Subfield: Thermodynamics

2. Chimney physics 

Introduction

Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$.

Assume that:

- the velocity of the gases in the furnace is negligibly small
- the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal
- the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible
- the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow:

$\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const,

where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height

- the change of the density of the gas is negligible throughout the chimney

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Task 1

Question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney.

Solution:
Let $p(z)$ denote the pressure of air at height $z$; then, according to one of the assumptions $p(z)=p(0)-\rho_{\text {Air }} g z$, where $p(0)$ is the atmospheric pressure at zero altitude.

Throughout the chimney the Bernoulli law applies, that is, we can write

$$
\frac{1}{2} \rho_{\text {Smoke }} \nu(z)^{2}+\rho_{\text {Smoke }} g z+p_{\text {Smoke }}(z)=\text { const. },
\tag{1}
$$

where $p_{\text {Smoke }}(z)$ is the pressure of smoke at height $z, \rho_{\text {Smoke }}$ is its density, and $v(z)$ denotes the velocity of smoke; here we have used the assumption that the density of smoke does not vary throughout the chimney. Now we apply this equation at two points, (i) in the furnace, that is at point $z=-\varepsilon$, where $\varepsilon$ is a negligibly small positive number, and (ii) at the top of the chimney where $z=h$ to obtain:

$$
\frac{1}{2} \rho_{\text {Smoke }} v(h)^{2}+\rho_{\text {Smoke }} g h+p_{\text {Smoke }}(h) \approx p_{\text {Smoke }}(-\varepsilon)
\tag{2}
$$

On the right hand side we have used the assumption that the velocity of gases in the furnace is negligible (and also $-\rho_{\text {Smoke }} g \varepsilon \approx 0$ ).

We are interested in the minimal height at which the chimney will operate. The pressure of smoke at the top of the chimney has to be equal or larger than the pressure of air at altitude $h$; for minimal height of the chimney we have $p_{\text {Smoke }}(h) \approx p(h)$. In the furnace we can use $p_{\text {Smoke }}(-\varepsilon) \approx p(0)$. The Bernoulli law applied in the furnace and at the top of the chimney [Eq. (2)] now reads

$$
\frac{1}{2} \rho_{\text {Smoke }} v(h)^{2}+\rho_{\text {Smoke }} g h+p(h) \approx p(0)
\tag{3}
$$

From this we get

$$
v(h)=\sqrt{2 g h\left(\frac{\rho_{\text {Air }}}{\rho_{\text {Smoke }}}-1\right)}
\tag{4}
$$

The chimney will be efficient if all of its products are released in the atmosphere, i.e.,



| $v(h) \geq \frac{B}{A}$, | $(5)$ |
| :--- | :--- |
| from which we have |  |
| $h \geq \frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{1}{\frac{\rho_{\text {Air }}}{\rho_{\text {Smoke }}}-1}$. | (6) |

We can treat the smoke in the furnace as an ideal gas (which is at atmospheric pressure $p(0)$ and temperature $\left.T_{\text {Smoke }}\right)$. If the air was at the same temperature and pressure it would have the same density according to our assumptions. We can use this to relate the ratio $\rho_{\text {Air }} / \rho_{\text {Smoke }}$ to $T_{\text {Smoke }} / T_{\text {Air }}$ that is,

| $\frac{\rho_{\text {Air }}}{\rho_{\text {Smoke }}}=\frac{T_{\text {Smoke }}}{T_{\text {Air }}}$, and finally | (7) |
| :--- | :--- |
| $h \geq \frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{T_{\text {Smoke }}-T_{\text {Air }}}=\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$. | (8) |

For minimal height of the chimney we use the equality sign.

Final answer: $\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$