Question 10.6: A person holds her hand out of an open car window while the ...

We put ourselves in the frame of the person’s hand so that the hand is still and the air moves with speed 65 mph (or 220 mph). We can visualize the airflow as sketched in Figure 10.17.

Note that there is a dividing streamline that impinges on the hand at a stagnation point. (Airflow either goes above this streamline, up and over the hand, or below it.) At this stagnation point, the air will be at its maximum pressure, the stagnation pressure. (Recall that pressure and velocity are inversely proportional.) If we assume that this stagnation streamline is level, so that gravitational effects are easily neglected, we can apply the Bernoulli equation on this streamline to find the stagnation pressure. The Bernoulli equation has the form

\left\lgroup P+\frac{1}{2}ρ  V^2\right\rgroup _{upstream} = \left\lgroup P+\frac{1}{2}ρ  V^2\right\rgroup _{SP} ,

or

P_{atm}+\frac{1}{2}ρ  V^2=P_{\max} .

Plugging in the atmospheric pressure, air density, and V = 65 mph, we have

P_{\max} =101,325  \textrm{ Pa}+\frac{1}{2}(1.23 \frac{\textrm{kg}}{\textrm{m}^3}) \left\lgroup 65 \textrm{ mph}\frac{0.447  \frac{\textrm{m}}{s}}{1 \textrm{ mph}} \right\rgroup ^2 = 101.844 kPa (abs)

= 520 Pa (gage).

If the car (and hence the air, in the frame of the hand) moves with speed V =220 mph,

P_{\max} =101,325  \textrm{ Pa}+\frac{1}{2}(1.23  \frac{\textrm{kg}}{\textrm{m}^3}) \left\lgroup 220 \textrm{ mph}\frac{0.447  \frac{\textrm{m}}{s}}{1 \textrm{ mph}} \right\rgroup ^2 = 107.28 kPa (abs)

= 5.95 kPa (gage).