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Ch. 11: Introduction to Compressible Flow

Ch. 11: Introduction to Compressible Flow. Focus on 1-dimensional, compressible, inviscid flows. Liquids,  = constant for us (1% increase in  for every 1.6 km deep). Air, 1% change for every 26 m deep. M = 0.3 ~ 5% = /; M = 0.3 ~ 100 m/s or 230 mph.

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Ch. 11: Introduction to Compressible Flow

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  1. Ch. 11: Introduction to Compressible Flow • Focus on 1-dimensional, compressible, inviscid flows • Liquids,  = constant for us (1% increase in  for every 1.6 km deep) • Air, 1% change for every 26 m deep • M = 0.3 ~ 5% = /; M = 0.3 ~ 100 m/s or 230 mph • Significant density changes imply significant compression • or expansion work on the gas, which can change T, e, s, … • Compressibility: fluid acceleration because of friction, fluid deceleration in a converging duct, fluid temperature decrease • with heating • Ideal Gas: p = RT (simple, good approximations for our engineering • applications, captures trends)

  2. Smits/A Physical Introduction To Fluid Mechanics

  3. Density gradients will affect how light is transmitted though medium (by affecting index of refraction). By applying the Gladstone-Dale formula it becomes evident that the shadowgraph is sensitive to changes in the 2nd derivative of the gas density. Strength of shock can be related to width of dark band. - Methods of Experimental Physics – Vol 18, Martin Deflection of light caused by shock compressed gas ahead of a sphere flying at supersonic speed.

  4. Thoughts on the increased complexity of incompressible flow

  5. GOVERNING EQUATIONS FOR NEWTONIAN FLUIDS INCOMPRESSIBLE /t + /xk(uk) = 0 becomesuk/xk= 0 uj/t + ukuk/xk = -p/xj+/xj( uk/xk)+/xi[(ui/xj+uj/xi)]+fj becomes uj/t + ukuj/xk = -p/xj + (2ui/xjxj) + fj 4 Equations: continuity and three momentum 4 Unknowns: p, u, v, w Know: , , fj

  6. GOVERNING EQUATIONS FOR NEWTONIAN FLUIDS COMPRESSIBLE /t + /xk(uk) = 0 uj/t + uk/xk = -p/xj +/xj( uk/xk) + /xk[(ui/xk + (uj/xi)] + fj p = p(,T) Thermal ~ p = RT e = e(,T) Caloric ~ e = CvT e/t + uke/xk = -puk/xj +/xj(k T/xj) + (uk/xk)2 + (ui/xk + uj/xi)(uk/xk) e is the internal energy 7 Equations: continuity, momentum(3), energy, thermal, state 7 Unknowns: p, u, v, w, e, T,  Know: , fj, , k

  7. Thoughts on the speed of sound

  8. COMPRESSIBLE FLOW front c2 = (p/)s

  9. COMPRESSIBLE FLOW front If fluid incompressible, gas would behave like solid body and move everywhere at piston speed. If pressure disturbance is small relative to p1 then “front” propagates at speed of sound. If large shock waves occur where speed, temperature, density and pressure change significantly across shock. (Speed of shock is between the speed of sound in the compressed and undisturbed gas.)

  10. ~ SPEED OF SOUND ~ Sound waves are pressure disturbances << ambient pressure. For loud noise: p ~ 1Pa whereas ambient pressure is 105 Pa Speed of sound: c2 = (p/)s Assumptions: ideal gas & isentropic p/k = const, or differentiating dp/k – pk -k-1 d= 0 dp/(p/cont) – (const k) k -k-1 d= 0 dp/p – kd/ = 0

  11. ~ SPEED OF SOUND ~ dp/p – kd/ = 0 dp/d = kp/ p = RT for ideal gas then c2 = kRT For 20oC and 1 atmosphere c = 343 m/s = 1126 ft/s = 768 mph

  12. M = 1 Dynamic Pressure  Static Pressure

  13. M = V/c M2 = V2/c2 = V2/kRT (ideal gas) p = RT (ideal gas) M2 = 2(1/2V2/kRT) = 2(1/2 V2/(kp/)) M2 = 2[1/2 V2/(kp)] ~ 1/2 V2/p M2 ~ dynamic pressure/ static pressure

  14. Thoughts on the speed of sound As related to the speed of the source

  15. Regimes of flow: • Acoustics – fluid velocities << c, speed of sound; • fractional changes in p, T and  are important. • (2) Incompressible flow – fluid velocities < c, speed of sound; • fractional changes in  are not significant; fractional changes • in p and T are very important • (3) Compressible flow (gas dynamics) – fluid velocities ~ c, • speed of sound; fractional changes in p, T and  are all • important.

  16. SUB SONIC SUPER Propagation of Sound Waves from a Moving Source

  17. Some Assumptions

  18. It is assumed that the system is always in equilibrium.

  19. It has been found by experiment that as long as the temperatures and pressures are not too extreme, the flow attains an instantaneous equilibrium. This continues to hold even inside shock waves. For all the flows examined here, all systems will be assumed to be in equilibrium at all times. p1,1, T1, s1, h1 p2,2, T2, s2, h2

  20. It is assumed that all gases obey ideal gas law: p =  RT Kelvin (or Rankine) Not gauge pressure R = Ru/Mm = 287.03 m2/(s2-K) = (N-m)/(kg-K) = J/(kg-K) R = 1716.4 ft2/(s2-R)

  21. Conservation of Energy (note that u is now the internal energy)

  22. FIRST LAW OF THERMODYNAMICS Q + W = E = (KE + PE + U) Q/m + W/m = E/m; q + w = u W = - pdV U, internal energy, is energy stored in molecular bonding forces and random molecular motion. (KE and PE we will ignore)

  23. Ideal gas is composed of point particles which exhibit perfect elastic collisions. Thus internal energy is a function of temperature only. U = f(T) Enthalpy, h, defined as: h = u + pv ; h = f(T) since h(T) = u(T) + RT

  24. Specific Heat for Ideal Gas dQ = mCv,pdT

  25. Specific heat is defined as the amount of heat required to raise the temperature of a unit mass of substance by1oK. Different for constant volume or pressure.

  26. Definition of heat capacity at constant volume: mCvdT = dQ or CvdT = dq dq + dw = du if Vol constant, w = -pdv = 0, then dq = du, Cv = du/dT “It can be shown that du = Cv dT even if volume not held constant!!” - pg 41, Thermal-Fluid Engineering, Warhaft

  27. Definition of heat capacity at constant pressure: mCpdT = dQ or Cp = dq/dT dq + dw = dq – pdv = du h = u + pv dh = du +pdv +vdp if pressure constant, dh = du + pdv = dq Cp = dh/dT ~ again can be shown to be true even if pressure is not constant!!

  28. Cp = dh/dT* Cv = du/dT* h = u + pv = u + RT* dh = du + RdT dh/dT = du/dT + R Cp – Cv = R* * IDEAL GAS

  29. cp/cv = k cp – cv = R cp/cp –cv/cp = R/cp 1 – 1/k = R/cp (1– 1/k)/R = 1/cp cp = R/(1-1/k) = kR/(k-1)

  30. cp/cv = k cp – cv = R cp/cv –cv/cv = R/cv k – 1 = R/cv (k – 1)/R = 1/cv cv = R/(k – 1)

  31. Another Assumption

  32. It is assumed that cp/cv is not a function of T calorically perfect For a perfect gas cp/cv = 1.4

  33. cp/cv = k is not a function of temperature k

  34. cp/cv = k = 1.4 for perfect gas

  35. The Second Law

  36. “The second law of thermodynamics can be stated in several ways, none of which is easy to understand.” – Smits, A Physical Introduction to Fluid Mechanics DEFINITION S = rev Q/T or dS = (Q/T)rev dq = du + pdv Tds = du + pdv = dh –vdp always true!

  37. DEFINITION S = rev Q/T or dS = (Q/T)rev Change in entropy intimately connected with the concept of reversibility – for a reversible, adiabatic process entropy remains constant. For any other process the entropy increases.

  38. What we can do with: Tds = du + pdv = dh –vdp + ideal gas and constant specific heats

  39. Cv = du/dT Ideal Gas Cp = dh/dT p = RT = (1/v)RT Tds = du + pdv = dh –vdp ds = du/T + RTdv/T ds = CvdT/T + (R/v)dv s2 – s1 = Cvln(T2/T1) + Rln(v2/v1) s2 – s1 = Cvln(T2/T1) - Rln(2/1)

  40. s2 – s1 = Cvln(T2/T1) - Rln(2/1) If isentropic s2 – s1 = 0 ln(T2/T1)Cv = ln(2/1)R Cp – Cv = R; R/Cv = k – 1 2/1 = (T2/T1)Cv/R = (T2/T1)1/(k-1) assumptions ISENROPIC & IDEAL GAS & constant cp, cv

  41. Cv = du/dT Ideal Gas Cp = dh/dT p = RT = (1/v)RT Tds = du + pdv = dh –vdp ds = dh/T - vdp ds = CpdT/T - (RT/[pT])dp s2 – s1 = Cpln(T2/T1) - Rln(p2/p1) s2 – s1 = Cpln(T2/T1) - Rln(p2/p1)

  42. s2 – s1 = Cpln(T2/T1) - Rln(p2/p1) If isentropic s2 – s1 = 0 ln(T2/T1)Cp = ln(p2/p1)R Cp – Cv = R; R/Cp = 1- 1/k p2/p1 = (T2/T1)Cp/R = (T2/T1)k/(k-1) assumptions ISENROPIC & IDEAL GAS & constant cp, cv

  43. Stagnation Reference (V=0)

  44. BE: 1-D, energy equation for adiabatic and no shaft or viscous work. (p2/2) + u2 + ½ V22 + gz2 = (p1/1) + u1 + ½ V12 + gz1 Definition: h = u + pv = u + p/; assume z2 = z1 h2 + ½ V22 = h1 + ½ V12 = ho + 0 Cp = dh/dT (ideal gas) ho – h1 = cp (To – T) = ½ V12 T0 = ½ V12/cp + T = T (1 + V2/[2cpT])

  45. T0 = ½ V12/cp + T = T (1 + V2/[2cpT]) cp = kR/(k-1) T0 = T (1 + V2/[2(kR/(k-1)]T]) T0 = T (1 + (k-1)V2/[2kRT]) c2 = kRT T0 = T (1 + (k-1)V2/[2c2]) V2/ c2 T0 = T (1 + [(k-1)/2] M2)

  46. To/T = 1 + {(k-1)/2} M2 STEADY, 1-D, ENERGY EQUATION FOR ADIABATIC FLOW OF A PERFECT GAS

  47. /o = (T/To)1/(k-1) To/T = 1 + {(k-1)/2} M2 /o = (1 + {(k-1)/2} M2 )1/(k-1) Ideal gas and isentropic and constant cp, cv (isentropic = adiabatic + reversible)

  48. p/p0 = (T/To)k/(k-1) To/T = 1 + {(k-1)/2} M2 p/p0 = (1 + {(k-1)/2} M2)k/(k-1) Ideal gas and isentropic and constant cp, cv (isentropic = adiabatic + reversible)

  49. QUIZ • When a fixed mass of air • is heated from 20oC to 100oC – • What is the change in enthalpy? • For a constant volume process, • what is the change in entropy? • For a constant pressure process, • what is the change in entropy? • For an isentropic process • what are the changes in p and ? • Compare speed of sound • for isentropic and isothermal conditions.

  50. h2 – h1 = Cp(T2- T1) • s2 – s1 = Cvln(T2/T1) • s2 – s1 = Cpln(T2/T1) • 100/ 20 = (T100/T20)2.5 • 2.5 = 1/(k-1) k = 1.4 for ideal gas • p100 / p20 = (T100/T20)3.5 • 3.5 = k/(k-1) k = 1.4 for ideal gas

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