Problem 3.4

Given:
\(\Delta p = \)30 Pa,
\( T= \) 30^oC = 303 K,
\(p = \) 101 kPa.

To calculate: \( d\rho, dT, dV \)

The problem description is schematically shown in Fig. 1.

Fig. 1: Schematic of the problem description

Placing the frame of reference on the sound wave, as shown on the right side of Fig. 1. The velocity at which the flow enters the wave is equal to the speed of sound,

\[ V=a=\sqrt{\gamma\,R\,T}=\sqrt{1.4\times287\times303}= 348.92\, \text{m/s}\ . \]

Also \( M=V/a=1 \).

There are possibly two methods of solving this problem.

Since the wave is a weak pressure wave with a very small jump in pressure, \( \Delta p/p\approx dp/p \). We can assume the flow through the wave to be isentropic. Therefore, we can use the differential relations,

\[ \frac{dp}{p}=-\gamma\,M^{2}\,\frac{dV}{V} \]

which we can solve for \( dV \) to be,

\[ dV=-\frac{V\,dp}{\gamma\,p\,M^{2}}=-\frac{348.92\times30}{1.4\times101\times10^{3}\times1^{2}} \]

\[ \boxed{dV= -0.0740282\, \text{m/s}}\ . \]

The change of temperature is given as,

\[ \frac{dT}{T}=-\left(\gamma-1\right)\,M^{2}\,\frac{dV}{V} \]

which can be solved for \( dT \) as,

\[ dT=-T\,\left(\gamma-1\right)\,M^{2}\,\frac{dV}{V}=-303\times\left(1.4-1\right)\times1^{2}\times\left(\frac{-0.0740282}{348.92}\right) \]

\[ \boxed{dT=0.02571425\,\text{K}}\ . \]

The change in density is given as,

\[ \frac{d\rho}{\rho}=-M^{2}\,\frac{dV}{V} \]

which can be solved for \( d\rho \) as,

\[ d\rho=-\rho\,M^{2}\,\frac{dV}{V} \]

using the equation of state for ideal gas equation,

\[ d\rho=-\frac{p}{R\,T}\,M^{2}\,\frac{dV}{V}=-\frac{101\times10^{3}}{287\times303}\times1^{2}\times\left(\frac{-0.0740282}{348.92}\right) \]

\[ \boxed{d\rho= 2.464\times10^{-4}\,\text{kg/m}^{3}}\ . \]

Let us name the two sides of the wave as 1 and 2 and shown in Fig. 1.

\( p_{1} = \) 101 kPa,

\( p_{2}=p_{1}+\Delta p=101\times10^{3}+30= 101030\,\text{Pa} \).

Using isentropic relation of ratios of pressure and temperature, we get,

\[ \frac{T_{2}}{T_{1}}=\left(\frac{p_{2}}{p_{1}}\right)^{\left(\gamma-1\right)/\gamma} \] \[ T_{2}=T_{1}\left(\frac{p_{2}}{p_{1}}\right)^{\left(\gamma-1\right)/\gamma} \] \[ T_{2}-T_{1}=T_{1}\left(\frac{p_{2}}{p_{1}}\right)^{\left(\gamma-1\right)/\gamma}-T_{1} \]

\[ \Delta T=303\times\left[\left(\frac{101030}{101000}\right)^{\left(1.4-1\right)/1.4}-1\right] \]

\[ \boxed{\Delta T=0.025712\,\text{K}}\ . \]

Using isentropic relation of ratios of pressure and density, we get,

\[ \frac{\rho_{2}}{\rho_{1}}=\left(\frac{p_{2}}{p_{1}}\right)^{1/\gamma} \] \[ \rho_{2}-\rho_{1}=\rho_{1}\left[\left(\frac{p_{2}}{p_{1}}\right)^{1/\gamma}-1\right] \] Using the equation of state for ideal gas, \[ \Delta\rho=\frac{p_{1}}{R\,T_{1}}\left[\left(\frac{p_{2}}{p_{1}}\right)^{1/\gamma}-1\right] \]

\[ \Delta\rho=\frac{101000}{287\times303}\left[\left(\frac{101030}{101000}\right)^{1/1.4}-1\right] \]

\[ \boxed{\Delta\rho=2.464\times10^{-4}\,\text{kg/m}^{3}}\ . \]

Using the conservation of energy equation for isentropic flow,

\[ c_{p}T_{1}+\frac{V_{1}^{2}}{2}=c_{p}T_{2}+\frac{V_{2}^{2}}{2} \] \[ V_{2}=\sqrt{V_{1}^{2}+2\,c_{p}\,\left(T_{1}-T_{2}\right)} \] \[ V_{2}=\sqrt{V_{1}^{2}-\frac{2\,\gamma\,R}{\gamma-1}\,\Delta T} \] \[ V_{2}-V_{1}=\sqrt{V_{1}^{2}-\frac{2\,\gamma\,R}{\gamma-1}\,\Delta T}-V_{1} \] \[ \Delta V=\sqrt{V_{1}^{2}-\frac{2\,\gamma\,R}{\gamma-1}\,\Delta T}-V_{1} \]

\[ \Delta V=\sqrt{348.92^{2}-\frac{2\times1.4\times287}{1.4-1}\times0.025712}-348.92 \]

\[ \boxed{\Delta V=-0.0740297\,\text{m/s}}\ . \]

Both the methods give same results, with minor differences because the differential equations assume infinitesimal changes.

Problem 3.4:

Solution: