We will calculate the input impedance of a cylindrical pipe with rigid walls, of length $L$ and cross-sectional area $S$. The pipe runs from $x=0$, where the impedance is to be determined, to $x=L$, where it has an open end. Actually, we will assume that $L$ includes the end correction for the open end, so the physical length of the pipe is really slightly shorter than that.
At $x=0$ we drive the pipe with a volume flow rate $Ve^{i \omega t}$. This will generate an outgoing sound wave with acoustic pressure $Ae^{i(\omega t – kx)}$ above ambient, where $A$ is a complex constant and $k=\omega/c$. There will also be a returning reflected wave with pressure $Be^{i(\omega t + kx)}$, where $B$ is another complex constant. Each of these travelling waves has an associated particle velocity $u(x,t)$, satisfying equation (7) from section 4.1.3:
$$-\rho_0 \dfrac{\partial u}{\partial t} = \dfrac{\partial p’}{\partial x} \tag{1}$$
where $\rho_0$ is the mean density of air. For a wave $e^{i(\omega t – kx)}$ this requires
$$u=\dfrac{1}{Z_0}e^{i(\omega t – kx)} \tag{2}$$
where $Z_0=\rho_0 c$ is the characteristic impedance of sound waves in air. The corresponding result for a left-travelling wave $e^{i(\omega t + kx)}$ is
$$u=-\dfrac{1}{Z_0}e^{i(\omega t + kx)} . \tag{3}$$
Now the total pressure $p’$ in the pipe is
$$p’=Ae^{i(\omega t – kx)}+Be^{i(\omega t + kx)} , \tag{4}$$
and the total volume flow rate is
$$v=\dfrac{AS}{Z_0}e^{i(\omega t – kx)}-\dfrac{BS}{Z_0}e^{i(\omega t + kx)} . \tag{5}$$
At $x=L$ we require $p’=0$ for the open end, so
$$Ae^{-ikL} +Be^{ikL} =0 \tag{6}$$
or
$$B=-Ae^{-2ikL} . \tag{7}$$
At $x=0$, the input impedance we want, $Z(\omega)$, is now given by the ratio of pressure to volume flow rate:
$$Z(\omega)=\dfrac{A+B}{(S/Z_0) (A-B)}=\dfrac{Z_0}{S}\dfrac{1-e^{-2ikL}}{1+e^{-2ikL}}=\dfrac{iZ_0}{S}\tan kL . \tag{8}$$
So far, we have assumed that there is no dissipation in the problem. As a final step, we can introduce a realistic level of damping by the simple expedient of allowing $k$ in equation (8) to become complex and to depend on frequency. There are three sources of energy dissipation for our sound waves in the pipe. One is the result of energy loss at the open end by sound radiation. The other two come from effects internal to the pipe, as the waves propagate. First, the particle velocity $u$, parallel to the axis of the pipe, results in some dissipation from viscous friction on the walls of the pipe.
The second effect comes from something we haven’t previously mentioned. When we derived the wave equation back in section 4.1.1, we used the approximation that at acoustic frequencies of interest the changes in density and pressure are adiabatic (or isentropic). This means that the density changes happen sufficiently rapidly that there is no change in total energy of a small element of air. The universal gas law governing the thermodynamics then requires that the temperature of the air also changes. When air (or any other gas) is compressed, it gets hotter; and when it is expanded, it gets cooler. This principle is used in refrigerators, for example.
So as a sound wave propagates along the pipe, there are small temperature fluctuations associated with it. There is thus a tendency for heat diffusion to occur from the warmer to the cooler regions, leading to energy dissipation. Furthermore, the pipe wall is a reservoir of heat, and it remains more or less at a constant temperature. So the air in contact with the walls will have a significant temperature gradient, and again heat diffusion will result.
The combination of these three energy dissipation mechanisms is quite complicated: see sections 8.2–8.5 of Fletcher and Rossing [1] for a detailed discussion. For our purposes we can use a simple approximation that these authors recommend as appropriate to most musical wind instruments. Assuming that the pipe is neither extremely narrow (i.e. it is not a capillary), nor extremely wide, the loss from sound radiation can be neglected, and the other two effects can be allowed for in a combined formula in which
$$k=\omega/c+i \alpha(\omega) \tag{9}$$
with
$$ \alpha \approx 1.2 \times 10^{-5} \sqrt{\omega}/a \tag{10}$$
where $a$ is the pipe radius.
We can deduce something else from the first part of equation (8). If our pipe were semi-infinite, so that no reflected wave ever came back, we would have $B=0$ and the input impedance would be simply $Z_0/S$. So in order to plot the impedance, it makes a lot of sense to normalise it by this value for an infinite tube. That is what was done in Fig. 13 of section 11.1, reproduced here as Fig. 1. We can now see that the mean trend of this decibel plot is very obviously the 0 dB line. This neatly illustrates Skudrzyk’s theorem (see section 5.3.2): the logarithmic mean of a driving-point impedance or admittance is given by the behaviour of the corresponding infinite system.
[1] Neville H Fletcher and Thomas D Rossing; “The physics of musical instruments”, Springer-Verlag (Second edition 1998)