How to derive this temperature-pressure-specific volume relationship?

Question

(for isentropic, adiabatic, ideal gas flow) $$\frac{T_x}{T_y} = \left(\frac{p_x}{p_y}\right)^{\frac{(\gamma-1)}{\gamma}} = \left(\frac{V_y}{V_x}\right)^{\gamma-1}$$ where $V$ is the specific volume, $x$ and $y$ are any 2 rocket nozzle axial sections (though I presume this relation holds true for many other things as well), and $\gamma$ is the specific heat capacity ratio $c_p/c_v$, numerically equivalent to the molar heat capacity ratio $(\bar{C}_p/\bar{C}_v)$.

Thanks! p.s. I may have just learnt the hard way that it is probably not the best idea to write a markdown-dense blurb on a phone ;)

Answer 1

For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression:

$$\mathrm dU = \delta Q + \delta W$$

$$\mathrm dU = - p\,\mathrm dV$$

This expression is pretty much useless however, in that you can't integrate it, since $T$, $V$, and $p$ are all constantly changing interdependently in the system. However if you assume the gas is ideal and making a few substitutions...

$$C_V\,\mathrm dT = \frac{-nRT}{V}\,\mathrm dV$$

$$C_V \frac{1}{T}\,\mathrm dT = -nR\frac{1}{V}\,\mathrm dV$$

$$C_V \int_{T_1}^{T_2} \frac{1}{T}\,\mathrm dT = -nR\int_{V_1}^{V_2} \frac{1}{V}\,\mathrm dV$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -nR \cdot \ln\left(\frac{V_2}{V_1}\right)$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -(C_p - C_V) \cdot \ln\left(\frac{V_2}{V_1}\right) = (C_p - C_V) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \left(\frac{C_p}{C_V}-1\right) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \ln\left(\frac{V_1}{V_2}\right)^{\gamma-1} ; \gamma = \frac{C_p}{C_V}$$

$$\frac{T_2}{T_1} = \frac{V_1}{V_2}^{\gamma-1}$$

From here it's just more cycling through variables; see if you can work the mathematics a bit so to get your second relationship.

Answer 2

From the open system (control volume) version of the first law of thermodynamics, between cross sections x and y, $$\dot{Q}-\dot{W}_s-\dot{m}\Delta h=0$$where $\dot{Q}$ is the rate of heat addition to the control volume, $\dot{W}_s$ is the rate of doing shaft work, $\dot{m}$ is the mass flow rate, and $\Delta h$ is the change in specific enthalpy between cross sections x and y. If the process is adiabatic and reversible, no shaft work is being done, and the gas is ideal, this equation reduces to $$dh=C_pdT=VdP=\frac{RT}{P}dP$$ along the path between the two cross sections, where V is the molar volume. This equation integrates to $$\frac{T_x}{T_y}=\left(\frac{P_x}{P_y}\right)^{(\gamma-1)/\gamma}$$