\left(-\frac{\dot{n}t}{n}\right)\right]$$

So, after some quick research I've found that $$\displaystyle A=8.712$$ and that a usual heater provides a maximum power of $$\displaystyle \dot{Q}=28$$ kW for 21 L/min of water flowing trough it. At this point I've made other hypothesis: that the ratio $$\displaystyle \frac{\dot{Q}}{\dot{n}}$$ is constant - since there must be a controller inside the heater in some way that it will not provide the maximum power when the flow rate is small.

All right, so I can finally write,
$$\displaystyle T(t)=T_0+K\left[1-\exp\left(-\frac{\dot{n}t}{n}\right)\right]$$
where $$\displaystyle K$$ is constant. Also, $$\displaystyle \lim_{t\to \infty} T(t)=T_0+K$$, which makes sense.

So, looking for the equation above I can tell that if I raise the mass flow rate (by opening the tap) it will take less time to the water reach some temperature.

I want to ask you: is there any mistake that I've committed? For me, the final result doesn't make sense - I was expecting that a lower flow rate would make my time waiting for shower decrease.

PS. I'm really sorry if you can't get something, I'm not a native speaker. Thanks everybody!

#### oz93666

Hello everybody!

I was thinking: if I change the mass flow rate (by opening the tap when I'm going to take a shower) the water will take more or less time to heat?
Hallo tom ... that's a very simple question ... shurely no need to get into all those mind numbing equations !!!

So the water is heated by an electrical heater , lets assume the power input to heater is constant ...

you take a shower , the water is hot ... now you open the tap more , the flow of water through the heater will now increase ... since the power input is the same the temperature must drop.

You ask "will the water take more or less time to heat?" ... the flow rate has increased through the heater it will take less time to heat

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#### studiot

Hello Tom and welcome.

Are you studying Transport Phenomena?

I agree with Oz that you are making the answer too complicated.
But I would add that you have made the information too simple.

You have not said enough about the supply conditions or the heater itself.
Is there a tank of hot water?
Or is this an 'instantaneous' or direct heater?

For a direct heater we have that these are nowhere near 100% efficient.

If the heater is constant power then the flow rate and therefore velocity will increase.
So a given parcel of water will spend less time in the heat exchanger so will emerge cooler.
So temperature of the heated water will drop.
However, by Newton's Law the rate of heat transfer will increase due to the increased temperature difference. Thus the efficiency will increase slightly.
This will not be enough to offset the faster flow.

If the heater is thermostatically controlled then the temperature of the water will not drop so the only way to increase the rate of heating is to increase the heater power (and therefore the element temperature).
The control system should make the increase in efficiency needed exactly balance the increase in water flow.

#### ChipB

PHF Helper
When I read this I thought the question was regarding how long it takes for hot water to flow from the heater to the shower, and whether running a tap in the bathroom would help the shower get warm faster. The obvious answer is yes - to get the shower to be warm you have to empty out all the cold water in the pipes, and its clearly quicker to do that if you run the hot water in both shower and sink at the same time - this assumes that all the fixtures are fed by a single hot water line from the heater to the bathroom. But upon reflection I'm guessing that the OP has a tankless water heater, in which case it's still best to run the water as quickly as possible until the hot water reaches the shower, but after that the question of whether you get hotter water running the shower at full force versus a slower rate depends on the insulation properties of the pipes and the rate of heat transfer into the water by the tankless heater.

One thing I would quibble with in the OP's calculations is the assumption that the rate of heat transfer into the water is proportional to the rate of water flow - most hot water heaters are designed to be either on or off; there is no "ramping" of heat flow that depends on water flow rate (what my old-time professors would call a "bang-bang" control system - the heater is either in or off, there is no middle setting, like the way a typical home thermostat calls for heat).

#### benit13

I want to ask you: is there any mistake that I've committed? For me, the final result doesn't make sense - I was expecting that a lower flow rate would make my time waiting for shower decrease.
I get the same result, but a different conclusion, because you need to consider the dependence of $$\displaystyle K$$ on $$\displaystyle \dot{n}$$.

If you rearrange the new formula for $$\displaystyle t$$ and substitute for $$\displaystyle K$$, you get

$$\displaystyle t = \frac{n}{\dot{n}} \ln \left(\frac{1}{1 - \frac{\dot{n}RA}{\dot{q}}\left(T - T_0\right)}\right)$$

This formula is the product of two factors, one which is proportional to $$\displaystyle 1/\dot{n}$$ and one which is proportional to the logarithm term. The two factors compete with each other to be the main factors affecting the result. When I plotted $$\displaystyle t$$ as a function of $$\displaystyle \dot{n}$$, I got a slowly increasing slope, so time to heat increases with increasing $$\displaystyle \dot{n}$$. This was because the logarithm term tends to win out with the choice of parameters I used, but I need to plot it with the particular parameters you're adopting to check whether your conclusion is consistent.