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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=0.8m, then the rate of flow is …..m^(3)s^(-1). |
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Answer» `0.0314` `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)` |
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