Steady isothermal flow of an ideal gas

Question

So I have a steady isothermal flow of an ideal gas through a smooth duct (no frictional losses) and need to compute the mass flow rate (per unit area) as a function of pressures at any two different arbitrary points, say 1 and 2. I have the following momentum equation in differential form:

\begin{equation} \rho vdv + dP = 0\end{equation} where $v$ is the gas the flow velocity and $P$ is static pressure. The mass flow rate per unit cross section $G$, can be calculated by integrating this equation between points 1 and 2. This is where it gets confusing. I do the integration by two ways:

1) Use the ideal gas equation $P = \rho RT$ right away and restructure the momentum equation: \begin{equation} vdv + RT\frac{dP}{P} = 0\end{equation}, integrate it between points 1 and 2 and arrive at:

\begin{equation} v_1^2-v_2^2 + 2RTln\frac{P_1}{P_2} = 0 \end{equation}

Since the flow is steady, I can write $G = \rho_1 v_1 = \rho_2 v_2$, again use the ideal gas law to write density in terms of gas pressure and finally arrive at the mass flow rate expression:

\begin{equation} G^2 = \frac{2ln\frac{P_2}{P_1}}{RT(\frac{1}{P_1^2}-\frac{1}{P_2^2})} \end{equation}

2) In another way of integrating (which is mathematically correct), I start by multiplying the original momentum equation by $\rho$ to get

\begin{equation} \rho^2 vdv + \frac{1}{RT}PdP = 0\end{equation} write $\rho^2v = G^2/v$, integrate between points 1 and 2 to arrive at

\begin{equation} G^2ln\frac{v_2}{v_1} = \frac{P_1^2 - P_2^2}{2RT} \end{equation} Using the ideal gas law the velocity ratio can be written as the pressure ratio to finally arrive at the mass flow rate equation

\begin{equation} G^2 = \frac{P_1^2 - P_2^2}{2RTln\frac{P_1}{P_2}} \end{equation}

Both the expressions are dimensionaly sound and I know that the second expression is the correct one. My question is, whats wrong with first expression.

Answer 1

The reason why you arrive at different solutions is that the assumptions in the assignment are inconsistent. One way to show this is to show in isothermal flow if ideal gas in straight duct of constant cross-section with no friction the gas has to have the same pressure everywhere; different pressures at the entrance and the exit $P_1, P_2$ are not possible.

To show this, let us take the Euler equation

$$\ \rho v \frac{dv}{dx} + \frac{dp}{dx} = 0, $$

and transform it into equation for $P(x)$ only. We express $v$ as $G/\rho$ and obtain

$$ \frac{d}{dx}\left(\frac{G^2}{\rho}\right) + \frac{dP}{dx} = 0. $$ From the state equation of ideal gas, we can express the density as a function of pressure and temperature:

$$ \rho = \frac{MP}{RT}, $$ where $M$ is molar mass of the gas and $R$ is the universal gas constant. Replacing $\rho$ in the last equation, we obtain

$$ \frac{d}{dx}\left(\frac{G^2 RT}{M} \frac{1}{P(x)} + P(x)\right) = 0. $$

Since the temperature $T$ is assumed constant, this equation implies that the pressure $P$ is constant as well, which contradicts the assumption about different pressures $P_1, P_2$.

In other words, if there is no friction, the gas will move inertially with the same pressure and velocity everywhere. The velocity can be any number so it cannot be deduced from the data in the assignment.

The assignment may have reasonable solution if we introduce friction into the model. The mathematically simplest way seems to be to add a negative constant force to the right-hand side of the Euler equation:

$$ \frac{d}{dx}\left(\frac{G^2 RT}{M} \frac{1}{P(x)} + P(x)\right) = -f. $$ This equation has sensible solution $P(x)$ for given pressures $P_1, P_2$ and will allow you to find $G$ and $v(x)$. However, physically constant friction is rather unrealistic model, especially if the friction is low, as then the gas velocity and its transversal gradient will increase rapidly along the duct. Since the Newton friction forces are proportional to transversal gradient, they should get stronger along the duct and thus damp the increase in velocity (but not completely, in reality the velocity should increase along the duct). More realistic approach would be to solve the Navier-Stokes equations for stationary isothermal flow, similarly to what one does in the derivation of the Poiseuille law for incompressible liquid flow in a pipe.

Answer 2

Apart from the physical meaning i think that there is a mathematics issue here. When you divide the original differential equation you result in an equivalent form. But when you have a condition related to the variables then things change and essentially you generate a different equation and thus different solution.

Answer 3

Have a look into Rayleigh line flow (Flow with heat addition or loss). I just spent some time refreshing my memory by reading a chapter from "Gas Dynamics" by James E. A. John. That book has a very good introduction to the topic. At a glance, Wikipedia seems to have a decent looking page on it as well. The discussion of Rayleigh line flow from these two sources is for perfect gases only. I suspect that some very interesting things might be possible for other equations of state. A key result for perfect gases, where the speed of sound $a = \sqrt{\gamma R T}$, is that temperature and Mach number (likewise speed and temperature since the speed of sound is only a function of temperature) $M = v/a$, are related by $$T=\frac{M^2}{(1+\gamma M^2)^2}\times constant$$ For a fixed temperature you can only hope to satisfy this equation with at most two Mach numbers (2nd order polynomial equation in $M^2$ with $M>0$ required). Which Rayleigh line we are dealing with all depends on the downstream and upstream pressure conditions that set up the flow. Consequently, the only valid isothermal solution for a perfect gas is $$v_1=v_2$$ $$P_1=P_2$$ $$\rho_1=\rho_2$$ $$\dot{Q}=0$$ where $\dot{Q}$ is the heat flux. I believe this is probably why you are able to find an unlimited number of what appear to be valid solutions to the problem, all of which satisfy the only real solution.