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ME 510 – Gas Dynamics Final Exam Spring 2009 NAME ME 510 – Gas Dynamics Final Exam Spring 2009 NAME: 1. (20 pts total – 2pts each) - Circle the most correct answer for the following questions. i. A normal shock propagated into still air travels with a speed (a) equal to the speed of sound in the still air (b) larger than the speed of sound in the still air (c) smaller than the speed of sound in the still air (d) all of the above are possible, depending on the air temperature ii. A perfect gas enters a frictionless, insulated passage with a supersonic speed. It is desired to increase the static pressure of the gas as it flow through the passage. The passage area should: (a) decrease (b) remain constant (c) increase (d) be converging-diverging iii. Which of the following is true for a Fanno flow (a) the Mach number always increases as one moves downstream (b) the static pressure always decreases as one moves downstream (c) the maximum length of the duct is the sonic length (d) none of the above iv. A characteristic curve is a curve (a) across which a variable, e.g. the velocity, is continuous, but the derivatives of that variable are indeterminate (b) the governing PDE can be reduced to an ordinary differential equation (c) along which disturbances in the flow propagate (d) all of the above v. For flow of a compressible fluid from a storage tank through a converging-diverging channel operating under choked conditions (a) the pressure at the exit will be equal to the sonic pressure (b) the flow in the diverging section must be supersonic (c) the mass flow rate through the channel cannot be increased by changing the storage tank conditions (d) none of the above Page 2 of 11 ME 510 – Gas Dynamics Final Exam Spring 2009 NAME: 1. (cont…) vi. The conditions across a normal shock (a) lie at the intersection of the Fanno and Rayleigh lines for the flow (b) have the same stagnation temperature (c) both (a) and (b) are true (d) both (a) and (b) are false vii. When using small perturbation theory, the boundary conditions for a flow (a) must be satisfied exactly to prevent unstable solutions (b) become trivial (this is the chief advantage of small perturbation analysis) (c) should be modified to be consistent with the small perturbation assumptions (d) must satisfy the no slip condition at a solid wall viii. Kelvin’s theorem states that for a flow with the following conditions, the circulation about a closed curve (or the vorticity contained within that curve) will remain constant. (a) adiabatic and no body forces (b) uniform flow and no body forces (c) inviscid and the pressure is a function of the density only (d) irrotational and an ideal gas ix. A Mach line (a) is a curve which is everywhere perpendicular to the stream lines in a subsonic flow (b) is a wave which is perpendicular to the stream lines in a supersonic flow (c) is perpendicular to the stream lines when the flow is sonic (d) has the same slope as an arbitrary oblique shock wave x. When heat is added to a compressible flow (a) the flow temperature will always increase (b) the Mach number will always increase (c) the entropy may decrease (d) the flow stagnation temperature will always increase Page 3 of 11 rayleigh_17 A converging‐diverging nozzle with a test section‐to‐throat area ratio of 3.0 supplies air to a supersonic wind tunnel. If there is moisture in the air, it is possible for the water vapor to condense during the expansion process if the local static temperature drops below the saturation temperature. In practice, this condensation process occurs very rapidly, leading to an almost discontinuous change in the flow properties (and thus is referred to as a “condensation shock”). Assume that the stagnation temperature of the air/water vapor mixture entering the nozzle is 600 K and that the mass fraction of water vapor in the stream is mH2O/mmix = 0.01 (the ratio of the mass of water vapor to the mass of the vapor air mixture). The saturation temperature for the air/water vapor mixture is 14 C and the heat of vaporization of water is 2470 kJ/kg (i.e. the heat released per unit mass of water when water vapor condenses to liquid water). You may assume that the air/vapor mixture behaves as a perfect gas and has the same flow properties as air ( = 1.4, R = 287 J/(kg.K)). condensation shock T0 = 600 K ATS/Athroat = 3.0 mH2O/mmix = 0.01 Athroat Acond ATS a. Determine the area ratio, Acond/Athroat, where the condensation shock occurs, i.e. the area ratio where the static temperature of the flow first drops below the saturation temperature of 14 C. b. Determine the test section Mach number when no condensation shock is present. c. Determine the test section Mach number when the condensation shock is present. [Hint: Be careful differentiating between mH20 and mmix.] d. Sketch the process with the condensation shock on a T‐s diagram. SOLUTION: Assume the flow is isentropic up to the point of the condensation shock. The Mach number at the saturation temperature of Tcond = (14 + 273) = 287 K may be found using: 1 Tcond 1 2 1Macond Macond = 2.335 where Tcond = 287 K and T0 = 600 K ( Tcond/T0 = 0.4783) (1) T0 2 The area ratio at this Mach number may be found using: 1 1 2 21 AAcond cond 1 1Ma 2 cond *1 Acond/Athroat = 2.264 (2) AAthroatMa cond 1 2 * Note that in the previous equation Athroat = A . The Mach number in the test section when no condensation shock is present may be found from the area ratio * ATS/Athroat = ATS/A = 3.0: 1 1 2 21 AATS TS 1 1Ma 2 TS *1 MaTS = 2.637 (3) AATSMa TS 1 2 Page 1 of 3 rayleigh_17 When the condensation shock is present, we must account for the heat released by the flow as the water vapor condenses from vapor to liquid. The rate at which heat is released into the flow is the mass flow rate of water multiplied by the heat of vaporization, hfg: qm H 2Ofg h (4) Thus, the flow through the condensation shock may be modeled as a Rayleigh flow. condensation T = 600 K shock 0 A /A = 3.0 TS throat mH2O/mmix = 0.01 1 2 Athroat Acond ATS The stagnation temperature change through the condensation shock is given from conservation of energy as: qcTT12P 02 01 (5) where q mh H 2Ofg q12 (6) mmmix mix Combine the previous two equations to get: mhH 2Ofgm HO2 h fg cTP 02 T 01 T 02 T 01 T02 = 624.1 K (or T02/T01 = 1.041) (7) mmcmix mix P * The Mach number just downstream of the condensation shock may be found using T02/T : TTT02 02 01 * ** T02/T0 = 0.7641 Macond,2 = 2.154 (from the Rayleigh flow relations) (8) TT00T01 where * T01/T0 = 0.7340 using Macond,1 = 2.335 and the Rayleigh flow relations. (9) The sonic area ratio corresponding to the downstream Mach number is: 1 1 2 21 Acond 1 1Ma 2 cond,2 * Acond/A2 = 1.926 (10) *1Ma A2 cond,2 1 2 The sonic area ratio for the test section is: AAAATS TS throat cond * ** ATS/A2 = 2.552 (11) AA22AAthroat cond The Mach number in the test section may be found from the sonic area ratio. 1 1 2 21 ATS 1 1Ma 2 TS *1 MaTS = 2.465 (12) A2 MaTS 1 2 Thus, we see that the Mach number in the test section decreases when there is a condensation front. It would be a good idea to de‐humidify the air before sending it through the wind tunnel. Page 2 of 3 rayleigh_17 Sketch the process on a T‐s diagram. p02 T T01 T02 p01 Rayleigh curve p2 T2 T1 p1 TTS pTS s Page 3 of 3 expansionfan_13 A two-dimensional double-wedge profile is at zero angle of attack in an air stream of Mach number 2.0. Ma∞ = 2.0 10 10 l Calculate the drag coefficient for the airfoil based on the chord length, l. SOLUTION: 2 1 23 12 3 12 Use the oblique shock relations to determine the conditions in region 2. Ma1 = 2.0, 12 = 10 Ma2 = 1.6405, 12 = 39.31, p2/p1 = 1.7066 (1) Determine the Mach number in region 3 using the Prandtl-Meyer angle. Ma2 = 1.6405 2 = 16.0574 (2) 3 = 2 + 23 = 36.0574 where 23 = 20 (3) 3 = 36.0574 Ma3 = 2.3717 (4) Determine the pressure ratio for region 3 using the isentropic relations. 1 2 1 p3 1Ma 2 3 p3/p2 = 0.3227 (5) p 1Ma 1 2 2 2 2 Note that: pp33p2 0.5507 ppp121 The drag coefficient on the object is given by: F c D (6) D 1 2 2 11Vl where 1 FpplD 23 2 tan (7) and ½l 11222p1 1 2211VRTp 11Ma 2 11 Ma (8) RT1 So that: pp 1 ltan 232 p2 p3 tan cD (9) 1 22pp 2 p11Ma11 Ma 1 cD = 0.036 using p2/p1 = 1.7066, p3/p1 = 0.5507, = 10, = 1.4, and Ma1 = 2.0 (10) Page 1 of 2 expansionfan_13 Now solve the problem using thin airfoil theory. 2 2 2 l dy dy cdxD (11) 2 l Ma 1 0 dxupper dx lower where tan 10 0 x 1 l dy 2 (12) 1 dx upper tan 10 lxl 2 dy 0 (13) dx lower Substitute and simplify.
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