A venturimeter has a pipe diameter of 4 ,cm a | vedclass

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(B) Given:Diameter of pipe $D_1 = 4 \,cm$,radius $r_1 = 2 \,cm$.Diameter of throat $D_2 = 2 \,cm$,radius $r_2 = 1 \,cm$.Velocity in pipe $V_1 = 10 \,m

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$7.5 	imes 10^5 \,Pa$

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(B) Given:Diameter of pipe $D_1 = 4 \,cm$,radius $r_1 = 2 \,cm$.Diameter of throat $D_2 = 2 \,cm$,radius $r_2 = 1 \,cm$.Velocity in pipe $V_1 = 10 \,m/s$.Density of water $ho = 1000 \,kg/m^3$.Step $1$: Using the equation of continuity $A_1 V_1 = A_2 V_2$:$\pi r_1^2 V_1 = \pi r_2^2 V_2$$(2)^2 	imes 10 = (1)^2 	imes V_2$$V_2 = 40 \,m/s$.Step $2$: Using Bernoulli's theorem for horizontal flow:$P_1 + \frac{1}{2} ho V_1^2 = P_2 + \frac{1}{2} ho V_2^2$$P_1 - P_2 = \frac{1}{2} ho (V_2^2 - V_1^2)$$P_1 - P_2 = \frac{1}{2} 	imes 1000 	imes (40^2 - 10^2)$$P_1 - P_2 = 500 	imes (1600 - 100)$$P_1 - P_2 = 500 	imes 1500 = 7.5 	imes 10^5 \,Pa$.

$A$ venturimeter has a pipe diameter of $4 \,cm$ and a throat diameter of $2 \,cm$. The velocity of water in the pipe section is $10 \,m/s$. The pressure drop between the pipe section and the throat section is (use density of water $= 1000 \,kg/m^3$):

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