EKC 212
FLUID FLOW FOR CHEMICAL
ENGINEERING
CHAPTER 9 (SOLUTION WIP EXERCISE):
AGITATION & MIXING OF LIQUIDS
Dr Mohd Azmier Ahmad
Tel: +60 (4) 5996459
Email: chazmier@eng.usm.my
1. Agitation is (i)………………….... while mixing is
(ii)………………………
Answer:
(i) an induced motion of a material in a specified way.
(ii) random distribution, into & through one another of two
or more initially separate phases
2a. Give 4 purposes of agitation of liquids
Answer:
¾ Suspending solid particles.
¾ Blending miscible liquids e.g. methyl alcohol & water.
¾ Dispersing gas through liquid in the form of small
bubbles.
¾ Promoting heat transfer between liquids & a coil/jacket.
2b. 4 purposes of suspension of solid
particles
Answer:
¾To produce a homogenous mixture for
feeding to a processing unit.
¾To promote growth of a crystalline product
from a supersaturated solution.
¾To dissolve the solids.
¾To catalyze a chemical reaction.
3. Sketch the typical diagram of
agitation vessel.
Answer:
4. Three main types of impellers are (i)……, (ii)……. & (iii) ………
Answer:
(i) Propellers; (ii) paddles & (iii) turbines
5. Sketch the schematic diagram of (a) axial flow & (b) radial flow.
Answer:
6. Sketch the flow pattern for
off-center propeller.
Answer:
7. Sketch the draft tube with flow
pattern in baffled tank for (a)
turbine & (b) propeller impeller.
Answer:
Flow pattern with off-center propeller
8. A flat-blade turbine with six blades is installed centrally in
a vertical tank. The tank is 3.6 m in diameter, the turbine is
1.2 m in diameter & is positioned 1.2 m from the bottom of
the tank. The turbine blades are 240mm wide. The tank is
filled to a depth of 3.6m with a solution of 50% caustic soda
at 65.6oC, which has a viscosity of 10.785 P and a density of
1498 kg/m3. The turbine is operated at 60 rpm. What power
will be required to operate the agitator if:(a)The tank was baffled & (b) The tank was unbaffled.
Answer:
n = 60rpm / 60 s = 1.0 r/s
Da = E = 1.2 m
µ = 10.785 P = 1.0785 kg/ms
1 poise = 0.1 kg/ms
(a) Baffled tank
N
RE
=
D a2 n ρ
μ
(1 . 2 )
2
=
(1 )( 1498 )
= 2000
1 . 0785
From Fig. 9.12, curve A for baffle, NP = 5.0
∴ P = N P n 3 Da5 ρ = (5)(1) 3 (1.2) 5 (1498) = 1.86 x10 4 W
(b) Unbaffled tank
From Fig 9.12, curve D, NP =2.0.
Froude number,
n 2 Da (1) 2 (1.2)
N Fr =
=
= 0.122
g
9.81
From Table 9.1, a & b are 1.0 & 40.0 respectively
a − log10 N Re 1.0 − log10 2000
m=
=
= −0.0575
b
40
So the corrected value of NP,
NP(Corr) = NP Χ NFrm = 2Χ0.122−0.0575 = 2.257
Thus power,
P = N P n3 Da5 ρ
= (2.257)(1)3 (1.2)5 (1498)
= 8413mN / s = 8413W
9. A propeller with three blades is installed centrally in a
vertical tank. The tank is 2.7 m in diameter, the propeller is
0.81 m in diameter & is positioned 0.81 m from the bottom of
the tank. The tank is filled to a depth of 2.7m with a caustic
soda solution, which has a viscosity of 1.5 cP and a density of
1498 kg/m3. The turbine is operated at 3.21 rpm. What
power will be required to operate the agitator if:- The tank
was baffled & (b) The tank was unbaffled.
Answer:
n = 3.21rpm / 60 s = 0.0535 r/s
Da = E = 0.81 m
µ = 1.5cP = 1.5x10-3 kg/ms
1 poise = 0.1 kg/ms
(a) Baffled tank
N
RE
=
D a2 n ρ
μ
(0 . 81 )
2
=
( 0 . 0535 )( 1498 )
= 3 . 51 x 10
1 . 5 x 10 − 3
4
From Fig. 9.13, NP = 0.9
∴ P = N P n 3 Da5 ρ = (0.9)(0.0535) 3 (0.81) 5 (1498) = 0.072W
(b) Unbaffled tank
From Fig 9.13, NP =0.58.
Froude number,
n 2 Da (0.0535) 2 (0.81)
N Fr =
=
= 2.36 x10 −4
g
9.81
From Table 9.1, a & b are 1.7 & 18.0 respectively
a − log10 N Re 1.7 − log10 35055
m=
=
= −0.158
b
18
So the corrected value of NP,
N P ( Corr ) = N P Χ N Frm = 0.58Χ(2.36 x10−4 ) −0.158 = 2.17
Thus power,
P = N P n3 Da5 ρ
= (2.17)(0.0535)3 (0.81)5 (1498)
= 0.173mN / s = 0.173W
10. An agitated vessel 0.45m in diameter contains a six-blade straight-blade
turbine 0.15m in diameter located 0.15m above the vessel floor & rotating at
54 rpm. It is proposed to use this vessel for neutralizing a dilute aqueous
solution of NaOH with a stoichiometrically equivalent quantity of
concentration nitric acid (HNO3). The final depth of liquid in the vessel is to
be 0.45m. Assuming that all the acid is added to the vessel at one time, how
long will it take for the neutralization to be completed for (a) baffled tank
and (b) unbaffled tank? (Given ρliquid = 997 kg/m3; μliquid = 9.82x10-2 Pa.s).
Answer:
(a) baffled tank
(b) unbaffled tank
Dt= 0.45 m, Da= E = 0.15 m, n = 0.9 rps,
∴ N RE =
Da2 nρ
μ
(
0.15) (0.9)(997)
=
= 205.6
2
9.82 x10−2
From Fig. 9.15, ntT = 170
Thus tT= 170/0.9 = 189s
From Fig. 9.15, ntT = 350
Thus tT= 350/0.9 = 389s