1. (20 pts total 2pts each) - Circle the most correct answer for the following questions.

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1 ME 50 Gas Dynamics Spring 009 Final Exam NME:. (0 pts total pts each) - Circle the most correct answer for the following questions. i. 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. 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. 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 of

2 ME 50 Gas Dynamics Spring 009 Final Exam NME:. (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. 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

3 rayleigh_7 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 ). ssume 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 m HO /m mix = 0.0 (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 4 C and the heat of vaporization of water is 470 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 ( =.4, R = 87 J/(kg.K)). T 0 = 600 K m HO /m mix = 0.0 condensation shock throat cond / throat = 3.0 a. Determine the area ratio, cond / throat, where the condensation shock occurs, i.e. the area ratio where the static temperature of the flow first drops below the saturation temperature of 4 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 m H0 and m mix.] d. Sketch the process with the condensation shock on a T s diagram. SOLUTION: ssume the flow is isentropic up to the point of the condensation shock. The Mach number at the saturation temperature of T cond = (4 + 73) = 87 K may be found using: Tcond Macond Ma cond =.335 where T cond = 87 K and T 0 = 600 K ( T cond /T 0 = ) () T0 The area ratio at this Mach number may be found using: cond cond Ma cond * throat Macond Note that in the previous equation throat = *. cond / throat =.64 () The Mach number in the test section when no condensation shock is present may be found from the area ratio / throat = / * = 3.0: Ma * Ma Ma =.637 (3) Page of 3

4 rayleigh_7 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, h fg : q m H Ohfg Thus, the flow through the condensation shock may be modeled as a Rayleigh flow. (4) T 0 = 600 K m HO /m mix = 0.0 condensation shock throat cond / throat = 3.0 The stagnation temperature change through the condensation shock is given from conservation of energy as: q cp T0 T0 (5) where q mh Ohfg q mmix mmix (6) Combine the previous two equations to get: m H Ohfg m h H O fg cp T0 T0 T0 T0 m m c T 0 = 64. K (or T 0 /T 0 =.04) (7) mix mix The Mach number just downstream of the condensation shock may be found using T 0 /T * : P T 0 T 0 T 0 * * T0 T0 T0 T 0 /T * 0 = Ma cond, =.54 (from the Rayleigh flow relations) (8) where T 0 /T * 0 = using Ma cond, =.335 and the Rayleigh flow relations. (9) The sonic area ratio corresponding to the downstream Mach number is: cond Ma cond, * Ma cond, The sonic area ratio for the test section is: throat cond * * throat cond cond / * =.96 (0) / * =.55 () The Mach number in the test section may be found from the sonic area ratio. Ma * Ma Ma =.465 () 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 of 3

5 rayleigh_7 Sketch the process on a T s diagram. T T 0 p 0 p 0 T 0 Rayleigh curve p T T p p T s Page 3 of 3

6 expansionfan_3 two-dimensional double-wedge profile is at zero angle of attack in an air stream of Mach number.0. Ma = l Calculate the drag coefficient for the airfoil based on the chord length, l. SOLUTION: 3 3 Use the oblique shock relations to determine the conditions in region. Ma =.0, = 0 Ma =.6405, = 39.3, p /p =.7066 () Determine the Mach number in region 3 using the Prandtl-Meyer angle. Ma =.6405 = () 3 = + 3 = where 3 = 0 (3) 3 = Ma 3 =.377 (4) Determine the pressure ratio for region 3 using the isentropic relations. p Ma 3 3 p Ma p 3 /p = 0.37 (5) Note that: p3 p3 p p p p The drag coefficient on the object is given by: FD Vl (6) where 3 tan FD p p l and ½l (7) p V RTMa p Ma RT (8) So that: p p3 ltan p p3 tan p Ma p p Ma (9) c D = using p /p =.7066, p 3 /p = , = 0, =.4, and Ma =.0 (0) Page of

7 expansionfan_3 Now solve the problem using thin airfoil theory. l dy dy dx () l Ma dx 0 upper dx lower where dy tan 0 0 x l dx () upper tan 0 l x l dy dx lower 0 (3) Substitute and simplify. l l tan 0 dx tan 0 dx l Ma 0 l l tan0 l Ma c D tan 0 Ma This result is nearly identical to the result found previously despite the fact that 0 is on the edge of being considered a small angle. (4) (5) (6) Page of

8 shock_tube_05 normal shock moves down an open-ended tube with a velocity of 45 m/s (with respect to the stationary air upstream of the shock). The ambient air pressure and temperature are 0 kpa (abs) and 5 C upstream of the shock wave. Determine the velocity, with respect to the ground, of the first and last expansion waves that move down the tube after reflection of the shock from the open end. V S = 45 m/s incident reflected open tube p = 0 kpa (abs) u,w/r/t ground =? u B,w/r/t ground =? SOLUTION: u S = 45 m/s p = 0 kpa (abs) T = (5 + 73) = 98 K p = 0 kpa (abs) t /(u 3 c 3 ) 3 pathline of an air particle /(u c ) /u /u 3 Note: The speeds, u, are measured with respect to the ground. /u S initially stationary air x The first and last reflected expansion waves are left running characteristics. Thus, the speed of these waves with respect to the ground will be: u u c (),w/r/t ground ub,w/r/t ground u3 c3 () The Mach number of the shock wave is: us us Ma S Ma S =.99 (3) c RT where u S = 45 m/s, =.4, R = 87 J/(kg.K), and T = 98 K. Page of

9 shock_tube_05 Using the normal shock relations, the Mach number and the velocity, temperature, and pressure ratios across the shock are: Ma Ma Ma = (relative to the shock) (4) Ma T Ma Ma T T /T =.76 (5) Ma p Ma p /p =.54 (6) p p p T T u 0 0 S Ma Ma u u S Ma Ma p /p 0 = (7) T /T 0 = (8) (u S u )/u S = u = 05.4 m/s (9) u u = 0 u S u u S u S shock moves in this frame of reference shock is stationary in this frame of reference The air velocity and speed of sound downstream of the shock wave are: c RT c = m/s (T = K) (0) p = 53. kpa (abs) () The shock wave reflects as an expansion fan from the free surface boundary in order to maintain a constant pressure boundary condition. The speed of the first expansion wave front relative to the ground is: u = u c u = -6.0 m/s (to the left) () The speed of sound in region 3 may be found from the pressure ratio and noting that the process in going from to 3 is isentropic: c3 p3 c 3 /c = c 3 = m/s (3) c p The velocity in region 3 may be found by noting that we re crossing left-running characteristic curves in going from region to region 3: u 3 u 3 3 c c c c u 3 = 0.7 m/s (4) Thus, the speed of the last expansion wave front relative to the ground is: u B = u 3 c 3 u B = m/s (to the left) (5) lternately, the velocity in region 3 may be found directly in terms of the pressure ratio using: u3 u p3 c p3 u3 u c c p u 3 = 0.7 m/s (6) p Page of

Introduction to Fluid Mechanics. Chapter 13 Compressible Flow. Fox, Pritchard, & McDonald

Introduction to Fluid Mechanics Chapter 13 Compressible Flow Main Topics Basic Equations for One-Dimensional Compressible Flow Isentropic Flow of an Ideal Gas Area Variation Flow in a Constant Area Duct