Glycerine of density 1.25 times 10^3 kg/m^3 | vedclass

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(A) Given: Density $ho = 1.25 	imes 10^3 \ kg/m^3$,$A_1 = 10 \ cm^2 = 10^{-3} \ m^2$,$A_2 = 5 \ cm^2 = 5 	imes 10^{-4} \ m^2$,$\Delta P = P_1 - P_

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$4 	imes 10^{-5} \ m^3/s$

Vedclass · Answer

(A) Given: Density $ho = 1.25 	imes 10^3 \ kg/m^3$,$A_1 = 10 \ cm^2 = 10^{-3} \ m^2$,$A_2 = 5 \ cm^2 = 5 	imes 10^{-4} \ m^2$,$\Delta P = P_1 - P_2 = 3 \ N/m^2$.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$$P_1 - P_2 = \frac{1}{2} ho (v_2^2 - v_1^2)$$3 = \frac{1}{2} 	imes 1.25 	imes 10^3 	imes (v_2^2 - v_1^2)$$v_2^2 - v_1^2 = \frac{6}{1.25 	imes 10^3} = 4.8 	imes 10^{-3} \dots (i)$From the equation of continuity,$A_1 v_1 = A_2 v_2$:$10 	imes 10^{-4} 	imes v_1 = 5 	imes 10^{-4} 	imes v_2 \Rightarrow v_2 = 2v_1 \dots (ii)$Substituting $(ii)$ into $(i)$:$(2v_1)^2 - v_1^2 = 4.8 	imes 10^{-3}$$3v_1^2 = 4.8 	imes 10^{-3} \Rightarrow v_1^2 = 1.6 	imes 10^{-3}$$v_1 = \sqrt{1.6 	imes 10^{-3}} \approx 0.04 \ m/s$Rate of flow $Q = A_1 v_1 = 10 	imes 10^{-4} 	imes 0.04 = 4 	imes 10^{-5} \ m^3/s$.

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