Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is 15 Nm ^ -2 . The area of cross-section at P and Q are 40

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 400

From the equation of continuity, the rate of flow is constant:

A_P v_P = A_Q v_Q

Given

A_P = 40 \text{ cm}^2

and

A_Q = 20 \text{ cm}^2

, we have:

40 v_P = 20 v_Q \Rightarrow v_Q = 2 v_P

Applying Bernoulli's equation for a horizontal pipe:

P_P + \dfrac{1}{2} \rho v_P^2 = P_Q + \dfrac{1}{2} \rho v_Q^2

P_P - P_Q = \dfrac{1}{2} \rho (v_Q^2 - v_P^2)

Substitute the given values (

P_P - P_Q = 15 \text{ N m}^{-2}

,

\rho = 1000 \text{ kg m}^{-3}

)

and v_Q = 2 v_P

:

15 = \dfrac{1}{2} \times 1000 \times ((2 v_P)^2 - v_P^2)

15 = 500 \times (4 v_P^2 - v_P^2)

15 = 500 \times 3 v_P^2

15 = 1500 v_P^2

v_P^2 = \dfrac{1}{100} \Rightarrow v_P = 0.1 \text{ m s}^{-1} = 10 \text{ cm s}^{-1}

The rate of flow of water is:

Q = A_P v_P = 40 \text{ cm}^2 \times 10 \text{ cm s}^{-1} = 400 \text{ cm}^3 \text{ s}^{-1}

Answer:

400

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