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grin007 [14]
3 years ago
7

Fluid Dynamics: How do I find gauge pressure of an air current at various points around a cylinder?A freestream air current of v

elocity 15m/s flows through a wind tunnel. The current hits a cylinder of diameter 19mm, and has a laminar boundary layer. The stagnation pressure in the test section of the wind tunnel is 1 atm. Determine the minimum and maximum gauge pressures of the flow around the cylinder.I'm given the equation Cp=2(P-Po)/(?V^2), where Cp is the pressure coefficient, Po is the upstream static pressure. My problem is largely confusion over what to do with the diameter of the cylinder and where it comes into play here.
Engineering
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

The answer is as given in the explanation.

Explanation:

The 1st thing to notice is the assumptions required. Thus as the diameter of the cylinder and the wind tunnel are given such that the difference is of the orders of the magnitude thus the assumptions as given below are validated.

  1. Flow is entirely laminar, there's no boundary layer release.
  2. Flow is streamlined, ie, it follows the geometrical path imposed by the curvature.

By D'alembert's paradox, "The net pressure drag exerted on a circular cylinder that moves in an inviscid fluid of large extent is identically zero".Just in the surface of the cylinder, the velocity profile can be given in the next equation:

V=2Usin\theta

And the pressure P on the surface of cylinder is given by Bernoulli's equation along the streamline through that point:

P=P_{_{\infty }}+\frac{1}{2}\rho U^{2}(1-4sin^{2}\Theta ))

where P_∞ is  Pressure at stagnation point, U is the velocity given, ρ is the density of the fluid (in this case air) and θ is the angle measured from the center of cylinder to the adjacent point where your pressure point will be determine.

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A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.
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What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

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Ns = 12 × 120 / π2

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Now we find the metal removal rate using the equation

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= 1 in

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