Page 233 - Bird R.B. Transport phenomena

§7.7 Use of the Macroscopic Balances for Unsteady-State Problems 217

EXAMPLE 7.7-1 An open cylinder of height Я and radius R is initially entirely filled with a liquid. At time t =
0 the liquid is allowed to drain out through a small hole of radius R at the bottom of the tank
Acceleration Effects in (see Fig. 7.7-1). o
Unsteady Flow from a
Cylindrical Tank (a) Find the efflux time by using the unsteady-state mass balance and by assuming Torri-
celli's equation (see Problem 3B.14) to describe the relation between efflux velocity and the in-
stantaneous height of the liquid.
(b) Find the efflux time using the unsteady-state mass and mechanical energy balances.
SOLUTION (a) We apply Eq. 7.1-2 to the system in Fig. 7.7-1, taking plane 1 to be at the top of the tank (so
that w ] = 0). If the instantaneous liquid height is h(t), then

2
y- (rrR hp) = - P V 2 (TTRI) (7.7-1)
at
Here we have assumed that the velocity profile at plane 2 is flat. According to Torricelli's
equation v = Vlgh, so that Eq. 7.7-1 becomes
2
2gh (7.7-2)
R
When this is integrated from t = 0 to t = f we get
efflux
x _ /2NH (7.7-3)
'efflux —
in which N = (R/R ) 4 » 1. This is effectively a quasi-steady-state solution, since we have
0
used the unsteady-state mass balance along with Torricelli's equation, which was derived for
a steady-state flow.
(b) We now use Eq. 7.7-1 and the mechanical energy balance in Eq. 7.4-2. In the latter, the terms
W m and E are identically zero, and we assume that E is negligibly small, since the velocity gra-
c
v
dients in the system will be small. We take the datum plane for the potential energy to be at the
bottom of the tank, so that Ф = gz 2 = 0; at plane 1 no liquid is entering, and therefore the poten-
2
tial energy term is not needed there. Since the top of the tank is open to the atmosphere and the
tank is discharging into the atmosphere, the pressure contributions cancel one another.
To get the total kinetic energy in the system at any time t, we have to know the velocity
of every fluid element in the tank. At every point in the tank, we assume that the fluid is mov-
ing downward at the same velocity, namely v (R /R) 2 SO that the kinetic energy per unit vol-
0
2
4
ume is everywhere lpvl(R /R) .
0
To get the total potential energy in the system at any time t, we have to integrate the po-
2
2
tential energy per unit volume pgz over the volume of fluid from 0 to h. This gives 7rR pg(lh ).
Therefore the mechanical energy balance in Eq. 7.4-2 becomes
2
j t [{irR h)(\pv ){R /Rf (7.7-4)
2
2
Q
2
From the unsteady-state mass balance, v = -(R/R ) (dh/dt). When this is inserted into Eq.
2 0
7.7-4 we get (after dividing by dh/dt)
(7.7-5)

Outlet of radius R Fig. 7.7-1. Flow out of a cylindrical tank.
o