451
Chapter 6 x Viscous Flow in Ducts
6.27 Let us attack Prob. 6.25 in symbolic
fashion, using Fig. P6.27. All parameters
are constant except the upper tank depth Z(t).
Find an expression for the flow rate Q(t) as
a function of Z(t). Set up a differential
equation, and solve for the time t0 to drain
the upper tank completely. Assume quasisteady laminar flow.
Solution: The energy equation of Prob. 6.25,
using symbols only, is combined with a
control-volume mass balance for the tank
to give the basic differential equation for Z(t):
Fig. P6.27
energy: h f
32PLV
Ugd 2
or:
S
4
h Z; mass balance:
D2
dZ
dt
S
4
d ªS 2
S
º
D Z d2 L »
«
dt ¬ 4
4
¼
d 2 V, where V
Q
S
4
d 2 V,
U gd 2
(h Z)
32P L
Separate the variables and integrate, combining all the constants into a single “C”:
Z
³
Zo
dZ
hZ
t
C³ dt, or: Z
(h Zo )e Ct h, where C
0
Tank drains completely when Z
0, at t 0
U gd 4
32 P LD2
Z ·
1 §
ln ¨ 1 o ¸
C ©
h¹
Ans.
Ans.