Calculating the temperature of flowing gas in copper tube

Question

A copper coil is immersed in a water bath. The water bath provides heat to heat the nitrogen gas when it passes through the copper coil. Nitrogen gas at room temperature is passed through the copper coil at a set flow rate. The heated nitrogen gas is then supplied to a drying vessel. I want to find the temperature of the nitrogen gas when it emerged from the outlet of the copper coil.

I used the heat conduction formula to calculate it but it is not the right method. $q = kA (T_1-T_2)/d$ where $q$ is the rate of heat transferred, $A$ is the area of the surface, $k$ is the thermal conductivity, $d$ is the thickness and $(T_1-T_2)$ is the temperature difference. But since the temperature in the copper coil is not constant because the temperature of the nitrogen gas will increase while it flows through the copper coil as it gains more heat, this approach is wrong. Is there a better approach to find the temperature of the nitrogen gas when it emerged from the copper coil?

Answer 1

Firstly, some assumptions:

1. water temperature $T_{\infty}$ is uniform and constant,
2. nitrogen gas flow is highly turbulent, plug flow. Because nitrogen viscosity is very low, that is a reasonable assumption,
3. nitrogen gas is heated by convection only. Longitunal conduction is negligle. This is a reasonable assumption because nitrogen's thermal conductivity is low,
4. thermal expansion of nitrogen gas is considered neglibible.

Now look at the heat balance of an element $\mathbf{d}x$, mass $\mathbf{d}m$: it receives a heat flux $\frac{\mathbf{d}q}{\mathbf{d}t}$, in accordance with Newton's cooling/heating Law:

$$\frac{\mathbf{d}q}{\mathbf{d}t}=2 \pi r_o R\mathbf{d}x[T_{\infty}-T(x)]\tag{1}$$

Where $R$ is the thermal resistance, I'll come back to it extensively below. $r_o$ is the outside radius of the pipe.

Now we also know that:

$$\mathbf{d}q=c_p\mathbf{d}m\mathbf{d}T(x)$$

Dividing both sides by $\mathbf{d}t$ and because $\frac{\mathbf{d}m}{\mathbf{d}t}=\dot{m}$, we get with $(1)$: $$c_p\dot{m}\mathbf{d}T(x)=2 \pi r_o R\mathbf{d}x{R}[T_{\infty}-T(x)]\tag{2}$$

where $c_p$ is the constant pressure heat capacity of the nitrogen gas and $\dot{m}$ its mass flow through the pipe.

Now for ease of reading, set:

$$\alpha=\frac{2\pi r_o R}{c_p\dot{m}}$$

And with $(2)$, rearrange slightly:

$$\frac{\mathbf{d}T(x)}{T_{\infty}-T(x)}=\alpha \mathbf{d}x\tag{3}$$

Integrating $(3)$ between $(0,L)$ and $(T_0,T(L))$ we get:

$$-\ln\Big[\frac{T_{\infty}-T(L)}{T_{\infty}-T_0}\Big]=\alpha L$$

Slight reworking then yields:

$$\boxed{T(L)=T_{\infty}-(T_{\infty}-T_0)e^{-\alpha L}}\tag{4}$$

Note that for $L \to +\infty$, $T(L) \to T_{\infty}$.

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Thermal Resistance:

A full treatise of the concept of themal resistance is outside the scope of this answer. May I suggest you read up on it here.

In our pipe problem we have three components to the thermal resistance:

1. outer convection zone: water to pipe ($R_1$),
2. conduction zone: heat conducting through circular pipe wall ($R_2$),
3. inner convection zone: pipe to nitrogen gas ($R_3$).

The overall thermal resistance $R$ is obtained from:

$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$$

where:

1. $R_1=h_o$, the outer (convection) heat transfer coefficient,
2. Conduction zone:

$R_2=\frac{1}{k_{Cu}}\ln\frac{r_o}{r_i}$

where the $r$ are outer and inner radii of the pipe and $k_{Cu}$ the thermal conductivity of copper.

3. $R_3=h_i$, the inner (convection) heat transfer coefficient.