How do I find pressure coefficient?

[gharrington44]

a. A very large reservoir of pressurized air feeds a contraction and forms a free jet of air of circular cross section and diameter 60 cm in a laboratory, where the ambient pressure is atmospheric. A Pitot static tube is arranged in the jet and attached to a vertical U-tube manometer containing alcohol. If the difference in levels recorded by the manometer is 65 mm, find the air speed.

I was able to find the air speed leading the contraction to be 28.8 m/s

b. The Pitot static tube is removed and replaced by a small body, which is equipped with static pressure holes in its surface. If the minimum gauge pressure recorded on the body is found to be -310 Pa, find the maximum velocity on the body.

I found the maximum velocity of the body to be 36.5 m/s.

c. State the values of pressure coefficient on the body (i) at the stagnation point, and (ii) at a point where the velocity achieves a value 50% above the value in the jet.

I have no idea how to approach this at all. I am assuming the stagnation point is in the center of the object where the air velocity is 0. But I am not familiar with a pressure coefficient or where to find where the velocity is half of what it should be.

 

[anathema]

Hello, thank you for your post. I can help you with the third part of your question.

The pressure coefficient is a dimensionless quantity that relates the local pressure at a specific point on a body to the free stream pressure. It is defined as Cp = (P-P0)/(1/2*rho*V^2), where P is the local pressure, P0 is the free stream pressure, rho is the density of the fluid, and V is the free stream velocity.

For part (i), at the stagnation point, the velocity is 0, so the pressure coefficient would be Cp = (P-P0)/(1/2*rho*0^2) = P/P0. This means that the pressure at the stagnation point is equal to the free stream pressure, and the pressure coefficient would be 1.

For part (ii), we need to find the point where the velocity is 50% higher than in the jet. We can use the Bernoulli's equation to relate the velocity at this point to the free stream velocity. The equation is P + 1/2*rho*V^2 = P0 + 1/2*rho*V0^2, where V0 is the free stream velocity. Solving for V, we get V = V0 * sqrt((P0-P)/P0).

Now, we can substitute this value of V into the pressure coefficient equation. Cp = (P-P0)/(1/2*rho*V0^2 * (P0-P)/P0) = 2*(P0-P)/P0. Since we know that the velocity at this point is 50% higher than in the jet, we can substitute V0 = 28.8 m/s * 1.5 = 43.2 m/s. Substituting this into the equation, we get Cp = 2*(P0-P)/P0 = 2*(P0-(-310 Pa))/(1/2*rho*43.2^2) = 0.015.

So, the pressure coefficient at a point where the velocity is 50% higher than in the jet is 0.015. I hope this helps. Let me know if you have any further questions.