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FLOW IN A CONSTANT-AREA DUCT WITH FRICTION

FLOW IN A CONSTANT-AREA DUCT WITH FRICTION. (not isentropic, BUT constant area, adiabatic). (short ducts/pipes; insulated ducts/pipes). Constant Area Duct Flow with Friction. friction. Quasi-one-dimensional flow affected by: no area change, friction , no heat transfer, no shock. C O N

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FLOW IN A CONSTANT-AREA DUCT WITH FRICTION

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  1. FLOW IN A CONSTANT-AREA DUCT WITH FRICTION (not isentropic, BUT constant area, adiabatic) (short ducts/pipes; insulated ducts/pipes)

  2. Constant Area Duct Flow with Friction friction Quasi-one-dimensional flow affected by: no area change, friction, no heat transfer, no shock

  3. C O N S T A N T A R E A • Governing Euations • Cons. of mass • Cons. of mom. • Cons. of energy • 2nd Law of Thermo. • (Ideal Gas/Const. cp,cv) • Eqs. of State • p = RT • h2-h1 = cp(T2 – T1) • s = cpln(T2/T1) • - Rln(p2/p1) A D I A B A T I C 12.3 F R I C T I O N CH {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, adiabatic = Fanno Flow A1= A2 RXonly friction No Q/dm term

  6. Constant area, adiabatic but friction If know: p1, 1, T1, s1, h1, V1 and Rx Can find: p2, 2, T2, s2, h2, V2 properties changed because of Rx

  7. (T-s curve)

  8. T-s diagram for Fanno Line Flow s2-s1 = cpln(T2/T1) – Rln(p2/p1) p = RT; p2/p1 = 2T2/(1T1); R = cp-cv s2-s1 = cpln(T2/T1) – Rln(p2/p1) = cpln(T2/T1) – [Rln(2/1) + (cp-cv)ln(T2/T1)] = – Rln(2/1) + cvln(T2/T1) 2V2 = 1V1; 2/1 = V1/V2 s2-s1 = cvln(T2/T1) – Rln(V1/V2)

  9. s2-s1 = = cvln(T2/T1) – Rln(V1/V2) Energy equation (adiabatic): h + V2/2 = ho; V = (2[ho – h])1/2 Ideal Gas & constant cp; h = cpT V = (2cp[To – T])1/2 -ln[V1/V2] = -(1/2)ln[(To-T1)/(To-T2)] = (1/2)ln[(To-T2)/(To-T1)] s2-s1 = cvln(T2/T1) + ½Rln [(To-T2)/(To-T1)] (p to to V to T)

  10. s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)] To Locus of possible states that can be obtained under the assumptions of Fanno flow: Constant area Adiabatic (ho = h1+V12/2 = cpTo) x T1, s1, V1, …

  11. TS curve properties direction ? F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  12. s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)] Note – can only move from left to right because s2> s1 non isentropic. (Friction, Rx, is what is changing states from 1 to 2 and it is not an isentropic process.)

  13. TS curve properties where is sonic ? F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  14. s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)] Properties at P Where ds/dT = 0

  15. s-s1 = cvln(T/T1) + ½Rln [(To-T2)/(To-T1)] d(s – s1) /dT= ds/dT = 0 ds/dT = cv/T+{(cp-cv)/2}[-1/(To-T)] = 0 1/T = {(k-1)/2}[1/(To-T)] T(k-1) = 2(To – T)

  16. T(k-1) = 2(To – T) h + V2/2 = cpT + V2/2 = ho = cpTo V = (2cp[To – T])1/2 2(To – T) = V2/cp T(k-1) = V2/cp

  17. T(k-1) = V2/cp at P V2 = cp T (k-1) = cp T (cp/cv- cv/cv) V2 = (cp/cv)T(cp-cv) V2 = kRT For ideal gas and ds = 0: c2 = kRT Therefore V = c at P, where ds/dT = 0 Sonic condition

  18. TS curve properties how does V change ? F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  19. What else can we say about Fanno Line? Energy equation: h + V2/2 = constant = ho=cpTo As h goes down, then V goes up; but h=cpT, so as T goes down V goes up; To = const T goes down so V goes up Subsonic ? Sonic Supersonic ? T goes up so V goes down Tds > 0

  20. TS curve properties how does M change ? F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  21. What else can we say about Fanno Line? What does M do? M = V/[kRT]1/2 T goes down; V goes up Subsonic M increasing Sonic Supersonic M decreasing h+V2/2 = ho h = cpT T goes up; V goes down

  22. Note – friction causes an increase in velocity in subsonic flow! Turns out that pressure dropping rapidly, making up for drag due to friction.

  23. TS curve properties how does  change F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  24. What else can we say about Fanno Line? What does  do? V goes up, then goes down Subsonic Sonic Supersonic V= constant V goes down,then goes up

  25. TS curve properties how does p change F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  26. What else can we say about Fanno Line? What does p do? T &  goes down, p goes down Subsonic Sonic Supersonic p =  R T T &  goes up, p goes up

  27. What else can we say about Fanno Line? in summary  V = constant T goes down; V goes up Subsonic p and  decreases Sonic Supersonic p and  increases p =  R T T goes up; V goes down

  28. TS curve properties how does o and pochange F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  29. What else can we say about Fanno Line? What do o and po do? po =oRTo Since To is a constant (so To1 = To2 = To) then po and o must change the same way.

  30. What else can we say about Fanno Line? What do o and po do? 1 so2 – so1 = cpln(To2/To1) – Rln(po2/po1) so2 – so1 = cpln(To2/To1) – Rln(o2/o1) 1 Since so2 > so1 then po2 and o2 must both decrease!

  31. TS curve properties (summary) F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  32. TS curve properties (critical length) F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  33. M<1 a M=0.2 M=0.5 b c ? a b c

  34. M1 < 1 flow is choked For subsonic flow can make adjustments upstream – mass flow decreases

  35. M1 > 1 For supersonic flow adjustments can not be made upstream – so have shock to reduce mass flow

  36. subsonic, supersonic, shock M<1 M<1 M<1 M>1 M>1 M>1

  37. Fanno Flow (examples) F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  38. Example ~ FIND Ve and Te

  39. Basic equations for constant area, adiabatic flow: V = constant Rx + p1A –p2A = (dm/dt)(V2 – V1) h1 + V12/2 = h2 + V22/2= ho {= constant} s2 > s1 p = RT h = h1 – h2 = cp T; {To = constant} s = s2 – s1 = cpln (T2/T1) –R ln(p2/p1) Local isentropic stagnation properties To/T = 1 + [(k-1)/2]M2 Given: adiabatic, constant area, choked (Me = 1), To = 25oC, Po = 101 kPa (abs) Find: Ve, Te; include Ts diagram

  40. Given: adiabatic, constant area, choked (Me = 1), To = 25oC, Po = 101 kPa (abs) Find: Ve,Te; include Ts diagram Computing equations: (1) To/Te = 1 + [(k-1)/2]Me2 (2) Ve = Mece = Me(kRTe)1/2

  41. ASSUMPTIONS / NOTES for EQUATIONS USED To/Te = 1 + [(k-1)/2]Me2 Equation for local isentropic stagnation property of ideal gas, so assume ideal gas Used the relation: To = constant from h + V2/2 = h0 = cpTo Assumed that cp is constant; adiabatic flow, P.E. = 0; 1-D flow (uniform at inlet), steady, dWs/dt = dWshear/dt = 0 Ve = Mece = Me(kRTe)1/2 Ideal gas (experimentally shown that sound wave propagates isentropically)

  42. (1) To/Te = 1 + [(k-1)/2]Me2; (2) Ve = Mece = Me(kRTe)1/2 To constant so at exit know To and Me so use (1) to solve for Te Given Me and having solved for Te can use (2) to compute Ve Te = 248K, Ve = 316 m/s

  43. T-s Diagram (Me = 1)

  44. Fanno Flow (examples) F R I C T I O N CH A D I A B A T I C 12.3 C O N S T A N T A R E A

  45. Example ~ constant mass flow ? Pmin, Vmax ? Where do they occur?

  46. Basic equations for constant area, adiabatic flow: V = constant Rx + p1A –p2A = (dm/dt)(V2 – V1) h1 + V12/2 = h2 + V22/2= ho {= constant} s2 > s1 p = RT h = h1 – h2 = cp T; {To = constant} s = s2 – s1 = cpln (T2/T1) –R ln(p2/p1) Local isentropic stagnation properties To/T = 1 + [(k-1)/2]M2 P2 V2 V1

  47. P2 V2 V1 Computing equations: (1) p = RT (2) dm/dt = VA (3) To/Te = 1 + [(k-1)/2]Me2 (2) Ve = Mece = Me(kRTe)1/2

  48. ASSUMPTIONS / NOTES for EQUATIONS USED p = RT Ideal gas ( point particles, non-interacting) dm/dt = VA Conservation of mass

  49. ASSUMPTIONS / NOTES for EQUATIONS USED To/Te = 1 + [(k-1)/2]Me2 Equation for local isentropic stagnation property of ideal gas Used the relation: To = constant from h + V2/2 = h0 = cpTo Assumed that cp is constant; adiabatic flow, P.E. = 0; 1-D flow (uniform at inlet), steady, dWs/dt = dWshear/dt = 0 Ve = Mece = Me(kRTe)1/2 Ideal gas (experimentally shown that sound wave propagates isentropically)

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