The maximum velocity of airflow in a pipe is dependent on several factors such as the diameter of the pipe, the pressure of the air, and the length of the pipe. In this case, we have a 2.5 inch diameter pipe and a stoic pressure of 14.7 psi. To calculate the maximum airflow, we can use the equation Q = (π/4) x D^2 x V, where Q is the volumetric flow rate, D is the pipe diameter, and V is the velocity of airflow.
Plugging in the values, we get Q = (π/4) x (2.5 inches)^2 x V. Converting the diameter to meters (2.5 inches = 0.0635 meters), we get Q = (π/4) x (0.0635 meters)^2 x V. Rearranging the equation to solve for V, we get V = (4 x Q)/π x D^2. Plugging in the given stoic pressure of 14.7 psi, we get V = (4 x 14.7 psi)/π x (0.0635 meters)^2. This gives us a maximum velocity of approximately 8.99 meters per second.
For the next part of the question, we need to determine the time it would take to compress the air inside the pipe to 20.7 psi. To do this, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Assuming the temperature remains constant, we can rearrange the equation to solve for time (t = (P2-P1) x V/(nR)), where P2 is the final pressure (20.7 psi) and P1 is the initial pressure (14.7 psi).
Plugging in the values, we get t = (20.7 psi - 14.7 psi) x (6 ft x 0.0635 meters/ft) / (n x 8.314 J/mol·K). Assuming the air inside the pipe is at standard temperature and pressure, we can use n = 1 mol. This gives us a time of approximately 0.084 seconds to compress the air inside the pipe to 20.7 psi.
Finally, to calculate the new airflow at the increased pressure,
