Bernoulli's Equation — Question 8

Question 8

Different streamlines can have different Bernoulli constants

An inviscid incompressible liquid rotates rigidly in a horizontal plane with velocity $v_\theta=\Omega r$. The flow is steady. Radial force balance gives $dp/dr=\rho\Omega^2r$. Let $p(0)=p_c$ and compare points at the same height.

1. Find $p(r)$ and the Bernoulli quantity $C(r)=p+\rho v^2/2$.

2. Why does variation of $C$ with radius not violate Bernoulli along a streamline? Identify the incorrect prediction from treating $C$ as one universal constant across the flow.

Question 8 - Solution

Different streamlines can have different Bernoulli constants

(a) Derivation and calculation

$$\frac{dp}{dr}=\rho\Omega^2r\Rightarrow p(r)-p_c=\int_0^r\rho\Omega^2r^{\prime}\,dr^{\prime}.$$$$\boxed{p(r)=p_c+\tfrac12\rho\Omega^2r^2}.$$$$\boxed{C(r)=p+\tfrac12\rho v^2=p_c+\rho\Omega^2r^2}.$$

(b) Result and interpretation

Each streamline has fixed radius $r$, so$$\left.\frac{dC}{ds}\right|_{\mathrm{streamline}}=0,\qquad\frac{dC}{dr}=2\rho\Omega^2r\ne0.$$A universal cross-streamline constant would incorrectly imply$$\frac{dp}{dr}=-\rho v\frac{dv}{dr}=-\rho\Omega^2r,$$$$\text{opposite to the required radial pressure gradient}.$$

Physical check

The flow has vorticity $2\Omega$; Bernoulli along each circular streamline does not require one constant for all circles.