Question

A viscous fluid is flowing through two horizontal pipes. They have the same length, although the radius of one pipe is three times as large as the other. The pressure difference P2 - P1 between the ends of each pipe is the same. If the volume flow rate in the narrower pipe is QA = 2.94 × 102 m3/s, what is the volume flow rate QB in the wider pipe?

Answer 1

The volume flow rate in the wider pipe is 2646 m^3/s.

Step-by-step explanation:

In fluid dynamics, the volume flow rate, Q, is given by Q = Av, where A is the cross-sectional area of the pipe and v is the velocity of the fluid. Based on the principle of continuity, the flow rate must be the same at all points along the pipe. Therefore, if the volume flow rate in the narrower pipe is QA, and the radius of the wider pipe is three times larger than the radius of the narrower pipe, then we can calculate the volume flow rate in the wider pipe, QB.

Since the cross-sectional area is directly proportional to the square of the radius, we have A1/A2 = (r1^2)/(r2^2), where r1 and r2 are the radii of the narrower and wider pipes, respectively. Given that r2 = 3r1, we substitute this into the equation to get A1/A2 = (r1^2)/(9r1^2) = 1/9. Since QA = Q1 = A1v1, we can write QA = Q1 = (1/9)QB, where QB is the volume flow rate in the wider pipe.

Therefore, QB = 9QA = 9 * 2.94 × 10^2 m^3/s = 2646 m^3/s.

Answer 2

The conservation of the mass of fluid through two sections (be they A1 and A2) of a conduit (pipe) or current tube establishes that the mass that enters is equal to the mass that exits. Mathematically the input flow must be the same as the output flow,

`Q_1 = Q_2`

The definition of flow is given by

`Q =VA`

Where

V = Velocity

A = Area

The units of the flow of flow are cubic meters per second, that is to say that if there is a continuity, the volume of input must be the same as that of output, what changes if the sections are modified are the proportions of speed.

In this way

`Q_A = Q_B`

`Q_B = 2.94*10^2 m^3/s`