This example shows how the Bernoulli equation is applied in a Pitot tube, which is being
used to measure the velocity of a boat, can also be used to measure the velocity of a
plane, but in this example it's for a boat as it's moving through water. Picture this
tube here, it's kind of like a manometer tube in a way, and it's attached to a boat, and
the boat is moving at a velocity of u through the water. So if we consider the captain of
the boats perspective then the tube is actually stationary, and the water is moving at a velocity
of u, so the water is all moving at a velocity of u right here. Now what happens is that
first moves at a velocity of u through this tube until it gets sort of stuck at this point,
and it turns up, and then its pushing a column of water up into this tube, which at a certain
point the pressure head due to this additional height above the ambient pressure, so this
height is a height of h, and this point right here we're going to call point 2, and this
is point 1, the pressure at point 2 is ambient, as is the pressure right above the surface
of the water, but this pressure head right here of rho*g*h is counterbalancing the speed
of the boat, the faster it moves the higher this moves up here, and so this point right
here is called a stagnation point because obviously for the fluid inside this tube to
not be moving the velocity really must be zero right there, but right on the other side
of it the velocity is u, because that's how fast the water is hitting this now stationary
column of water. This tube is at a depth of d below the surface, and so the pressure here
at point 1 is equal to P(ambient), plus rho*g*d, and the pressure here at point 2 is P(ambient),
because both point 2 and the surface of the water are exposed to the ambient air. So if
we apply the Bernoulli equation to this fluid then u(1)^2/2 + P(1)/rho + gz(1) is equal
to u(2)^2/2 + P(2)/rho + gz(2). So we'll call point 1 at height of 0, which makes point
2 at a height of d + h. P(1) is equal to P(2) plus rho*g*d, because of what we said earlier,
P(2) is ambient, and P(1) is ambient, plus rho*g*d. u(2) is zero because it's not moving
at this point, and while it's not moving just to the right of 0.1 the fluid is trying to
move at a velocity of u, just to the left of point 1, and so u(1) is really the velocity
of the boat, and so that's u. We can simplify this equation a little bit by saying u(1)^2/2
+ P(1)/rho is equal to g*(d+h), plugging in for z(2), plus P(2)/rho, and because of the
fact that we explained above that P(1) is equal to P(2), plus rho*g*d we can plug in
right here for P(1) and rewrite this equation on the left hand side as u(1)^2/2 plus, I
am plugging in for P(1) right here, P(2)/rho + g*d. Now, taking this side of the equation
and setting it equal to this side of the equation I see that I get a g*d on each side, u(1)^2/2,
plus g*d. I have a P(2)/rho on each side, those cancel out, as do the g*d's, I can solve
for u(1) as u(1). I am going to take the square root of both sides, u(1) is equal to the square
root of 2*g*h. The column of fluid in the Pitot tube above the surface of the water
translates to the velocity of the boat. The kinetic energy of the water right before it
enters the Pitot tube is converting to pressure energy at the top surface of the fluid in
the Pitot tube. So we can measure a height or, obviously that wouldn't work very well
for error on an airplane you could use a pressure transducer to measure the pressure that is
required to counteract the velocity at which the Pitot tube is moving through the fluid,
or visa-versa, the fluid is moving over the Pitot tube.