Rate of water temperature change

Goal:

I have a pipe containing water. Solving for $k$ in Newton's Law of Cooling, I want to detect whether or not fresh water is flowing through the pipe. Assumption: The closer the empirical $k$ is to $\alpha = \frac{hA}{mc}$, the less flow present and vice versa.

Specs:

Pipe material: Iron

Length: 100 cm

$h \approx 20\:\mathrm{Wm^{-2}K^{-1}}$

$c_{water}=42000\:\mathrm{Jkg^{-1}K^{-1}}$

Sample 1

Outer diameter: 3.81 cm

Inner diameter: unknown

$A=\pi D L=0.12\:\mathrm{m^2}$

$m=\rho V=\rho \frac{\pi D^2}{4}L=1.14\:\mathrm{kg}$

With no fresh water flow, using $\alpha = \frac{hA}{mc}$ = .0005

Empirical Data: $T_1$ = 294.43, $T_2$ = 295.15, $T_{\infty}$ = 299.43, $t$ = 316 sec

$$k=\frac{1}{-t} \times \ln\Big(\frac{T_2 - T_{\infty}}{T_1 - T_{\infty}}\Big)$$

$$k = .0005$$

This is reaching ambient temperature as environmentally expected, indicating no fresh flow.

However, $T_1$ = 303.15, $T_2$ = 299.43, $T_{\infty}$ = 303.43, $t$ = 330 sec

$$k = -.0081$$

Indicating an external factor (i.e. fresh water flow) effecting the natural rate of temperature change.

Sample 2

Outer diameter: 5.08 cm

Inner diameter: unknown

$A=\pi D L=0.06\:\mathrm{m^2}$

$m=\rho V=\rho \frac{\pi D^2}{4}L=2.03\:\mathrm{kg}$

With no fresh water flow, using $\alpha = \frac{hA}{mc}$ = .0004

Empirical Data where fresh water is flowing: $T_1$ = 294.43, $T_2$ = 295.15, $T_{\infty}$ = 300.29, $t$ = 482 sec

$$k = .0003$$

Question

I am trying to understand how the resulting $k$ in sample 2 is so similar to the no-flow $k$ of .0004.

• It is not clear to me what you are asking. Please could you provide a full description (and diagram) of your experiment and your calculations? What assumptions have you made? What measurements have you made? Is the flow rate the same in both samples 1 & 2? How do you know? It is particularly difficult to understand (1) how you have been able to measure water temperature but not flow rate, and (2) why the unknown inner diameter can be assumed to make no difference. – sammy gerbil Sep 4 '16 at 22:00 We assume the pipe thickness to be negligible (as it is 'unknown') and no radial temperature gradients to exist. In addition plug flow is assumed.

We consider an infinitesimal element of length $dx$ and apply Newton's cooling law:

$$\frac{dQ}{dt}=hdA[T(x)-T_{\infty}]=\pi Dh[T(x)-T_{\infty}]dx$$

If the infinitesimal element changes temperature then:

$$dQ=-dmcdT(x)$$ So:

$$\frac{dQ}{dt}=-\frac{dm}{dt}cdT(x)$$ $$\frac{dQ}{dt}=-\dot{m}cdT(x)$$

Where $\dot{m}=\frac{dm}{dt}$ is the mass throughput flowing through the pipe.

This gives us the identity:

$$-\dot{m}cdT(x)=\pi Dh[T(x)-T_{\infty}]dx$$ $$\frac{dT(x)}{T(x)-T_{\infty}}=-\frac{\pi Dh}{\dot{m}c}dx$$

Call:

$$k=\frac{\pi Dh}{\dot{m}c}$$ $$\int_{T_1}^{T_2}\frac{dT(x)}{T(x)-T_{\infty}}=-k\int_0^Ldx$$ $$\ln\Big[\frac{T_2-T_{\infty}}{T_1-T_{\infty}}\Big]=-kL$$ $$\large{T_2=T_{\infty}+[T_1-T_{\infty}]e^{-kL}}$$

I am trying to understand how the resulting $k$ in sample 2 is so similar to the no-flow $k$ of .0004.

One cannot treat the problem the same way when there's no water flow. For $\dot{m}=0 \implies k \to +\infty$! That is of course not allowed.

The cases with and without water flow must be treated differently.

• I haven't yet worked through the whole response. However, at the time of gathering the temperature data, I do not yet know if there is flow. This is what I am trying to determine based on Newton's Law. How else do you advise determining the level of flow based on the temperatures? – Cindy Aug 29 '16 at 19:55
• Additionally, it appears on the diagram as $T_1$ is the initial temperature at location x, whereas $T_2$ is the exiting temp at location y. However, that is not the case. In our scenario, the are BOTH at location x where $T_2$ is the temperature after $t$ amount of time. – Cindy Aug 29 '16 at 19:59
• I rated your answer! – Cindy Aug 29 '16 at 20:00
• Do you suggest an alternate method of calculating rate of flow based on temperatures? – Cindy Aug 29 '16 at 20:07
• A flowmeter? :-) Using temperatures is really awkward and inaccurate. – Gert Aug 29 '16 at 20:22