p0646 - 6.46. CHAPTER 6, PROBLEM 46 741 6.46 Chapter 6,...

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Unformatted text preview: 6.46. CHAPTER 6, PROBLEM 46 741 6.46 Chapter 6, Problem 46 Problem: A two-dimensional channel of width H = 3h has two slots of width h as shown. Fluid is injected at the indicated velocities through the lower slot at an angle φ to the horizontal and normal to the flow direction from the upper slot. The fluid is incompressible with density ρ and body forces can be neglected. Velocity and pressure can be assumed constant across the channel. The value of λ is 4/3. (a) Derive an equation governing mass for the channel. (b) Derive an equation governing x momentum for the channel. (c) Now, using Bernoulli’s equation, eliminate p1 and p2 from the momentum equation. (d) If U2 = 4U1 , what is the angle φ? ........... ...... ........... ........... ........... ........... ........... ........... ........... ..................................................................................................................................................................................................... ..................................................................................................................................................................................................... ................................................................................................................................................................... .................................................................................................................................................................................. ........... ...... ........... ........... ........... ........... ........... ........... .. .. .. .................................................................................................................................................................... .............................. ......................................................................................................................................... . ....... .. . . ................................................................................................................................................................... ............................................................... h λV U1 p1 y U2 p2 H x V . . . .. ..... . .......................................................................................................................................................... ............................................................................................ ............ ....... ....... ... ........... ... ............. ................................................................................................................................................................... ......................................................................................................................................................................................................... . . . . ... .. ... . . .... .... ........ . ........ . . ......................... ... ................................................................................................................................................................ ................................................................................................................................................................................................ ............................................................................................................................................................. ................................................................................................................................................................................................. . ............. .. .............. .............. .............. . ............ .............. ................. ......... .. ......................................................... .. . ........................................................................... ................................... . . ..................................................................... ......................... ............. .... .. .. ....................................................................... .......................................................... ....... .... . . . ... . .... ................................................................... φ h Solution: We select the stationary control volume shown in the figure. (a) Mass Principle: For steady, incompressible flow, the mass principle simplifies to S ρ u · n dS = 0 Noting that the lower slot width as measured along the x axis is h/ sin φ, we find ρ (−3U1 h) + ρ (−V sin φ h/ sin φ) + ρ (−4V h/3) + ρ (3U2 h) = 0 In et Lower s ot Upper s ot Simplifying, the mass principle for the channel tells us V= 9 (U2 − U1 ) 7 Out et 742 CHAPTER 6. CONTROL-VOLUME METHOD (b) x-Momentum Principle: For the x-momentum principle, we need no information about the pressure in the slots. This is true as the orientation of the control volume yields no net pressure-force component in the x direction from the slots. Thus, S ρ u(u · n)dS = −i · p n dS S So, noting that no x momentum is carried into the control volume from the upper slot, expanding the closed-surface integrals yields, (ρ U1 ) (−3U1 h) + (ρV cos φ) (−V sin φ h/ sin φ) + (ρ U2 ) (3U2 h) = (p1 − p2 ) (3h) Inlet Lower slot Outlet which can be rearranged to read 1 p1 − p2 2 2 U2 − U1 − V 2 cos φ = 3 ρ (c) Bernoulli’s equation tells us that 1 1 2 2 p1 + ρ U1 = p2 + ρ U2 2 2 p1 − p2 1 2 2 = U2 − U1 ρ 2 =⇒ So, the momentum equation developed in Part (b) becomes 1 1 2 2 2 U2 − U1 − V 2 cos φ = U 2 − U1 3 22 Simplifying, we find 2 2 2 U2 − U1 = V 2 cos φ 3 (d) If U2 = 4U1 , then the result from Part (a) yields V= 9 27 (4U1 − U1 ) = U1 7 7 Therefore, using the result from Part (c), we have 2 2 16U1 − U1 = 2 27 U1 37 2 cos φ =⇒ cos φ = 245 >1 162 Hence, no angle φ exists for these conditions. This tells us that Bernoulli’s equation does not hold for U2 = 4U1 . A straightforward calculation shows that we must have U2 > 4.31U1 in order to have cos φ < 1. ...
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