Water is flowing at 3 m/s in a circular pipe. If the diameter of the pipe decreases to four-ninth its former value, what is the velocity of the water downstream?
Answer in units of m/s.
The law of continuity states that
Area1*flow1=Area2*flow2
To find the velocity of water downstream when the diameter of the pipe changes, we can use the principle of continuity. According to the principle of continuity, the product of the cross-sectional area of a pipe and the velocity of the fluid flowing through it remains constant, assuming no leakage or change in density.
The formula for the principle of continuity is:
A1V1 = A2V2
Where A1 and V1 are the original cross-sectional area and velocity, respectively, and A2 and V2 are the new cross-sectional area and velocity, respectively.
In this case, the diameter decreases to four-ninth its former value (which means the radius becomes two-thirds of its original value). Therefore, the new cross-sectional area (A2) is given by:
A2 = π(R2)^2
= π[(2/3)R1]^2
Since the relationship between radius and diameter is R = d/2, we can rewrite A2 as:
A2 = π[(2/3)(D1/2)]^2
= π[4/9(D1^2)]
= (4/9)π(D1^2)
Now, substituting the values into the principle of continuity equation:
A1V1 = A2V2
(V1)(πD1^2) = (4/9)(πD1^2)V2
V2 = (V1)(9/4)
Given that the velocity of water flowing through the circular pipe is 3 m/s (V1 = 3 m/s), we can calculate V2 as:
V2 = (3 m/s)(9/4)
= 27/4
= 6.75 m/s
Therefore, the velocity of the water downstream is 6.75 m/s.