Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2016 / Paper 1 / 3

Let $e=c_{v}T$ be the specific internal energy of a calorically perfect gas and define its specific gas constant $\mathcal{R}=R^{\ast }/\mu$, so $p=\mathcal{R}\rho T$. A reversible adiabatic change has no heat input and obeysUsing $c_p-c_v=\mathcal{R}$ and $\gamma =c_p/c_v$ givesIntegration yields $p/{\rho }^{\gamma }=K$, or $\boxed{p=K{\rho }^{\gamma }}$, with $K$ fixed on that isentrope. Equivalently $s-s_{\mathrm{ref}}=c_v\ln [(p/{\rho }^{\gamma })/K_{\mathrm{ref}}]$, so $K$ specifies entropy as well as composition. In a smooth nondissipative flow this means $D(p{\rho }^{-\gamma })/Dt=0$ along each parcel. A single common $K$ throughout the fluid additionally requires a homentropic flow. No heat transfer alone does not forbid entropy production: different sides of an adiabatic shock generally have different $K$ values. Constant composition and heat capacities are needed for the fixed exponent used here.

Write $\mathbf{\omega }=\nabla \times \mathbf{u}$ and use the vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$. Taking the curl of the momentum equation eliminates all pure gradients, including gravity and the $\nabla (\nabla \cdot \mathbf{u})$ viscous contribution. Before invoking barotropy, the pressure term givesThis is baroclinic vorticity generation. For a barotropic fluid, $p=p(\rho )$, so $\nabla p$ is parallel to $\nabla \rho$ and this source vanishes. Constant $\nu$ permits the curl and Laplacian to commute. The result isNo incompressibility assumption was used. Since $\nabla \cdot \mathbf{\omega }=0$, an equivalent form isThe terms represent vortex stretching/tilting, compression/dilatation and viscous diffusion. In particular, removing the baroclinic source does not make the vorticity of a three-dimensional parcel constant. A spatially varying entropy-dependent $K$ in $p=K{\rho }^{\gamma }$ need not define a single barotropic law; the common-pressure-law assumption matters.

Use the idealized fully developed, laminar cylindrical-flow model implied by the fixed cross-section and zero velocity at its sides. Take the jet length to be $R$, as for a central source reaching the cloud boundary; if the source is off-centre, the actual path length must replace $R$. Let $s$ denote distance from the jet axis and write the axial velocity as $u(s)$. Then $(\mathbf{u}\cdot \nabla )\mathbf{u}=0$, the pressure gradient is $dp/dz=-p_0/R$, and the axial Navier-Stokes equation reduces toRegularity at $s=0$ excludes a logarithmic term. The imposed no-slip boundary condition $u(r_0)=0$ gives the Hagen-Poiseuille flow profileThe kinetic energy crossing each area element per unit time is $(\rho u^2/2)udA$, so the velocity must be cubed before integration. ThusConsequentlyThe mean velocity is $\bar{u}=U_{\max }/2$. Replacing the profile by $\bar{u}$ in $\frac{1}{2}\rho \pi r_0^2{\bar{u}}^3$ underestimates the flux by two; the kinetic energy correction factor is two for this profile. The radius dependence follows from $u\propto r_0^2$ and cross-sectional area $\propto r_0^2$.

The specified $P$ is outgoing bulk kinetic-energy flux, not the pressure-work rate $p_{0}Q$, which scales as $r_0^4$. In a fully developed section the profile is independent of $z$ and pressure work is dissipated as heat; entrance acceleration and energy supply establish the profile. A turbulent or freely entraining astrophysical jet need not have this no-slip parabolic structure. The model assumes a long enough, laminar confined region and neglects entrance losses and gravity.