FLOW IN A CONSTANT-AREA DUCT WITH HEAT EXCHANGE

1. FLOW IN A CONSTANT-AREA DUCT WITH HEAT EXCHANGE (Combustion Chambers, Heat Exchangers)

2. Frictionless Flow in a Constant Area Duct with Heat Exchange Q/dm h1, s1, h2, s2, Rx= 0 Quasi-one-dimensional flow affected by: area change, friction, heat transfer, shock

3. N O F R I C T I O CH 12-4 H E A T E X C H A N G E C O N S T A N T A R E A • Governing Equations • Cons. of mass • Cons. of mom. • Cons. of energy • 2nd Law of Thermo. • Ideal Gas/Const. cp,cv • p = RT • h2-h1 = cp(T2 – T1) • s = cpln(T2/T1) • - Rln(p2/p1) {1-D, Steady, FBx=0 only pressure work}

4. Quasi-One-Dimensional, Steady, FBx = 0, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas*, cv, cp is constant Cons. Of Mass Cons. of Momentum Cons. of Energy Property relations for ideal gas with cv and cp constant 2nd Law of Thermodynamics

5. Constant area, frictionless, heat exchange = Rayleigh Flow No Rx

6. Constant area, frictionless, heat exchange = Rayleigh Flow Can find: p2, 2, T2, s2, h2, V2 If know: p1, 1, T1, s1, h1, V1 and Q/dm

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8. TS curve H E A T E X C H A N G E N O F R I C T I O N C O N S T A N T A R E A

9. Frictional, Constant Area, Adiabatic Flow Frictionless, Constant Area with Heat Transfer ? T Rayleigh Line Isentropic Flow s Isentropic Flow Fanno Line dA0 No Frictional, Changing Area, Adiabatic Flow

10. s2-s1 = cpln(T1/T2)-Rln(p2/p1) Need p2/p1 in terms of T2 and T1 After manipulation eqs 12.30a – 12.30g T s

11. Rayleigh Line For the same mass flow, each point on the curve corresponds to a different value of q added or taken away. T x s

12. FLASHBACK - FANNO LINE, ADIABATIC & CONSTANT AREA BUT FRICTION To h1 + V12/2 = h2 + V22/2 hO1 = hO2 cpTO1 = cpTO2 TO1 = TO2 x T1, s1 s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)]

13. RALEIGH LINE, NOTADIABATIC & CONSTANT AREA BUT NO FRICTION q+h1+V12/2 = h2+V22/2 q = hO2 – hO1 q = cp(TO2-TO1) The effect of heat addition is to directly change the stagnation (total) temperature of the flow

14. TS curve properties H E A T E X C H A N G E where is sonic ? N O F R I C T I O N C O N S T A N T A R E A

15. T Rayleigh Line B dT/ds = 0 s Properties: at A – highest s at B – highest T A ds/dT = 0

16. Want differential form of governing equations. V = ( +  )(V +  V) • pA – (p+ p)A = ( +  ) A(V +  V)2 - AV2 • p A= V A(V +  V) - AV2 • p A= V A  V • dp/ = -VdV

17. Momentum: dp/ = -VdV Ideal gas: p = RT dp = Rd(T) + RTd() dp/p = dT/T+ d/ Continuity: V = constant d/ + dV/V = 0

18. ds/dT = 0 dp/ = -VdV Tds = du + pdv (1.10a) du = d(h-pv) = dh – pdv –vdp Tds = dh – vdp = dh – dp/ Ideal gas: dh = cpdT Tds = cpdT – dp/ Tds = cpdT +VdV ds/dT = cp/T + (V/T)(dV/dT)

19. ds/dT = cp/T + (V/T)(dV/dT) Momentum: dp/ = -VdV Ideal gas: dp/p =d/ + dT/T Continuity: d/ + dV/V = 0 -VdV/p = d/ + dT/T p = RT -VdV/(RT)= d/ + dT/T -VdV/(RT)=d/+ dT/T -VdV/(RT)=-dV/V+ dT /T dV( 1/V – V/[RT]) = dT/T dV/dT = (1/T)/(1/V –V/RT) = 1 / (T/V – V/R)

20. ds/dT = cp/T + (V/T)(dV/dT) dV/dT = 1 / (T/V – V/R) ds/dT = cp/T + (V/T) ( 1/ (T/V – V/R)) ds/dT = 0 = cp/T + (VA/T) ( 1/ (T/VA – VA/R)) -cp = VA / ([T/VA –VA/R]) -cpT/VA + cpVA/R = VA VA (1 – cp/R) = -cpT/VA VA2 = -cpT / (1 - cp/R) = cpT / (cp/R - 1) VA2 = cpRT / (cp – R) R = cp – cv; cp/cv = k VA2 = kRTso MA = VA/(kRT)1/2 = 1

21. At A ds/dt = 0 A T s MA = 1

22. At B ds/dt = 0 B T s MB = ?

23. dT/ds = 1/ (ds/dT) dT/ds = 1/{cp/T + (V/T) ( 1/ (T/V – V/R))} 0 = (TB /VB - VB /R)/[cp/VB – cpVB/(RTB) + VB/TB] 0 = TB /VB - VB /R TB /VB = VB /R VB = (RTB)1/2 MB = VB/ (kRTB) = (RTB)1/2 / (kRTB)1/2 MB = (1/k)1/2 (~ 0.715 < 1)

24. At B ds/dt = 0 MB= (1/k)1/2 MB < 1 B T s Subsonic on top. Supersonic on bottom. What happens as you add heat? subtract heat?

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26. TS curve properties H E A T E X C H A N G E how does s change with q ? N O F R I C T I O N C O N S T A N T A R E A

27. T S = rev Q/T s Entropy, s, always increases with heating and decreases with cooling. If continue to heat at A, can’t stay on Rayleigh line since s must increase, so mass flow must change, get new (lower mass flow) Rayleigh line.

28. EFFECTS OF HEATING / COOLING ON FLUID PROPERTIES FOR RAYLEIGH FLOW Heat addition increases disorder and hence always increases entropy, whereas cooling decreases disorder and hence decreases entropy. Independent of Ma. cp ( To2 – To1 ) Heat addition increases stagnation temperature, cooling decreases stagnation temperature. Independent of Ma.

29. TS curve properties H E A T E X C H A N G E how does V change with q ? N O F R I C T I O N C O N S T A N T A R E A

30. To learn more about Rayleigh flow we need to also consider the energy equation. How does velocity change with Q? How does temperature change with Q? How does Mach number change with Q?

31. q = h + dh + (V + dV)2/2 – h – V2/2 q = dh + 2VdV/2 = dh + VdV Ideal gas: dh = cpdT q = cpdT + VdV q/(cpT) = dT/T + VdV/(Tcp) Ideal gas: cp = Rk/(k-1) q/(cpT) = dT/T + (k-1)VdV/(RkT) q/(cpT) = dV/V{[V/dV][dT/T] + (k-1)V2/(RkT)} dV/V = {q/(cpT)} {[V/dV][dT/T] + (k-1)V2/(RkT)}-1

32. dV/V = {q/(cpT)} {[V/dV][dT/T] + (k-1)V2/(RkT)}-1 dT/dV = T/V – V/R (slide 19) [V/dV][dT/T] + (k-1)V2/(RkT) = (V/T)[T/V – V/R] + V2/(RT) – V2/(kRT) = (V/T)[T/V – V/R + V/R] – Ma = 1 - Ma dV/V = {q/(cpT)}[1 – Ma]-1

33. RAYLEIGH – LINE FLOW ~0.715 dV/V = {q/(cpT)}[1 – Ma]-1 A B s is + s is - Ts diagram for frictionless flow in a constant-area duct with heat exchange

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35. TS curve properties H E A T E X C H A N G E how does T change with q ? N O F R I C T I O N C O N S T A N T A R E A

36. To learn more about Rayleigh flow we need to also consider the energy equation. How does velocity change with Q? How does temperature change with Q? How does Mach number change with Q?

37. RAYLEIGH – LINE FLOW A B ? How does T change with q ? Ts diagram for frictionless flow in a constant-area duct with heat exchange

38. RAYLEIGH – LINE FLOW T increasing with q A B Note, between A and B heating the fluid results in reducing the temperature! Not surprising if consider stagnation temperature and fluid velocity changes in this region. Ts diagram for frictionless flow in a constant-area duct with heat exchange

39. (from Rayleigh Line) For supersonic flow, temperature increases with heating and decreases with cooling. For subsonic flow and M < 1/(k)1/2, temperature increases with heating and decreases with cooling. For subsonic flow and 1/(k)1/2 < M < 1, temperature decreases with heating and increases with cooling.

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41. TS curve properties H E A T E X C H A N G E how does M change with q ? N O F R I C T I O N C O N S T A N T A R E A

42. To learn more about Rayleigh flow we need to also consider the energy equation. M = V/c How does velocity change with Q? How does temperature change with Q? How does Mach number change with Q?

43. (from Rayleigh Line) For supersonic flow, M decreases with heating and increases with cooling. For subsonic flow, M increases with heating and decreases with cooling. < 1 M = V/(kRT)1/2

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45. TS curve properties H E A T E X C H A N G E how does p change with q ? N O F R I C T I O N C O N S T A N T A R E A

46. To investigate how p depends on heat transfer, first go back to: (1st and 2nd laws, ideal gas, constant specific heats) s2 – s1 = cpln(T2/T1) – Rln(p2/p1) Keeping p constant T = Toe(s-so)/cp

48. (from Rayleigh Line) For supersonic flow, pressure increases with heating and decreases with cooling. For subsonic flow, pressure decreases with heating and increases with cooling.

49. (from Rayleigh Line) For supersonic flow, stagnation pressure decreases with heating and increases with cooling. For subsonic flow, stagnation pressure decreases with heating and increases with cooling.

50. For supersonic flow, pressure increases with heating and decreases with cooling. For subsonic flow, pressure decreases with heating and increases with cooling. + p1A + (dm/dt)V1 = p2A + (dm/dt)V2 For supersonic flow, velocity decreases with heating and increases with cooling. For subsonic flow, velocity increases with heating and decreases with cooling.