Question
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 $40{\text{ }cm}^{2}$ and $20{\text{ }cm}^{2}$, respectively. The rate of flow of water through the pipe, in ${cm}^{3}{\text{ }s}^{- 1}$, is: [Take density of water $= 1000\text{ }kg{\text{ }m}^{- 3}$ ]
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Solution
Sol. [Key Concept: Streamline flow — equation of continuity]
$A_{P}V_{P} = A_{Q}V_{Q}$
$${40V_{P} = 20V_{Q},\ V_{Q} = 2V_{P} } {Using\ Bernoulli's\ equation: } \begin{matrix} & P_{P} - P_{Q} = \frac{1}{2}\rho\left( V_{Q}^{2} - V_{P}^{2} \right) \\ & 15 = \frac{1}{2} \times 10^{3}\left( 4V_{P}^{2} - V_{P}^{2} \right) \\ & V_{P} = \sqrt{\frac{15}{1500}} = 0.1\text{ }m/s \\ & V_{Q} = 0.2\text{ }m/s \\ & \text{~}\text{Rate}\text{~} = A_{P}V_{P} = 40 \times 10^{- 4} \times 10^{- 1}{\text{ }m}^{3}/s \\ & \ = 4 \times 10^{- 4}{\text{ }m}^{3}/s \\ & \ = 4 \times 10^{- 4} \times 10^{6}{\text{ }cm}^{3}/s \\ & \ = 400{\text{ }cm}^{3}/s \end{matrix}$$
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