Problem: 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 4.80x10-2 m2; at point 3, where the water is discharged, it is 1.60x10-2 m2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.What is the gauge pressure pgauge at point 2? 

FREE Expert Solution

Applying Bernoulli's principle to points 2 and 3:

P2+12ρv22+h2ρg=P0+12ρv32+h3ρg

We're looking for P2.

We can solve for P2 as follows:

P2-P0=12ρv32-12ρv22+ρgh3-ρgh2

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Problem Details

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 4.80x10-2 m2; at point 3, where the water is discharged, it is 1.60x10-2 m2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.

What is the gauge pressure pgauge at point 2? 

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Bernoulli's Equation concept. If you need more Bernoulli's Equation practice, you can also practice Bernoulli's Equation practice problems.

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Based on our data, we think this problem is relevant for Professor Stavola's class at LEHIGH.