# Bernoulli’s equation for rotational flow

Consider fluid moving in a circular path. Higher pressure must be exerted from the outside, towards the centre of rotation, in order to provide the centripetal force. That is, some outside pressure force must be available to prevent the particles moving in a straight Une. This suggests that the pressure is growing in magnitude as the radius increases, and a corollary is that the velocity of flow must fall as the distance from the centre increases.

With a segmental element at P(r, 6) where the velocity is qt only and the pressure p, the pressures on the sides will be shown as in Fig. 3.26 and the resultant pressure thrust inwards is

which reduces to

(3.53)

This must provide the centripetal force = mass x centripetal acceleration

= pr8r66q^/r

Equating (3.53) and (3.54):

dp _ pgj

dr r

The rate of change of total pressure H is

dH = d{p + pq) = dr dr

and substituting for Eqn (3.55):

Now for this system (1 /r)(dqn/d9) is zero since the streamlines are circular and therefore the vorticity is (q{jr) + (dq{jdr) from Eqn (2.79), giving

(3.56)

## Leave a reply