dse +; thus dp0 −, or p0 decreases. A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 94 dse +; thus dp0 −, or p0 decreases. For cooling, dse −; thus dp0 +, or p0 increases. In practice, the latter condition is difficult to achieve because the friction that is inevitably present introduces a greater drop in stagnation pressure than the rise created by the cooling process, unless the cooling is done by vaporization of an injected liquid. 6.4 WORKING EQUATIONS FOR PERFECT GASES By this time you should have a good idea of the property changes that are occurring in both subsonic and supersonic Rayleigh flow. Remember that we can progress along a Rayleigh line in either direction, depending on whether the heat is being added to or removed from the system. We now proceed to develop relations between properties at arbitrary sections. Recall that we want these working equations to be expressed in terms of Mach numbers and the specific heat ratio. To obtain explicit relations, we assume the fluid to be a perfect gas.  Momentum We start with the momentum equation developed since this will lead directly to a pressure ratio: p + GV = canst. or this can be written as p + pV2 = canst. Substitute for density from the equation of state: p RT and for the velocity V2 = M2a2 = M2 yRT p =

Show that equation becomes A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 95 Show that equation becomes If we apply this between two arbitrary points, we have which can be solved for  Continuity From previous section we have ρV = G = constant Again, if we introduce the perfect gas equation of state together with the definition of Mach number and sonic velocity, Written between two points, this gives us which can be solved for the temperature ratio: The introduction of the pressure ratio results in the following working equation for static temperatures:

A course in Gas Dynamics…………………………………. …. …Lecturer: Dr A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 96 The density relation can easily be obtained from equations and the perfect gas equation of state:  Stagnation Conditions This is the first flow that we have examined in which the stagnation enthalpy does not remain constant. Thus we must seek a stagnation temperature ratio for use with perfect gases. We know that y - 1 2 If we write this for each location and then divide one equation by the other, we will have 1 + fy - 1l M2 T01 T1 1 + fy - 1l M2 Since we already have solved for the static temperature ratio, this can immediately be written as 2 2 2 y - 1 2 T01 1 + yM2 M1 1 + fy - 1l M2 Similarly, we can obtain an expression for the stagnation pressure ratio, since we know that y p0 = p (1 + 2 M ) which means that T0 = T (1 + M ) 2 T02 T2 ( 2 2 = 2 1 T02 1 + yM1 M2 1 + f l M2 ( 2 = 2 2 2 1 y - 1 (y-1) 2

A course in Gas Dynamics…………………………………. …. …Lecturer: Dr A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 97 y p02 = p2 (1 + r 2 l M2 2 1 Substitution for the pressure ratio yields p02 = 1 + yM1 1 + r 2 l M2 1  Energy h01 + q = h02 For perfect gases we express enthalpy as h = cpT which can also be applied to the stagnation conditions h0 = cpT0 Thus the energy equation can be written as cpT01 + q = cpT02 or q = cp(T02 - T01) Note carefully that q = &'∆)* ≠ &'∆) 6.5 REFERENCE STATE ANDTHE RAYLEIGH TABLE The equations developed in Section 6.4 provide the means of predicting properties at one location if sufficient information is known concerning a Rayleigh flow system. Although the relations are y - 1 2 (y-1) p01 p1 1 + ry - 1l M2 y - 1 2 (y-1) 2 p01 1 + yM2 2 1 + ry - 1l M2 2

We introduce still another (∗) reference state defined as before, in A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 98 straightforward, their use is frequently cumbersome and thus we turn to techniques used previously that greatly simplify problem solution. We introduce still another (∗) reference state defined as before, in that the Mach number of unity must be reached by some particular process. In this case we imagine that the Rayleigh flow is continued (i.e., more heat is added) until the velocity reaches sonic. Figure 6.6 shows a T –s diagram for subsonic Rayleigh flow with heat addition. A sketch of the physical system is also shown. If we imagine that more heat is added, the entropy continues to increase and we will eventually reach the limiting point where sonic velocity exists. The dashed lines show a hypothetical duct in which the additional heat transfer takes place. At the end we reach the ∗ reference point for Rayleigh flow.

Figure 6.6 The ∗ reference for Rayleigh flow. A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 99 Figure 6.6 The ∗ reference for Rayleigh flow. We now rewrite the working equations in terms of the Rayleigh flow ∗ reference condition. Consider first: Let point 2 be any arbitrary point in the flow system and let its Rayleigh ∗ condition be point 1. Then and equation becomes

A course in Gas Dynamics…………………………………. …. …Lecturer: Dr A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 100 6.6 APPLICATIONS The procedure for solving Rayleigh flow problems is quite similar to the approach used for Fanno flow except that the tie between the two locations in Rayleigh flow is determined by heat transfer considerations rather than by duct friction. The recommended steps are, therefore, as follows: 1. Sketch the physical situation (including the hypothetical ∗ reference point). 2. Label sections where conditions are known or desired. 3. List all given information with units. 4. Determine the unknown Mach number. 5. Calculate the additional properties desired.

Table 6.1Fluid Property Variation for Rayleigh Flow A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 101 Table 6.1Fluid Property Variation for Rayleigh Flow 6.7 CORRELATION WITH SHOCKS We can now picture a supersonic Rayleigh flow followed by a normal shock, with additional heat transfer taking place subsonically. Such a situation is shown in Figure 6.7. Note that the shock merely jumps the flow from the supersonic branch to the subsonic branch of the same Rayleigh line. This also brings to light another reason why the supersonic stagnation curve must lie above the subsonic stagnation curve. If this were not so, a shock would exhibit a decrease in entropy, which is not correct.

Figure 6.7 Combination of Rayleigh flow and normal shock. A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 102 Figure 6.7 Combination of Rayleigh flow and normal shock. Example Air enters a constant-area duct with a Mach number of 1.6, a temperature of 200 K, and a pressure of 0.56 bar (as shown in figure). After some heat transfer a normal shock occurs, whereupon the area is reduced as shown. At the exit the Mach number is found to be 1.0 and the pressure is 1.20 bar. Compute the amount and direction of heat transfer.

q = cp (T02 - T01) = (1005)(226 - 302) = -76.38 k]/kg A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi 103 It is not known whether a heating or cooling process is involved. The flow from 3 to 4 is isentropic; thus p04 1 4 Note that point 3 is on the same Rayleigh line as point 1 and this permits us to compute M2 through the use of the Rayleigh table p03 p03 p1 p01 2.2714 1 01 From the Rayleigh table we find M3 = 0.481 and from the shock table, M2 = 2.906. Now we can compute the stagnation temperatures: T01 1 1 T02 T* 1 01 and the heat transfer: q = cp (T02 - T01) = (1005)(226 - 302) = -76.38 k]/kg The minus sign indicates a cooling process that is consistent with the Mach number’s increase from 1.60 to 2.906. p03 p04 = = p4 = (0.5283) (1.2) = 2.2714 bar p = = ( ) (0.2353)(1.1756) = 1.1220 p* p p p* 0.56 T01 = T1 = (0.6614) (200) = 302 K T T02 = T01 = (0.6629) ( ) (302) = 226 K 0.8842 T* T