Motivation

In traditional compressible flow classes there is very little discussion about the speed of sound outside the ideal gas. The author thinks that this approach has many shortcomings. In a recent consultation an engineerdesign a industrial system that contains converting diverging nozzle with filter to remove small particles from air. The engineer was well aware of the calculation of the nozzle. Thus, the engineer was able to predict that was a chocking point. Yet, the engineer was not ware of the effect of particles on the speed of sound. Hence, the actual flow rate was only half of his prediction. As it will shown in this chapter, the particles can, in some situations, reduces the speed of sound by almost as half. With the ``new'' knowledge from the consultation the calculations were within the range of acceptable results.

The above situation is not unique in the industry. It should be expected that engineers know how to manage this situation of non pure substances (like clean air). The fact that the engineer knows about the chocking is great but it is not enough for today's sophisticated industry. In this chapter an introductory discussion is given about different situations which can appear the industry in regards to speed of sound

Introduction

$\begin{figure}\centerline{\includegraphics {cont/sound/illustrations/piston}} \end{figure}$

The people had recognized for several hundred years that sound is a variation of pressure. The ears sense the variations by frequency and magnitude which are transferred to the brain which translates to voice. Thus, it raises the question: what is the speed of the small disturbance travel in a ``quiet'' medium. This velocity is referred to as the speed of sound. To answer this question consider a piston moving from the left to the right at a relatively small velocity (see Figure ). The information that the piston is moving passes thorough a single ``pressure pulse.'' It is assumed that if the velocity of the piston is infinitesimally small, the pulse will be infinitesimally small. Thus, the pressure and density can be assumed to be continuous

$\begin{figure}\centerline{\includegraphics {cont/sound/illustrations/stationaryWave}} \end{figure}$

In the control volume it is convenient to look at a control volume which is attached to a pressure pulse. Applying the mass balance yields

$\displaystyle \rho c = (\rho + d\rho)(c-dU)$

(3.1)

or when the higher term $dUd\rho$ is neglected yields

$\displaystyle \rho d U = c d\rho \Longrightarrow dU = {c d \rho \over \rho}$

(3.2)

From the energy equation (Bernoulli's equation), assuming isentropic flow and neglecting the gravity results

$\displaystyle { ( c - dU)^2 - c^{2} \over 2} + {dP \over \rho} = 0$

(3.3)

neglecting second term (

) yield

$\displaystyle -c dU + {dP \over \rho} = 0$

(3.4)

Substituting the expression for

from equation (3.2) into equation (3.4) yields

$\displaystyle c^{2} \left( { d\rho \over \rho } \right) = {dP \over \rho} \Longrightarrow c^2 = {dP \over d\rho}$

(3.5)

An expression is needed to represent the right hand side of equation (3.5). For an ideal gas,

is a function of two independent variables. Here, it is considered that $P= P(\rho, s)$ where

is the entropy. The full differential of the pressure can be expressed as follows:

$\displaystyle dP = \left. {\Pxy{P}{\rho} } \right\vert _{s} d\rho + \left. {\Pxy{P}{s} } \right\vert _{\rho} ds$

(3.6)

In the derivations for the speed of sound it was assumed that the flow is isentropic, therefore it can be written

$\displaystyle \Dxy{P}{\rho} = \left. {\Pxy{P}{\rho} } \right\vert _{s}$

(3.7)

Note that the equation (3.5) can be obtained by utilizing the momentum equation instead of the energy equation.

The momentum equation written for the control volume shown in Figure (3.2) is

$\displaystyle \overbrace{(P + dP) - P}^{\sum F} = \overbrace{(\rho +d\rho)(c - dU)^{2} - \rho c^2}^ {\int_{cs} U(\rho U dA)}$

(3.8)

Neglecting all the relative small terms results in

$\displaystyle dP = (\rho + d\rho) \left( c^{2} - \cancelto{\sim 0}{2cdU} + \cancelto{\sim 0}{\rule{20pt}{0pt} dU^{2}}\qquad\;\right) - \rho c^{2}$

(3.9)

$\displaystyle dP = c ^{2} d \rho$

(3.10)

This yields the same equation as (3.5).

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 $T_B \longrightarrow T_A$ .

Compressible Flow and Gas Dynamic

الجمعة، 10 أغسطس 2012

Speed of Sound

Motivation

Speed of sound in ideal and perfect gases

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