1 / 56

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

wynter-graves
Download Presentation

FLOW IN A CONSTANT-AREA DUCT WITH FRICTION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related