Question

At a certain point in a horizontal pipeline, the water's speed is 3.50 m/s and the gauge pressure is 2.30×104 Pa .

Part A

Find the gauge pressure at a second point in the line if the cross-sectional area at the second point is twice that at the first.

Please shwo all your work.

Answer 1

his question is about Bernoulli's principle - the total energy of the water in the pipe remains unchanged, as an application of the principle of conservation of energy. It is assumed that the flow is non-turbulent and non-viscous so that the only forms of energy are gravitational potential energy, pressure energy, and kinetic energy, and the sum of these is a constant at all points on the pipe. Since the pipe is horizontal the heights of the 2 points mentioned are the same, so the change in gravitational PE between the points is zero; only KE and pressure energy change.

Consider 1L of water at the first point. Its mass is 1kg and its KE is 0.5*3.5^2 = 6.125J. Its pressure energy is pressure*volume = 2.30×10^4*10^-3 = 23J. The total energy is 29.125J.

At the 2nd point, the cross sectional area is twice that at the 1st point, so the speed (V) of the water halved; its KE per litre is one quarter, namely 6.125/4 = 1.531J. The pressure energy is 23 - 1.531 = 21.469J and the pressure (P) is therefore 21.469*10^3Pa

You asked for an explanation, so I have broken the reasoning down into simple steps. The solution could be reached through solving this equation, which is Bernoulli's equation in terms of energy per unit volume with the gravitational terms omitted:

P1 + 0.5*V1^2*d = P2 + 0.5*V2^2*d

where d is the density of water and subscripts 1 & 2 refer to the 2 locations.