Water flows steadily from an open tank as shown in the figure. ( Figure 1 ) The
ID: 2254930 • Letter: W
Question
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 2.00m . The cross-sectional area at point 2 is 4.8010?2m2 ; at point 3, where the water is discharged, it is 1.6010?2m2 . The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.
A) Assuming that Bernoulli's equation applies, compute the volume of water
?V that flows across the exit of the pipe in 1.00s . In other words, find the discharge rate ?V/?t.
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 2.00m . The cross-sectional area at point 2 is 4.80 times 10?2m2 ; at point 3, where the water is discharged, it is 1.60 times 10?2m2 . The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe Assuming that Bernoulli's equation applies, compute the volume of water ?V that flows across the exit of the pipe in 1.00s . In other words, find the discharge rate ?V/?t.Explanation / Answer
v2 = 12.52 m/s
A1*v1 = A2*v2
V1 = (A2/A1)*V2
= (1.6/4.8)*12.52
= 4.173 m/s
P + 0.5*rho*v1^2 = po + 0.5*rho*v2^2
P - P0 = 0.5*rho*(v2^2-v1^2)
Pgauge = 0.5*1000*(12.52^2 - 4.173^2)
Pgauge = 69668 pa
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