Water flows in a horizontal tube as shown in | vedclass

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(B) Given: $A_1 = 30\, cm^2 = 30 	imes 10^{-4}\, m^2$,$A_2 = 15\, cm^2 = 15 	imes 10^{-4}\, m^2$,$\Delta P = P_1 - P_2 = 600\, N/m^2$,$ho = 1000\,

Vedclass · Accepted Answer

$\frac{6}{\sqrt{10}} 	imes 10^{-3}\, m^3/s$

Vedclass · Answer

(B) Given: $A_1 = 30\, cm^2 = 30 	imes 10^{-4}\, m^2$,$A_2 = 15\, cm^2 = 15 	imes 10^{-4}\, m^2$,$\Delta P = P_1 - P_2 = 600\, N/m^2$,$ho = 1000\, kg/m^3$.From the equation of continuity,$A_1 v_1 = A_2 v_2$,so $v_2 = (A_1/A_2) v_1 = (30/15) v_1 = 2 v_1$.Using Bernoulli's equation for horizontal flow: $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$.$P_1 - P_2 = \frac{1}{2} ho (v_2^2 - v_1^2) = \frac{1}{2} ho ((2v_1)^2 - v_1^2) = \frac{1}{2} ho (3v_1^2)$.$600 = \frac{1}{2} 	imes 1000 	imes 3 v_1^2 = 1500 v_1^2$.$v_1^2 = 600 / 1500 = 6/15 = 0.4 = 4/10$.$v_1 = \sqrt{4/10} = 2/\sqrt{10}\, m/s$.The rate of flow $Q = A_1 v_1 = (30 	imes 10^{-4}) 	imes (2/\sqrt{10}) = 60 	imes 10^{-4} / \sqrt{10} = \frac{6}{\sqrt{10}} 	imes 10^{-3}\, m^3/s$.

Water flows in a horizontal tube as shown in the figure. The pressure of water changes by $600\, N/m^2$ between $A$ and $B$ where the cross-sectional areas are $30\, cm^2$ and $15\, cm^2$ respectively. Find the rate of flow of water through the tube.

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