A pitot tube is mounted in a main pipe of diameter 40 cm. If the difference in pressure indicated by the gauge ie 6cm of water column, calculate the volume of air passing through the main pipe in two minutes.

Explanation

Given, `r=(40 cm)/(2) =20 cm =0.2 m`
`h=6cm =0.06m`
The given situation is shown below:

At A, the plane of aperture is parallel to the direction of air flow, so velocity of air flow at A in the pipe (say v). At B, the stane of aperture is perpendicular to the direction of air flow. So, velocity of air entering the tube is reduced to zero.
Applying Bernoulli.s theorem at points A and B, we get `p_(A)+1/2 pv_(A)^(2)=p_(B)+1/2 pv_(B)^(2)`
Here,` v_(A)=v and v_(B)=0`
`1/2 pv^(2)=p_(B)-p_(A)-hpg`
(U-tube contains water)
`rArr v=sqrt(2gh) =sqrt(2 xx 9.8 xx 0.06) [therefore h=0.06m]`
=1.08m/s
Volume of air flowing per second,
`Q=av=pir^(2) v`
`=3.14 xx (0.2)^(2) xx 0.832`
`=0.1045m^(3)//s`
Volume of air passing in 2 mins is
`v=Qt=0.1045 xx 2 xx 60`
`=12.54 m^(3)`