# Outlet temperature of gas through a pipe

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?

• Ambient temperature outside the pipe. Lots of engineering sites have calculators as mentioned in the link. physics.stackexchange.com/q/91792 – Farcher Oct 31 '16 at 14:53

Although this is a difficult problem to model, a simple lumped thermal analysis can bring some understanding and an approximate solution. 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:

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

• Numerical evaluation added. – Gert Nov 1 '16 at 1:54