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III. Long Answer Questions · Q12

Q.State and prove Bernoulli's theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.

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Step 1. Statement. Bernoulli's theorem states that for an incompressible, non-viscous fluid in streamlined flow, the sum of pressure energy, kinetic energy and potential energy per unit mass remains constant: Pρ+12v2+gh=constant\dfrac{P}{\rho}+\tfrac12v^2+gh=\text{constant}.

Step 2. Setup. Consider a liquid flowing through a pipe from A (cross-sectional area aAa_A, velocity vAv_A, pressure PAP_A, height hAh_A) to B (aBa_B, vBv_B, PBP_B, hBh_B). A mass m of liquid enters at A in time t, equal to the mass that leaves at B in the same time (by the equation of continuity).

Step 3. Energy at A. The pressure energy of this mass at A is EPA=PAV=PAρmE_{PA}=P_AV=\dfrac{P_A}{\rho}m (writing V=m/ρV=m/\rho); its potential energy is PEA=mghA\text{PE}_A=mgh_A; its kinetic energy is KEA=12mvA2\text{KE}_A=\tfrac12mv_A^2. Total: EA=PAρm+12mvA2+mghAE_A=\dfrac{P_A}{\rho}m+\tfrac12mv_A^2+mgh_A.

Step 4. Energy at B. By identical reasoning, EB=PBρm+12mvB2+mghBE_B=\dfrac{P_B}{\rho}m+\tfrac12mv_B^2+mgh_B.

Step 5. Conservation of energy. For an ideal (non-viscous) fluid, no energy is lost between A and B, so EA=EBE_A=E_B. Dividing through by m: PAρ+12vA2+ghA=PBρ+12vB2+ghB\dfrac{P_A}{\rho}+\tfrac12v_A^2+gh_A=\dfrac{P_B}{\rho}+\tfrac12v_B^2+gh_B -- since A and B were arbitrary points along the flow, this proves Pρ+12v2+gh=constant\dfrac{P}{\rho}+\tfrac12v^2+gh=\text{constant} everywhere along a streamline. …

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