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I have a pipe, Mica, 2 meters in length. Inner Diameter 8mm, and Outer Diameter 10mm. Thermal conductivity of the pipe is 0.528 $\mathrm{\frac{W}{m^\circ C}}$.

I have a gas inlet temperature of 1100°C. The mass flow rate of the gas through the pipe is 0.0325 kg/hour.

As the flow is very slow, I assume that there will be heat loss to the pipe.

How can I determine the outlet temperature?

In this enough information?

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  • $\begingroup$ Ambient temperature outside the pipe. Lots of engineering sites have calculators as mentioned in the link. physics.stackexchange.com/q/91792 $\endgroup$ – Farcher Oct 31 '16 at 14:53
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Although this is a difficult problem to model, a simple lumped thermal analysis can bring some understanding and an approximate solution.

Cooling tube

We'll consider the material flow through the pipe to be plug flow and study the temperature of a small mass element $dx$ travelling down the pipe.

Using Newton's cooling/heating law and considering convective heat losses only, we can write:

$$\frac{dQ}{dt}=u(T-T_{\infty})dA,$$

where the LHS is the heat flux leaving the element, $u$ the overall heat transfer coefficient, $dA$ the surface area of the element, $T$ its temperature and $T_{\infty}$ the ambient temperature. Developing a little, we get:

$$\frac{dQ}{dt}=\pi Du(T-T_{\infty})dx,\tag{1}$$

where $D$ is pipe outer diameter.

As the element has lost heat:

$$dQ=-dmc_pdT$$

$c_p$ is the heat capacity of the gas. Dividing both sides by $dt$ and with $\dot{m}=\frac{dm}{dt}$ gives:

$$\frac{dQ}{dt}=-\dot{m}c_pdT,\tag{2}$$

where $\dot{m}$ is the mass throughput of the gas.

Using the identity $(1)=(2)$, we get a simple differential equation:

$$-\dot{m}c_pdT=\pi Du(T-T_{\infty})dx$$

$$\frac{dT}{T-T_{\infty}}=-\frac{\pi Du}{\dot{m}c_p}dx=-\alpha dx,\tag{3}$$

where:

$$\alpha=\frac{\pi Du}{\dot{m}c_p}$$

Integrating $(3)$ between $0, T_1$ and $L, T_2$ gives:

$$\ln\frac{T_2-T_{\infty}}{T_1-T_{\infty}}=-\alpha L,\tag{4}$$

where $T_1$ and $T_2$ are the incoming and outgoing temperatures of the gas, respectively and $L$ is the pipe length. From $(4)$, $T_2$ can easily be extracted.

The overall heat transfer coefficient $u$ can be estimated from:

$$\frac1u\approx\frac{1}{h_1}+\frac{\theta}{k}+\frac{1}{h_2},$$

where $h_1$ is the convection heat transfer coefficient gas/mica, $k$ the thermal conductivity of mica, $\theta$ the wall thickness and $h_2$ is the convection heat transfer coefficient mica/air.

Limitations of the model:

At high temperature, convection is not the only heat loss mode: radiation loss will also be important. This can be rectified by adding a radiative heat loss function to $(1)$. With Stefan-Boltzmann the loss function would be:

$$\sigma\epsilon(T^4-T^4_{\infty})dA$$

Secondly, assuming plug flow in the case of a low speed low viscosity fluid like a hot gass isn't very realistic. Assuming laminar flow requires a far more demanding mathematical approach though.

Numerical evaluation:

$(4)$ reworks to:

$$T_2=T_{\infty}+(T_1-T_{\infty})e^{-\alpha L}$$

To estimate $\alpha$, I used:

$u\approx 17\:\mathrm{Wm^{-1}K^{-1}}$, based on OP data and literature values.

$c_p=1000\:\mathrm{Jkg^{-1}K^{-1}}$

$\dot{m}=0.000009\:\mathrm{kg/s}$

$D=0.010\:\mathrm{m}$

Which gives an estimate of $\alpha\approx 70\:\mathrm{m^{-1}}$.

With $T_1=1100\:\mathrm{C}$ and $T_{\infty}=20\:\mathrm{C}$, with $L=1\:\mathrm{m}$, I get:

$$T_2\approx 20\:\mathrm{C}$$

So over $1\:\mathrm{m}$, the gas would have cooled down completely.

For other pipelengths $x$, evaluate as:

$$T_2=20+1080e^{-70x}$$

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  • $\begingroup$ Numerical evaluation added. $\endgroup$ – Gert Nov 1 '16 at 1:54

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