Applied Gas Dynamics Flow With Friction and Heat Transfer

Transcription

1 Applied Gas Dynamics Flow With Friction and Heat Transfer Ethirajan Rathakrishnan Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 1 / 121

2 Introduction So far, we have discussed about compressible flow of gases in ducts, where changes in flow properties were brought about solely by area change, i.e. where effects of viscosity are neglected. But, in a real flow situation like stationary power plants, aircraft propulsion engines, highvacuum technology, transport of natural gas in long pipe lines, transport of fluids in chemical process plants, and various types of flow systems, the high-speed flow travels through passages of sufficient length, the effects of viscosity (friction) cannot be neglected. In many practical flow situations, friction may even have a decisive effect on the resultant flow characteristics. The inclusion of friction terms in the equations of motion makes the analysis of the problem far more complex. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 2 / 121

3 Like area-change and friction, energy effect such as external heat exchange, combustion or moisture condensation can also produce continuous changes in the state of flowing stream. Here n our discussion on energy effect, we will consider processes involving change in the stagnation temperature or the stagnation enthalpy of a gas stream which flows at constant area and without frictional effects. Though a process involving simple T 0 change is difficult to achieve in practice, many useful conclusions of practical significance may be drawn by analyzing the process of simple T 0 change. Flow where the energy effect is the sole cause for change of state is termed Rayleigh flow. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 3 / 121

4 In this chapter, we intend to discuss flows with friction alone and simple T 0 change alone as the primary factors causing change of states from a simple one-dimensional point of view. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 4 / 121

5 Flow in Constant Area Duct with Friction Consider one-dimensional steady flow of a perfect gas, with constant specific heats, through a constant-area duct. Also, let there be neither external heat exchange nor external shaft work and differences in elevation produce negligible changes as compared with frictional effects. The flow with the above-mentioned conditions, namely, adiabatic flow with no external work, is called Fanno line flow. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 5 / 121

6 Let the wall friction (due to viscosity) be the chief factor bringing about changes in fluid properties, for the adiabatic compressible flow through ducts of constant-area under consideration. The energy equation of steady flow under the above assumptions may be written [Eq. (1.39)] as h + V 2 2 = h 0 (8.1) where h and V are respectively the corresponding values of the enthalpy and velocity at an arbitrary section of the duct and h 0 (the stagnation enthalpy) has a constant value for all sections of the duct. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 6 / 121

7 By equation of continuity, ṁ A = ρv = G (8.2) where ρ is the density at the section where V and h are measured and G is called the mass velocity, which has a constant value for all sections of the duct. Combining Eqs. (8.1) and (8.2), we get the equation of the Fanno line in terms of the enthalpy and density as h = h 0 G2 2ρ 2 (8.3) Because h 0 and G are constants for a given flow, Eq. (8.3) defines a relation between the local density and the local enthalpy. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 7 / 121

8 Large G This relation defines families of curves (the particular curve depending of the choice of the parameters G and h 0 ) in the plane of any two thermodynamic variables; in Figure 8.1, this relation is shown graphically in the h-v plane, for a single value of h 0 and for several values of G. Such curves are, in general, called Fanno lines. h Lines of constant entropy h 0 Small G Medium G v = 1/ρ Figure 8.1 Fanno lines on h-v plane. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 8 / 121

9 The Fanno Line For a pure substance, s = s(h,ρ) That is, the entropy is determined by the enthalpy and density. The curves of Figure 8.1 may thus be transferred to the enthalpy-entropy diagram, giving the familiar Fanno curves of Figure 8.2. For all substances so far investigated, the Fanno curves have the general shape shown in Figure 8.2. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 9 / 121

10 The three curves shown in Figure 8.2 have the same stagnation enthalpy but different mass flow rates per unit area. h p 0a = p 0b p a p 0a = p 0b a p a = p b h 0 Large G b p b Small G Figure 8.2 Fanno lines on h s diagram. s Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 10 / 121

11 We know by the Second Law of Thermodynamics that for an adiabatic flow, the entropy may increase but cannot decrease. Thus, in Figure 8.2, the path of states along any one of the Fanno curves must be towards the right. Therefore, if the flow at some point in the duct is subsonic (point a of Figure 8.2), the effect of friction will be to increase the velocity and Mach number and to decrease the enthalpy and pressure of the stream. On the other hand, if the flow is initially supersonic (point b of Figure 8.2), the effect of friction will be to decrease the velocity and Mach number and to increase the enthalpy and pressure of the stream. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 11 / 121

12 A subsonic flow may therefore never become supersonic and a supersonic flow may never become subsonic, unless a discontinuity is present. Thus we observe that, as in the case of isentropic flow, the qualitative character of the flow is markably influenced by the flow speed, i.e. whether it is subsonic or supersonic. The limiting pressure, beyond which the entropy would decrease, occurs at Mach number unity and is denoted by p. The asterisk here denotes the state where M = 1, for the particular case of adiabatic flow through ducts of constant area. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 12 / 121

13 From Figure 8.2 it is seen that the isentropic stagnation pressure is reduced as a result of friction, irrespective of whether the flow is subsonic or supersonic. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 13 / 121

14 Adiabatic, Constant-Area Flow of a Perfect Gas In this section, the fluid is assumed to be perfect so as to make the analytical treatment of the problem simpler. Further, with this assumption, it becomes possible to draw broad conclusions which would not be otherwise apparent. The aim here is to express the variations in flow characteristics along the length of a duct of constant area in analytical form. This requires the introduction of momentum equation, with a term accounting for the frictional forces acting on the control volume, since the rate of change of flow properties depends on the amount of friction. In Chapter 1, the isentropic relations were derived by writing the various physical relations for two sections a finite distance apart. To demonstrate another method of approach, let us carry out the present analysis in differential form. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 14 / 121

15 Select an infinitesimal control volume as shown in Figure 8.3. In the figure, τ w is the shear stress due to friction, acting on the wall of the duct. τ w Control surface p ρ T M p + dp ρ + dρ T + dt M + dm dx Figure 8.3 Control surface for analysis of adiabatic, constant-area flow. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 15 / 121

16 For a perfect gas, This may also be expressed as p = ρrt dp p = dρ ρ + dt T (8.4) By definition of the Mach number, M 2 = V 2 γrt This gives dm 2 M 2 = dv 2 V 2 dt T (8.5) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 16 / 121

17 The energy equation for a perfect gas gives c p dt + d V 2 = 0 2 Dividing throughout by c p T and using the definition of Mach number, we get dt T + γ 1 M 2 dv 2 2 V 2 = 0 (8.6) The continuity equation [Eq. (8.2)] is G = ṁ A = ρv Noting that G is a constant, this equation can be expressed as dρ ρ + 1 dv 2 2 V 2 = 0 (8.7) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 17 / 121

18 Referring to Figure 8.3, the momentum balance gives A dp τ w da w = ṁ dv where A is the cross-sectional area of duct and da w is the wetted wall surface area of the duct over which the shear stress τ w acts. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 18 / 121

19 Definition of Friction Coefficient The coefficient of drag, or the coefficient of friction, as it is generally referred to for flow in ducts, is defined as f f Wall shear stress Dynamic pressure head of the stream τ w 1 2 ρv 2 Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 19 / 121

20 It is a common practice in such analysis to use a parameter called hydraulic diameter D, defined as D D 4 (cross-sectional area) wetted perimeter 4A da w dx = 4 A da w dx The advantage of using hydraulic diameter is that the equations in terms of hydraulic diameter are valid even for ducts with noncircular crosssection. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 20 / 121

21 Introducing the above f and D along with continuity equation into the momentum equation, we get dp 4 f ρv 2 2 dx D = ṁ A dv = ρ V 2 dv V Dividing throughout by p and expressing ρv 2 as γpm 2, we obtain dp p + γm2 dx 4f 2 D + γm2 2 dv 2 V 2 = 0 (8.8) The isentropic stagnation pressure p 0 is given by Eq. (1.73) as p 0 = p 1 + γ 1 M 2 γ/(γ 1) 2 Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 21 / 121

22 i.e. dp 0 p 0 = dp p + γm 2 /2 dm2 1 + γ 1 M 2 M 2 (8.9) 2 Now, we may define a new parameter called impulse function F as F pa + ρav 2 = pa(1 + γm 2 ) After noting that A is a constant, this may be expressed in differential form in terms of p and M, as df F = dp p + γm2 dm γm 2 M 2 (8.10) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 22 / 121

23 Effects of Wall Friction on Fluid Properties Now we see that the simultaneous algebraic Eqs. (8.4)-(8.10) relate eight differential variables: dp p, dρ ρ, dt T, dm 2 M 2 dv 2 V 2, dp 0 p 0, df F, 4f dx D The physical phenomenon causing changes in flow properties is the viscous friction. Hence, we choose the variable 4fdx/D as independent. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 23 / 121

24 Now, solving the seven equations as simultaneous equations for the remaining seven variables, we can obtain dm 2 M 2 = γm2 dp p = γm2 1 + γ 1 M M 2 4f dx D 1 + (γ 1)M 2 2(1 M 2 ) dv V = γm 2 dx 2(1 M 2 4f ) D 4f dx D (8.11) (8.12) (8.13) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 24 / 121

25 dt T = 1 da γ(γ 1)M4 dx = 2 a 2(1 M 2 4f ) D dρ ρ = γm2 dx 2(1 M 2 4f ) D dp 0 p 0 = γm2 2 4f dx D df F = γm 2 dx 2(1 + γm 2 4f ) D (8.14) (8.15) (8.16) (8.17) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 25 / 121

26 Second Law of Thermodynamics For an adiabatic flow, the stagnation temperature is invariant. Therefore, from Eq. (1.57), the entropy change can be expressed as ds = γ 1 dp 0 (8.18) c p γ p 0 Substituting for dp 0 /p 0, from Eq. (8.16), we have ds (γ 1)M2 = 4f dx c p 2 D (8.19) The Second Law of Thermodynamics states that the entropy should not decrease in an adiabatic flow process. Therefore, from Eq. (8.19), it follows that the friction coefficient f must always be a positive quantity, since by convention dx in Eq. (8.19) is positive in the direction of flow. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 26 / 121

27 In Figure 8.3 we have marked the shear stress in a direction opposite to that of the flow. Since f must always be positive, we conclude that the shear stress must always act in a direction opposite to the flow, as marked in the figure. From Eqs. (8.11)-(8.17), for flow parameters in terms of friction coefficient f, it may be summarized that, Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 27 / 121

28 For subsonic inlet flow, the effect of friction on the downstream flow is such that: (a) Pressure p decreases (b) Mach number M increases (c) Velocity V increases (d) Temperature T decreases (e) Density ρ decreases (f) Stagnation pressure p 0 decreases (g) Impulse function F decreases. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 28 / 121

29 For supersonic inlet flow, the effect of friction on the downstream flow is such that: (a) Pressure p increases (b) Mach number M decreases (c) Velocity V decreases (d) Temperature T increases (e) Density ρ increases (f) Stagnation pressure p 0 decreases (g) Impulse function F decreases. From the above summary we may observe that the friction has the net effect of accelerating a subsonic stream and causes a rise in static pressure at supersonic speeds. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 29 / 121

30 Working Relations Equations (8.11)-(8.19) can be integrated to result in formulae suitable for practical calculations. Let the Mach number be the independent variable for this purpose. Then Eq. (8.11) may be rearranged to give Lmax 0 4f dx 1 D = 1 M 2 M 2 γm γ 1 dm 2 M 2 2 where the integration limits are taken as the section where the Mach number is M and where x is arbitrarily set equal to zero, and as the section where Mach number is unity and x is the maximum possible length of duct, L max. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 30 / 121

31 On integration, the above equation yields 4f L max D = 1 M2 γm 2 + γ + 1 2γ ln (γ + 1)M γ 1 2 M 2 (8.20) where f is the mean friction coefficient with respect to duct length defined by f = 1 Lmax f dx L max 0 Equation (8.20) gives the maximum value of 4f (L/D) corresponding to any initial Mach number M. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 31 / 121

32 Because 4f(L max /D) is a function only of M, the duct length required for the flow to pass from a given initial Mach number M 1 to a given final Mach number M 2 is obtained from the expression 4f L ( D = 4f L ) max D M 1 ( 4f L ) max (8.21) D M 2 Similarly, the local flow properties can be found in terms of local Mach number. Indicating the properties at M = 1 as superscripted with asterisk and integrating between the duct sections with Mach number M and 1, we obtain [from Eqs. (8.12)-(8.17) and (8.19)] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 32 / 121

33 p p = 1 M V V = M γ γ 1 2 γ γ 1 2 T T = a2 a = γ γ 1 ρ ρ = V V = 1 M γ 1 2 γ + 1 M 2 M 2 1/2 1/2 (8.22) (8.23) M 2 (8.24) M 2 1/2 (8.25) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 33 / 121

34 p 0 p 0 = 1 M γ 1 2 γ + 1 M 2 (γ+1)/2(γ 1) (8.26) F F = 1 + γm 2 [ M 2(γ + 1) 1 + γ 1 ] 1/2 (8.27) M 2 2 s s = ln M 2 (γ + 1) c p 2M γ 1 M 2 2 (γ+1)/(2γ) (8.28) We know that the quantities marked with asterisk in theses equations are constants for a given adiabatic constant-area flow. Therefore, they may be regarded as convenient reference values for normalizing the equations. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 34 / 121

35 To find the change in a flow property, say the density, between sections of the duct where the Mach numbers are M 1 and M 2, we set ρ ρ 2 ρ M2 = ρ 1 ρ ρ M1 where ρ ρ is the value on the right hand side of Eq. (8.25) corre- M1 sponding to M 1, and so on. The variation of dimensionless ratios given by Eqs. (8.20) and (8.22)- (8.27) with Mach number is tabulated in Table 4 in the Appendix. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 35 / 121

36 Example 8.1 Atmospheric air at kpa and 300 K is drawn through frictionless bell-mouth entrance into a 3 m long tube having a 0.05 m diameter. The average friction coefficient f = 0.005, for the tube. The system is perfectly insulated. (a) Find the maximum mass flow rate and the range of backpressure that will produce this flow. (b) What is the exit pressure required to produce 90% of the maximum mass flow rate and what will be the stagnation pressure and the velocity at the exit for that mass flow rate? Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 36 / 121

37 Solution (a) The mass flow rate will be maximum for choked flow conditions. For choked flow, For 4f Lmax D 4f L max D = = = 1.2, from Fanno flow table (Table 4 of the Appendix), M 1 = 0.485, p 1 p = , T 1 T = From isentropic tables (see Table 1 of the Appendix) for M 1 = 0.485, p 1 p 0 = 0.851, T 1 T 0 = Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 37 / 121

38 Therefore, p 1 = (0.851)(101.35) = 86.25kPa T 1 = (0.955)(300) = K p = T = p = 39.07kPa T = 250 K ρ = p = kg/m3 RT a = γrt = = m/s Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 38 / 121

39 The maximum flow rate is ṁ = ρ a A = π 4 (0.05)2 ṁ = kg/s The range of backpressure (p b ) that would produce this flow is p b p (39.07 kpa) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 39 / 121

40 (b) 90% of ṁ is ṁ = = kg/s p 1 p 0 G = ṁ A = kg/(s m2 ) ρ 1 V 1 = G = ṁ γ A, p 1M 1 = G RT 1 R T0 γrt0 G G = γ p 0 γ T 0 T 1 M 1 = p 0 = = Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 40 / 121

41 Solving this for M 1, by trial and error method, we get M 1 = 0.42 From isentropic table, for M 1 = 0.42, By Eq. (8.21), p 1 p 0 = 0.886, p 1 = 89.8kPa T 1 T 0 = 0.966, T 1 = K 4f L ( D = 1.2 = 4f L ) ( max 4f L ) max D M 1 D M 2 Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 41 / 121

42 Using Eq. (8.20) or the Fanno flow table, we get for M 1 = 0.42, ( 4f L ) max = D M 1 Hence, ( 1.2 = f L ) max D M 2 i.e. ( 4f L ) max = D M 2 For this value 4f(L max /D) from Fanno flow table, M 2 = Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 42 / 121

43 Now, using Fanno table (a), for M 1 = 0.42, For M 2 = 0.541, p 1 p = 2.563, T 1 T = 1.159, p 01 p 0 p 2 p = 1.96, T 2 T = 1.134, p 02 p 0 = = Therefore, p 2 = p 2 p p p 1 p 1 = kpa T 2 = T 2 T T T 1 T 1 = K Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 43 / 121

44 p 02 = p 02 p 0 p 0 p 01 p 01 = kpa a 2 = γrt 2 = m/s The exit velocity is V 2 = M 2 a 2 = = m/s Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 44 / 121

45 Example 8.2 A straight pipe of 0.05 m diameter is attached to a large air reservoir at 1.38 MPa and 310 K. The exit of the pipe is open to atmosphere. Assuming adiabatic flow with an average friction coefficient of 0.005, calculate the pipe length necessary to obtain a mass flow rate of 2.25 kg/s. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 45 / 121

46 Solution Let the subscripts 1 and 2 refer to conditions at entry and exit of the pipe, respectively. Given By Eq. (8.2), the mass velocity is Also, ṁ = 2.25 kg/s G = ṁ A = 2.25 (π/4)(0.05) 2 = kg/(m 2 s) G = p 1 M 1 γ RT 1 1/2 = p 2 M 2 γ RT 2 1/2 = constant Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 46 / 121

47 Now the relation can be rewritten as p 1 M 1 γ RT 1 1/2 = G 1/2 p 1 T 0 M 1 = R 1/2 G T 1/2 0 γrt0 = p 0 T 1 γ p 0 γ G p 0 (i) = = Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 47 / 121

48 The left hand side of Eq. (i) has three parameters and out of them the pressure ratio and temperature ratio are uniquely related to Mach number. By trials we can solve this equation as follows: Let M 1 = Then L.H.S. = = L.H.S. < R.H.S. Let M 1 = Then L.H.S. = = L.H.S. > R.H.S. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 48 / 121

49 Hence, M 1 lies between 0.21 and For M 1 = 0.213, L.H.S. is nearly equal to R.H.S.. Therefore, M 1 = can be taken as the correct solution. For this value of M 1, p 1 p 0 = 0.969, p 1 p = 5.12 Thus, p 1 = MPa p = kpa For M 1 = 0.213, from Eq. (8.20), 4f L max D = Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 49 / 121

50 Therefore, L max = = m Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 50 / 121

51 Flow With Heating or Cooling in Ducts So far we have considered only the effect of area change and friction on gas flow process. From one-dimensional aspect, there is yet another effect producing continuous changes in the state of flowing stream and this third factor is called energy effect. External heat exchange, combustion, or moisture condensation are examples of energy effects. In the discussion on the effects of area change on flow state, we considered the process to be isentropic with frictional and energy effects absent. In Section 8.2 we dealt with the effects of wall friction in the absence of area change and energy effects; the corresponding process is described by the Fanno curve and may aptly be termed simple friction. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 51 / 121

52 In this section, we discuss processes involving change in the stagnation temperature or the stagnation enthalpy of a gas stream which flows at constant area and without frictional effects. Though a process involving simple T 0 change is difficult to achieve in practice, many useful conclusions of practical significance may be drawn by analyzing the process of simple T 0 change. These conclusions can be expected to have a higher degree of validity when the departures from the assumptions of the model are small. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 52 / 121

53 Governing Equations For flow of gas through a constant-area duct without friction, the momentum equation may be written as p + ρv 2 = F A = constant (8.29) By continuity, ρv = ṁ = G = constant (8.30) A Combining Eqs. (8.29) and (8.30), we get p + G2 ρ = F A (8.31) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 53 / 121

54 For constant values of G and F/A, Eq. (8.31) defines a unique relation between pressure and density, called Rayleigh line. Since both the enthalpy h and entropy s are functions of p and ρ, Eq. (8.31) may be used for representing Rayleigh line on the h-s diagram, as illustrated in Figure 8.4. In general, most of the fluids in practical use have Rayleigh curves of the general form shown in Figure 8.4. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 54 / 121

55 h h 01 p 01 Heating Cooling g Rayleigh, M < 1 p Isentropic 1 Fanno p 1 Cooling Heating Rayleigh M > 1 Figure 8.4 Rayleigh curve for simple T 0 -change. s Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 55 / 121

56 The portion of the Rayleigh curve above the point of maximum entropy usually corresponds to subsonic flow and the portion below the maximum entropy point corresponds to supersonic flow. The process of simple heating is thermodynamically reversible; therefore, heat addition should correspond to an entropy increase and heat rejection must correspond to an entropy decrease. 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. Thus, like friction, heat addition also always tends to make the Mach number in the duct approach unity. Cooling causes the Mach number to change always in the direction away from unity. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 56 / 121

57 For heat addition at either subsonic or supersonic speeds, the amount of heat input cannot be greater than that for which the leaving Mach number is unity. If the heat addition is too large, the flow will be choked, i.e. the initial Mach number will be reduced to a magnitude which is consistent with the specified amount of heat input. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 57 / 121

58 Simple-Heating Relations for a Perfect Gas As in Section 8.3, we shall describe flow of a perfect gas through a constant-area duct. Let there be no friction. Consider the control volume shown in Figure 8.5. Heat Control surface T 01 T 02 p 1 ρ 1 p 2 V 1 V 2 ρ 2 T 1 M 1 T 2 M Figure 8.5 Control volume for Rayleigh flow. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 58 / 121

59 For the flow through constant-area duct, by continuity, ρ 2 ρ 1 = V 1 V 2 (8.32) The momentum balance in the absence of friction gives p 1 p 2 = ṁ A (V 2 V 1 ) But ṁ/a = ρv and ρv 2 = γpm 2 (for perfect gas). Using these relations, the above momentum equation may be rewritten as p 2 p 1 = 1 + γm γm 2 2 (8.33) By the state equation, p 2 p 1 = ρ 2T 2 ρ 1 T 1 (8.34) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 59 / 121

60 The Mach number ratio between states 1 and 2 can be expressed as M 2 = V 2a 1 = V 2 T 1 (8.35) M 1 V 1 a 2 V 1 T 2 Similarly, for impulse function, F 2 = p 2(1 + γm2 2) F 1 p 1 (1 + γm1 2) Using, Eq. (8.33), we get F 2 = 1 (8.36) F 1 The isentropic stagnation pressure ratio is given by Eq. (1.73) as p 02 p 01 = p 2 p γ γ 1 2 M 2 2 M 2 1 γ/(γ 1) γ/(γ 1) (8.37) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 60 / 121

61 The entropy change may be found from Eq. (1.59) as s 2 s 1 c p T = ln 2 /T 1 (γ 1)/γ p 2 /p 1 (8.38) So far, we have seen the relation between the parameters at two different states of the process. All these changes are brought about due to changes in stagnation temperature. That is, the rate of change of stream properties along Rayleigh lines is a function of the rate of change of stagnation temperature. From the energy relation, the stagnation temperature T 0 is T 0 = T + V 2 = T 1 + V 2 2c p 2c p T = T 1 + γ 1 M 2 2 Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 61 / 121

62 Therefore, γ 1 T 02 = T 1 + M T 01 T γ 1 (8.39) M1 2 2 For the process involving only heat exchange, the change in stagnation temperature is a direct measure of the amount of heat transfer. If Q is the heat added to the control volume, then by the energy equation, Q = c p (T 2 T 1 ) + V 2 2 V = c p (T 02 T 01 ) (8.40) Equations (8.32)-(8.34) may be combined to result in T 2 T 1 = M2 2 M 2 1 ( 1 + γm 2 1 ) 2 ( 1 + γm 2 2 ) 2 (8.41) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 62 / 121

63 Using Eq. (8.41) in Eq. (8.39), we get T 02 T 01 = M2 2 M 2 1 ( 1 + γm 2 1 ) 2 ( 1 + γm 2 2 ) γ γ 1 2 M 2 2 M 2 1 (8.42) Equations (8.41) and (8.42) express the static and stagnation temperature ratios between states 1 and 2 in terms of Mach numbers at these states. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 63 / 121

64 Following the same procedure as that in Section 8.3, we can get the following normalized expressions (working formulae) for the present flow process involving only heat transfer: T T = (γ + 1)2 M 2 ( ) 1 + γm 2 2 (8.43) T 0 T0 = 2(γ + 1)M γ 1 M 2 2 ( ) 1 + γm 2 2 (8.44) V V = ρ ρ = (γ + 1)M2 1 + γm 2 (8.45) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 64 / 121

65 p 0 p0 p p = γ γm 2 (8.46) = γ γ 1 γ/(γ 1) M γm γ (8.47) s s = ln M 2 γ + 1 c p 1 + γm 2 (γ+1)/γ (8.48) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 65 / 121

66 Also, T 02 T 01 = (T 0/T 0 ) M 2 (T 0 /T 0 ) M 1 (8.49) where (T 0 /T 0 ) M 2 is given by Eq. (8.44) and so on. The variation of the dimensionless ratios given by Eqs. (8.43)-(8.47) with Mach number is tabulated in Table 5 in the Appendix. From our discussions on Rayleigh flow and the properties relations the physical trends associated with flow with simple T 0 -change may be summarized as follows: Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 66 / 121

67 For subsonic flow (M 1 < 1), when heat is added: (a) Pressure decreases, p 2 < p 1 (b) Mach number increases, M 2 > M 1 (c) Velocity increases, V 2 > V 1 (d) Temperature increases for M 1 < γ 1/2 and temperature decreases for M 1 > γ 1/2 (e) Total temperature increases, T 02 > T 01 (f) Total pressure decreases, p 02 < p 01. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 67 / 121

68 For supersonic flow (M 1 > 1), when heat is added: (a) Pressure increases, p 2 > p 1 (b) Mach number decreases, M 2 < M 1 (c) Velocity decreases, V 2 < V 1 (d) Temperature increases, T 2 > T 1 (e) Total temperature increases, T 02 > T 01 (f) Total pressure decreases, p 02 < p 01. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 68 / 121

69 Note that for subsonic flow when heat is added, the temperature increases for M 1 < γ 1/2 and decreases for M 1 > γ 1/2. This is due to the fact that the value of T/T goes through a maximum at M = 1/ γ, corresponding to point g on Figure 8.4. In the case of air, therefore, the values of M between 0.85 and 1, heat addition results in decrease of stream temperature and heat rejection results in increase of stream temperature. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 69 / 121

70 Example 8.3 Air at standard sea level conditions enters the tube shown in Figure E8.3, at Mach 0.68 and reaches a value of Mach 0.25 at the exit of the diffuser (station B). The entrance area is 1 m 2. (a) Assuming no dissipative losses in the diffuser, show that the area at station B is 2.16 m 2. Will the area be larger or smaller if losses are present? (b) Show that, assuming no losses, the static pressure at B is kpa and the density is 1.48 kg/m 3. If losses are present, will the stagnation pressure rise or fall from station A to station B? Will the stagnation density rise or fall? Give reason for your answer. (c) Heat is added at Mach 0.25 between stations B and C until thermal choking occurs. Show that the heat added is N-m/kg and the stagnation temperature at station C is 1225 K. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 70 / 121

71 Heat addition M = 0.68 A B Figure E8.3 C Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 71 / 121

72 Solution Given, p A = kpa, M A = 0.68 T A = 288 K, M B = 0.25 ρ A = kg/m 3, A A = 1 m 2 (a) From isentropic flow table, A A A = for M A = 0.68 A B A = for M B = 0.25 Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 72 / 121

73 Therefore, A B = A B A A A A A A = = m 2 If losses were present, the area would have been larger since where ṁ = Γ γrt0 p 0 f(m)a Γ = γ 2 γ + 1 (γ+1)/2(γ 1) ṁ, T 0, M being constants, p 0 decreases with losses; hence A B has to increase. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 73 / 121

74 (b) From isentropic table, p A p 0 = , p B p 0 = , ρ A ρ 0 = for M A = 0.68 ρ B ρ 0 = for M B = 0.25 Therefore, p B = p B p 0 p 0 p A p A = kpa ρ B = ρ B ρ 0 ρ 0 ρ A ρ A = 1.48 kg/m 3 s = R ln p 0A p 0B for s > 0, p 0A > p 0B Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 74 / 121

75 Hence, if losses were present, the stagnation pressure would have decreased from A to B: p 0A ρ 0A = RT 0A, p 0B ρ 0B = RT 0B ρ 0B ρ 0A p 0A p 0B = 1 since T 0A = T 0B ρ 0B ρ 0A = p 0B p 0A < 1 Therefore, the stagnation density falls from A to B. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 75 / 121

76 (c) For M A = 0.68, from isentropic tables, we have Hence, For thermal choking, from Table 5, T A T 0A = T 0A = K = T 0B T 0B T 0 = at M B = 0.25 Therefore, T0 = K = T 0C Heat added is given by Eq. (8.40) as Q = c p (T 0C T 0B ) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 76 / 121

77 Also, Thus, c p = γ γ 1 R = m2 /s 2 K Q = ( ) = N-m/kg Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 77 / 121

78 Summary From our discussions in Chapters 1-8, it is clear that the change of state in flow properties is achieved by three means: (a) with area change, treating the fluid to be inviscid and passage to be frictionless, (b) with friction, considering the heat transfer between the surrounding and system to be negligible and (c) with heat transfer, assuming the fluid to be inviscid and passage to be frictionless. These three types of flows are called isentropic flow, frictional or Fanno type flow and Rayleigh type flow, respectively. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 78 / 121

79 All gas dynamic problems encountered in practice can be classified under these three flow processes, of course with the assumptions mentioned. Although it is impossible to have a flow process which is purely isentropic or Fanno type or Rayleigh type, in practice it is justified in assuming so, since the results obtained from these processes prove to be accurate enough for most of the practical situations in gas dynamics. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 79 / 121

80 Flows in which wall friction is the chief factor bringing about changes in flow properties, assuming that no heat is transferred to or from the fluid stream are termed Fanno type flow. When the ducts are short, the flow is approximately adiabatic. However, when the ducts are extremely long, as in the case of natural-gas pipe lines, there is sufficient area for heat transfer to make the flow nonadiabatic and approximately isothermal. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 80 / 121

81 Considering one-dimensional steady flow of a perfect gas through a constant area duct, with the assumption that there is no external heat exchange and external shaft work and differences in elevation produce negligible changes as compared to frictional effects, we can write h = h 0 G2 2ρ 2 (8.3) where h and h 0 are the static and stagnation enthalpy, ρ is density and G is mass velocity. This equation shows that for a given initial condition, the relation between the local density ρ and local enthalpy h is fixed. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 81 / 121

82 This implies that the relation between any two properties of the flowing gas is also fixed. Thus all the states that satisfy Eq. (8.3) can be plotted on h s diagram. The locus of these states on such a diagram is called the Fanno line. Figure 8.2 shows such lines for a certain value of h 0. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 82 / 121

83 The friction coefficient f is defined as f = Wall shear stress Dynamic pressure head of the stream The hydraulic diameter D is defined as D = 4 (cross-sectional area) wetted perimeter The advantage of using hydraulic diameter is that the equations, in terms of hydraulic diameter, are valid even for ducts with non-circular cross-section. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 83 / 121

84 The maximum length of the duct required for the flow to choke for a given initial Mach number is given by 4f L max D = 1 M2 γm 2 + γ + 1 2γ ln (γ + 1)M ( γ 1 2 ) (8.20) M 2 where f is the mean friction coefficient with respect to duct length, defined by f = 1 Lmax f dx L max 0 Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 84 / 121

85 The duct length required for the flow to pass from a given initial Mach number M 1 to a given final Mach number M 2 can be obtained from the expression 4f L D = 4f L max D M 1 4f L max D (8.21) M 2 Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 85 / 121

86 At the maximum entropy point of the Fanno curve the velocity is sonic velocity. For subsonic flow, the enthalpy decreases as the velocity increases in the direction of flow. For supersonic flow, the enthalpy increases as the velocity decreases in the direction of flow. Thus the upper part of the Fanno line represents the state of subsonic flow, while the lower part of the line represents the states of supersonic flow. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 86 / 121

87 The physical significance of the point of maximum entropy may be illustrated by considering the flow in a pipe with friction. If in such a case both the initial pressure and the discharge pressure are maintained constant, there is a maximum pipe length that we can use for a given mass flow rate. However, the variation of pressure, velocity, entropy, etc. of the fluid as a function of pipe length can be predicted if the friction coefficient for the pipe is known. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 87 / 121

88 From the discussion on Fanno flow it is clear that friction always drives the Mach number toward unity, decelerating a supersonic flow and accelerating a subsonic flow. In Figure 8.2, which is a Mollier diagram of one-dimensional flow with friction, the above mentioned effect of friction on Mach number is emphasized. For any given initial Mach number, for a certain value of L the flow becomes sonic. For this condition the flow is said to be choked, since any further increase in L is not possible without causing a drastic change of the inlet conditions. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 88 / 121

89 For instance, if the inlet conditions were achieved by expansion through a supersonic nozzle, and if L were larger than that allowed for attaining Mach 1 at the exit, then a normal shock would form inside the supersonic nozzle and the duct inlet conditions would suddenly become subsonic. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 89 / 121

90 It is important to note that friction always causes the total pressure to decrease whether the inlet flow is subsonic or supersonic. Further, unlike the Rayleigh curve for flow with heating or cooling, the upper and lower portions of the Fanno curve cannot be traversed by the same onedimensional flow. In other words, it is not possible to first decelerate a supersonic flow to sonic condition by friction, and then further retard it to subsonic speeds also by friction, since such a subsonic deceleration violates the second law of thermodynamics. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 90 / 121

91 A process involving changes in the stagnation enthalpy or stagnation temperature of a gas stream which flows at constant area and without frictional effects is called a process with Simple To-change. In this process, energy effects such as external heat exchange, combustion, or moisture condensation is the prime parameter causing changes in the state of a flowing gas. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 91 / 121

92 For fixed values of the flow per unit area and the impulse function per unit area, a unique relation between the pressure and the density is defined as p + G2 ρ = F (8.31) A This equation is called the Rayleigh line relation. Since both the enthalpy and entropy are functions of pressure and density, it follows that Eq. (8.31) may be used for representing the Rayleigh line on the enthalpyentropy diagram, as shown in Figure 8.4. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 92 / 121

93 From a physical point of view the changes in stream properties are due primarily to changes in stagnation temperature. That is, the rate of change of stream properties along a Rayleigh line is a function of the rate of change of stagnation temperature. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 93 / 121

94 The stagnation temperature corresponding to a given state is that temperature which the stream would assume if it were adiabatically decelerated to zero velocity. The ratio of stagnation temperatures at sections 1 and 2 in terms of Mach numbers at these sections, for a Rayleigh flow, can be expressed as T 02 T 01 = M2 2 M 2 1 ( 1 + γm 2 1 ) 2 ( 1 + γm 2 2 ) γ γ 1 2 M 2 2 M 2 1 (8.42) Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 94 / 121

95 It is important to note that heat addition always drives the Mach numbers toward 1, accelerating a subsonic flow and decelerating a supersonic flow. This is emphasized on the Rayleigh curve in Figure 8.4. Heating always acts to reduce the stagnation pressure, irrespective of whether the speed is subsonic or supersonic. Increase in stagnation pressure, on the other hand, may be obtained at either subsonic or supersonic speeds by a cooling process which reduces to stagnation temperature; in practice this is difficult because other effects are always present which tend to reduce the stagnation pressure. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 95 / 121

96 Finally, it is extremely important to realize that we have considered only simple types of flow to study the flow processes in which only a simple independent parameter was allowed to change, e.g. isentropic flow in Chapter 2 where effects of area change alone was considered; Fanno and Rayleigh flows in this chapter where the effects of friction alone and the effects of changes in stagnation temperature alone, respectively, have been considered. But in many practical problems of interest these effects occur simultaneously and in addition, there may be present such other phenomena as chemical reaction, change of phase, injection or withdrawal of gases, and changes in molecular weight and specific heats. Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 96 / 121

97 Rocket nozzles, ram jets, combustion chambers of gas turbine engines, moving flame fronts, moisture condensation shocks, injectors and ejectors, detonation waves, and heat exchangers are typical examples of flow passages in which simultaneous effects are present. For solving such flows all the effects associated with such processes must be taken into account simultaneously. For solving such flows, readers are encouraged to consult books specializing on such topics, see, for instance Shapiro (1953). Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 97 / 121

98 Exercise Problems 8.1 The stagnation chamber of a wind tunnel is connected to a high pressure air reservoir by a long pipe of 100 mm diameter, with a friction coefficient of If the static pressure ratio between the reservoir and the stagnation chamber is 10 and the reservoir static pressure is MPa, how long can the pipe be without choking? Assume the flow to be adiabatic, subsonic, one-dimensional. [Ans. L max = m] 8.2 Air at a pressure of 350 kpa and 300 K is to be transported at the rate of kg/s over a distance of 600 m through a pipe. The final pressure is to be at least 140 kpa. Assuming isothermal flow and the average friction coefficient of the pipe f = 0.004, determine the minimum pipe diameter. [Ans m] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 98 / 121

99 8.3 With an experimental rig comprising a convergent-divergent nozzle attached to a smooth round tube, the following data were measured with the aim of measuring friction coefficient for the supersonic flow of air: Stagnation pressure and temperature upstream of the nozzle are p 0 = 6.73 MPa and T 0 = 312 K; throat diameter = m; diameter of nozzle exit and tube D = m; pressure of streams at stations x 1 /D = 1.75 and x 2 /D = from the tube inlet: p 1 = 238 kpa and p 2 = 485 kpa. Calculate the average friction coefficient between stations x 1 and x 2. Assume that the flow to the throat of the nozzle is isentropic and that the flow in the entire system is adiabatic. [Ans ] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 99 / 121

100 8.4 An isentropic nozzle having an area ratio of 2, discharges air into an insulated pipe of length L and diameter D. The nozzle is supplied at 700 kpa and 300 K and the duct discharges into a space where the pressure is 280 kpa. Calculate the 4f L/D of the pipe and mass flow rate per unit area in the pipe for the cases where a normal shock stands: (a) at the nozzle throat, (b) at the nozzle exit plane and (c) at the duct exit plane. [Ans. (a)4.8576, kg/(m 2 s), (b) , kg/(m 2 s), (c) , kg/(m 2 s)] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 100 / 121

101 8.5 A gaseous mixture of air and fuel enters a ramjet combustion chamber with a velocity of m/s, at a static temperature and pressure of K and kpa. The heat of reaction H of the fuel-air mixture is kj/kg. Assuming that the working fluid has the same thermodynamic properties as air before and after combustion, the friction is negligible and the cross-sectional area of the combustion chamber is constant, calculate: (a) the stagnation temperature after combustion, (b) the Mach number after combustion, (c) the final static temperature, (d) the loss in stagnation pressure due to heat addition, (e) the entropy change, (f) the final velocity of combustion mixture and (g) the maximum heat of reaction for which flow with the specified initial conditions can be maintained. [Hint: H = c p (T 02 T 01 )] [Ans. (a) K, (b) 0.68, (c) K, (d) 8.53 kpa, (e) J/(kg K), (f) m/s, (g) kj/kg] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 101 / 121

102 8.6 Air flows adiabatically through a duct of diameter 20 mm. At a station 1 in the duct, M 1 = 0.2, p 1 = 5 atm and T 1 = 300 K. Compute p 2, T 2, V 2 and p 02 at a station 2 where M 2 = 0.5. [Ans kpa, 288 K, m/s, kpa] 8.7 Air flows through a perfectly insulated square tube of cross-section 0.1 m by 0.1 m. At a section 1 inside the tube, M 1 = 0.2, T 1 = 72 C and p 1 = 2 atm. At a downstream section 2, M 2 = Determine the mass flow rate through the tube and the drag force acting on the duct between sections 1 and 2. [Ans kg/s, N] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 102 / 121

103 8.8 Carbon dioxide gas enters an insulated circular tube of length-todiameter ratio 50. At the entrance, the flow velocity is 195 m/s and temperature is 310 K. If the flow at the tube exit is choked, determine the average friction factor of the tube. [Ans ] 8.9 Air flows through a pipe of 25 mm diameter and 51 m length. The conditions at the exit of the pipe are M 2 = 0.8, p 2 = 1 atm and T 2 = 270 K. Assuming adiabatic one-dimensional flow, calculate M 1, p 1 and T 1 at the pipe entrance. Take the local friction coefficient to be [Ans. 0.13, 6.56 atm, K] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 103 / 121

104 8.10 Air enters a perfectly insulated tube of 5 cm diameter with a stagnation state at p 0 = 135 kpa and T 0 = 359 K. The velocity at the entrance is V 1 = 135 m/s. If the average friction factor is 0.02, determine, (a) the minimum length of the duct required for the flow to choke and (b) the mass flow rate and the stagnation pressure at the exit if the tube length is 0.6 m. [Ans. (a) 1.99 m, (b) kg/s, kpa] 8.11 Hydrogen enters an insulated tube of 25 mm diameter with V 1 = 200 m/s, p 1 = 250 kpa and T 1 = 303 K. What is the length of the tube required for this flow to choke? Determine the exit pressure. The friction factor of the tube f = [Ans m, kpa] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 104 / 121

105 8.12 An air stream flowing out of a convergent nozzle at 200 m/s and 30 C is made to enter an insulated pipe of diameter 20 mm. Determine the length of the pipe at which the flow will become sonic if the average friction factor is [Ans cm] 8.13 Methane gas flows in a commercial steel pipe of 25 mm diameter. At the inlet p 1 = 1.0 MPa, T 1 = 320 K and V 1 = 25 m/s. Determine the velocity and pressure at the pipe length at which the flow just chokes. Treat the flow to be adiabatic. For Methane R = J/(kg K) and the viscosity coefficient µ = kg/(m s) at the given inlet condition. Take the average friction coefficient to be f = [Ans m/s, kpa] Applied Gas Dynamics, John Wiley & Sons (Asia) Pte Ltd c 2010 Ethirajan Rathakrishnan 105 / 121