p06122

# p06122 -

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Unformatted text preview: 6.122. CHAPTER 6, PROBLEM 122 933 6.122 Chapter 6, Problem 122 Problem: A hover craft has two air compressors, neither of which draws any air from the surroundings. One compressor supplies a large number of small hover jets with vertical velocity downward of magnitude we that provide a mass-flow rate m1 = ρwe Ah ˙ as shown, where ρ is air density and Ah is the total area of the hover jets. The other compressor supplies a horizontal jet with velocity relative to the hover craft, ue = −ue i and mass-flow rate m2 = ρue Ae , where Ae is the area of the jet. The flow ˙ is incompressible and flow through the compressors is steady as observed from within the hover craft. Ignoring both pressure variations around the hover craft’s surface and buoyancy effects on the air flow, determine the equations of motion for mass and ˙ x momentum. Express your answers for dM/dt and M dU/dt in terms of m1 , m2 and ˙ any pertinent velocities, where M is the mass of the hover craft and the air remaining in the compressors. .............. ................ .... ................ ..... . ... .... .... ... .. .... ... . . ... .. .. .. ... .. .. .. .. .. .. . . . . . . . x ue m2 U .................................................................... .................................................................... . ................................................................ ..................................................................................................................................................... . .................................................................................................................................................. ............ .... .................. .... .... ...................................................................................................................................................................................................................... . . .. . . . . . . .. . .. .... . . . . ..................... ...... ..................................................................................................... . ....................................................................................................................................................................... . ........................................................................................................................................................................................ e 1 ......................................................................................................................................................................................................................... .................................................................................................................................................................................................. . .. .. . ... .. . . . . . . ..... . . .. . .. . .. . .. . . . . . . . . ..... ....... . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .... .... ...... .. .. ..................................................................................................................................................................................... . . . ........................................................................................... .. . . .. .. .. .................................................................................................... . . . ......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... . ... .................................................................................................................................................................................................................................................................................................................................................................................................................... .......................................................................................................................................................................................................................................................................................................................................................................................................................................... ... . . . . . . . . . . .. . .. . ............. ...... ... . .... .. ........... ... ... ..................................... ...... ............................................................................................ . .. wm Solution: Let the control volume be coincident with the surface of the hover craft. It remains fixed on the hover craft, and is thus accelerating. Also, let the positive x direction be to the right. Mass Principle: In general, we know that d ρ dV + ρ ure · n dS = 0 dt V S By definition, the mass of the hover craft, including the air in the compressors, is M and the only place fluid crosses the control-volume surface is through the jets. Thus, M≡ and ρ dV V S ρ ure · n dS = ρ (we Ah ) + ρ (ue Ae ) = m1 + m2 ˙ ˙ Substituting into the mass-principle, we have dM = − (m1 + m2 ) ˙ ˙ dt x-Momentum Principle: The x-momentum principle tells us that d dt ρu dV + V S ρu ure · n dS = −i · S (p − pa ) n dS 934 CHAPTER 6. CONTROL-VOLUME METHOD The absolute velocity is u = U i + urel where urel is the velocity relative to the hover craft. Assuming the flow in the air jets is steady as seen by an observer on the hover craft, the unsteady term is d dt ρu dV = V d dt ρ U dV + V d dt ρurel dV V Because the flow in the jet is steady, the last integral on the right-hand side vanishes. Thus, d d d ρu dV = ρ U dV = (MU ) dt V dt V dt The only part of the control-volume boundary across which momentum passes is the jets. Hence, S ρu urel · n dS = (ρ U ) (we Ah ) + ρ (U − ue ) (ue Ae ) = U m1 + (U − ue ) m2 ˙ ˙ Vertical jet Horizontal jet Note that the vertical jets carry x momentum, proportional to ρ U , out of the control volume. Finally, we can ignore pressure variations. Consequently, the net pressure integral is zero and the momentum equation simplifies to d (MU ) + U m1 + (U − ue ) m2 = 0 ˙ ˙ dt =⇒ d (MU ) + U (m1 + m2 ) − ue m2 = 0 ˙ ˙ ˙ dt We can expand the first term on the left-hand side of the momentum equation according to dU dM dU d (MU ) = M +U =M − U (m1 + m2 ) ˙ ˙ dt dt dt dt where we make use of the mass principle to eliminate dM/dt. So, the momentum equation becomes dU − U (m1 + m2 ) + U (m1 + m2 ) − ue m2 = 0 ˙ M ˙ ˙ ˙ ˙ dt Therefore, we arrive at the following differential equation for the velocity. M dU = m2 ue ˙ dt ...
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