Water is flowing in a non viscous tube as shown in the diagram .The diameter of point A and point B are 0.5 m and 0.1 m respectively . The pressure difference between point A and B are `DeltaP=0.8`m, then the rate of flow is …..`m^(3)s^(-1)`.

Options

A. `0.0314`
B. `0.00314`
C. `0.000314`
D. `0.0000314`

Answer

c

Explanation

From Bernoulli.s equation,
`P_(1)+(1)/(2)rhov_(1)^(2)=P_(2)+(1)/(2)rhov_(2)^(2)(because` Tube is horizontal)
`thereforeP_(1)-P_(2)=(1)/(2)rho(v_(2)^(2)-v_(1)^(2))`
but according to equation of continuity
`A_(1)v_(1)=A_(2)v_(2)`= rate of flow Q
`thereforeQ=A_(1)v_(1)rArrv_(1)=(Q)/(A_(1))`
and `Q=A_(2)v_(2)rArrv_(2)=(Q)/(A_(2))`
`therefore` Equation (1),
`P_(1)-P_(2)=(1)/(2)rho[(Q^(2))/(A_(2)^(2))-(Q^(2))/(A_(1)^(2))]`
`therefore(2(P_(1)-P_(2)))/(rho)=Q^(2)[(A_(1)^(2)-A_(2)^(2))/(A_(1)^(2)A_(2)^(2))]`
`thereforeQ^(2)=(A_(1)^(2)A_(2)^(2)[2(P_(1)-P_(2))])/(rhoA_(1)^(2)-A_(2)^(2))`
`thereforeQ^(2)=A_(1)A_(2)sqrt((2(P_(1)-P_(1)))/(rho(A_(1)^(2)-A_(2)^(2))))`

`A_(1)=pi(0.25)^(2)=0.19625m^(2)andA_(1)^(2)=0.0385m^(4)`
`A_(2)=pi(0.05)^(2)=0.00785m^(2)andA_(2)^(2)=6.16xx10^(-5)m^(4)`
`DeltaP=P_(1)-P_(2)=0.8m` `thereforeQ=0.19625xx0.00785xxsqrt((2xx0.8)/(10^(3)xx(0.03844)))`
`thereforeQ=0.00154xxsqrt(0.0416)`
`thereforeQ=0.00154xx0.0204`
`thereforeQ=0.000314m^(3)s^(-1)`