Question 6 is the answer I need
1 answer:
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Answer:
discharge rate (Q) = 0.2005 m^{3} / s
Explanation:
if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:
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 1.60x10-2m. 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.
solution:
time = 1 s
elevation of point 1 (z1) = 10 m
elevation of point 2 (z2) = 2 m
elevation of point 3 (z3) = 2 m
cross section area of point 2 = 4.8 x 10^{2} m
cross section area of point 3 = 1.6 x 10^{2} m
g
acceleration due to gravity (g) = 9.8 m/s^{2}
find the discharge rate at point 3 which is the exit pipe.
discharge rate (Q) = A3 x V3
where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below
pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes
V3 =
V3 =
V3 = 12.53 m/s
discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53
discharge rate (Q) = 0.2005 m^{3} / s