Water is flowing at 6 m/s in a circular pipe.
If the diameter of the pipe decreases to five-seventh its former value, what is the velocity
of the water downstream?
assuming the same flow:
V1A1=V2A2
V2=6*A1/A2=6(7/5)^2 so flow changes inversly as the square of diameter change.
11.76
To find the velocity of the water downstream when the diameter of the pipe decreases, we can use the principle of continuity. According to the principle of continuity, the mass flow rate of a fluid remains constant in a closed pipe system, assuming incompressible fluid and steady flow.
The equation for the principle of continuity is:
A1V1 = A2V2
Where:
A1 and V1 are the initial cross-sectional area and velocity of the water.
A2 and V2 are the final cross-sectional area and velocity of the water.
In this case, the diameter of the pipe decreases to five-sevenths its former value. We know that the area of a circle is given by:
A = πr^2
Where r is the radius of the circle.
Let's assume the initial diameter of the pipe is D and the final diameter is D/7*5.
The initial radius, r1, will be D/2, and the final radius, r2, will be (D/7*5)/2.
The initial cross-sectional area, A1, will be π(D/2)^2, and the final cross-sectional area, A2, will be π((D/7*5)/2)^2.
Now, we can substitute these values into the principle of continuity equation:
A1V1 = A2V2
π(D/2)^2 * V1 = π((D/7*5)/2)^2 * V2
Simplifying the equation, we get:
(D/2)^2 * V1 = ((D/7*5)/2)^2 * V2
(1/4)D^2 * V1 = (1/4)((D/7*5)^2) * V2
Simplifying further:
D^2 * V1 = (D/49)(25) * V2
Simplifying again:
49D^2 * V1 = 25D^2 * V2
Now, we can cancel out D^2 from both sides:
49V1 = 25V2
Finally, we can solve for V2:
V2 = (49/25) * V1
Given that the initial velocity, V1, is 6 m/s, we can substitute this value into the equation to find the velocity downstream, V2:
V2 = (49/25) * 6
V2 = (49/25) * 6
V2 = 11.76 m/s
Therefore, the velocity of the water downstream, when the diameter of the pipe decreases to five-sevenths its former value, is approximately 11.76 m/s.
To find the velocity of water downstream when the diameter of the pipe decreases, we can use the principle of conservation of mass. In a closed system like a pipe, the mass of water entering the pipe should be the same as the mass of water exiting the pipe.
The equation for conservation of mass is:
A1 * V1 = A2 * V2
Where:
A1 is the cross-sectional area of the pipe before the diameter change
V1 is the velocity of the water before the diameter change
A2 is the cross-sectional area of the pipe after the diameter change
V2 is the velocity of the water after the diameter change
Since the pipe is circular, the cross-sectional area of the pipe is given by:
A = π * r^2
Where:
π is a mathematical constant approximately equal to 3.14159
r is the radius of the pipe
Let's assume the initial diameter of the pipe is D1, and the final diameter is D2. The relationship between the diameter and radius is:
D = 2 * r
So, when the diameter decreases to five-sevenths (5/7) of its former value, we can write:
D2 = (5/7) * D1
From this, we can solve for r2:
r2 = (1/2) * D2 = (1/2) * (5/7) * D1 = (5/14) * D1
Now, we can calculate the cross-sectional areas A1 and A2:
A1 = π * r1^2 = π * (D1/2)^2 = π * (D1^2/4)
A2 = π * r2^2 = π * ((5/14) * D1)^2 = π * (25/196) * D1^2
As we know the velocity before the diameter change (V1) is 6 m/s, we can substitute these values into the conservation of mass equation to solve for V2:
A1 * V1 = A2 * V2
π * (D1^2/4) * 6 = π * (25/196) * D1^2 * V2
Canceling out the common terms and simplifying the equation:
(25/196) * D1^2 * V2 = (1/4) * D1^2 * 6
V2 = (1/4) * 6 * (196/25)
V2 = 6 * 196 / (4 * 25)
V2 = 1176 / 100
V2 = 11.76 m/s
Therefore, the velocity of the water downstream when the diameter of the pipe decreases to five-sevenths its former value is 11.76 m/s.