91Ó°ÊÓ

For an incompressible liquid, the density \(\rho\) is constant and independent of pressure. Since \(\mu\), \(c_p\), and \(k\) are also independent of pressure, both the kinematic viscosity \(v = \frac{\mu}{\rho}\) and thermal diffusivity \(\alpha = \frac{k}{\rho c_p}\) will be constant and independent of pressure for an incompressible liquid. For an ideal gas, the density \(\rho\) depends on the pressure and temperature according to the ideal gas equation: \(\rho = \frac{P}{RT}\), where \(P\) is the pressure, \(R\) is the specific gas constant, and \(T\) is the temperature. As the pressure increases, the density will increase proportionally. Therefore, for an ideal gas: 1. The kinematic viscosity v will be inversely proportional to the pressure: \(v \propto \frac{1}{P}\). 2. The thermal diffusivity \(\alpha\) will be proportional to the pressure: \(\alpha \propto P\).

Using the relationship for thermal diffusivity \(\alpha = \frac{k}{\rho c_p}\), we can calculate the values of \(\alpha\) for given pressure values. Assume that air behaves like an ideal gas at the given temperature and pressure conditions. At 350 K and 1 atm (101325 Pa), the properties of air are as follows: - Specific gas constant, \(R = 287 \, \mathrm{J/(kg \cdot K)}\) - Density, \(\rho = \frac{P}{RT} = \frac{101325}{287 \cdot 350} = 1.177 \, \mathrm{kg / m^3}\) - Thermal conductivity, \(k = 0.03 \, \mathrm{W/(m \cdot K)}\) - Specific heat, \(c_p = 1005 \, \mathrm{J/(kg \cdot K)}\) Now, we can find the thermal diffusivity of air at 1 atm: \[ \alpha_{1 \, \mathrm{atm}} = \frac{k}{\rho c_p} = \frac{0.03}{1.177 \cdot 1005} = 2.54\times10^{-5} \, \mathrm{m^2/s} \] Now, we can find the thermal diffusivity of air at 5 and 10 atm using the proportionality relationship between \(\alpha\) and \(P\) for ideal gases: \[ \alpha_{5 \, \mathrm{atm}} = 5\alpha_{1 \, \mathrm{atm}} = 1.27\times10^{-4} \, \mathrm{m^2/s} \] \[ \alpha_{10 \, \mathrm{atm}} = 10\alpha_{1 \, \mathrm{atm}} = 2.54\times10^{-4} \, \mathrm{m^2/s} \]

To find the distance of transition, \(x\), we will use the transition Reynolds number equation \(Re_x = \frac{\rho u_s x}{\mu}\). Using the given transition Reynolds number \(Re_x = 5\times10^5\), we can solve for the distance \(x\) at each pressure: As \(\rho\) and \(\mu\) are dependent on pressure, we can first find \(\mu\) for each condition using the relationship \(v = \frac{\mu}{\rho}\), then calculate the distance of transition: At 1 atm: - Dynamic viscosity, \(\mu = \rho v = 1.177 \cdot 1.516\times10^{-5} = 1.785\times10^{-5} \, \mathrm{Pa \cdot s}\) - \(x_{1 \, \mathrm{atm}} = \frac{\mu \, Re_x}{\rho u_s} = \frac{1.785\times10^{-5} \cdot 5\times10^5}{1.177 \cdot 2} = 3.78 \, \mathrm{m}\) At 5 atm: - Density is 5 times higher, \(\rho = 5 \cdot 1.177 = 5.885 \, \mathrm{kg / m^3}\) - Dynamic viscosity, \(\mu = \rho v = 5.885 \cdot 0.303\times10^{-5} = 1.785\times10^{-5} \, \mathrm{Pa \cdot s}\) (remains unchanged) - \(x_{5 \, \mathrm{atm}} = \frac{\mu \, Re_x}{\rho u_s} = \frac{1.785\times10^{-5} \cdot 5\times10^5}{5.885 \cdot 2} = 0.756 \, \mathrm{m}\) At 10 atm: - Density is 10 times higher, \(\rho = 10 \cdot 1.177 = 11.77 \, \mathrm{kg / m^3}\) - Dynamic viscosity, \(\mu = \rho v = 11.77 \cdot 0.1518\times10^{-5} = 1.785\times10^{-5} \, \mathrm{Pa \cdot s}\) (remains unchanged) - \(x_{10 \, \mathrm{atm}} = \frac{\mu \, Re_x}{\rho u_s} = \frac{1.785\times10^{-5} \cdot 5\times10^5}{11.77 \cdot 2} = 0.378 \, \mathrm{m}\)