Question

Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.65 x 10° Pa and the pipe radius is 2.60 cm. At Pa and the pipe radius is 1.20 cm. the higher point located at y2.50 m, the pressure is 1.29 x 10 P (a) Find the speed of flow in the lower section 1.77 Your response differs from the correct answer by more than 10 %. Double check your calculations. m/s (b) Find the speed of flow in the upper section. 8.30 Your response differs from the correct answer by more than 10 %. Double check your calculations. m/s (c) Find the volume flow rate through the pipe. 3.76e-3 Your response differs from the correct answer by more than 10%. Double check your calculations, m /s

Answer 1

Solution

From equation of continuity

A₁V₁=A₂V₂..................................................(i)

A₁ is the area of the bottom section

V₁is the velocity of the bottom section

A₂ is the area of upper section

V₂ is the volume of upper section

Now, by bernoullis equation

+ |- 3 | + เพิ่ง

པ བརླ|ས P9pg

– 2P1 – 2P2 – 2pgh. – 03 – cum pg

putting the value of equation ......(i)

| 2(R-B- pgh.) P9 4S

_ 2(P1 – P2 – pgh:) _ A - A, pg

=> _ 2(P1 – P2 - pgh.) A - A} – A Pg

putting the given values from the question we get

= V1 = 0.11m/s

Therefore

*0.11m/s = 0.52m/s

Flow rate = A1V1 = 2.33e^-4 m³/s

If any doubt feel free to comment