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State and prove Bernoulli's theorem for a flow of incompressible, non-viscous, and streamlined flow or fluid. |
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Answer» Solution :Bernoulli's theorem According to Bernoulli's theorem the sum of pressure energy, kinetic energy, and potential energy per UNIT mass of an imcompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically, `(P)/(rho) + (1)/(2)v^(2) + gh` = constant This is KNOWNAS Bernoulli's equation. Proof :  Let us consider a flow of liquid through a pipe AB as shown in figure. 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 leving B in the same time. Let `a_(A), v_(A)` and `P_(A)` be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively. Let the force exerted by the liquid at A is `F_(A) = P_(A)a_(A)` Distance travelled by the liquid in time t is `d = v_(A) t` Therefore, the work done is `W = F_(A)d = P_(A) a_(A) v_(A) t` But `a_(A)v_(A) t = a_(A) d = V`, volume 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)` Pressure energy per unit mass at `A = ("Pressure energy")/("mass") = (P_(A)V)/(m) = (P_(A))/((m)/(V)) = (P_(A))/(rho)` Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is `E_(PA) = P_(A)V = P_(A)V xx ((m)/(m)) = m(P_(A))/(rho)` Potential energy of the liquid at A, `PE_(A) = MG h_(A)`, Due to the flow of liquid, the kinetic energy of the liquid at A, `KE_(A) = (1)/(mv_(A)^(2)` Therefore, the TOTAL energy due to the flow of liquid at A, `E_(A) = E_(PA) + KE_(A) + PE_(A)` `E_(A) = m(P_(A))/(rho) + (1)/(2)m v_(A)^(2) + mg h_(A)` Similarly, let `a_(B), v_(B)`, and `P_(B)` be the area of cross section of the tube, velocity of the liquid, and pressure exerted by the liquid at B. Calculating the total energy at `E_(B)`, we get `E_(B) = m(P_(B))/(rho) + (1)/(2)mv_(B)^(2) + mg h_(B)` From the low of conservation of energy, `E_(A) = E_(B)` `m_(P_(A))/(rho) + (1)/(2) mv_(A)^(2) + mgh_(A) = m(P_(B))/(rho) + (1)/(2) mv_(B)^(2) + mgh_(B)` `(P_(A))/(rho) + (1)/(2)v_(A)^(2) + gh_(A) = (P_(B))/(rho) + (1)/(2) v_(B)^(2) + gh_(B)` = constant Thus, the above equation can be written as `(P)/(rho G) + (1)/(2)(v^(2))/(g) + h` = constant. |
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