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Calculate the rate of flow of glycerine ...

Calculate the rate of flow of glycerine of density 1.20 `xx 10^(3) kg//m^(3)` through the conical section of a pipe. If the radii of its ends are 0.1 m and 0.02 m and the pressure drop across the length is 10 `N/m^(2)`.

Text Solution

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According to continuity equation,
`A_(1)V_(1) = A_(2)V_(2)`
`rArr (V_(2))/(V_(1))=(A_(1))/(A_(2))=(pi r_(1)^(2))/(pi r_(2)^(2))=(r_(1)^(2))/(r_(2)^(2))`
`rArr (V_(2))/(V_(1))=((r_(1))/(r_(2)))^(2)=((0.10)/(0.02))^(2)=5^(2)=25" "...(i)`
and according to Bernoulli's equation
`P_(1)+(1)/(2)rho v_(1)^(2)+rho gh_(1)=P_(2)+(1)/(2)rho v_(2)^(2)+rho gh_(2)`
Here pipe is horizontal. So `h_(1) = h_(2)`
`rArr P_(1)+(1)/(2) rho v_(1)^(2) + rho gh_(1) = P_(2) + (1)/(2)rho v_(2)^(2) + rho gh_(1)`
`rArr P_(1) + (1)/(2) rho v_(1)^(2)=P_(2)+(1)/(2)rho v_(2)^(2)`
`rArr (1)/(2) rho v_(2)^(2) - (1)/(2) rho v_(1)^(2) = P_(1)-P_(2)`
`rArr v_(2)^(2)-v_(1)^(2)=(2(P_(1)P_(2)))/(rho)=(2xx10)/(1.20 xx 10^(3))=(20)/(1.20 xx 10^(3))`
`rArr v_(2)^(2) - v_(1)^(2) = (20)/(1200) = (2)/(120) = (1)/(60) = 16.7 xx 10^(-3)" "...(ii)`
Substituting the value of `V_(2)` from equation (i) in (ii)
`(25 V_(1))^(2) - V_(1)^(2) = 16.7 xx 10^(-3)`
`rArr 624 V_(1)^(2) = 16.7 xx 10^(-3)`
`rArr v_(1)^(2) = (167 xx 10^(-4))/(624)`
`rArr V_(1) = sqrt((167)/(624))xx 10^(-2)m//sec`
So, rate of flow through the tube
`R=A_(1)V_(1)=A_(2)V_(2) = pi xx (0.1)^(2) xx sqrt((167)/(624)) xx 10^(-2)`
`=pi xx sqrt((167)/(624))xx 10^(-4)`
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