Fanno flow.pdf

MECH 430 – FLUIDS 2Flow of a compressible fluid in a constant area duct with friction For an incompressible fluid where the density remains constant, friction results in a pressure drop in the flow direction. For a compressible flow, viscous dissipation heats up the gas resulting in a density change. Thus, the flow velocity and pressure can increase or decrease depending on the rate of expansion due to density decrease and the convective mass flux through the cross section of the pipe. As in the case of area change, friction has opposite effect if the flow is subsonic or supersonic. Let us first review the steady flow of an incompressible fluid in constant area pipe with friction (which results in a wall shear stress W w ). Referring to the sketch below: L p + dp p Ww The conservation of mass for steady flow gives U u=constant and since U constant, the flow velocity is constant throughout the length of the duct. The momentum equation gives: pA   p  dpA  W w Aw where A is the area of the duct ( A ( Aw 0 SD2 for a circular duct) and Aw is the wetted area 4 SDL for a circular pipe). Rearranging the momentum equation gives: dp W w Aw A 4 Ww L D where we have assumed a circular pipe for convenience. The wall shear stress is given by: In general.) This parabolic velocity profile gives the wall shear stress as: Ww § wu · ¸ © wr ¹ r P¨  R 4 Pu R  8Pu D The negative sign denotes the direction of W w that gives the parabolic profile and we have already considered the appropriate sign (i. we write:  dp L where Re D p1  p2 L 4W w D 32 Pu D2 64 § Uu 2 · 1 ¨ ¸ Re D © 2 ¹ D UuD is the Reynolds number based on the pipe diameter. The velocity profile for steady laminar incompressible flow in a circular pipe is given by: u u § § r · 2· 2¨ 1  ¨ ¸ ¸ © © R¹ ¹ where u is the mean flow velocity (i. Thus. we P define a friction factor “f” as: f 4W w u2 U 2 where f  Re D . direction) of W w when we write the momentum equation. the classical Hagen-Poiseuille Flow.e.e.§ wu · ¸ © wr ¹ r R § wu · where P is the coefficient of viscosity and ¨ ¸ © wr ¹ r R Ww P¨ is the velocity gradient at the wall (if we assume a laminar flow). For the present case of laminar Poiseuille Flow: 2 . is determined from experiments. (Moody diagram where f is given for a wide range of Reynolds number and wall roughness). Compressible Flow (Fanno Flow) For steady compressible flow in a constant area. i.64 Re D f We may write the pressure drop as: p1  p2 L 64 Re D § Uu 2 · 1 ¸¸ ¨¨ © 2 ¹D f Uu 2 D 2 For incompressible flow. we need not worry about the energy equation since U constant and the problem is a dynamic problem with pressure forces balancing the friction forces. . adiabatic duct with friction as illustrated below. The conservation of momentum can be written as: 3 1.: U P+dp U  dU T+dT u+du u p T Ww dx The conservation of mass gives: m dU U UuA constant  du u 0 since dA=0 for a constant area duct.e. The viscous heating is at the expense of the kinetic energy decrease in the flow but since there is no heat transfer. the flow velocity is uniform across the cross section of the duct. For adiabatic (not isentropic) flow where h u2 2 H0 cons tan t we get dh+udu=0 4 . the energy equation is given by: u2 h 2 ho constant Note that there is no work done by the viscous stress W w since the velocity at the wall vanishes.dpA  W w Dsx mdu 2. We should consider the flow velocity as an averaged value to permit the presence of a velocity gradient for the viscous stress to occur. where S and A are the perimeter and the area of the cross-section of the tube. the viscous heating remains in the flow. If the flow is adiabatic. respectively. there will be no shear stress at the wall. The no slip (i.e. In the present assumption of a quasi one dimensional flow. In an inviscid flow. Defining a friction factor: Ww f U u2 2 and a hydraulic diameter “DH” by: 4 * area wetted perimeter DH 4A S Equation 2 becomes: Uudu  dp  Uu 2 4 fdx DH 2 3. u=0) condition at the wall generates a velocity profile resulting in viscous dissipation. Since h=cpT. the energy equation can be written as: and c 2 dT du  J  1. and c p JR J 1 JRT . URT p and hence dU dp p U  dT T 5. 5 and 6. we can get an expression for du u JM 2 4 fdx 21  M . 4. we obtain an expression for as p U dp p du JM 2  JM  2 u 2 § 4 fdx · ¨ ¸ © DH ¹ From Equations 1. Rewriting Equation 3 as: Uudu  dp Uu 2 4 fdx  p 2 p DH p and noting that c 2 dp Jp . M 2 T u The equation of state for a perfect gas is: 0 4. 2 From Equation 1. DH 6. du as u 7. we see that: dU U  du u  JM 2 4 fdx 21  M 2 . we get dT T   J  1. DH From Equation 4. . M 2 5 du u 8. and substituting Equation 7 into the above yields -J J -1. M 2 . M 4 4 fdx 2 1. DH dT T 9. Substituting Equations 8 and 9 into Equation 5. gives: dp p  .  JM 2 1   J  1. M 2 4 fdx DH 21  M 2 . dM M du 1 dT  u 2 T And substituting Equations 7 and 9 into the above. gives: dM M § © J 1 · M 2¸ ¹ 4 fdx 2 DH 21  M 2 . thus c To get an expression for the Mach number. u . we note M dM M and since c 2 JRT . du dc  u c 1 dT 2 T Hence. dc c 10. JM 2 ¨ 1  11. For the variation of the stagnation pressure along the pipe. J po p J  1 2 · J 1 § M ¸ ¨1 © ¹ 2 and differentiating. Thus. we get: dpo po dp  p 1 6 JM 2 J 1 2 M2 dM M . we note that the definition of the stagnation pressure is the pressure one would obtain if we decelerate a flow isentropically to zero velocity. Dh The variation of entropy along the pipe can be found from: Tds = dh-vdp with h cpT JRT .The variation of p and M along the pipe is given by Equations 10 and 11. we get: dpo po  JM 2 4 fdx 2 12. pv=RT. the above becomes: J 1 dT J  1 dp  T J p ds cp dT dp and . thus we obtain for the change in T p entropy along the pipe the following expression: Equations 9 and 10 give the variation of ds cp ds R or since c p  J  1. thus. the various thermodynamic states can either increase or 7 . J § J  1· JM 2 4 fdx ¨ ¸ © J ¹ 2 DH JM 2 4 f 2 DH dx  dpo po 13. U . R . all the equations contain the term (1-M2) in the respectively. Except for R po denominator. M. M>1 and (1-M2) <0. Thus. P. the entropy increase can be correlated to the stagnation pressure loss due to viscous dissipation. T. Equations 7-13 give the variation of u. and (1-M2)>0 or supersonic. po and s along the pipe dpo ds and . depending on whether the flow is subsonic M<1. As in the case of the normal shock. e. Integration of the equations To get the variation of the various state variables along the duct. The sign + or – denotes whether the variable increases or decreases along the pipe (i. The table below summarizes the effect of friction on the fluid and thermodynamic states. the differential equations have to be integrated. The reverse happens for supersonic flow when the rate of convection (characterized by the flow velocity) dominates over the rate of expansion. The important one to start is the variation of the Mach number with distance given by Equation 11. both subsonic and supersonic flows indicate the same behavior of loss in stagnation pressure and increase in entropy from irreversible viscous dissipation. a decrease in temperature and pressure due to the expansion. increased value of x or positive dx). Rewriting it in the following form: 4 fdx DH 1  M 2 . for the stagnation pressure and entropy. for subsonic flow.decrease due to friction. Thus. Variable Velocity Mach Number Pressure Temperature Density Stagnation pressure Entropy Subsonic Supersonic + + + + + + + u M p T U po s The difference between subsonic and supersonic flow is a result of the competition between the rate of expansion due to viscous heating of the flow and the rate of mass convected through the cross section of the pipe. However. the rate of expansion (characterized by the sound speed) dominates. and the density decreases resulting in an increase in the flow velocity. The above can be expanded as: 8 .dM 2 § © JM 4 ¨ 1  J 1 2 · M 2¸ ¹ we can integrate the above using the method of partial fraction. if we have a pipe of length L and inlet Mach number M1. and if we want to know the Mach number at L. we assume the Mach number at x=0 is M and at x=L*. the Mach number is M. we integrate the above from 0<x<L* and 0<M2<1 (i. the friction factor is a function of the Reynolds number and varies along the duct as the fluid properties change. So.J 1 dM 2 dM 2 J  1 dM 2 J  1 2   JM 4 2J M 2 2J § J  1 2 · ¨1  M ¸ © ¹ 2 4 fdx DH Since friction always drives the flow towards M=1. M=1 and the flow is choked. M=1.e. we do the following: M2 M1 L*M2 L LM1 Referring to the sketch above: L L* M1  L* M 2 or 9 . The above equation says that if we have a duct where at x=0. then at x=L*. In general. However. We then obtain: J 1 4 fL* DH ª J  1 2 º 2J 2 M « » 1 M 2 « »  ln JM 2 « ¨§ 1  J  1 M 2 ¸· » «¬ © ¹ »¼ 2 14. we shall assume an average value for the friction factor from 0<x<L*. Equation 14 is computed for a range of Mach numbers and tabulated (Fanno flow tables). etc.4 fL* M1 4 fL* M 2  DH DH 4 fL DH 4 fL* M1 We know “L” and “M1”. . etc. we can integrate the other equations for . we can look up the table for L such that corresponds DH to M 1 . T. U . If we want to know the other flow variables dp dT like p. we can find the value for: * 1 4 fL* M 2 DH 4 fL* M 2 4 fL   DH DH We can then find the value of M2 corresponding to the value of 4 fL* M 2 . Then. From Equations 10 and 11. DH Equation 14 gives “M” as a function of L*. However. we can find the other variables. we write:  . it is p T more convenient to express them in terms of the Mach number so that once “M” at a section is known.  1   J  1. M 2 JM 2 § 4 fdx · ¸ ¨ © DH ¹ 21  M 2 . dp p dM M JM 2 § J  1 2· M ¸ ¨1 ¹ § 4 fdx · © 2 2 ¸ ¨ © DH ¹ 1  M 2 . and simplifying the above yields: dp p  .  1   J  1. the above can be written as: dp p J 1 dM 2 dM 2 2   2M 2 § J 1 2 · M ¸ 2 ¨1  2 © ¹ 10 . M 2 dM 2 J  1 2· M 2 § M ¸ 2¨ 1  © ¹ 2 Using partial fraction. Similarly.which integrates to yield: p p* 1 M § 2¨ 1  © J 1 J 1 2 15. we can find expressions for the other variables in terms of the Mach number as follows: T T* u u* § c· ¨ *¸ ©c ¹ M 2 § 2¨ 1  © J 1 J 1 2 J  1. · M ¸ ¹ 2 The above equation gives the pressure “p” at “M’ and p = p* when M=1. 17. J  1 2· § 2¨ 1  M ¸ © ¹ 2 J  1 2· § 2¨ 1  M ¸ © ¹ 2 J 1 U U* u u po po * J  1 2 · º 2J 1. ª § 2¨ 1  M ¸» « ¹ 1 © 2 « » M« J  1. J 1 s  s* cp ª º « » J  1. · M ¸ ¹ 2 1 M 18. » «¬ »¼ * 16. 2« » ln M J  1 2· » « 2§ «¬ 2 M ¨© 1  2 M ¸¹ »¼ 11 19. J 1 2J 20. . e. The Fanno Line The adiabatic flow of a compressible fluid in a constant area duct with friction satisfies the conservation of mass and energy i.Equations 15 to 20 are also computed for various Mach numbers and tabulated together 4 fL* in the Fanno tables for convenience in solving problems of compressible flow with DH in ducts with friction. i.: dh  vdp Tds we get: dh J  1 dp  h J p ds cp and from the equation of state p dp p URT . From the Tds equation. we get: dU U  dT T Hence.e. Uu constant dU U h  u2 2 du u ho 0 constant dh  udu 0 From these two equations. 12 dU U  dh h . we can obtain an expression for the variation of the enthalpy (or equivalently the temperature) and the entropy along the duct. we write: dU  U du u  du 2 2u 2 From the energy equation. we note that: u2 2 and du 2 2  dh d ho  h .dh J  1 § dU dh · ¨   ¸¸ J ¨© U h h ¹ ds cp 1 dh J  1 dU  J h J U From the conservation of mass. Thus. since ho h o h constant. we may write: dU  U d ho  h . 2ho  h . and the equation for the entropy becomes: 1 dh J  1 d ho  h .  J h 2J ho  h . ds cp We can integrate the above as following: ds s* c p ³ s dh J  1 ho  h d  ho  h .  h* h 2J ³ho  h*  ho  h . we can eliminate ho in the above equation and obtain the equation for 2 the Fanno Line as: 13 . 1 J³ h and obtain: s  s* cp Since ho J 1 1 J ln h J  1 § ho  h ·  ln¨¨ ¸* 2J h* © ho  H¹ h* . For any given value of M. we get: udu  Tds From the conservation of mass Uu 1 U dp constant . friction will drive the flow towards sonic conditions M=1 where the entropy is a maximum. Alternately. p . The Fanno line gives the variation of the enthalpy as a function of the entropy for a given mass flow rate in and stagnation enthalpy ho (or equivalently h*). we write: dh  Tds and from the energy equation dh 1 U dp udu. one ds could differentiate Equation 20 and equate 0 and solve for M. The lower branch is for supersonic flow and the upper branch is for subsonic flow. To prove that entropy is a maximum when the flow becomes sonic. we do dM the following: from the Tds equation. we get : Udu  udU 0 If we express U s.s  s* cp J 1 J 1 º 1 ª J § 2 · 2J § J  1 2J h h § · · ¸¸ ¨ ln «¨ * ¸ ¨¨  *¸ » «© h ¹ © J  1 ¹ © 2 h ¹ » »¼ «¬ 21. p . . we can write: § wU · § wU · dU s. . ¨ ¸ ds  ¨¨ ¸¸ dp © ws ¹ p © wp ¹ s 1 § wU · ¨ ¸ ds  2 dp c © ws ¹ p since the sound speed “c” is defined by: c2 § wp · ¨ ¸ © wU ¹ s Replacing dU in the continuity equation by the above expression yields : Udu  udU Udu  u § wU · dp  u ¨ ¸ ds 2 c © ws ¹ p 14 0 22. we get:  Uudu  UTds dp and replacing dp in Equation 22 by the above and rearranging gives: § UT § wU · · ¨ ¸ ¸ © ws ¹ p ¸¹ ©c Udu 1  M 2 .From the Tds equation. we write: dh ds 1 1 R dp  T p dh From the energy equation.  uds¨¨ 2 0 23.e. Another alternate expression for the Fanno Line that involves the Mach number can be obtained as follows: From the Tds equation. as M o 1 . i. The bracketed term is always positive. wU  0 as ds>0 at constant “p”).: Since UT 2 pT § wU · ¨ ¸ ! 0 c 2 © ws ¹ p Thus. thus “s” corresponds to an extremum (maximum) value when M=1. we get: dh  du 2 2 0 And the conservation of mass gives: dU U  du u  15 1 du 2 2 u2 . we write: ds dh Rdp  T p and rearranging.e. i. § wU · is always positive and ¨ ¸ is always negative (since density decreases c © ws ¹ p when heat is added at constant pressure and heat added means entropy increases. ds must vanish. we can express dh as:  du dh And using the equation of state p 2 u2 2 dU U URT . dh u 2 And solving for the slope U  p 1  J  1. we can express: dU U dp dT  p T Hence. M 2 2 u . we get: dh dp dh We can now write dU dh ds 1 p· 1 R§ p  ¨ 2  J  1. dh as: ds dh ds or § dp dT · ¸ u 2 ¨¨  T ¸¹ © p dp . When M=1.M 2 2 ¸ u ¹ T p©u JM 2T M 2 1 The above is an alternate form of the equation for the Fanno Line and it is clear that for dh subsonic flow where M<1. i. ds dh Fanno Line. o f and s o max . ds To plot a Fanno Line. For M>1. the locus of states is represented by the lower branch of the Line. we first note that the Fanno Line involves the conservation of mass and energy. 16 .e.  0 which corresponds to the upper branch of the Fanno ds dh ! 0 . thus the value of h* can be found for a given stagnation enthalpy “ho”. the Mach number at the jet exit is given by the isentropic relationship J po p § J  1 2 · J 1 M ¸ ¨1  2 © ¹ If we now connect a pipe to the exit of the converging nozzle. that h* J 1 Effect of duct length and back pressure The flow inside a duct with friction is influenced by the duct length and the back pressure of the exit of the duct. We assume the inlet to the duct is from a converging nozzle connected to a large plenum chamber where the stagnation pressure and temperature is “po” and “To”.Uu m u2 h 2 constant ho constant Thus. the exit pressure must be the same as the back pressure. Consider first just the nozzle. we get: h Where v 1 U m 2v 2 2 ho constant 24. then the exit of the nozzle corresponds to the entrance to the duct. 17 . Thus. The system is shown schematically in the sketch below. Note 2 ho . For a subsonic jet. Combining conservation of energy and mass. And Equation 21 gives “s” as a function of “h”. Consider first the case of subsonic flow in the duct. we pick different values for U and compute h from Equation 24. So. for a given exit pressure. given m and ho . a Fanno Line corresponds to a value for “ m ” and “ ho ”. is the specific value. Let the end of the duct be connected to a large plenum chamber where the pressure can be controlled. For this case. the exit pressure pe=pB . value of is less than the value DH DH When we lower the back pressure pB. The 4 fL 4 fL* corresponding to the inlet Mach number M1. then pe=p*=pB and Me=1.stagnation conditions x Pe plenum chamber PB Po. the flow is similar but the inlet Mach number to the duct (hence the mass flow rate) is increased. When the back pressure is reduced to a value where the flow is choked at the exit. the 18 . To Pe a x Pe = PB b x Pe = PB c x Pe = P* = PB x Pe = P* PB < P* ho x Me x Me Me = 1 x a b c  decreasing m For curve “a”. the flow is subsonic throughout the duct. After choking occurs. Again. The limiting condition will be when a normal shock occurs just at the duct exit so that the pressure downstream of the normal shock matches the back pressure pB. Thus. are all determined from the isentropic flow relationships (look up the tables). the under expanded sonic jet will undergo further expansion downstream of the duct exit to adjust to the ambient pressure. Similarly. For high pB. i. then the inlet Mach number will have to decrease to permit choking to occur in the longer length of pipe. the shock moves into the duct and becomes stronger since the Mach number is higher. Consider first a duct length less the critical value for the given inlet Mach number M1. when * pe=p >pB. the exit Mach number is still supersonic and the flow in the duct is supersonic throughout.e. the flow is subsonic downstream of the duct and for a subsonic flow. the flow is chocked at the exit.e. a normal shock will eventually be formed inside the duct. When pB increases beyond this limiting value. When pB>pe. pe<p since the flow is not choked at the exit and pe<p > > B . thus the exit pressure will have to be equaled to the ambient pressure. the inlet conditions to the pipe can adjust to the different values of the back pressure (or exit pressure) and the length of the duct. This is the case provided the exit pressure of the duct is greater than the back pressure pB in the plenum * chamber. For supersonic flow. If the duct is very long. curve “c”.e. If pB=p*.duct length corresponds to the critical value L* for the inlet Mach number (curve “c”). then A the exit Mach number. If the back pressure is further decreased. i. a normal shock will be formed inside the duct. Once a normal shock occurs inside the duct. but not significantly greater. if the duct length L>L* for the inlet Mach number. Once the area ratio 1* of the nozzle is fixed. For a sub critical duct length. then conditions inside the duct remains unchanged and adjustment to ambient pressure occurs downstream of the jet exit via oblique shock waves. pB<p*. the exit pressure must match the ambient pressure. DH Dh the length of the pipe is increased beyond this critical length. for subsonic flow. Downstream of the shock is subsonic. If pB=p*. pressure. then the jet exit pressure matches the ambient pressure and the jet is parallel. further adjustment occurs 19 . The three cases of “a” “b” and “c” are also illustrated by the 4 fL 4 fL* ) if Fanno Lines on the h-s diagram. i. The under expanded jet will adjust to ambient pressure via expansion waves. the same kind of events as described earlier for the case of L<L* occurs. Thus. Mach number at the duct inlet is supersonic and conditions downstream in the duct cannot influence the inlet condition unless a shock wave is driven back into the nozzle. pe=p*. no change occurs within the duct and further expansion occurs downstream of the duct exit to permit the jet pressure to adjust to the ambient pressure. The location of the shock is such that this condition is satisfied. etc. The position of the shock is such that the exit pressure matches the ambient pressure. pe=pB. The above discussion is for the case when the duct length is less than the critical value corresponding to the inlet Mach number. flow rate. the inlet of the duct must be connected to the stagnation tank A via a converging diverging nozzle. When pB increased further. then pe=p* and Me=1. (i. When the back pressure is increased to the exit pressure. If the duct length is the critical value.e. the oblique shock becomes stronger by increasing its angle to the flow. then the maximum Mach number the subsonic flow downstream of the shock inside the duct can accelerate to is M=1. there is only one location where the shock inside the duct is located. For pe=p*. the process is illustrated in the h-s diagram below. ho M< 1 x x isentropic flow in nozzle 20 x Me . For supersonic inlet flow and a very high pB so that a shock is driven inside the duct.downstream of the duct exit. Pe = PB normal shock transition M> 1 .