Fanno FlowGas Dynamics 1 Simple frictional flow ( Fanno Flow ) Adiabatic frictional flow in a constant-area duct * The Flow of a compressible fluid in a duct is Always accompanied by :- Variation in the cross –sectional area of the duct * Heat transfer Friction The above parameters contribute to changes the flow properties. * Although, it is difficult in many cases to separate the effects of each of these parameters , yet in order to provide an insight into the effect of friction , adiabatic flow thermally insulated in a constant area duct is analyzed in this chapter. Gas Dynamics * The entropy of the system still increase because of friction …. - Friction is associated with the turbulence and viscous shear of molecules of the gas . - Irreversibility associated with friction causes a decrease in the stagnation pressure , steady 1-D flow as a perfect gas . Assumptions: - Adiabatic frictional - flow in a constant cross- sectional area duct . Gas Dynamics V 1 V + dV τw 2 ρ V X X + dX ρ + dρ V + dV W D 2g D 2g Fanning coefficient of friction Darcy-weisbech coefficient of friction 2 .Momentum equation B = 4A DH = wetted perimeter DH = hydraulic diameter V X ρV V + dV 1 τw 2 X + dX AP − A( P + dP) − τ w Bdx = ρAV (V + dV − V ) − AdP − τ w Bdx = ρAVdV 1 2 τ w = F ( ρV ) 2 2 V + dV 1 ρ + dρ 1 ρV 2 2 Dynamic pressure F is the conventional Fanning friction factor ( circular duct ) F for flow over as the drag coefficient hloss in pipe Gas Dynamics LV LV = 4 FF = f D. 1 ( eq. roughness of boundaries ) The friction factor for turbulent in smooth ducts is the Von-Karman Nikurdse formula Gas Dynamics 1 = −0.From eq.8 + 2 log10 (Re 4 f ) 4F .1 / A) − AdP − τ w Bdx = ρAVdV1 VdV= dV2/2 B dP + τ w dX + ρVdV = 0 A 4 F ρV 2 ρV 2 dV 2 dP + dX + =0 2 DH 2 2 V 2 ρV 2 dV 2 ρVdV = 2 V2 ρV 2 τw = F 2 4A DH = B F = f( Reynold’s number . dh + VdV = 0 2 Mach number 2 2 2 Gas Dynamics dM dV dT = 2 − 2 M V T V M = γRT 5 6 2nd law of thermodynamics ds ≥ 0 7 .Perfect gas law Continuity Eq. P = ρRT m = ρV =Constant A dρ dV ho V2 = h+ = Constant 2 ρ + V =0 dP dρ dT = + P ρ T 3 4 Energy Eq. Energy Eq. T . S & (τ DH Gas Dynamics ds ≥ 0 Variable ) . ρ . Mach number 2nd law of thermodynamics 6 equation dP dρ dT = + P ρ T dρ dV + =0 ρ V 2 2 dh + VdV = 0 dM dV dT = 2 − 2 M V T 4 FdX M . P. V .4 F ρV 2 ρV 2 dV 2 dP + dX + =0 2 DH 2 2 V 2 3 4 5 6 7 Momentum equation Perfect gas law Continuity Eq. From eq.5 dV C P dT = − 2 2 Gas Dynamics dh + VdV = 0 2 dT − dV − (γ − 1)dV γ − 1 2 dV = = =− M 2 T 2CPT 2γRT 2 V 2 2 9 . 2 Dividing by 2 2 P dP 4 F ρV ρV dV + dX + =0 2 P DH 2 P 2P V therefore 2 4 F ρV 2 ρV 2 dV 2 dP + dX + =0 2 DH 2 2 V 2 2 γ PV 2 ρV = = γPM 2 γ RT dP 4 F γM 2 γM 2 dV 2 + dX + =0 2 P DH 2 2 V 8 5 from _ Eq. (6) dV dM γ − 1 2 dV − =− M 2 2 2 V M 2 V 2 2 2 2 2 dV 1 dM 10 = * 2 2 γ − 1 V M 2 1+ M 2 dρ dV dP dV dT + = 0 (4) From eq.(9) with eq. 9 Gas Dynamics dP dV γ − 1 2 dV =− − M 2 P V 2 V 2 dV dV = 2 V 2V 2 . 3 & 4 = − + (3). ρ V P V T From eq. = − T 2CPT 2γRT 2 V2 M2 V2 T (6) Combining eq.dT − dV 2 − (γ − 1)dV 2 γ − 1 2 dV 2 dM 2 dV 2 dT = = =− M (9). 8 can 2 V P be replaced to give 2 dP 4 F γM 2 γM 2 dV 2 + dX + = 0 (8) 2 P DH 2 2 V dV 2 1 γ − 1 2 γM 2 dV 2 4 F γM 2 (− − M )+ + dX = 0 2 2 V 2 2 2 V DH 2 From Eq.dP dV 1 γ −1 2 11 = 2 (− − M ) P V 2 2 2 dV dP The term and the term in eq. 10 dV 2 1 dM 2 = * 2 (10) 2 γ − 1 V 1+ M2 M 2 2 1 1 dM 2 4 F γ M 2 2 [ ](−1 − (γ − 1) M + γM ) + dX = 0 2 2 1 + γ −1 M 2 M DH 2 2 Gas Dynamics . 1 1 dM 2 4 F γM 2 2 2 [ ](−1 − (γ − 1) M + γM ) + dX = 0 2 γ − 1 2 1+ DH 2 M2 M 2 4 FdX (1 − M ) dM = * 2 γ −1 2 DH M 2 γ M (1 + M ) 2 2 2 dM = 2 M 2 γM (1 + 2 γ −1 2 2 (1 − M ) M ) 2 4 FdX . DH M>1 12 M<1 Gas Dynamics dM +ve dM -ve . = 2 γ −1 2 V ρ 2(1 − M ) DH M (1 + M ) 2 M<1 Gas Dynamics M>1 dV -ve ρ +ve dV +ve ρ -ve .dV 2 = 2 V From continuity eq. ( eq. DH 12 13 dV dρ γM 4 FdX dM =− = . 10 & 12 1 dM 2 * (10) γ −1 2 2 M 1+ M 2 γ −1 2 2 dV 2 2VdV 2dV = = (11) 2 2 V V V dM 2 = 2 M γM (1 + 2 (1 − M 2 ) 2 M ) 4 FdX . 4 ) & from eqs. 11 & 13 dP γM 2 (1 + (γ − 1)M 2 ) 4FdX (1 + (γ − 1)M 2 ) =− . = −(γ − 1)( 2 γ −1 2 T 2(1 − M ) DH (1 + M ) 2 15 4 Gas Dynamics . 3.From eqs. = −[ ]dM 2 (γ − 1) 2 P 2(1 − M ) DH M (1 + M ) 14 2 From eqs.13& 14 dT γ (γ − 1)M 4FdX (MdM ) =− . CP 2 DH . 14 & 15 & CP = γ −1 (15) Then ds = γRM 2 2 ds ≥ 0 Gas Dynamics 4 FdX R(1 − M ) dM . = − ( γ − 1 )( γ −1 T 2(1 − M 2 ) DH (1 + M 2) 2 γR From eqs. = −[ ]dM (14) 2 ( γ − 1 ) P 2(1 − M ) DH M (1 + M 2) 2 4 dT γ (γ − 1) M 4 FdX ( MdM ) =− .Entropy changes are determined from dT dP ds = C P −R T P dP γM 2 (1 + (γ − 1) M 2 ) 4 FdX (1 + (γ − 1) M 2 ) =− . = γ −1 2 M DH M (1 + M ) 2 2 16 ∴F 2 Is positive ds (γ − 1) M 4 FdX = . R(1 − M 2 ) dM ds = γ −1 2 M M (1 + M ) 2 (16) Also as a result of frictional flow .For subsonic flow the Mach number increases M <1 (1 − M ) + ve 2 ds = +ve M >1 Gas Dynamics ∴ dM = +ve 2 .For supersonic flow the Mach number decreases ds = +ve (1 − M ) − ve ∴ dM = −ve . = γ −1 2 M P0 2 DH M (1 + M ) 2 17 . 12 & 14 then dM = 2 M 2 γ −1 P0 = P(1 + M ) 2 2 γ 2 γ −1 2 γM 2 (1 + γ −1 2 (1 − M 2 ) M 2) . 4 FdX (12). DH dP γM 2 (1 + (γ − 1) M 2 ) 4 FdX (1 + (γ − 1) M 2 ) =− .The stagnation pressure can be calculated from dPO dP γ M / 2 dM ∴ = + 2 γ − 1 P0 P 1+ 2 M M 2 From eqs. = −[ ]dM (14) 2 (γ − 1) 2 P 2(1 − M ) DH M (1 + M ) 2 2 2 Gas Dynamics dP0 − γ M 4 FdX − (1 − M ) dM = . = (17) γ − 1 P 2 DH 0 M (1 + M 2) M 2 .For supersonic flow dM = −ve Gas Dynamics M >1 ∴ dPO = −ve ds = +ve .For subsonic flow M <1 (1 − M ) + ve 2 dM = +ve ∴ dPO = −ve ds = +ve (1 − M ) − ve 2 .dP0 − γ M 2 4 FdX − (1 − M 2 ) dM = . γ −1 2 γ M 2 (1 + M ) dM 4 FdX 2 = . D = γ −1 2 M 0 H M (1 + M ) Gas Dynamics M<1 dM dV dρ dP dT dS dPo ho +ve +ve -ve -ve -ve +ve -ve M>1 -ve -ve +ve +ve +ve +ve -ve 2 Constant . 2 M 2(1 − M ) DH 2 dV d ρ γ M 4 FdX =− = . D = γ −1 2 M H M ( 1 + M ) 2 2 2 dP − γ M 4 FdX − ( 1 − M ) dM O P = 2 . 2 P = − 2(1 − M ) DH 4 dT γ (γ − 1)M 4 FdX =− . 2 V ρ 2(1 − M ) DH dP γM 2 (1 + (γ − 1) M 2 ) 4 FdX . 2 T 2(1 − M ) DH γRM 2 4 FdX R(1 − M 2 ) dM ds = 2 . Po2 =Po*. V/V*. 4 FdX DH dV dρ γM 2 4 FdX =− = . M. P 2(1 − M 2 ) DH ds = γRM 2 4 FdX 2 . ρ/ρ*. then L2 =L*. D H R (1 − M 2 ) dM = γ −1 M (1 + M 2) M 2 Gas Dynamics . ρ2 =ρ*. P/P*. If M2=1. 4FL*/D. Po/Po*. V ρ 2(1 − M 2 ) DH dT γ (γ − 1) M 4 4 FdX =− . P2 =P*. and S2 =S* dM M γM 2 (1 + = γ −1 2 2(1 − M 2 ) M 2) . V2 =V*.Fanno table relations. M=M2 at distance L. T 2(1 − M 2 ) DH dPO − γ M 2 4 FdX − (1 − M 2 ) dM = . T2 =T*. (S-S*)/Cp By separating the variables and by establishing the limits M=M1 at the duct entrance. = γ −1 2 M P 2 DH 0 M (1 + M ) 2 dP γM 2 (1 + (γ − 1) M 2 ) 4 FdX =− . T/T*. Energy eq. Eq.The Fanno Line * It is defined by the From Continuity eq. of state dρ V = 2CP (T0 − T ) dV =− ρ V dV d (TO − T ) ∴ = V 2(TO − T ) The change of entropy for the perfect gas d (TO − T ) dT ds = Cv −R T 2(TO − T ) Gas Dynamics dT dρ ds = Cv −R T ρ R = γ −1 Cv 19 ds dT γ − 1 d (TO − T ) ∴ = − Cv T 2 2(TO − T ) . (5) Continuity eq. (4) From Energy eq. this curve is right. hO = const. TO h. the change of entropy may be described in terms Of temperature for a given value of T0 (To = constant ) and m. 19 is integrated . M <1 M >1 V2 /2 M =1 h Fanno Line Gas Dynamics S S = Max S S = Max Prove that.ds dT γ − 1 d (TO − T ) ∴ = − Cv T 2 2(TO − T ) h. hO . T (19) Eq. TO hO . The Fanno Line equation is explained as a function of Mach . T hO = const./ A=constant . of state dρ ρ = dP dT − P T ∴ dh = V 2 ( dP P P = 2+ dh V C PT Gas Dynamics V 2 dP (1 + )dh = V CPT P 21 From 20 & 21 dh 1 = 1 R R dS − 2 − T V C PT γR CP = γ −1 .The Fanno Line equation is a function of Mach From the 1st & 2nd law of thermodynamics dh 1 = 1 R dP ds − T P dh dρ 2 ds = dh dP −R T P 20 ∴ dh ∝ dT Continuity dh = V dρ ρ + dV = 0. V Energy dh + VdV = 0 2 dh = −VdV dP dT − ) P T dP C P dT dh = V ( − ) P C PT 2 ρ Eq. S S = Max . or M1 <1 then Po -ve. TO h.dh V 2CPT V 2γRT = 2 = 2 dS V CP − RC PT − V R (γ − 1)[V 2CP − RC PT − V 2 R] dh 1 = 1 R R dS − 2 − T V C PT γR CP = γ −1 V 2 = M 2γRT dh γTM = 2 dS M − 1 At M<1 dh/dS -ve 2 22 dS ≥ 0 This equation describe fanno line as a function on Mach ∴ dh -ve dT –ve ∴ dV +ve At M>1 dh/dS +ve dS ≥ 0 ∴ dh +ve dT +ve ∴ dV -ve At M=1 dh/dS h which corresponding to the state of maximum entropy This means that. A Fanno line moving only in the direction from left to right M1 >1. Gas Dynamics hO . T hO = const./A increase Smax decease. M <1 M >1 V2 /2 M =1 h If m. It was noted that L*.the maximum length of duct which doesn’t cause choking. 1 Flow M1 L1-2 (L*) M1 2 M2 (L*) M2 * M=1 L1-2 Gas Dynamics = (L*) M1 - (L*) M2 Multiply the equation by (4F / DH) (4FL1-2)/DH = ((4F L*)/ DH) M1- ((4F L*)/ DH) M2 . Fanno flow in a constant cross-sectional area duct proceeded by an isentropic nozzle isentropic nozzle Flow 1 2 P/Po 3 Fanno flow ( Ld) Back pressure Pb e M2 <1 The flow in nozzle& duct are affected by Ld & Pb X Gas Dynamics . by decreasing To and/or increasing Po at the nozzle inlet. . In this case Me=1.M2<1 the gas will accelerate in the duct owing to friction and pressure decrease Me <1 or 1 depend on Pb & Ld If Me =1 (m. but Pe would be higher./A)max choked condition G*=(m. /A) max = Po To γ R [2/(γ+1)] (γ+1)/(2(γ-1)) Gas Dynamics To increase mass. the gas will decelerate in the duct due to friction. if no shock occur. max is determined by the throat area of the nozzle and also by the area of the duct Gas Dynamics . In this case (Me=1) . the flow will approach Me=1 directly.Changes in duct length do not affect the mass flow rate M2>1. m. /A)4 ./A)3 >(m./A)2> (m. L2d) Fanno flow L1d 1 2 h ho 3 4 2 P4* M=1 3 * P2*isen P3 4 P/Po1 Gas Dynamics S (m.Effect of increasing duct length isentropic nozzle ( (M duct entr <1) Po ho=const. the shock will be at the throat of the nozzle.(M duct entr >1) At certain length. if discontinuity occurs. Then. the flow is subsonic at all point. the shock will position itself further upstream. the M=1 at exit duct crosssectional area Subsonic conditions can also be attained at exit area. a shock appears in the duct If the duct length is increased further. there will be no shock at all. If the duct is very long. If [ Ld > Ld* at Me =1 ] . Gas Dynamics . Beyond that length. 1 Pback<Pexit a Pback/Po1 1 ho=constant P/Po 2 h a Pa Pb Pc X 1 a b M=1 c Gas Dynamics S .Effect of reducing back pressure (Mach at duct inlet ≤1) c b m. there are 5 cases:- L12 < L1* Design case Pe=Pb Expansion wave Shock wave stand somewhere N.S.W stand inside the at duct exit duct Oblique shock wave * M1 >1 M=1 Pe=Pb Gas Dynamics Pb>Pe Pb<Pe P /Po= Pb/Po X The flow in nozzle& duct are affected by Ld & Pb .Effect of reducing back pressure (Mach at duct inlet >1) Supersonic nozzle Flow 1 P/Po 2 Pb Fanno flow critical length L1* In the duct L12 <L*.