A water pipe tapers down from an initial radius of R 1 = 0.22 m to a final radiu

Question

A water pipe tapers down from an initial radius of R1 = 0.22 m to a final radius of R2 = 0.1 m. The water flows at a velocity v1 = 0.87 m/s in the larger section of pipe. What is the volume flow rate of the water? What is the velocity of the water in the smaller section? Using this water supply, how long would it take to fill up a swimming pool with a volume of V = 157 m3? (give your answer in minutes) The water pressure in the center of the larger section of the pipe is P1 = 263380 Pa. Assume the density of water is 103 kg/m3. What is the pressure in the center of the smaller section of the pipe?

Explanation / Answer

R1 = 0.22 m

R2 = 0.1 m

v1 = 0.87 m/s

volume flow rate = Q = A1v1 = R12v1 = 3.14*0.22*0.87 = 0.601 m^3/s

By equation of continuity,

Q = A1v1= A2v2

v2 = v1(A1/A2) = v1(R12/ R22) = 0.87*(0.22^2/0.1^2) = 4.21 m/s

volume flow rate = Q = V/t

t= V/Q = 157/0.601 = 261.23s = 4.35 min

By Burnolli’s principle,

P2+KE2+PE2= P1+KE1+PE1

P2+1/2v2^2+gh2 = P1+1/2v1^2+gh1

Since pope is horizontal h1=h2

P2+1/2v2^2 = P1+1/2v1^2

P2+1/2v2^2 = P1+1/2v1^2

P2 +1/2*103*4.21^2 = 263380 +1/2*103*0.87^2   => P2= 162506 Pa

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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