Saved Bookmarks
| 1. |
Prove Bernoulli 's Principle .OR Obtain Bernoulli's equation for steady incompressible irrotational and viscous liquid. |
|
Answer» Solution :A pipe is shown in figure. At point B of left end of pipe Cross sectional area `=A_(1)`. Pressure EXERTED on fluid `P_(1)=(F_(1))/(A_(1))`.  At point D , of right end of pipe , Fluid speed `=v_(2)` Cross esctional area `=A_(2)` Pressure exerted on the fluid `P_(2)=(F_(2))/(A_(2))`. Force `F_(1)=P_(1)A_(1)` exerted onfluid at point B, covers distance `v_(1)Deltat` and reaches at point C. the work done on fluid `W_(1)=` (Force)(displacement) `W_(1)=P_(1)A_(1)v_(1)Deltat` `thereforeW_(1)=P_(1)(DeltaV_(1))` ....(1) where `DeltaV=A_(1)v_(1)Deltat=` volume of fluid) Fluid at point d, covers distance `v_(2)Deltat` and reaches at point E, the work done by the fluid. `W_(2)=-P_(2)A_(2)v_(2)Deltat` `thereforeW_(2)=-P_(2)DeltaV` ....(2) where `DeltaV=A_(2)v_(2)Deltat`= volume of fluid) According to equation of continuity both fluids have same volume. Total work done on the fluid `W=W_(1)+W_(2)` `W=P_(1)DeltaV-P_(2)DeltaV` `W=(P_(1)-P_(2))DeltaV` ....(3) Part of this work goes into changing the kinetic energy of the fluid and part goes into changing the gravitational potential energy. `W=DeltaK+DeltaU` .... (4) The mass flowing fluid of density `RHO` at TIME `Deltat`. `Deltam="Volume"xx"Density"` `Deltam=rho(DeltaV)` ...(5) This liquid when moves from B to D its velocity becomes `v_(1)` to `v_(2)` . Hence change in its kinetic energy `DeltaK=(1)/(2)m(v_(2)^(2)-v_(1)^(2))` `DeltaK=(1)/(2)rhoDeltaV(v_(2)^(2)-v_(1)^(2))`.....(6) When fluid of mass `Deltam` move from B to D , it posses HEIGHT from `h_(1)` to `h_(2)(h_(2)gth_(1))` . Hence its potential energy is botained as below `DeltaU=mg(h_(2))-mg(h_(1))` `DeltaU=mg(h_(2)-h_(1))` `DeltaU=(rhoDeltaV)g(h_(2)-h_(1))`....(7) The values of W , `DeltaK,DeltaU` in equation (4) from equation (3),(6),(7) `(P_(1)-P_(2))DeltaV=(1)/(2)rhoDeltaV(v_(2)^(2)-v_(1)^(2))+(rhoDeltaV)g(h_(2)-h_(1))` Dividing each by `DeltaV` `P_(1)-P_(2)=(1)/(2)rho(v_(2)^(2)-v_(1)^(2))+rhogh_(2)-h_(1))` `thereforeP_(1)-P_(2)=(1)/(2)rhov_(2)^(2)-(1)/(2)rhov_(1)^(2)+rhogh_(2)-rhogh_(1)` `thereforeP_(1)+(1)/(2)rho_(1)^(2)+rhogh_(1)=P_(2)+(1)/(2)rhov_(2)^(2)+rhogh_(2)` ....(8) Equation (8)is called Bernoulli.s equation In general `P+(1)/(2)rhov^(2)+rhogh=` constant ....(9) In words the Bernoulli.s equation may be stated as follows : As moving along a streamline the sum of the pressure (P) , the kineticenergy per unit volume `((1)/(2)rhov^(2))` and the potential energy per unit volume `(rhogh)` remains a constant. |
|