Chapter 18

Q. 18.6

A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the car were traveling at 130 mph?

Given: Speed of airflow past hand; standard atmospheric conditions.
Find: Maximum pressure on hand.
Assume: Flow is steady and incompressible, with negligible viscous effects; air has constant, uniform density, equal to its tabulated value at 20°C (1.23 kg/m³). Standard atmospheric pressure p_{atm} = 101.325 kPa.


Verified Solution

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 130 mph). We can visualize the airflow as sketched below:

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(p+\frac{1}{2} \rho V^2\right)_{\text {upstream }}=\left(p+\frac{1}{2} \rho V^2\right)_{\mathrm{SP}}{}^{\prime}


p_{\mathrm{atm}}+\frac{1}{2} \rho V^2=p_{\max }

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

\begin{aligned} p_{\max } & =101,325 \mathrm{~Pa}+\frac{1}{2}\left(1.23 \mathrm{~kg} / \mathrm{m}^3\right)\left(65 \mathrm{~mph} \frac{0.447 \mathrm{~m} / \mathrm{s}}{1 \mathrm{~mph}}\right)^2=101.8 \mathrm{~kPa} \text { (abs) } \\ & =520 \mathrm{~Pa} \text { (gage) } \end{aligned}

If the car (and hence the air, in the frame of the hand) moves with speed V = 130 mph:

\begin{aligned} p_{\max } & =101,325 \mathrm{~Pa}+\frac{1}{2}\left(1.23 \mathrm{~kg} / \mathrm{m}^3\right)\left(130 \mathrm{~mph} \frac{0.447 \mathrm{~m} / \mathrm{s}}{1 \mathrm{~mph}}\right)^2=103.4 \mathrm{~kPa} \text { (abs) } \\ & =2.08 \mathrm{~kPa} \text { (gage) } \end{aligned}