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Lesson 32. SOLVING NUMERICAL
Lesson 32
SOLVING NUMERICAL
Example 1
A flat-blade turbine with six blades is installed centrally in a vertical tank. The tank is 6 ft (1.83 m) in diameter; the turbine is 2 ft (0.61m) in diameter and is positioned 2 ft (0.61m) from the bottom of the tank. The turbine blades are 5 in (127 mm) wide. The tank is filled to a depth of 6 ft (1.83 m) with a solution of 50 percent caustic soda, at 150°F (65.6°C), which has a viscosity of 12 cP and a density of 93.5 lb/ft^{3} (1498 kg/m^{3}). The turbine is operated at 90 r/ min. The tank is baffled. What power will be required to operate the mixer ?
Fig. 32.1 Curve turbine
Solution
Curve A in Fig1. applies under the conditions of this problem.
S_{1} = D_{a} / D_{t} ; S_{2} = E / D_{t} ; S_{3} = L / D_{a} ; S_{4} = W / D_{a} ; S_{5} = J / D_{t} and S_{6} = H / D_{t}
The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.
Fig. 32.2 Dimension of turbine agitator
The Reynolds number is calculated. The quantities for substitution are, in consistent units,
${D}_{a}=2\cdot \text{ft}$
$\text{}$$n=\frac{90}{60}=\mathrm{1.5\; r/s}$
μ = 12 x 6.72 x 10^{-4} = 8.06 x 10^{-3} lb/ft-s
ρ = 93.5 lb/ft^{3} g= 32.17 ft/s2
${N}_{}$_{Rc}
The power $P=\frac{5.8\times 93.5\times {\mathrm{(1.5}}^{3}\times {\mathrm{(2}}^{5}}{32.17}=$
What would the power requirement be in the vessel described in example 1. If the tank were unbaffled?
Fig. 32.3 Flat-blade turbine
Curve D of fig.2 now applies. Since the dashed portion of the curve must be used, the Froude number is a factor ; its effect is calculated as follows:
${N}_{\text{FR}}=\frac{{n}^{2}\times {D}_{a}}{g}=\frac{{1.5}^{2}\times 2}{32.17}=0.14$
Constants a and b $m=\frac{}{}$
Figure |
Line |
a |
b |
1 |
D |
1.0 |
40.0 |
2 |
B |
1.7 |
18.0 |
The constants a and b for substitution into Eq. $m=\frac{a-{\text{log}}_{10}\times {N}_{\text{Re}}}{b}$ are
a = 1.0 and b = 40.0 from eq. $m=\frac{a-{\text{log}}_{10}\times {N}_{\text{Re}}}{b}$
$\left(\right)$ $m=\frac{1-{\text{log}}_{10}\times \text{69,600}}{40}=-0.096$0.096
The power number read from fig 1, curve D, for N _{RC }= 69,600, is 1.07; the corrected value of N_{p} is 1.07 x 0.14 ^{-0.096} = 1.29. Thus, from $P=\frac{{N}_{p}\times {n}^{3}\times {{D}_{a}}^{5}\times \rho}{{g}_{c}}$ Eq.
$P=\frac{}{}$
The power requirement is 406/550 = 0.74 hp (0.55 kW).
It is usually not good practice to operate an unbaffled tank under these conditions of agitation.
Example 3
The mixer of example1. is to be used to mix a rubber-latex compound having a viscosity of 1200 P and a density of 70 lb / ft^{3} (1120 kg/m^{3}). What power will be required?
Solution
The Reynolds number is now $$
This is well within the range of laminar flow. ${N}_{p}=\frac{{K}_{}}{}$_{ }
Values of constants K_{L} and K_{T} in Eqs. and N_{P} = K_{T} are given in the table
below for baffled tanks having four baffles at tank diameter
Type of impeller |
K_{L} |
K_{T} |
Propeller, three blades |
||
Pitch 1.0^{1} |
41 |
0.32 |
Pitch 1.5^{2} |
55 |
0.87 |
Turbine |
||
Six-blade disk ^{2} (S_{3} = 0.25, S_{4} = 0.2) |
65 |
5.75 |
Six curved blades^{1} (S_{4} = 0.2) |
70 |
4.80 |
Six pitched blades ^{3} (45°,S_{4} = 0.2) |
- |
1.63 |
Four pitched baldes^{2} (45°,S_{4} = 0.2) |
44.5 |
1.27 |
Flat paddle, two baldes^{1}(S_{4} = 0.2) |
36.5 |
1.70 |
Anchor ^{2} |
300 |
0.35 |
K_{L} = 65 ; from Eq. , N_{P} = 65 / 5.2 = 12.5, and
$P=\frac{12.5\times 70\times {1.5}^{3}\times {2}^{5}}{32.17}=$The power required is 2938 / 550 = 5.34 hp (3.99 kW). This power requirement is independent of whether or not the tank is baffled. There is no reason for baffles in a mixer operated at low Reynolds numbers, as vortex formation does not occur under such conditions.
Note that a 10,000-fold increase in viscosity increases the power by only about 60 percent over that required by the baffled tank operating on the low-viscosity liquid.