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 find the pressure at the narrow end of the segment, you can use the Bernoulli equation along with the fact that the velocity at the exit is 100 times the velocity at the entrance. Here's a step-by-step explanation:
1. Convert the given cross-sectional areas from cm^2 to m^2. The larger end has an area of 50 cm^2, which is equal to 0.005 m^2. The narrow end has an area of 0.5 cm^2, which is equal to 0.00005 m^2.
2. Convert the given pressure from Pa to N/m^2. The pressure at the larger end is 1.26 * 10^5 Pa.
3. Use the equation of continuity to relate the velocities at the entrance and the exit of the pipe. According to the continuity equation, the product of the cross-sectional area and velocity remains constant along a streamline. Therefore, we have:
A_in * V_in = A_out * V_out
Substituting the given values, we get:
0.005 m^2 * V_in = 0.00005 m^2 * (100 * V_in)
Simplifying the equation, we find:
V_out = 100 * V_in
This means that the velocity at the exit is 100 times the velocity at the entrance.
4. Now, we can use the Bernoulli equation to find the pressure at the narrow end. The Bernoulli equation states that the total energy per unit volume of a fluid remains constant along a streamline. Mathematically, it can be written as:
P + (1/2) * ρ * V^2 + ρ * g * h = constant
Where P is the pressure, ρ is the density of the fluid, V is the velocity of the fluid, g is the acceleration due to gravity, and h is the height of the fluid.
Since we are dealing with a horizontal pipe, the height difference (h) is zero. Therefore, the equation simplifies to:
P + (1/2) * ρ * V^2 = constant
Using the subscript "1" for the larger end and "2" for the narrow end, we can rewrite the Bernoulli equation as:
P_1 + (1/2) * ρ * V_1^2 = P_2 + (1/2) * ρ * V_2^2
Substituting the known values, we get:
1.26 * 10^5 N/m^2 + (1/2) * ρ * (0.029 m/s)^2 = P_2 + (1/2) * ρ * (100 * 0.029 m/s)^2
Expanding and simplifying, the equation becomes:
1.26 * 10^5 N/m^2 + 0.5 * ρ * 0.000841 m^2/s^2 = P_2 + 0.5 * ρ * 841 m^2/s^2
Since the density (ρ) of the fluid is not given, we can cancel it out from both sides of the equation. This gives:
1.26 * 10^5 N/m^2 + 0.5 * 0.000841 m^2/s^2 = P_2 + 0.5 * 841 m^2/s^2
0.63 * 10^5 N/m^2 = P_2 + 0.5 * 841 m^2/s^2
Subtracting 0.5 * 841 m^2/s^2 from both sides, we get:
0.63 * 10^5 N/m^2 - 0.5 * 841 m^2/s^2 = P_2
Finally, calculate the pressure at the narrow end:
P_2 ≈ 0.63 * 10^5 N/m^2 - 0.5 * 841 m^2/s^2
Simplifying the expression, we find:
P_2 ≈ 0.63 * 10^5 N/m^2 - 0.5 * 841 N/m^2
P_2 ≈ 0.63 * 10^5 N/m^2 - 420.5 N/m^2
P_2 ≈ 0.63 * 10^5 N/m^2 - 0.004205 * 10^5 N/m^2
P_2 ≈ 0.625795 * 10^5 N/m^2
Therefore, the pressure at the narrow end of the segment is approximately 0.625795 * 10^5 Pa, or 6.26 * 10^4 Pa.