Solution%20to%20HW#5

Solution%20to%20HW#5 - EGM6812/FluidMechanicsI Homework#5...

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EGM 6812/ Fluid Mechanics I Homework # 5      Due Oct. 19, 2009 1. Consider the energy equation of the fluid in the form of Dh Dp Dt Dt ρ - = -∇⋅ Φ q + For perfect gas, h = c p T. If heat conduction and viscous dissipation are neglected, show that for an ideal gas, the above equation can be reduced to 0 D p Dt γ = (so that / p =  constant along a pathline) Solution: Negelecting heat conduction and viscous dissipation, we have 0 Dh Dp Dt Dt - = Since h=c p T, we have  0 p DT Dp c Dt Dt - = Since T=p/( ρ R), the above becomes  0 p c Dp p D Dp R Dt Dt Dt - - = . Since 1 p v c c R R - = , we have  1 1 0 p v c c D Dp R Dt R p Dt - = . That is  ln ln 0 D D p Dt Dt - =  or  ( 29 ln / 0 D p Dt = . This is equivalent to   0 D p Dt =
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2. Using the vorticity dynamic equation for an inviscid flow to explain the initial stage of the formation of a cyclone
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Solution%20to%20HW#5 - EGM6812/FluidMechanicsI Homework#5...

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