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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,

namely

$\displaystyle P= constant\times \rho^{k}$

(3.11)

and hence

$\displaystyle c = \sqrt{\Dxy{P}{\rho}} = k \times constant \times \rho^{k-1}$	$\displaystyle =k \times {\overbrace{ constant \times \rho^k}^{P} \over \rho} \hfill$
	$\displaystyle = k \times {P \over \rho }$	(3.12)

Remember that $P / \rho$ is defined for an ideal gas as

, and equation (3.12) can be written as

$\displaystyle c = \sqrt{ k R T}$

(3.13)

$\begin{examl} Calculate the speed of sound in water vapor at $20 [bar]$ and $350 \celsius$, (a) utilizes the steam table (b) assuming ideal gas. \end{examl}$
Solution

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

$\displaystyle c = \sqrt{\Delta P \over \Delta \rho}_{s=constant}$

(3.14)

and $350 \celsius$ : s = 6.9563 $\left[ kJ \over K\; kg\right]$ $\rho$ = 6.61376 $\left[ kg \over m^3 \right]$
At

and $350 \celsius$ : s = 7.0100 $\left[ kJ \over K\; kg\right]$ $\rho$ = 6.46956 $\left[ kg \over m^3 \right]$
At

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 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

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

(3.15)

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

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

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

$\begin{examl} \index{speed of sound!linear temperature} The temperature in the a... ...d to travel from point \lq\lq A'' to point \lq\lq B'' under this assumption.? \end{examl}$
Solution

The temperature is denoted at ``A'' as

and temperature in ``B'' is

. The distance between ``A'' and ``B'' is denoted as

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

Where the distance

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

is equal to

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

integration of the above equation yields

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

(3.16)

For assumption of constant temperature the time is

$\displaystyle t = { h \over \sqrt{kR\bar{T}} }$

(3.17)

Hence the correction factor

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

(3.18)

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

Next: Speed of Sound in Up: Speed of Sound Previous: Introduction Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21

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