Ls] FLOW THROUGH CONSTANT AREA DUCTS 3.41. FLOWIN CONSTANT AREA DUCTS WITH HEAT TRANSFER [RAYLEIGH FLOW] 3.4.4 Introduction Flow in a constant area duct with heat transfer and without friction is known as Rayleigh flow. Inthe case of combustion chambers, regenerators, heat exchangers and intercoolers the fluid flow takes place with heat transfer. In such 2 flow, the following assumptions are made 1 2 3 4 s (One dimensional steady flow Flow takes place in constant area section. ‘There is no frietion ‘The gas is perfect. “Absence of work transfer across the boundaries. 3.1.2 Rayleigh Line (or) Curve Flow in @ constant area duct with heat transfer and without friction is described by a curve known as Rayleigh line (or) curve.4.2 Gas Dynamics and Jet Propulsion Toh Constant enthalpy may) R Maximum entropy ‘Temperature or enthalpy Constant entropy = Rayleigh Entropy () Fig. 3.1A Rayleigh line or curve We know that Mass flow rate, m= pAc 2 Bev = G- Bape where G > Mass flow density = Grpe Momentum equation is given by p+ pc? = constant ___Flow in constant area ducts with heat transfer [Rayleigh Flow) 3.3 eg Substitute o= @ = ptpx constant o 2 pe Scant => | p+G2v= constant [-vspecific volume, » Equation (3.2) may be used for representing Rayleigh line on the bh-s diagram, as illustrated in fig3.1 In general, most of the fluids in practical use have Rayleigh curves of the general form shown in fig.3.1 The portion of the Rayleigh curve above the point of maximum entropy usually represents subsonic flow (M < 1) and the portion below the ‘maximum entropy point represents supersonic flow (M-> 1). ‘An entropy increases due to heat addition and entropy decreases due to heat rejection, Therefore, the Mach number is increased by heating and decreased by cooling at subsonic speeds. On the other hand, the Mach number is decreased by heating and increased by cooling at supersonic speeds. Therefore, like friction, heat addition also tends to make the Mach ‘number in the duet approach unity. Cooling causes the Mach number to ‘change in the direction away from unity. 3.1.3. Fundamental Equations ‘The following fundamental equations will be used to determine the variation of flow parameters in Rayleigh flows.3-4 Gas Dynamics andl Jet Propulsion 3 Continuity equation We know that ‘Mass low rate, m= pyAye1 = D2A2ey For constant area duct Ay = Ay = m= pie ~ pre > pier = P2e where 4 - Velocity of fluid at inlet ~ mis 2 - velocity of fluid at outlet — mis 154 > Density of Mud at inlet — kg/m? pa > Density of Mud at outlet — kg/m? 3.1.8. Momentum equation Momentum equation between State (1) and State (2) is given by P\A + me; = PyA + me PyA~PyA~ mep—me, > Pr PYA = mMey~ 1) > (P)- PA = mey~ me, = pyAey * 2 pyACy* 6 t =pAc] 2 oad = pyAee = Prd; cee ee Flow in constant area ducts with heat transfer(Rayleigh Flow) 3.5. = @-P) = meF - eye? PioPa = prt pM?r = pl+ 7M?) 3.1.6. Mach number = huge ag- PL xmyexa? RTy RT t Pax Myx RTz Px My? IRT) RT) RT) = Mal SIRT = pM? Y —ppM?Y = pat PMY ~ mit 1M) Pa Pr 1+ YM? * aM? acne ‘The Mach number atthe two states are36 Gas Dynamics ond Jet Propulsion Flow in constant area ducts with heat transfer{ Rayleigh Flow] 3.7 3.1.9. Stagnation pressure ‘Stagnation pressure ~ Mach number relation is given by 3.4.7. Energy The heat transfer during the proces is given by Qa me (To - Toi) where > Mass flow rate ~ kg/s cp > Specific heat t constant pressure ~ sik K Ty Stagnation enthalpy at State,2 To, > Stagnation enthalpy at State, 1 3.4.8, Impulse function Impulse function, F = [1+ 1 M2}p FL + (+ YM?) By = [1+ Y M23] py Ft YM? opp Ferm? Ph G7) Peg Ley M 2 By 15M? 66)3.8 Gas Dynamics and Jet Propulsion Ler? 3.4.10 Static temperature From Equation (3.5), we know that ey My My er oe - Ba YM? - G8) | Flow in constant area ducts with heat transfer[ Rayleigh Flow] 3.9 BSH 7 1, Pe peel eeu Te T, Pe 2 [s,m t) Ty LM Ty M, = n[ 2 LM 2M, 2 ou. Pe3.10 Gas Dynamics and Jet Propulsion 3.1.11 Stagnation temperature (To) ‘Stagnation Temperature ~ Mach number relation is given by M2 deyMpy TMP aerM,ey Th. Maem? “Ty MpaHM,2? Perea reece Flow in constant area ducts with heat transfer [Rayleigh Flove] 3.11 3.1.42 Density ‘We know that fu oe Rr, 7 RT Pe Po Rhy erpraeae RT Pa me TL or my Py OT fy PMP eT PPL Maer)? 1 _ From Equation no. (3. 10] Tt atm? emmy fh From Equation no. eo] 1 Pa ME 17M? -@.12) PL My? 1+YM,? ‘We know that cy 2 M21 M2) 7 ear ans {From equation (3.3)]3.120_Gas Dynamics and Jet Propulsion Flow in constant area ducts with friction [Fanno flow] 3.121 3.2 FLOW IN CONSTANT AREA DUCTS WITH FRICTION AND WITHOUT HEAT TRANSFER (FANNO FLOW) a G—Mass flow density 3.2.4. Introduction e-velocity of fuid Flow in a constant area duct with friction and without heat transfer Density of fluid and work transfer is known as Fanno flow, aaa In the previons chapter, frictionless flow in a constant area duct = G=pe was discussed. But in many engineering applications where the effect @ of friction may not be neglected. in this chapter, the frictional fees | HEC phenomena will be discussed, in a simplified manner. similar leigh flow, in a es i a Rayleigh flow, the following assumptions are made Stagnation enthalpy, iy = A+ ; s 1. One dimensional steady flow. Sate ae 2 2. Flow takes place in constant sectional area, = a+ = 3, There is no heat transfer, > : i > ha hy = (G23) 4... The gas is perfect with constant specific heats, 2 Absence of work transfer across the boundaries, Density (p) isa function of entropy and enthalpy. 3.2.2, Fanno Line (or) Curve aad pm fis.) Flow in a constant area duct with friction and without heat transfes Substitute p value in Equation no(3.23) is described by a curve is known as Fanno line or Fanno curve. ct > =hy= fp ‘We know that [6 MP Mass flow rate, m = pAc 2 eee cece hehy- yy (628) m fear » Tape Equation (3.23) or (3.24) may be used for representing fenno line an the f-s diagram as shown in fig 3.3,3.122 Gas Dynamics and Jet Propulsion Fig. 3.3 A Fanno line or curve ‘The curve consists of two branches AB and BC. At point B the flow is sonic ie, M= 1. The flow A to B is subsonic (M < 1) and C to B is supersonic (M> 1). In subsonic flow region (A to B), the effect of friction will increase the velocity and Mach number and to decrease the enthalpy and pressure of the gas, In supersonic flow region (C10 B), the effect of friction will decrease the velocity and Mach number and to increase the enthalpy and pressure of the gas. ‘We know by the second law of Thermedynamics that for an adiabatic flow, the entropy may increase but cannot decrease. So the processes in the direction B to A and B to C are not possible because they lead to ‘decrease in entropy. ‘The three fanno curves for different values of the mass flow density (G)isshown infig3.4 Flow in constant area ducts with friction [Fanno flow] 3.123, Fanno curves for various values of the mass flow density Fig 3.4 3.2.3 Fanno flow equations We know that Mass flow density, = pe > n(@)=Mnfpo} In@)=inp+ine Differentiating, é oe [+ Mass flow density & G= constant) c3.124 Gas Dynamics and Jet Propulsion [ec@y=2ede ~ ale] ~~ (3.25) = de 2 afc] 22 Ges equation Pa s pot = poo kr = Inf) = InfoRT} = Infp] = Info] +la{R]+Mn(T] Differentiating oe ieee eee [s. R= constant] ~ (3.26) Le on ViRT, Sn) = infe}-mfyRT) = inet] [ny + nR+ nT Inf?) = Info?) tn {T] Flow in constant area ducts with friction [Fanno flow] 3.125 Differentiating G27 R= constant Stagnation enthalpy, hy h+ V2 c® = b+ Ae = by= constant Differentiating as 20M. 2 > GAT +ede~0 > qiteald]-o te bog ah= 67] = th. Dears wee-0 = WE.2. yaano oa = xs wect=o I V7 Multiply throughout by (1-1) : } = eh Beary T 2, Hl yee, =o, > wT? 2 ere [eo M=%)3.126 Gas Dynamics and Jet Propulsion ay 7 act ME. (8.28) Fanning’s co-efficient of skin friction () Dynamic head t Ao Tw Ya pot 6.29) = fr, = tee ‘The area of the duct is given by dA, = perimeter « length GA, = pede - ‘The hydraulic mean diameter ofthe duct is given by 4A Deccepl aA [EeeseD Substituting perimeter p vale in equation (3.30) = 6.30) irea ducts with friction [Fanno flow] 3.127 The momentum equation between state (1) and state (2) is given by PA+me™(p + dp) A+m(c+de) Considering scar stress PA+me= p+ dp) Atm (c+ de) #Ty dAy => PA+me=pA+ Adp+me+mde+ T,,. dA, = Adp+mde+T,,dA,=0 = [inde=—[adp+T, 4A] (6.32) Substituting T,,and dA, values in Equation (3.32) = mide == [Adp + € = Yo pot $A dy] = ----633)3.128 Gas Dynamics and Jet Propulsion Flow in constant area ducts with friction [Fanno flow] 3.129 Divided by p. 1 2 plone Ax constant pte pe eet Fig. 3.5 Varlation of flow properties in Fano flow ‘We know that, pac n Bae ae ~ pede po? atc?) een +s d[o2] = 2ede ae [ dle} =2ede} = weeate Differentiating 22 fe pe2 = pM] fate? Seo. ™ gg Medel ----G.34) 7 A 2 Substitute = de value in equation (3.33) : a > @ yea? dlc?) 0 22 TMP dle), 4 YPM, ge Ot og 3 me agence D PP3.130 Gas Dynamics and Jet Propulsion dy 1a Po ; - & Po ‘We know that Impulse function, F = pA (1+¥M2) 2 In{F) = In[pA + YNP)} = In{F) = Inp+ In A+ in(4¥MP) Differentiating a FP |, ortam?y 14M? = Flo 3.2.4 Solution of fanno flow equations A In this section, the differential flow parameters $2, 9, op. a7 gee 0, ays a¢ ae expe a tons of Ya MP ety FromBqution 27), weknow thi ae deh at Mt ee From equation (3.28), we know that, ~~~ =(.38) From equation (3.25), we know that few. dle) P 208 1 2 constant area ducts with friction [Fanno flow] 3.131[From Equation no (3.38)} From equation (3.28), we know that a 7 From equation (3.26), we know that 4p, @ , aT Pop T surstiae ant as on usin 9a G40) ~“ + a-vae] i dM? 7, ot Mf 2) : ] [isa aw P 7 f etgie] From Equation (3.36), we know that ce Po e sue 2 rte Fomenton 034 ----G.41)3.134 Gas Dynamics and Jet Propulsion Flow in constant area ducts with friction [Fanno flow] 3.135 ute 2. 2 : : dy | A G-DM | ae 12 i Substtte SP, SS values [Fam equation no.) and (3.38)] n2| i Le GDM owe we afi (3.42) _ UG DMAyM Lave From equation (3.37), we know that j 7 afi eee ee Me dM ea Fooitym? MP svsine 2 eke on stin 34 a oat eee - Ferm? ow? From equation (3.35), we know dp, 1M a? far] fo MP dP YM [gp Se) Bina ree nena ple3.248 Gas Dynamics and Jet Propulsion 9 Show that 2 ge aM’ D we Solution [Refer Section 3.2.4) 3.2.18 TWO MARK QUESTIONS AND ANSWERS 1. What is Fanno flow? Flow in a constant area duct with friction and without heat transfer is known as fenno flov: 2. What are she assumptions made for fanno flow? (Apr'99~ MU] 1. One dimensional steady flow 2. Flow takes place in constant sectional area. 3, There is no heat transfer 4. The gas is perfect with constant specific heats 3. Differentiate Fanno flow and Rayleigh flow. [MU~ Apr'03 & Apr'2000 & Anna Univ May'05 } Rayleigh flow Flow in a constant area duct with heat transfer and without fiction is known as Rayleigh flow. Fanno flow Flow in a constant area duct with friction and without hest transfer is known as Fanno flow. Flow in constant area ducts with friction [Fanno flow]3.249 4 Give the fanno flow in h~ 5 digram. Show various Mach number regions and write the fanno flow equation. [MU- 02198] Fanno flow equation hein 5. Explain chocking in Fanno flow. ‘[MU~ Apr'2000} ina Fanno flow, subsonic flow region, the effect of friction will increase the velocity and Mach number and to decrease the enthalpy and pressure of the gas. In Supersonic flow region, the effect of friction will decrease the velécity and Mach number and to inerease the enthalpy and pressure of the gas. In both cases entropy increases upto limiting state where the Mach hnumber is one (M=1), So the mass flow rate is maximum at M = 1 and itis constant afterwards. At this point flow ig said to be chocked flow, (Refer figure 3.2.1)3.250 Gas Dynamies and Jet Propulsion 6. Explain the difference between Fanno flow and Isothermal flow. [Anna Univ— Dec 2003 & Dec ‘04] Fanno Flow Isothermal Flow 1. Flow in a constant area duct | Flow in a constant area duct with friction and without | with ftiction and heat transfer is heat transfer is known as | known as Isothermal flow. Fanno flow. . Static temperature is not | Static temperature remains constant. constant, 7. Write down the ratio of velocities between any two sections in terms of their Mach numbers in a Fanuo flow. [Anna Univ-~ May 2004] wel ye]? o M {ita M 2-#| 8 Write down the ratio of density between any two sections in terms of their Mach number in a Fano flow. [Anna Univ ~ May 2004) Flow in constant area ducts with friction [Fanno flow]3.251 9. Write down the ratlo of pressure betveen any two section in terms of their Mach number in a Fanno flow. 10, What are the three equation governing Fanno process? [MU Apr'98] 1. Energy equation 2, Continuity equation 3, Equation of state ., Give the expression to find increase in entropy for Fanno flow. (MU Apr'96] te 20-1) 12, Write down the expression for the temperature ratio between two ‘sections in terms of Mach numbers for flow in a constant area duct with friction. [MSU- Apr'97]4.252 Gas Dynamics and Jet Propulsion 13. Give two practical examples where the fanno flow oceurs. [MSU- Nov'95] 1. Flow ia ae breathing engines 2, Flow in refrigeration and airconditioning 3. Flow of fluids in long pipes. 14, What is Rayleigh line and Fanno line. Rayleigh line Flow in ¢ constant area duct with beat transfer end without friction is described by a curve is known as Rayleigh line (or) Rayleigh [MU-Apr'97] Fanno line Flow in a constant area duct with friction and without heat transfer is described by a curve is known as Fanno line (or) Fanno curve. 15, White down the expression forthe length of duct in terms ofthe tv0 Mach numbers Mand M fora flow through a constant aren duct wlth the influence of friction. [MKU- 4pr'96] Fit D Is, at -(| D D Im 16, Air at py = 10 bar and Ty = 400 K is supplied to « 50 mm diameter ‘pipe, the friction factor for the pipe surface is 0.002. If the Mach number changes from 3 at entry to 1 at the exit. Determine the length of the pipe. (MSU Apr'96] Solution Refer Fanno flow table for My =3 and ¥ = 1.4 Flow in constant area duets with friction [Fanno flow]3.253 Refer Fanno flow table for My = 1 and y= 14 afl oy D We know that 0.522 ~0 522*D at 0,522 x 0.050 4x 0.002 L=326m 17, Define fanning's coefficient of skin friction It is the ratio between wall shear stress and dynamic head Dynamic head