Question 6.4: An aircraft is to fly at a height of 9 km (where the tempera...
An aircraft is to fly at a height of 9 km (where the temperature and pressure are -45^{\circ} \mathrm{C} and 30.2 \mathrm{kPa} , respectively) at 400 \mathrm{~m} / \mathrm{s} . \mathrm{A} 1 / 20^{\text {th }} scale model is tested in a pressurized wind-tunnel in which the air is at 15^{\circ} \mathrm{C} . For complete dynamic similarity what pressure and velocity should be used in the wind tunnel? \text { (For air, } \mu \propto T^{3 / 2} /(T+117), E_{s}=\gamma p, p= \rho R T where the temperature T is in kelvin, \gamma is the ratio of specific heats).
We find from
Eq. (6.24) that for complete dynamic similarity the Reynolds number, Re and Mach number, Ma for the model must be the same with those of the prototype. From the equality of Mach number Ma,
\frac{F}{\rho V^{2} l^{2}}=\phi\left(\frac{\rho V l}{\mu}, \frac{V}{a}\right) (6.24)
\frac{V_{m}}{a_{m}}=\frac{V_{p}}{a_{p}}Or V_{m}=V_{p} \frac{a_{m}}{a_{p}}
=V_{p} \sqrt{\frac{E_{s_{m}} / \rho_{m}}{E_{s_{p}} / \rho_{p}}}
(Subscripts m and p refer to the model and prototype, respectively.)
=V_{p} \sqrt{\frac{\gamma p_{m}}{\gamma p_{p}} \cdot \frac{\rho_{p}}{\rho_{m}}}
(Since a=\sqrt{E_{s} / \rho}, and E=\gamma p)
Again from the equation of state,
\frac{p_{m}}{\rho_{m}} \frac{\rho_{p}}{p_{p}}=\frac{T_{m}}{T_{p}}Hence, V_{m}=V_{p} \sqrt{\frac{T_{m}}{T_{p}}}
=400 m / s \sqrt{\frac{(273.15+15)}{(273.15-45)}}
= 450 m/s
From the equality of Reynolds number,
\frac{\rho_{m} V_{m} l_{m}}{\mu_{m}}=\frac{\rho_{p} V_{p} l_{p}}{\mu_{p}}Or \frac{\rho_{m}}{\rho_{p}}=\frac{V_{p}}{V_{m}} \cdot \frac{l_{p}}{l_{m}} \cdot \frac{\mu_{m}}{\mu_{p}}
Or \frac{p_{m}}{p_{p}}=\frac{V_{p}}{V_{m}} \cdot \frac{l_{p}}{l_{m}} \cdot \frac{T_{m}}{T_{p}}, \frac{\mu_{m}}{\mu_{p}}
=\frac{V_{p}}{V_{m}} \cdot \frac{l_{p}}{l_{m}}\left[\frac{T_{m}}{T_{p}}\right]^{5 / 2}\left[\frac{T_{p}+117}{T_{m}+117}\right]
[Since \mu \propto T^{3 / 2} /(T+117) ]
Therefore p_{m}=30.2 kPa \left(\frac{400}{450}\right)(20)\left(\frac{273.15+15}{273.15-45}\right)^{5 / 2}\left(\frac{273.15-45+117}{273.15+15+117}\right)
= 821 kPa