Chapter 14435P14.71Note: Variation of atmospheric pressure with altitude is included inthis solution. Because of the small distances involved, this effect isunimportant in the final answers.(a)Consider the pressure at points A and B in part (b) of thefigure:Using the left tube: PPghgLhAaw=++−atmρafwhere thesecond term is due to the variation of air pressure withaltitude.Using the right tube: gLBatm0But Pascal’s principle says that AB=.Therefore,PgLhgLhawatmatm+=++−ρρ0orwawhL−=−bg0, givingw=−−FHGIKJ=−−FHGIKJ=01000 7501000 129500125...cmcm(b)Consider part (c) of the diagram showing the situationwhen the air flow over the left tube equalizes the fluidlevels in the two tubes. First, apply Bernoulli’s equation topoints A and B yyvvvAB===,,and0This gives: PvgyPgyAa aABaaB++=++1212022and since =, this reduces to:
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