Problem 2.6

Given:
isothermal air flow \( \implies dT/dx=0 \),
\( V = \) 200 m/s,
\( T = \) 25^oC = 298 K,
\( p = \) 120 kPa,
\( dV/dx = \) -0.3 \( V \).

To calculate: \( dp/dx \), \( ds/dx \), \( d\rho/dx \).

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The schematic diagram of the problem description is shown in Fig. 1.

The density at the section can be calculated using the equation of state for an ideal gas as,

\[ \rho=\frac{p}{R\,T}=\frac{120\times10^{3}}{287\times298}= 1.4030821\,\text{kg/m}^{3} \]

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Using the conservation of momentum (Euler equation) in the horizontal direction, we get,

\[ \frac{1}{\rho}\frac{dp}{dx}+V\,\frac{dV}{dx}=0 \]

which can be solved for pressure gradient as,

\[ \frac{dp}{dx}=-\rho\,V\,\frac{dV}{dx} \]

Substituting the know gradient of velocity, we get,

\[ \frac{dp}{dx}=0.3\rho\,V^{2}=0.3\times1.4030821\times200^{2} \]

\[ \boxed{\frac{dp}{dx}= 16836.9852\,\text{Pa/m}}\ . \]

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Using the conservation of energy equation in its differential form, we get,

\[ c_{p}\cancelto{\text{isothermal flow}}{\frac{dT}{dx}}+V\,\frac{dV}{dx}-\frac{dq}{dx}=0 \]

\[ \frac{dq}{dx}=V\,\left(-0.3\,V\right)=-0.3\times200^{2}=-12000\,\text{(J/kg)/m} \]

Using the definition of change in entropy,

\[ \frac{ds}{dx}=\frac{1}{T}\,\frac{dq}{dx}=\frac{1}{298}\times\left(-12000\right) \]

\[ \boxed{\frac{ds}{dx}=-40.268456\,\text{(J/kg-K)/m}}\ . \]

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Differentiating the equation of state for an ideal gas,

\[ \frac{1}{p}\frac{dp}{dx}-\frac{1}{\rho}\frac{d\rho}{dx}-\frac{1}{T}\cancelto{\text{isothermal flow}}{\frac{dT}{dx}}=0 \] \[ \frac{d\rho}{dx}=\frac{\rho}{p}\frac{dp}{dx} \]

\[ \frac{d\rho}{dx}=\frac{1.4030821}{120\times10^{3}}\times16836.9852 \]

\[ \boxed{\frac{d\rho}{dx}=0.1968639\,\text{(kg/m}^{3}\text{)/m}}\ . \]

Problem 2.6:

Solution: