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State Bernoulli's theorem. |
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Answer» Solution :Bernoulli theorem: According to bernoulli the sum of pressure energy kinetic energy and ponteniel energey per unit mass of an incompressible, non viscous fluid in a streamlined flow remains a constatnt, Mathmatically, Proof: Let us consinder a flow of liquid through a pipe AB. Let V be the VOLUME Of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time . Lrt `a_(A),v_(A)` and `P_(2)` br the area of cross section of the TUBE, velocity of the liquid and pressure EXERTED by the liquid at A respetivily. Let the force exerted by the liquid ata is `F_(2)=P_(A) a_(A)` Distance travlled by theliquid in theyime t is `d=v_(A)t` Therefore the work done is `W=F_(2)d=P_(A)a_(A)v_(A)t` But `a_(A)v_(A)t=a_(A)d=V`, Volime of the liquid ENTERING at A. Thus the work done is the pressure energy (at A), `W=F_(A)d=P_(A)V` Pressure energy per unit volume at `A=("Pressure Energy")/("Volume")=(P_(A)V)/(V)=P_(A)` Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is `EP_(A)=P_(A)V=xx(m/m)=m(P_(A))/(rho)` Due to the flow of liquid, the kinectic energy of the liquid at A, `KE_(A)=(1)/(2)mv_(A)^(2)` Therefore the total energey due to the flow of liquid at A, `E_(A)=EP_(A)+KE_(A)+PE_(A)` `E_(A)=m(P_(A))/(rho)+(1)/(2)mv_(A)^(2)+mg h_(A)` Similarly, let `a_(B), v_(B) and P_(B)` be the area of cross section of the tube , velocityof the liquidand pressure exerted by the liquid at B. Calculating the total energy at `F_(B)`, we get `E_(B)=m(P_(B))/(rho)+(1)/(2)mv_(B)^(2)+mg h_(B)``m(P_(A))/(rho)+(1)/(2)mV_(A)^(2)+mgh_(A)=m(P_(B))/(rho)+(1)/(2)mv_(B)^(2)+mgh_(B)` `m(P_(A))/(rho)+(1)/(2)mV_(A)^(2)+mgh_(A)=m(P_(B))/(rho)+(1)/(2)v_(B)^(2)+gh_(B)=`Constant Thus the above equation can be written as `(P)/(rhog)+(1)/(2)(v^(2))/(g)+h= `Constant  |
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