Question

A 6.6 cm diameter pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 30.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answer 1

Now using bernoulli's theorem for horizontal pipes:

P1 + 0.5*rho*V1^2 = P2 + 0.5*rho*V2^2

P1 - P2 = 0.5*rho*(V2^2 - V1^2)

(V2^2 - V1^2) = 2*(P1 - P2)/rho

Using known values:

P1 = 30 kPa & P2 = 24 kPa

Density of water = rho = 1000 kg/m^3

(V2^2 - V1^2) = 2*(30 - 24)*10^3/1000

V2^2 - V1^2 = 12

Now Volume flow rate is given by:

Q = A*V

A = pi*d^2/4

V2^2 - V1^2 = (Q/A2)^2 - (Q/A1)^2

V2^2 - V1^2 = (4Q/pi)^2*(1/d2^4 - 1/d1^4) = 12

(1/d2^4 - 1/d1^4) = 1/0.04^4 - 1/0.066^4 = 337923.34

Q^2 = 12*pi^2/(16*337923.34)

Q^2 = 2.19*10^-5

Q = sqrt (2.19*10^-5) = 4.68*10^-3 m^3/sec

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