- #31
jack action
Science Advisor
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Dan Thomas said:
The Bernoulli equation (adapted for this problem) states the following:
$$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$
This means that along a streamline:
{The static pressure + the dynamic pressure} at state 1 equals {The static pressure + the dynamic pressure} at state 2.
The static pressure is the atmospheric pressure when there is no wind.
For example, in a wind tunnel when the air doesn't move the static pressure ##p_1## is, say, 15 psi. Obviously, the dynamic pressure is 0 since the wind speed is zero. This is state 1. Total pressure is thus 15 + 0 = 15 psi.
Then somebody starts the fan and the wind speed increases to 100 mph. This is state 2. The dynamic pressure is now at 0.17 psi, as calculated in previous posts. This means that the new static pressure is now 15 psi - 0.17 psi = 14.83 psi, such that:
$$\{15\ psi + 0\ psi\}_1 = \{14.83\ psi + 0.17\ psi\}_2$$
That is in a closed wind tunnel. In the atmosphere that is so vast, a gust of wind will not be in a solid tunnel but surrounded by stagnant air. This would be like if our wind tunnel walls were made of stretchy rubber, like a ballon. This means that the walls would be subjected to the outside pressure and thus, the static pressure will always remain at 15 psi because the wall will move and compress the air in the tunnel until the inside pressure equals the outside pressure. In such a case, with a wind speed of 100 mph, the total pressure will be 15 psi + 0.17 psi = 15.17 psi.
If the atmospheric pressure would be 30 psi, it wouldn't change anything to the dynamic pressure, which would remain at 0.17 psi. Only the total pressure would change.