# Law for tap water temperature

I was wondering if anyone put together a law to describe the rising temperature of the water coming out of a tap.

The setup is fairly simple: there's a water tank at temperature T, a metal tube of length L connected to it and a tap at the end where temperature is measured. The water flows at P l/s.

Given that the metal tube is at room temperature initially, what law describes the temperature of the water at any instant? What is the limit temperature of the water?

Thanks.

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Which water temperature? The tank, or the outlet water temperature? –  Herb Nov 3 '10 at 14:24
the temperature at the tap. –  Sklivvz Nov 3 '10 at 14:41
Physics is just amazing, it can describe something as small as this to the lives of huge stars that last billion of years, to what proton is made of. –  Jonathan. Nov 3 '10 at 16:27
Uhg. Last spring I tries to do an internally consistent model of a forced fluid thermal control/redistribution system that was being considered for a experiment I participate it. The problem is ugly. The proper answer depends on the Reynold's number of the flow and several other dimensionless constants that I hadn't even heard of until I started reading for the project. The engineers who do this stuff regularly use a mixture of well known designs, empirical constants and rules of thumb and then over build them. –  dmckee Nov 20 '10 at 2:09

We can consider the following model: a tube of constant temperature $T_e$ of lenght L, radius $r$ where water is flowing uniformly at a speed $v$ (that you can obtain from your flow $P$).

A "slice" of water travels an interval $dx$ in a duration $dt = \frac{dx}{v}$.

The tube will contribute to the "heating" of the water by $\frac{dQ}{dt} = (T-T_e) k 2 \pi r dx$ where $k$ is the conductivity and where we use a very simple model (in particular for the radius, we do not distinguish external and internal radii).

During this interval the temperature $T(x)$ of the water will vary by $dT = -\frac{dQ}{c \rho dV}$ where $C$ is the heat capacity at constant pressure of water, and where $dV = 2 \pi r dx$.

Replacing we have $\frac{dT}{T-T_e}=-\frac{k}{\rho C v} dx$ whose solution, if the temperature in the tank (ie x = 0) is $T_t$ :

$T(x) = (T_t - T_e) e^{(-\alpha x)}+T_e$ where $\alpha = \frac{k}{\rho C v}$.

Depending on the lenght of the tube you have the temperature at the tap.

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Agree - this is the steady-state solution for classical heat exchanger problem. The original question implied that the temperature at the tap was changing over time, which it isn't. Or the question needs to be rephrased somewhat and both @Cedric and I are over-complicating the solution considerably. –  Herb Nov 3 '10 at 15:14
@Herb: As you said, to consider the transient solution, we need a model for the tube and the surrounding environment. –  Cedric H. Nov 3 '10 at 15:18
Aye, and the math there is a bit more hairy. By the way, nice job with the use of MathJax - I think you're missing a grouping in the e^{-\alpha x} term, though (or I've missed a step in the integration, which is possible/likely). –  Herb Nov 3 '10 at 15:29
This is too simple; water in a tube follows Poiseuille flow, so some insulating layer could form. Also there will be a turbulent transition in the system. –  mbq Nov 3 '10 at 15:48
Issues/simplifications: We never calculated / worried about the Reynolds number, so we don't know if we have laminar or turbulent flow, we assume that tank of water is insulated from the tube and the water inside always remains at the initial temperature, and that the outside of the tube is at constant temperature over its length. –  Herb Nov 3 '10 at 21:59