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Water is flowing in a non viscous tube a...

Water is flowing in a non viscous tube as shown in the diagram. The diametre at a point A and B are 0.5m and 0.1m respectively. The pressure difference between points A & B are `triangleP=0.8` then find out the rate of flow:

Text Solution

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The correct Answer is:
`Q=A_(1)A_(2)sqrt((2(P_(1)-P_(2)))/(rho(A_(1)^(2)-A_(2)^(2))))`
`P_(1)+(1)/(2)rhoV_(1)^(2)=P_(2)+(1)/(2)rho V_(1)^(2)" "Q=A_(1)V_(1)=A_(2)V`
`P_(1)-P_(2)=+(1)/(2)rho(V_(1)^(2)-V_(1)^(2))`
`P_(1)-P_(2)=rho[(Q^(2))/(A_(2)^(2))-(Q^(2))/(A_(1)^(2))]`
`2(P_(1)-P_(2))=rho[(A_(1)^(2)-A_(2)^(2))/(A_(1)^(2)A_(2)^(2))]Q^2`
`Q=A_(1)A_(2) sqrt((2(P_(1)-P_(2)))/(rho(A_1^2-A_2^2)))`
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