assuming the same flow:
V1A1=V2A2
V2=6*A1/A2=6(7/5)^2 so flow changes inversly as the square of diameter change.
If the diameter of the pipe decreases to five-seventh its former value, what is the velocity
of the water downstream?
V1A1=V2A2
V2=6*A1/A2=6(7/5)^2 so flow changes inversly as the square of diameter change.
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.
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:
Ï€ 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.