A pipe of radius 6mm is connected to another pipe of radius 9mm. If water flows in the wider pipe at a speed of 2m/s,what is the speed in the narrow pipe?
The gallons per minute is the same for both
so V1 A1 = V2 A2
or
V2 = V1 * (A1/A2)
= 2 (9^2 / 6^2) = 2 * 81/36
To determine the speed of water in the narrow pipe, we can use the principle of continuity, which states that the volume flow rate is constant at different points in a fluid system.
The volume flow rate (Q) is given by the equation Q = A × v, where A is the cross-sectional area of the pipe and v is the velocity of the fluid.
Let's denote the cross-sectional area of the wider pipe as A1 and the cross-sectional area of the narrow pipe as A2. Since the pipes are connected, the volume flow rates should be the same in both pipes.
To find the speed in the narrow pipe, we can rearrange the equation as v2 = Q / A2, where v2 is the velocity in the narrow pipe.
Now, we need to determine the relationship between the cross-sectional areas of the two pipes. The cross-sectional area of a pipe is proportional to the square of its radius.
Therefore, we can write A1 / A2 = (r1)^2 / (r2)^2, where r1 and r2 are the radii of the wider and narrow pipes, respectively.
Given that the radius of the wider pipe (r1) is 9 mm and the radius of the narrow pipe (r2) is 6 mm, we can substitute these values into the equation:
A1 / A2 = (9 mm)^2 / (6 mm)^2
= 81 / 36
= 9 / 4.
Now, since the volume flow rate is constant, we know that A1 × v1 = A2 × v2. So, we can write:
A1 × v1 = A2 × v2,
9/4 × v1 = 1 × v2,
v2 = (9/4) × v1.
Given that the speed in the wider pipe (v1) is 2 m/s, we can substitute this value into the equation:
v2 = (9/4) × 2
= 9/2
= 4.5 m/s.
Therefore, the speed in the narrow pipe is 4.5 m/s.