Please show work! Problem 13.43 Part A Water is flowing in a pipe with a varying
ID: 1792505 • Letter: P
Question
Please show work!
Problem 13.43 Part A Water is flowing in a pipe with a varying cross- sectional area, and at all points the water completely fills the pipe. At point 1, the cross- sectional area of the pipe is 7.30x10 m and the magnitude of the fluid velocity is 2.70 m/s What is the fluid speed at points in the pipe where the cross- sectional area is 0.107 m? 0 m/s Submit My Answers Give Up Part B What is the fluid speed at points in the pipe where the cross- sectional area is 4.00x10 m? A2 0 m/s Submit My Answers Give Up Provide Feedback ContinueExplanation / Answer
If we assume that the water is incompressible (a good approximation), then the volumetric flow rate (e.g., in cubic meters per second) is constant throughout the pipe.
The volumetric flow rate (or flux) is simply the volume of water passing through a cross sectional area of the pipe in a unit amount of time, and is given by the cross sectional area of the pipe multiplied by the linear speed of the flow (measured perpendicular to the cross section): F = v*A. This flow rate is constant throughout the pipe.
If you know the flow rate, speed and area at one point in the pipe (call these F, v1, and A1), then the speed at a point that has a different cross-sectional area is given by:
F = v1*A1 = v2*A2
v2 = v1*(A1/A2)
We know that v1 = 2.70 m/s and A1 = 7.30*10^-2 m^2, so at the point where the pipe has a cross-sectional area of 0.107 m^2, the speed is:
v2 = (2.70 m/s)*(7.30*10^-2)/(0.107) = 1.842 m/s
At the point where the pipe has a cross-sectional area of 4.00*10^-2 m^2, the speed is:
v2 = (2.70 m/s)*(7.30*10^-2)/(4.00*10^-2) = 4.927 m/s