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• 05-05-2013 12:02am
Registered Users, Registered Users 2 Posts: 1,213 ✭✭✭

I have my college exams next week, i've most of the questions learned but im stuck on the below formula. Can anyone help..the answer is shown I just don't know how he arrived at that answer?

I've attached the file....thanks

• Moderators, Science, Health & Environment Moderators Posts: 18,213 Mod ✭✭✭✭

This guy explains his answer to a similar question

When you sub in the density of mercury (13594kg/m^3) the answer comes out at 179m/s which is close enough to 180 for them to say 180.

I could, of course be completely wrong!

• Moderators, Arts Moderators Posts: 10,518 Mod ✭✭✭✭

You need to use Bernoulli's equation:

$P_{1} + \frac{1}{2}\rho_{1} V_{1}^2 + \rho_{1} g h_{1} = P_{2} + \frac{1}{2}\rho_{2} V_{2}^2 + \rho_{2} g h_{2}$

This equation relates the hydrostatic properties of a system at two different points. You need to look at the problem and identify points where you can remove some terms, where you know the state of the system is the same, or where you know enough terms to solve the equation.

In this case the velocity of the air results in pressure differential between the total (stagnation) and static pressure ports which causes the change in the height of the Mercury.

See #2 here for the solution. You can figure out the pressures at points A and B and equate the two since they are actually equal according to Bernoulli's equation.

You can bring the air velocity into the mix by, again, using Bernoulli's equation to write an expression for the pressure differential across the pitot static probe.
Combine your set of equations to solve for the velocity and get rid of the pressure terms. Now the velocity is given in terms of just the densities, the head and gravity.

Also, don't simply 'learn' the questions :mad:. The whole point is that you understand the concepts - there is nothing worse than a student who just regurgitates content blindly.