Show that for Rayleigh flow, the maximum amount of heat that may be added to the gas is given by: \\[\frac{q_{\max }}{c_{p} T_{1}}=\frac{\left(\mathrm{Ma}_{1}^{2}-1\right)^{2}}{2(k+1) \mathrm{Ma}_{1}^{2}}\\]

Short Answer

Expert verified
The given formula represents the maximum amount of heat addition that can be made to achieve sonic conditions (Ma=1) under Rayleigh Flow and can be derived by understanding the relationships underlying the perfectly gas and the heat addition process.

Step by step solution

01

Recognize the fundamental relationships

The fundamental governing equations for one-dimensional isentropic flow of a perfect gas can be expressed as: \[T T_{0}=\frac{T_{0}}{T}=\frac{1+k M a^{2}}{2}\] and \[M a^{2}=\frac{2}{k-1}\left(\frac{T_{0}}{T}-1\right)\] where T is the absolute static temperature, T0 is the total (or stagnation) temperature, Ma is the Mach number, and k is the specific heat ratio.
02

Understand the Rayleigh Line and Heat addition

In Rayleigh flow, the addition of heat along the flow moves the state towards the Rayleigh line and eventually towards sonic conditions. Also realize that the maximum heat addition would correspond the flow state reaching Mach number equal to 1 (Ma=1).
03

Calculate the Maximum Heat Addition

To set up an equation for maximum heat addition (q_max), we need to understand that the process starts from the initial Mach number (Ma1), passes through multiple states upon heat addition and ends up at Ma=1, representing maximum heat. For the flow to reach the sonic state upon heat addition, we need to go via the Rayleigh line and can thus obtain the heat addition as: \[ \frac{q_{\max }}{c_{p} T_{1}}=T_{01}-T_{1}\] where T01 is the total temperature at the sonic state upon maximum heat addition.
04

Substitute Values in Equation

Substitute T01 using the equation from Step 1 and solving it appropriately to eventually result into the required form: \[\frac{q_{\max }}{c_{p} T_{1}}=\frac{(Ma_{1}^{2}-1)^{2}}{2(k+1) Ma_{1}^{2}}\]

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