Chapter 3 Steady 1D Flow of a Perfect Gas. Steady: Properties and other quantities do not vary with time.

Transcription

1 Chapter 3 Steady 1D Flow of a Perfect Gas Steady: Properties and other quantities do not vary with time. 1D: Only one spatial coordinate is needed to describe the dependent variables u, T, etc. Differential-form governing equations are ODEs. 3.1 General 1D flow of a perfect gas Governing equation for 1D, steady flow of a perfect gas 1D flow through a CV(differential in flow dir.): (Fig. 3.1) (Uniform properties are assumed at a flow cross section) 1

2 Continuity: ua = constant or Thus, -> Momentum: where c is the duct circumference Energy: Equation of state and other thermodynamic relations : P = RT, dh = c p dt, c p c v = R, c p /c v =, etc. [Example] Integral-form governing eqs Important concepts and definitions (1) Speed of sound : speed of a weak compression or expansion wave traveling in a medium. a = p/ s In a perfect gas, a = ( RT) 1/2 In an incompressible medium, a -> infinity (2) Mach number : M = u/a (3) Stagnation state and sonic state (i) Stagnation state(o) : Consider a hypothetical process in which a fluid element at a given state is decelerated to zero velocity and zero potential energy : 2

3 For the hypothetical process w/o PE change : For constant specific heats : In terms of M and : For a perfect gas undergoing an isentropic process : Thus, and 3

4 (a) h o or T o : a measure of the total energy possessed by the fluid element at a given state (the starting state of the hypothetical process) h o or T o = constant if a flow is steady, adiabatic and no work transfer. (b) p o : a measure of the high-grade energy possessed by the fluid element. p o = constant for an adiabatic, no-work transfer flow without flow losses (e.g., friction, unrestrained expansion, shock, etc.) Flow cross sectional area A : For 1D flow, Using the result of energy equation and the M and a definitions, we obtain : Thus the mass flow rate can be expressed as : (ii) Sonic state(*) : similar to the definition of the stagnation state, but the fluid element is accelerated or decelerated to Mach one in a reversible, adiabatic process without work transfer. 4

5 3.2 Isentropic and non isentropic flows Recall the thermodynamic relation for entropy change : 5

6 Replacing T and p in the above equation by stagnation properties, Irreversibility effects (friction, shock, etc.) on s, p o, and A* : 6

7 Since the flow is steady without heat and work transfer with its surroundings -> Total energy remains unchanged : h o1 = h o2 = h o3 = h o4 and T o1 = T o2 = T o3 = T o4 Flow is isentropic between 1 and 2 and between 3 and 4, but irreversible between 2 and 3. Thus, s 1 = s 2 < s 3 = s 4 High-grade energy is converted to low-grade energy in process 2 to 3 -> p o1 = p o2 > p o3 = p o4 7

8 The flow is (internally) heated in the above energy conversion process. As a result, A 1 * = A 2 * < A 3 * = A 4 * 3.3 Isentropic flow with area change For 1D, steady reversible and adiabatic flow w/o work transfer, p o, T o, and A* remain unchanged. The equations introduced in Sec. 3.1 and Fig. 3.3 allow p/p o, T/T o, A/A*, etc. to be determined from local M. [Example] Solve for downstream properties from (a) the continuity, momentum, equations; (b) the isentropic flow diagram(fig. 3.3) or tables (App. A) Variation in flow cross-sectional area 8

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10 Conclusions : (1) A throat is needed to accelerate an isentropic flow from subsonic to supersonic flow. (2) Different nozzle and diffuser contours for M > and < Converging-only nozzles 10

11 3.3.3 Converging-diverging nozzles 11

12 3.4 Shocks Normal shocks Normal compression and expansion waves (wave front perpendicular to the flow direction). [Example] A piston-cylinder device (1) Generation of a weak compression wave : At time t = 0, the piston is given an incremental velocity du (say, du = 10-6 m/s) to the left. 12

13 u = 0 p a Wave front du p + dp + d du p + dp Pressure p If the frame of reference is traveling with the wave front, the flow becomes a steady flow from the left to the right. u = a p Wave front Control volume u = a - du p + dp + d Governing equations for 1D constant-area flow of a perfect gas through the control volume : Continuity : a = ( + d ) (a - du) du ~ a d / Momentum : Aa[(a - du) - a] = pa - (p + dp)a du ~ dp/( a) Combining the two equations to eliminate du, a 2 = dp/d where a = speed of sound Since the disturbance is infinitesimal if du ~ 0, the process is reversible and adiabatic (i.e., isentropic). 13

14 Thus, a 2 = p/ s For a perfect gas, pv = constant and pv = RT -> a 2 = RT (2) Formation of a shock wave : p u = 0 p Wave 1 (generated at t1) a1 p + dp 1 a 2 du1 p + dp1 + d 1 Wave 2 (generated at t2) du + du 1 2 p + dp1+ dp + d 1+ d 2 2 p + dp + dp 1 2 Pressure (3) Generation of expansion waves : 14

15 p u = 0 p Wave 1 (generated at t1) a1 a 2 - du -du du 2 p - dp 1 d 1 Wave 2 (generated at t2) p - dp 1- dp d 1- d 2 2 Wave 1 (generated at t1) p - dp 1 Wave 2 (generated at t2) p - dp - dp 1 2 Pressure u = 0 p At time t3 Wave 1 (generated at t1) Wave 2 (generated at t2) u = 0 p At St time t4 [Examples] Stationary NS, moving NS, CD nozzle operation Oblique shocks (1) Generation of a Mach wave (an oblique wave since the wave front is not perpendicular to the flow) 15

16 (2) Generation of an oblique shock [Example] A supersonic flow over a corner Gradual compression or expansion in the vicinity of the corner : 16

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18 Equations for an oblique shock. Oblique - shock geometry : 18

19 (i) Mass and momentum transfer normal to the shock wave : 1 u 1n = 2 u 2n p u 1n 2 = p u 2n 2 -> pressure increases due to deceleration. (ii) Momentum transfer tangent to the shock wave :... -> u 1t = u 2t = u t -> tangential velocity component is constant across the shock. Thus, an oblique shock can be considered as a tangential velocity u t superimposed on a normal shock : (iii) Energy equation : h 1 + u 1 2 /2 = h 2 + u 2 2 /2 -> h 1 + u 1n 2 /2 = h 2 + u 2n 2 /2 Combining the above equations and using the velocity- Mach number-wave angle relations to eliminate all properties and velocity components, we obtain the -M relation : 19

20 If upstream Mach number M 1 and deflection angle are known : -> determine shock angle from the -M relation. -> determine M n,1 from M n,1 = M 1 sin > use the normal shock results to find M n,2, p 2 /p 1, T 2 /T 1, p o2 /p o1 -> determine M 2 from M n,2,, and... etc. The oblique shock diagrams for = 1.4 (Figs to 12) For a given deflection angle : (i) < max : 2 shock angles : one corresponds to a weak shock, and the other a strong shock (depending on the downstream pressure) (ii) = max : 1 shock angle (iii) > max : No solution exists for a straight oblique shock. A detached, curved shock will occur. 3.5 Prandtl-Meyer Expansion Waves 20

21 (1) Consider the first expansion wave : The relation between the change in flow direction and the upstream Mach number is : (2) The deflection angle of the expansion fan is To carry out the integral on the RHS, du/u must be expressed as a function of M : 21

22 (3) Prandtl-Meyer function : Let LHS = (M) in the above equation and RHS is integrated from M = 1 to an arbitrary M value -> A table can be constructed for P-M flows. The deflection angle for a supersonic flow expanding from M 1 to M 2 can be determined from : = (M 2 ) - (M 1 ). 3.6 Compressible flow analyses using FLUENT. Compressible flow through a CD nozzle, supersonic missiles, etc. 22

23 Appendix A : Compressible- flow tables for = 1.4 (posted on the course website) (1) Isentropic flow tables. (2) Normal shock table. (3) Prandtl-Meyer flow table. Appendix B : Matlab codes for CH. 3 (posted on the course website) (1) Isentro : for 1D isentropic flows (Multiple solutions possible) (2) NormalShock : for 1D Normal shocks (3) PrandtlMeyer : Generation of a Prandtl-Meyer flow table. 23