Question

If the pressure reading of your pitot tube is xv.0 mm Hg at a speed of 200 km/h, what volition it exist at 700 km/h at the same altitude?

Solution Video

OpenStax College Physics Solution, Chapter 12, Problem 19 (Bug & Exercises) (3:14)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The way this question is worded where it says the pressure reading of U-Pattou Tube is blah blah blah is a scrap confusing considering pattou tubes don't measure pressure, instead they measure speed of the air going past these tubes here. And they measure that speed using pressure differences that arise in these 2 sides of this pipe here but this plenty itself is non a measurement of the air pressure level, it's only an indication of the air speed. So, what this question should exist proverb is what will the pressure be exterior given a pattou tube telling you lot that the air speed is 700 kilometers per hour versus 200 kilometers per hour? Then this is 15 millimeter of Mercury pressure reading did not come up from the pattou tube, it's the 200 kilometers per hour that's what came from the pattou tube. So hither'southward Bernoulli's principle and we're at constant altitude of course. So the plus rho gh terms disappeared from both sides because there is no difference in height. So we take the pressure in the start example which is 15 millimeters in Mercury plus one half times the density of air times initial speed of 200 Kilometers in an hour equals the pressure in the 2d case which we want to find plus one half density of air times the speed in the second example squared. So we'll solve this for P2 by subtracting this term from both sides and so factor out the one half rho, common factor betwixt these two terms. And nosotros end up with this line. Then the pressure is going to be the initial pressure level plus i one-half times density times the difference in the squares of the velocities, so that'south 15 millimeters of Mercury times 133 pascals per millimeter of Mercury plus i one-half times 1.29 Kilograms per cubic meter density of air times 200 kilometers per hours squared minus 700 kilometers per 60 minutes squared. And what I've washed here is a bit unusual. Unremarkably, I would convert this kilometers per hour hither to meters per second, but I just felt similar doing things a little differently. And then, I'm taking the squares of the kilometer per hours and we take to multiply by the conversion cistron here. And so that's one hour for 2,600 seconds times 1000 meters for every kilometer and we square that because we're converting this difference which volition accept units of kilometers squared per hours squared. If you didn't like this matter or this method, then just convert the 200 kilometers per hr into meters per 2d before substituting it and likewise with the 700. And so we end with this milli pascals and we multiply information technology with one millimeter of Mercury for every 133 pascals and become negative 153 millimeters of Mercury will exist the pressure when the air speed is 700 kilometers per hour.

Solutions for problems in affiliate 12