Speed of sound in ideal and perfect gases

The speed of sound can be obtained easily for the equation of state for an ideal gas (also perfect gas as a sub set) because of a simple mathematical expression. The pressure for an ideal gas can be expressed as a simple function of density, $\rho$ , and a function ``molecular structure'' or ratio of specific heats, $k$ namely

$$P= constant\times \rho^{k}$$ (3.11)

and hence

$$c = \sqrt{\Dxy{P}{\rho}} = k \times constant \times \rho^{k-1} =k \times {\overbrace{ constant \times \rho^k}^{P} \over \rho} \hfill$$

$$= k \times {P \over \rho }$$ (3.12)

Remember that $P / \rho$ is defined for an ideal gas as $RT$ , and equation (3.12) can be written as

$$c = \sqrt{ k R T}$$ (3.13)

Solution

The solution can be estimated by using the data from steam table^3.3

$$c = \sqrt{\Delta P \over \Delta \rho}_{s=constant}$$ (3.14)

At $20[bar]$ and $350 \celsius$ : s = 6.9563 $\left[ kJ \over K\; kg\right]$ $\rho$ = 6.61376 $\left[ kg \over m^3 \right]$
At $18[bar]$ and $350 \celsius$ : s = 7.0100 $\left[ kJ \over K\; kg\right]$ $\rho$ = 6.46956 $\left[ kg \over m^3 \right]$
At $18[bar]$ and $300 \celsius$ : s = 6.8226 $\left[ kJ \over K\; kg\right]$ $\rho$ = 7.13216 $\left[ kg \over m^3 \right]$

After interpretation of the temperature:
At $18[bar]$ and $335.7 \celsius$ : s $\sim$ 6.9563 $\left[ kJ \over K\; kg\right]$ $\rho \sim$ 6.94199 $\left[ kg \over m^3 \right]$
and substituting into the equation yields

$$c = \sqrt{ 200000 \over 0.32823} = 780.5 \left[ m \over sec \right]$$ (3.15)

for ideal gas assumption (data taken from Van Wylen and Sontag, Classical Thermodynamics, table A 8.)

$$c = \sqrt{kRT} \sim \sqrt{ 1.327 \times 461 \times (350 + 273)} \sim 771.5 \left[ m \over sec \right]$$

Note that a better approximation can be done with a steam table, and it will be part of the future program (Potto-GDC).

Solution

The temperature is denoted at ``A'' as $T_A$ and temperature in ``B'' is $T_B$ . The distance between ``A'' and ``B'' is denoted as $h$ .

$$T = (T_B - T_A) {x \over h} + T_A$$

Where the distance $x$ is the variable distance. It should be noted that velocity is provided as a function of the distance and not the time (another reverse problem). For an infinitesimal time $dt$ is equal to

$$d t = {d x \over \sqrt{kRT(x)} } = {dx \over \sqrt{ kRT_A \left( {(T_B - T_A) x \over T_A h} +1 \right) } }$$

integration of the above equation yields

$$t = {2 h T_A \over 3 \sqrt{kRT_A} \; (T_B - T_A) } \left( \left(T_B \over T_A\right)^{3\over 2}- 1 \right)$$ (3.16)

For assumption of constant temperature the time is

$$t = { h \over \sqrt{kR\bar{T}} }$$ (3.17)

Hence the correction factor

$${t_{corrected} \over t } = \sqrt{T_A \over \bar{T}} {2 \over 3} { T_A \over ( T_B - T_A )} \left( \left(T_B \over T_A\right)^{3\over 2}- 1 \right)$$ (3.18)

This correction factor approaches one when $T_B \longrightarrow T_A$ .