Extrema of Rayleigh Line Consider the T-s diagram of Rayleigh flow, as shown in Fig. 12–54. Using the differential forms of the conservation equations and property relations, show that the Mach number is Maa = 1 at the point of maximum entropy (point a), Mab = 1/√k and at the point of maximum

Question

Extrema of Rayleigh Line

Consider the T-s diagram of Rayleigh flow, as shown in Fig. 12–54. Using the differential forms of the conservation equations and property relations, show that the Mach number is [latex] M a_{a}=1 [/latex] at the point of maximum entropy (point a), [latex] M a_{b}=1 / \sqrt{k} [/latex] and at the point of maximum temperature (point b).

Accepted Answer

It is to be shown that [latex] M a_{a}=1 [/latex] at the point of maximum entropy and [latex] Ma _{b}=1 / \sqrt{k} [/latex] at the point of maximum temperature on the Rayleigh line.
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid.
Analysis The differential forms of the continuity (ρV = constant), momentum [rearranged as P + (ρV)V = constant], ideal gas (P = ρRT), and enthalpy change [latex] \left(\Delta h=c_{p} \Delta T ight) [/latex] equations can be expressed as

[latex] ho V= ext { constant } ightarrow ho d V+V d ho=0 \quad ightarrow \quad \frac{d ho}{ ho}=-\frac{d V}{V} [/latex] (1)

[latex] P+( ho V) V= ext { constant } \quad ightarrow \quad d P+( ho V) d V=0 \quad ightarrow \quad \frac{d P}{d V}=- ho V [/latex] (2)

[latex] P= ho R T \quad ightarrow \quad d P= ho R d T+R T d ho \quad ightarrow \quad \frac{d P}{P}=\frac{d T}{T}+\frac{d ho}{ ho} [/latex] (3)

The differential form of the entropy change relation (Eq. 12–40) of an ideal gas with constant specific heats is

[latex] d s=c_{p} \frac{d T}{T}-R \frac{d P}{P} [/latex] (4)

[latex] s_{2}-s_{1}=c_{P} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}} [/latex] (12–40)

Substituting Eq. 3 into Eq. 4 gives

[latex] d s=c_{p} \frac{d T}{T}-R\left(\frac{d T}{T}+\frac{d ho}{ ho} ight)=\left(c_{p}-R ight) \frac{d T}{T}-R \frac{d ho}{ ho}=\frac{R}{k-1} \frac{d T}{T}-R \frac{d ho}{ ho} [/latex] (5)

since

[latex] c_{p}-R=c_{v} ightarrow \quad k c_{v}-R=c_{v} ightarrow \quad c_{v}=R /(k-1) [/latex]

Dividing both sides of Eq. 5 by dT and combining with Eq. 1,

[latex] \frac{d s}{d T}=\frac{R}{T(k-1)}+\frac{R}{V} \frac{d V}{d T} [/latex] (6)

Dividing Eq. 3 by dV and combining it with Eqs. 1 and 2 give, after rearranging,

[latex] \frac{d T}{d V}=\frac{T}{V}-\frac{V}{R} [/latex] (7)

Substituting Eq. 7 into Eq. 6 and rearranging,

[latex] \frac{d s}{d T}=\frac{R}{T(k-1)}+\frac{R}{T-V^{2} / R}=\frac{R\left(k R T-V^{2} ight)}{T(k-1)\left(R T-V^{2} ight)} [/latex] (8)

Setting ds/dT = 0 and solving the resulting equation R(kRT - V²) = 0 for V give the velocity at point a to be

[latex] V_{a}=\sqrt{k R T_{a}} \quad ext { and } \quad Ma _{a}=\frac{V_{a}}{c_{a}}=\frac{\sqrt{k R T_{a}}}{\sqrt{k R T_{a}}}=1 [/latex] (9)

Therefore, sonic conditions exist at point a, and thus the Mach number is 1.
Setting [latex] d T / d s=(d s / d T)^{-1}=0 [/latex] and solving the resulting equation T(k - 1) × (RT - V²) = 0 for velocity at point b give

[latex] V_{b}=\sqrt{R T_{b}} \quad ext { and } \quad Ma _{b}=\frac{V_{b}}{c_{b}}=\frac{\sqrt{R T_{b}}}{\sqrt{k R T_{b}}}=\frac{1}{\sqrt{k}} [/latex] (10)

Therefore, the Mach number at point b is [latex] M a_{b}=1 / \sqrt{k} [/latex]. For air, k = 1.4 and thus [latex] Ma _{b}=0.845 [/latex].
Discussion Note that in Rayleigh flow, sonic conditions are reached as the entropy reaches its maximum value, and maximum temperature occurs during subsonic flow.

Question 12.13: Extrema of Rayleigh Line Consider the T-s diagram of Rayleig...

Related Answered Questions

Exit Conditions of Fanno Flow in a Duct Air enters a 27-m-long 5-cm-diameter adiabatic duct at V1 = 85 m/s, T1 = 450 K, and P1 = 220 kPa (Fig. 12–67). The average friction factor for the duct is estimated to be 0.023. Determine the Mach number at the duct exit and the mass flow rate of air. ...

Verified Answer:

Choked Fanno Flow in a Duct Air enters a 3-cm-diameter smooth adiabatic duct at Ma1 = 0.4, T1 = 300 K, and P1 = 150 kPa (Fig. 12–66). If the Mach number at the duct exit is 1, determine the duct length and temperature, pressure, and velocity at the duct exit. Also determine the percentage of ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Gas Flow through a Converging Nozzle Nitrogen enters a duct with varying flow area at T1 = 400 K, P1 = 100 kPa, and Ma1 = 0.3. Assuming steady isentropic flow, determine T2, P2, and Ma2 at a location where the flow area has been reduced by 20 percent. ...

Verified Answer:

Effect of Back Pressure on Mass Flow Rate Air at 1 MPa and 600°C enters a converging nozzle, shown in Fig. 12–24, with a velocity of 150 m/s. Determine the mass flow rate through the nozzle for a nozzle throat area of 50 cm² when the back pressure is (a) 0.7 MPa and (b) 0.4 MPa. ...

Verified Answer:

Critical Temperature and Pressure in Gas Flow Calculate the critical pressure and temperature of carbon dioxide for the flow conditions described in Example 12–3 (Fig. 12–19). ...

Verified Answer:

Verified Answer:

Effect of Heat Transfer on Flow Velocity Starting with the differential form of the energy equation, show that the flow velocity increases with heat addition in subsonic Rayleigh flow, but decreases in supersonic Rayleigh flow. ...

Verified Answer: