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State and prove Bernoulli's theorem. |
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Answer» Solution :1st part: N/A 2nd part: Let an incompressible, non-viscous LIQUID entering into the cross -sectional area `A_1` at point A with VELOCITY `V_1` and coming out a height `h_2` at point B with velocity `v_2` [Fig. 3.52]. The value of the potential energy (PE) and kinetic energy (KE) are increased , since `h_2 gt h_1 and v_2 gt v_1` . This is done by the pressure at the TIME of WORK done on the liquid. If `p_1 and p_2` are the pressure at A and B, for a small displacement at A and B, the work done on the liquid at A. `Deltax_1=p_1 A_1v_1 Delta t` So, net work done by pressure `=(p_1-p_2)A_0v_0 Delta t` [ where `A_1 v_1=A_2v_2=A_0v_0]` From conservation of energy, `(p_1-p_2)A_0 v_0 Delta t =` CHANGE in KE + change in PE or, `(p_1 -p_2) A_0v_0 Delta t =1/2A_0v_0 Delta t rho (v_2^2-v_1^2)+A_0 v_0 rho Delta tg (h_2-h_1)` or, `p_1-p_2=rhog(h_2-h_1)+(rho)/(2)(v_2^2-v_1^2)` or, `p_1+rho gh_1+(rho)/(2)v_1^2=p_2+rho gh_2+(rho)/(2)v_2^2` `therefore (p)/(rhog)+h+(v^2)/(2g) =` constant 
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