Question

For compressible flow, which one of the following is true?

• A. \(\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{average}}>\ln\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{local}}\)
• B. \(\ln\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{average}}>\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{local}}\)
• C. \(\ln\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{average}}<\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{local}}\)
• D. None of the above

Hint:

Remember: for positive numbers, \(\ln(x)<x - 1\) when \(x>1\).

Answer 1

The Correct Option is C

In compressible flow, we often encounter changes in density, necessitating a clear understanding of how average and local density ratios compare. Let's analyze the given statements mathematically:

Consider the options:

1. \(\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{average}}>\ln\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{local}}\)
2. \(\ln\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{average}}>\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{local}}\)
3. \(\ln\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{average}}<\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{local}}\)
4. None of the above

For compressible flows:

• The natural logarithm function, \(\ln(x)\), where \(x > 0\), is used to calculate the local effect using continuous data points.
• The density ratio \(\dfrac{\rho_2}{\rho_1}\) represents an instantaneous or pointwise ratio.
• The average effect, commonly processed as a natural log of many local density changes over a specific region, tends to be smoother and less intense than pointwise measurements.

Thus, the comparison that holds true is that the logarithm of the average density ratio is typically less than the local density ratio at any specific point in a compressible flow due to the smoothing nature of the averaging process over non-linear logarithmic transformations.

This reasoning leads us to option:

\(\ln\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{average}}<\left( \dfrac{\rho_2}{\rho_1} \right)_{\text{local}}\)