CHEMICAL ENGINEERING

90 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Ifhis the distance of the water level below the maximum depth of 1 m, then the head
above the orifice is equal to 1 hand:

GDCDA 0 

√

[2g 1 h]

When the tank contains 0.5 m of water, the flowrate is given by:

GD 0. 6 ð/ 4  0. 04 ^2 ð 1000

√

 2 ð 9. 81 ð 0. 5 D 2 .36 kg/s

The input to the tank is stated to be proportional toh, and when the tank is half full
the inflow is equal to the outflow,

or: 2. 36 DKð 0. 5 andKD 4 .72 kg/ms

Thus the inflowD 4. 72 hkg/s and the outflowDCDA 0 

p

2 g

p

 1 hkg/s.

The net rate of fillingD 4. 72 hCDA 0 

p

2 g

p

 1 hD 4. 72 h 3. 34

p

 1 h

Time to fill the tankD(mass of water/rate of filling)
D 1 ð 1. 5 ð 0. 45  0. 25 ð 1000 /rateD 300 /rate

The time to fill from 0.25 to 0.45 m above the bottom of the tank is then:

timeD

∫ 0. 75

0. 55

300dh

4. 72 h 3. 34

p

 1 h

This integral is most easily solved graphically as shown in Fig. 6c, where the area under
the curveD 0 .233 s/m and the timeD 300 ð 0. 233 D70 s.

3.0

2.5

2.0

1.5

1.0

0.5

0

0.55 0.55 0.65 0.70 0.70

h (m)

Area under curve = 0.233 s/m

1/(4.72

−3.34

√^1

−h

)

Figure 6c.