A horizontal pipe contains water at a pressure of 120kPa flowing with a speed of 1.4m/s . The pipe narrows to one-half its original diameter.
a) what is the speed of the water?
b) what is the pressure of the water?
I tried using the continuity equation A1V1=A2V2,
then V2= (A1/A2)V1 and I know that A1= pi (d1/2)^2 but I can't figure out how to find the diameter and so that I can get the radius to solve the area.
Help please?
To find the diameter of the narrowed pipe, we can use the fact that the pipe narrows to one-half its original diameter. Let's assume the original diameter is d1. Then the diameter of the narrowed pipe will be half of d1, so it is d2 = d1/2.
Now, let's solve the problem step by step:
a) Speed of the water: Using the continuity equation A1V1 = A2V2, we can rewrite it as V2 = (A1/A2) * V1.
Since A1 = π * (d1/2)^2 and A2 = π * (d2/2)^2, we can substitute these values to get V2 = (π * (d1/2)^2) / (π * ((d1/2)/2)^2) * V1.
Simplifying this expression, we have V2 = (d1^2) / (d1^2 / 4) * V1.
The d1^2 terms cancel out, so V2 = (4 * V1).
Therefore, the speed of the water in the narrowed pipe is 4 times the original speed: V2 = 4 * 1.4 m/s = 5.6 m/s.
b) Pressure of the water: To find the pressure of the water, we can use Bernoulli's equation, which states that the total energy of a fluid system remains constant along a streamline.
Bernoulli's equation is given by P1 + 0.5 * ρ * V1^2 + ρ * g * h1 = P2 + 0.5 * ρ * V2^2 + ρ * g * h2.
In this equation, P1 and P2 are the pressures at points 1 and 2 respectively, ρ is the density of the fluid, V1 and V2 are the velocities at points 1 and 2 respectively, g is the acceleration due to gravity, and h1 and h2 are the heights at points 1 and 2 respectively.
Assuming the pipe is horizontal, the heights h1 and h2 are equal, and ρ * g * h1 can be ignored.
Plugging in the known values: P1 + 0.5 * ρ * V1^2 = P2 + 0.5 * ρ * V2^2.
Since the density ρ is the same for both situations, we can cancel it out.
Thus, P1 + 0.5 * V1^2 = P2 + 0.5 * V2^2.
Now, substituting the given values: 120 kPa + 0.5 * 1.4^2 = P2 + 0.5 * 5.6^2.
Calculating, we have 120 kPa + 1.96 = P2 + 15.68.
Rearranging the equation, we get P2 = 120 kPa + 1.96 - 15.68.
Simplifying, P2 = 106.28 kPa.
Therefore, the pressure of the water in the narrowed pipe is 106.28 kPa.