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 105 Pa and the pipe radius is 2.70 cm. At the higher point located at y = 2.50 m, the pressure is 1.27 x 105 Pa and the pipe radius is 1.30 cm. P (a) Find the speed of flow in the lower section. m/s (b) Find the speed of flow in the upper section. m/s (c) Find the volume flow rate through the pipe. m3/s

Answer 1

Part A.

Using continuity equation at both ends

Q1 = Q2

A1*V1 = A2*V2

A = Area of pipe = pi*r^2

V2 = V1*(A1/A2) = V1*(r1/r2)^2

Since given that r1 = 2.70 cm and r2 =1.30 cm

V2 = V1*(2.70/1.30)^2 = 4.3136*V1

Now using bernoulli's theorem

P1 + 0.5rho*V1^2 + rho*g*h1 = P2 + 0.5*rho*V2^2 + rho*g*h2

0.5*rho*(V2^2 - V1^2) = P1 - P2 + rho*g*(h1 - h2)

V2^2 - V1^2 = [2*(P1 - P2)/rho] + 2*g*(h1 - h2)

Given that:

P1 = pressure at lower section = 1.65*10^5 Pa

P2 = pressure at higher section = 1.27*10^5 Pa

rho = density of fluid = 1000 kg/m^3

h1 - h2 = -2.50 m (See that h2 is at higher point)

So, Using above values:

(4.3136*V1)^2 - V1^2 = [2*(1.65 - 1.27)*10^5/1000] + 2*9.81*(-2.50) = 26.95

V1^2*(4.3136^2 - 1) = 26.95

17.607*V1^2 = 26.95

V1 = sqrt (26.95/17.607) = 1.23719 m/sec

V1 = 1.24 m/sec = flow of pipe in lower section

Part B.

V2 = 4.3136*V1

V2 = 4.3136*1.23719 = 5.33674 m/sec

V2 = 5.34 m/sec = flow of pipe in higher section

Part C.

Volume flow rate is given by:

Q = V*A = V*pi*r^2

Using V1 and r1

Q = 1.23719*pi*0.0270^2

Q = Volume flow rate = 2.83*10^-3 m^3/sec

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