An open-circuit wind tunnel draws in air from the atmosphere through a nozzle. I

Question

An open-circuit wind tunnel draws in air from the atmosphere through a nozzle. In the teat section. where the flow is straight and uniform, a static pressure tap is drilled into the wall. A manometer connected to the tap shows that the static pressure in the test section is 45 iron of water below atmospheric pressure. Assume that the outside air is incompressible, at 20degreeC, and at 100 kPa absolute. What is the air speed, in miles per hour, in the wind tunnel test section? Note the use of mmo f water as a pressure unit.

Explanation / Answer

Static pressure in test section, P2 = 45 mm of water below atmospheric.
Converting it in terms of absolute kPa pressure, we get, P2 = - [(45/1000)*1000*9.81]/1000 + 100 = 99.559 kPa.

Using Bernoulli's theorem, P1 + 1/2 V12 = P2 + 1/2 V22

Taking V1 = 0, and putting values, (100 - 99.559) * 103 = 1/2 * 1.189 * V22

This gives V2 = 27.24 m/s = 60.93 miles/hour.

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