Water flows steadily from an open tank as shown in the figure. (Figure 1) The el

Question

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0 meters, and the elevation of points 2 and 3 is 2.00 meters. The cross-sectional area at point 2 is 0.0480 square meters; at point 3, where the water is discharged, it is 0.0160 square meters. 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 discharge rate dVdt.

Express your answer in cubic meters per second.

dV/dt = m^3/s

Explanation / Answer

Discharge at 2 =discharge at 3

From bernauli principle

Total Head at 1 = Total head at 3

Pressure head at1 + velocity head at 1 + datum head at 1 =Pressure head at3 + velocity head at 3+ datum head at 3

0( free to enviournment ) + 0( free to enviournment ) + 10 = 0( free to enviournment ) + V^2/2g + 2

V=(2g* (10-8))^0.5

V=12.52 m/s (velocity at 3)

discharge = area at 3 (A) * velocity (V)

= 0.016* 12.52

= 0.2004 m^3 / second

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.