A horizontal segment of pipe tapers from a cross sectional area of 50 cm2 to 0.5 cm2. The pressure at the larger end of the pipe is 1.26 105 Pa and the speed is 0.029 m/s. What is the pressure at the narrow end of the segment? Pa
Please help!
physics - drwls, Monday, April 9, 2012 at 8:09pm
Due to the flow contraction, the velocity at the exit (Vout) will be 100 times the velocity at the entrance (Vin). That results from the incompresible-flow continuity equation.
The pressure at the entrance, combined with the Bernoulli Equation, Vin and Vout, will tell you the pressure at the exit.
You should use a ^ before exponents, when typing equations and scientific-notation numbers. I am sure you mean 10^5 and not 105.
Can someone or drwls please give a more detailed explaination please. I don't understand it still i keep getting it wrong.
To solve this problem, we can use the principles of fluid mechanics, specifically the continuity equation and Bernoulli's equation.
First, let's understand the continuity equation. It states that the mass flow rate of an incompressible fluid remains constant in a closed system. Mathematically, it can be expressed as:
A1 * V1 = A2 * V2
Where A1 and A2 are the cross-sectional areas of the pipe at points 1 and 2, and V1 and V2 are the velocities of the fluid at those points.
In this problem, A1 = 50 cm^2 and A2 = 0.5 cm^2. We are also given that V1 = 0.029 m/s. We need to find V2.
Using the continuity equation, we can solve for V2:
50 cm^2 * 0.029 m/s = 0.5 cm^2 * V2
Simplifying the equation:
V2 = (50 cm^2 * 0.029 m/s) / 0.5 cm^2
Now, let's apply Bernoulli's equation. It states that the total mechanical energy of a fluid (in this case, the sum of pressure energy, kinetic energy, and potential energy) remains constant along a streamline of steady flow.
Bernoulli's equation can be expressed as:
P1 + 1/2 * rho * V1^2 + rho * g * h1 = P2 + 1/2 * rho * V2^2 + rho * g * h2
Where P1 and P2 are the pressures, V1 and V2 are the velocities, rho is the density of the fluid, g is the acceleration due to gravity, h1 and h2 are the heights of the fluid above a reference point at positions 1 and 2.
In this problem, we are given P1 = 1.26 * 10^5 Pa, V1 = 0.029 m/s, and we need to find P2.
Since the problem is horizontal, we can neglect the height terms. Also, since the fluid is incompressible, rho can be considered constant.
Simplifying the Bernoulli's equation:
P1 + 1/2 * rho * V1^2 = P2 + 1/2 * rho * V2^2
Since we have the values of P1, V1, and V2, we can solve for P2:
P2 = P1 + 1/2 * rho * V1^2 - 1/2 * rho * V2^2
Plug in the values and calculate P2.