A liquid of density 750 kgm-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1=1.2×10−2m2 to A2=A12. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is __________________ ×10−3m3s−1.

A2=A12P1−P2=4500PaP1+12ρV12+ρgh=P2+12ρV22+ρghP1−P2=12ρV22−V12 .....(1)And A1V1=A2V2⇒ V2=2V1 .....(2)4500=12×750×3V12V1=2m/sVolume flow rate =A1V1=24×10−3m3s−1

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A liquid of density 750 kgm^-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from $A_{1} = 1 .2 \times 10^{- 2} m^{2}$ to $A_{2} = \frac{A_{1}}{2}$ . The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is __________________ $\times 10^{- 3} m^{3} s^{- 1}$ .

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Detailed Solution

$\begin{array}{l} A_{2} = \frac{A_{1}}{2} \\ P_{1} - P_{2} = 4500 Pa \\ P_{1} + \frac{1}{2} {ρV}_{1}^{2} + ρgh = P_{2} + \frac{1}{2} {ρV}_{2}^{2} + ρgh \\ P_{1} - P_{2} = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}) . . . . . (1) \end{array}$