final_prac_sol

# final_prac_sol - 2.25 ADVANCED FLUID MECHANICS FINAL EXAM...

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2.25 ADVANCED FLUID MECHANICS Fall 2004 FINAL EXAM SOLUTIONS Each problem is graded out of 20 points (40 points total) Please see Prof Sonin with grading queries regarding Question 1 Please see Prof McKinley with grading queries regarding Question 2 SOLUTION: PROBLEM 1 (Ain Sonin) PART A From the definitions of the lift and drag coefficients in the problem statement, R R ± 1 Lift = 2 ² Ldr = 2 ² c L 2 ³ ( ´ r ) 2 bdr = c L ³´ 2 bR 3 (1) 3 0 0 PART B Equating lift to the total helicopter weight Mg that is kept aloft, we obtain, using Part A, that the required rotation rate is 3 Mg ± = ² c L bR 3 (2) PART C Apply the steady-state angular momentum theorem, ± rV ² V rn dA = ( T z ) CV , (3) CS to a control volume that envelops the rotor blades and jet engines and “cuts” through the frictionless bearing and the exit planes of the two jet engines. At the jet exit planes, the - direction gas velocity relative to the inertial reference frame of the ground is V = R ± V 0 , and V = V 0 . The torque due to aerodynamic drag may be deduced from the formula for drag given in rn the problem statement: R 1 ( T z ) CV = ± 2 ² c D 2 ( r ) 2 brdr = ± c D ²³ 2 bR 4 . (4) 4 0 Inserting this into (3), and using (2) for the angular rotation rate, we cast (3) into the simple form

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2.25 Fall 2004 Final Examination Solutions V Mg 0 = ± R + 3 c D (5) 4 c L ² 0 V 0 A 0 where A 0 is the exit area of a jet. Equation (5) reduces to a quadratic equation for V 0 , which can be solved straightforwardly. Putting the equation in the form (5) is useful since it shows at a glance that the torque due to aerodynamic drag becomes negligible in the limit 3 c D Mg << R (6) 4 c L 0 V 0 A 0 PART D The power required for hovering flight is the torque exerted on the two rotor blades —Eq. (4)—multiplied by . NOTE: The above solution accounts for the dynamic effects of the jet exit velocity but neglects the effects of the inflow into the jet engines. This is a reasonable approximation: the hot exhaust gases are much less dense than the cool inlet air. Mass conservation implies that, with inlet and exit areas of the same order of magnitude, both the velocity and the momentum flux will be much higher at the exit than in the cool gas at the inlet. Problem 2: Lubricated Pipelining ( Gareth McKinley ) Model the fluid motion as flow in a ²,µ 2 water cylindrical pipe of radius R with a core of thickness R 1 consisting of viscous liquid 1 R 1 oil with viscosity µ 1 surrounded by a shell of water or other low viscosity fluid) of Q o oil R thickness = R ± R 1 that is density matched (so that 1 = 2 = ) with viscosity µ 2 < 1 . The interfacial tension between the two liquids is denoted . The average velocity of the oil through the pipe is denoted v o = Q o R 1 2 a)
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final_prac_sol - 2.25 ADVANCED FLUID MECHANICS FINAL EXAM...

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