## FANDOM

159 Pages

We consider a steady, isentropic, laminar fluid flow through a pipe whose cross sectional area changes continuously with time. We assume the flow is always along the axis of the pipe, and disregard motion perpendicular to the axis. In the next sections we derive relations between the area and the hydrodynamic variables for an ideal gas.

## Newtonian Edit

In steady state, the conservation of momentum reduces to

$v \frac{\partial v}{\partial x} +\frac{1}{\rho} \frac{\partial P}{\partial x} = 0$

where $v$ is the velocity, $\rho$ is the density, $P$ is the pressure and $x$ is the position along the pipe. We recall that the specific enthalpy is given by

$dw = T dS + \frac{1}{\rho} dP$

Hence, for an isentropic flow

$\frac{\partial}{\partial x} \left( \frac{1}{2} v^2 + w \right) = 0$

So the total enthalpy is conserved along the pipe. For an ideal gas the enthalpy is given by

$w = \frac{\gamma}{\gamma -1} \frac{P}{\rho}$

where $\gamma$ is the adiabatic index. Hence the following quantity is conserved along the pipe

$\frac{1}{2} v^2 + \frac{\gamma}{\gamma -1} \frac{P}{\rho} = W_t$

The mass current is also constant along the pipe

$J = A \rho v$

Plugging into the conservation of total momentum yields a relation between the pressure, velocity and cross sectional area

$\frac{1}{2} v^2 + \frac{\gamma}{\gamma -1} \frac{A P v}{J} = W_t$

If the flow is adiabatic, then $P = S \rho^{\gamma}$, when $S$ is a constant the represents the entropy. Under this assumption the equation reduces to

$\frac{1}{2} v^2 + \frac{\gamma}{\gamma-1} \frac{S}{v^2} \left( \frac{J}{A v} \right)^{\gamma-1} = W_t$

## Relativistic Edit

The particle current is constant throughout the pipe

$J = A n \beta \gamma$

where $n$ is the number density in the rest frame, $\beta = \frac{v}{c}$ is the dimensionless velocity and $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ is the Lorentz factor. Another quantity which is conserved is the energy current

$L = A \gamma^2 \beta w$

where $w = e + P$ is the relativistic enthalpy density (in the rest frame), $P$ is the pressure in the rest frame and $e$ is the energy density in the rest frame. By dividing the two conserved quantities we get a third conserved quantity

$\frac{L}{J} = \gamma \frac{w}{n}$

In case of an ideal gas with an adiabatic index $\eta$

$e = n m c^2 +\frac{P}{\eta -1}$

and

$w = n m c^2 + \frac{\eta}{\eta -1} P$

Eliminating $n$, we obtain

$\frac{L}{J} = \gamma \left( m c^2 + \frac{\eta}{\eta-1} \frac{P A \beta \gamma }{J} \right)$

It can be shown that this expression reduces to the Newtonian result in the limit $\beta \ll 1$.