A liquid of density 750,kg,m^{-3} flows smoot | vedclass

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(C) Given:Density $ho = 750\,kg\,m^{-3}$$A_{1} = 1.2 	imes 10^{-2}\,m^{2}$$A_{2} = \frac{A_{1}}{2} = 0.6 	imes 10^{-2}\,m^{2}$Pressure difference

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$24$

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(C) Given:Density $ho = 750\,kg\,m^{-3}$$A_{1} = 1.2 	imes 10^{-2}\,m^{2}$$A_{2} = \frac{A_{1}}{2} = 0.6 	imes 10^{-2}\,m^{2}$Pressure difference $\Delta P = P_{1} - P_{2} = 4500\,Pa$Using the Equation of Continuity:$A_{1}V_{1} = A_{2}V_{2}$$A_{1}V_{1} = (A_{1}/2)V_{2} \Rightarrow V_{2} = 2V_{1}$Using Bernoulli's Equation for a horizontal pipe $(h_{1} = h_{2})$:$P_{1} + \frac{1}{2}ho V_{1}^{2} = P_{2} + \frac{1}{2}ho V_{2}^{2}$$P_{1} - P_{2} = \frac{1}{2}ho(V_{2}^{2} - V_{1}^{2})$$4500 = \frac{1}{2} 	imes 750 	imes ((2V_{1})^{2} - V_{1}^{2})$$4500 = 375 	imes (4V_{1}^{2} - V_{1}^{2})$$4500 = 375 	imes 3V_{1}^{2}$$4500 = 1125V_{1}^{2}$$V_{1}^{2} = 4 \Rightarrow V_{1} = 2\,m\,s^{-1}$Rate of flow (Volume flow rate) $Q = A_{1}V_{1}$$Q = (1.2 	imes 10^{-2}) 	imes 2 = 2.4 	imes 10^{-2}\,m^{3}\,s^{-1}$$Q = 24 	imes 10^{-3}\,m^{3}\,s^{-1}$Thus,the rate of flow is $24 	imes 10^{-3}\,m^{3}\,s^{-1}$.

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