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The velocity of liquid flowing through a tube at certain distance from the axis of tube |
Answer» Solution :Construction:It consists of two wider tubes A and A' (with cross sectional area A) connected by a narrow tube B (with cross sectional area a). A MANOMETER in the form of U-tube is also attached between the wide and narrow tubes as shown in Figure. The manometer contains a liquid of density `'rho_(m)'`. Theory:Let `P_(1)` be the pressure of the fluid at the wider region of the tube A. Let us ASSUME that the fluid of density `'rho'` flows from the PIPE with speed `'v_(1)'` and into the narrow region, its speed increases to `'v_(2)'`. According to the Bernoulli's equation, this increases in speed is accompanied by a DECREASE in the fluid pressure `P_(2)` at the narrow region of the tube B. Hence, the pressure difference between the tubes A and B is noted by measuring the height difference `(DeltaP=P_(1)-P_(2))` between the surfaces of the manometer liquid. From the equation of continuity, we can say that `Av_(1)=av_(2)` which means that `v_(2)=(A)/(a)v_(1)` Using Bernoulli's equation `P_(1)+rho(v_(1)^(2))/(2)=P_(2)+rho(v_(2)^(2))/(2)=P_(2)+rho(1)/(2)((A)/(a)v_(1))^(2)` From the above equation, the pressure difference `DeltaP=P_(1)-P_(2)=rho(v_(1)^(2))/(2)((A^(2)-a^(2)))/(a^2)` Thus, the speed of flow of fluid at the wide end of the tube A `v_(1)^(2)=(2(DeltaP)a^2)/(rho(a^(2)-A^(2)))` `rArrv_(1)=sqrt((2(DeltaP)a^(2))/(rho(a^(2)-A^(2))))` |
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