CHEMICAL ENGINEERING

20 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

PROBLEM 3.3

A cylindrical tank, 5 m in diameter, discharges through a mild steel pipe 90 m long and
230 mm diameter connected to the base of the tank. Find the time taken for the water
level in the tank to drop from 3 m to 1 m above the bottom. The viscosity of water is
1mNs/m^2.

Solution

If at any time the depth of water in the tank ishand levels 1 and 2 are the liquid levels
in the tank and the pipe outlet respectively, then the energy balance equation states that:

u^2 / 2 CgzCvP 2 P 1 CFD 0

In this example, P 1 DP 2 Datmospheric pressure and vP 2 P 1 D0. Also u 1 /u 2 D
 0. 23 / 5 ^2 D 0 .0021 so thatu 1 may be neglected. The energy balance equation then
becomes:
u^2 / 2 hgC 4 R/u^2 l/du^2 D 0

The last term is obtained from equation 3.19 andzDh.
Substituting the known data:

u^2 / 2  9. 81 hC 4 R/u^2  90 / 0. 23 u^2 D 0

or: u^2  19. 62 hC 3130 R/u^2 u^2 D 0

from which: uD 4. 43

p

h/

√

[1C 3130 R/u^2 ]

In falling from a heighthtohdh,

the quantity of water dischargedD/ 4  52 dhD 19 .63dhm^3.

Volumetric flow rateD/ 4  0. 23 ^2 uD 0. 0415 uD 0. 184

p

h/

√

[1C 3130 R/u^2 ], and
the time taken for the level to fall fromhtohdhis:

 19. 63

0. 184

dh

p

h

√

[1C 3130 R/u^2 ]D 106. 7 h^0.^5

√

[1C 3130 R/u^2 ]dh

∴the time taken for the level to fall from 3 m to 1 m is:

tD 106. 7

√

[1C 3130 R/u^2 ]

∫ 1

3

h^0.^5 dh

R/u^2 depends upon the Reynolds number which will fall as the level in the tank falls
and upon the roughness of the pipeewhich is not specified in this example. The pressure
drop along the pipeDhgD 4 Rl/dN/m^2 andRDhgd/ 4 l.
From equation 3.23:

R/u^2 Re^2 DRd^2 / ^2 Dh^2 gd^3 / 4 l ^2

Dhð 10002 ð 9. 81 ð 0. 233 / 4 ð 90 ð 10 ^6 D 3. 315 ð 108 h

Thus ashvaries from 3 m to 1 m,R/u^2 Re^2 varies from (9. 95 ð 108 )to(3. 315 ð
108 .)