Water flows steadily from an open tank as shown in the figure.
(Figure 1) The elevation...
Water flows steadily from an open tank as shown in the figure.
(Figure 1) The elevation of point 1 is 10.0m , and the elevation of
points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is
4.80x10^-2m ; at point 3, where the water is discharged, it is
1.60?10^?2m. The cross-sectional area of the tank is very large
compared with the cross-sectional area of the pipe. Part A Assuming
that Bernoulli's equation applies, compute the volume of water
DeltaV that flows across the exit of the pipe in 1.00 s . In other
words, find the discharge rate \Delta V/Delta t. Express your
answer numerically in cubic meters per second.
Solutions
Expert Solution
Concepts and reason
The concepts required to solve the given questions is the speed of efflux and the continuity equation.
Initially, calculate the discharge area. Later, use the continuity equation. Finally, by using the continuity equation calculate the gauge pressure at point 2.
Fundamentals
The expression for the speed of efflux is as follows:
v=2gh
Here, g is the acceleration due to gravity and h is the height.
The height of the efflux is as follows:
h=y1−y2
Here, y1 is the elevation of the point 1 and y2 is the elevation at point 2.
Therefore, the speed of efflux is as follows:
v=2g(y1−y2)
The expression for the discharge charge is as follows:
Q3=v3A3
Here, v3 is the speed of efflux and A3 is the area at point 3, where the water is discharged.
The expression for the equation of continuity is as follows:
A2v2=A3v3
Here, A2 is the area at point 2 and v2 is the speed of efflux.
The expression to calculate the gauge pressure at point 2 is as follows:
P2=21ρ(v32−v22)
Here, ρ is the volume charge density.
Substitute 9.8m/s2 for g, 10.0 m for y1 , and 2.00 m for y2 in the equation v=2g(y1−y2) .
v=2(9.8m/s2)(10.0m−2.00m)=12.52m/s
Now, calculate the discharge charge using the equation Q3=v3A3 .
Substitute 12.52 m/s for v3 and 0.0160m2 for A3 in the equation Q3=v3A3 .
Q3=(12.52m/s)(0.0160m2)=0.20m3/s
Rearrange the equation P2=21ρ(v32−v22) as follows:
Now, substitute 1000kg/m3 for ρ , 9.8m/s2 for g, 10.0 m for y1 , 2.00 m for y3 , 0.0160m2 for A3 , and 0.0480m2 for A2 in the equation P2=ρ(g(y1−y3))(1−(A2A3)2) .
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Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 4.80 × 10-2m2; at point 3, where the water is discharged, it is 1.60 × 10-2m2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.Part BWhat is the gauge pressure Pgauge at point...
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