Water flowing out of a horizontal pipe emerges through a nozzle. The radius of the pipe...

Question

Question

Water flowing out of a horizontal pipe emerges through a nozzle. The radius of the pipe is 1.7 cm, and the radius of the nozz

Answer 1

Answer #1

By equation of continuity we can find the speed of flow through nozzle,

Ap * Vp = An * Vn

where Ap and An are the area of pipe and nozzle respectively. and Vp and Vn are the velocity of water in the pipe and nozzle respectively.

using known values,

Vn = (Rp²/Rn²)*Vp

Vn = 9.39 m/s

Now we can use equation of continuity to find the pressure,

Pp + (1/2)*d*Vp² = Pn + (1/2)*d*Vn²

At exit through nozzle pressure will be same as atmospheric pressure,

Pp = (1.01 * 10^5) + (0.5 * 1000 * 9.39²) - (0.5 * 1000 * 0.78²)

Pp = (1.01 * 10^5) + (0.441 * 10^5) + (0.003 * 10^5)

Pp = 1.454 * 10^5 Pascal.

Thus the pressure in the pipe is 1.454 * 10^5 Pa.

Answer 2

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