Problem 8.1 Problem: A rapid-mixing basin is to be designed for a water coagulation plant, and the design flow for the basin is 4.0 MGD. The basin is to be square with depth equal to 1.25 times the -1 O width. The velocity gradient is to be 900 s (at 50 F), and the detention time is 30 sec. Determine: a) The basin dimensions if increments of 1 in. are used. b) The input horsepower. c) The impeller speed if a vane-disc impeller with six flat blades is employed and the tank is baffled. The impeller diameter is to be 50% of the basin width. Approach: First, we solve for the volume of the tank by using the flow rate and the detention time. We then use this volume to compute basin dimensions. Then, we use the velocity gradient and the kinematic viscosity to obtain power. Lastly, we use the knowledge of impeller relations to obtain impeller speed. Variables: θ = detention time G = velocity gradient Q = flow rate µ = absolute viscosity of water n = impeller velocity
W = Power imparted
x = length and width of basin d = depth of basin P = Power input D = diameter of impeller = Volume of basin = water density
∀
Solution: a) First, we need to calculate the volume of the basin. We accomplish this through the following:
% 4×10 ˧Iˬ J 1 ˤI˳ 1 ℎ˯ 1 ˭˩ 1 ˦ˮ ∀ = ˝× = 1 ˤI˳ %× 24 ℎ˯J × 60 ˭˩ × 60 J˥ × 7.48 ˧Iˬ ×30 J˥ˤJ =185.68˦ˮ ∀ = ˲$ × ˤ ˤ = 1.25˲ ∴ ∀ = 1.25˲% 185.1.628˦ˮ5 % =5.3 ˦˥˥ˮ ≈5˦ˮ−4˩ ˲ = 1.∀25 = 185. ˤ = 1.25˲ = 1.25 × 5.3 ˦˥˥ˮ = 6.675 ˦˥˥ˮ ≈ 6˦ˮ − 8˩
We now take this volume and equate it to the following:
Now, solving for depth:
Summary: Basin Dimensions: 5ft-4in x 5ft-4in x 6ft-8in
˦ˮ−ˬ % ˣ=˙$ ×={900J#$ ×{2.73×10'=22.113 J˥−˦ˮ ˦ˮ−ˬ% × {5.33$ × {6.67 =4193 ˦ˮ−ˬ ˜=ˣ×∀=22. 1 13 J˥−˦ˮ J˥ J˥ ˜=4193 ˦ˮ−ˬ × J˥ 550 ˦ˮ−ˬ ×ℎ=7.62ℎ ˦ ˮ−ˬ 4193 ˜ ={H˖'#/% ={5.75×{2.667˦ˮJ˥' ×1.936#/% =1.41 ˥˰J˥ × 601 ˭˩J˥ =84.5˭
b) First, we need to calculate the power imparted:
Now, we can solve for the total power:
Lastly, we convert to horsepower:
c) For this, we simply need the following equation:
Problem 8.3: Problem: A flocculation basin is to be designed for a water coagulation plant, and the design flow is 13.0 MGD. The basin is to be a cross-flow horizontal-shaft, paddle-wheel type with a -1 0 mean velocity gradient of 26.7 s (at 50 F), a detention time of 45 min, and a GT value from 50,000 to 100,000. Tapered flocculation is to be provided, and three compartments of equal depth in series are to be used, as shown in Figure 8.19(b). The compartments are to be separated by slotted, redwood baffle fences, and the basin floor is level. The G values are to be 50, 20, and -1 10s . The flocculation basin is to have a width of 90ft to adjoin the settling basin. The paddle wheels are to have blades with a 6 in width and a length of 10ft. The outside blades should clear the floor by 1 ft and be 1 ft below the water surface. There are to be six blades per paddle wheel, and the blades should have a spacing of 1 ft. Adjacent paddle wheels should have a clear spacing of 30 to 36 in. between blades. The wall clearance is 12 to 18 in. Determine: a) The basin Dimensions b) The paddle-wheel design c) The power to be imparted to the water in each compartment and the total power required for the basin. d) The range in rotational speed for each compartment if 1:4 variable speed drives are employed. Approach: To begin, we’ll solve for the GT value. Also, we’ll solve for the dimensions of the basin using a volume equation compared with the flow. From this, we will assume square compartments in profile (depth = length for each profile). This will give us our dimensions. Using those dimensions, we can solve for the paddle wheel design relatively easily with the restrictions given. From here, we can use the viscosity of the water, the separate G values and the volume of each compartment to solve for the power (in hp) required for each compartment. We
then simply add those up to obtain total power. Lastly, we can use relations of water speed versus rotational speed to obtain the rotations in rpms (the range). Variables: θ = detention time G = velocity gradient Q = flow rate d’ = diameter of paddle wheel = Volume of basin
∀
L = Length of Basin
x = width of basin d = depth of basin µ = absolute viscosity of water v = blade velocity relative to water P = Power imparted GT = GT value
Solution: a) To begin, we need to check that the GT value will fall in the appropriate range (50,000100,000):
˙ˠ=˙×= 26.J˥7 ×45˭˩× 601 ˭˩J˥ =72,090 ,≤,≤, → Ϳͻ! % 13. 0 ˙˖ 1 ℎ˯ 1 ˦ˮ ∀ =˝= 24 ℎ˯J × 60 ˭˩ ×45˭˩× 7.48 ˧Iˬ =5. 43×10& ˦ˮ%
Next, we compute the volume of the entire basin:
We already know the width of the basin, so we can divide that out to obtain the profile area of the basin:
& % ∀ 5. 4 3×10 ˦ˮ ˤ×= ˲ = 90 ˦ˮ =603.46˦ˮ$ ˤ×3ˤ=603.46˦ˮ$ → ˤ=14˦ˮ−2˩ Iˤ =42˦ˮ−6˩ ∀
To obtain that each of profiles of the compartments is a square, we assume that per compartment, length = depth. Therefore:
Final basin dimensions: 3 L = 42’6” x = 90’ d = 14’2” = 54187.5 ft
b) If we assume a six-blade paddle wheel design and that both clearances from the top and bottom of 1 ft, we will try and get our D1 to match accordingly.
˖# =ˤ−2{1˦ˮ −212 ˤF=14. 17˦ˮ−2{1˦ˮ −212 ×0.5˦ˮF=11.67˦ˮ ≈.XY
We also know that each of the other diameters is equal to twice the spading between blades plus the width of each blade multiplied by two.
{1˦ˮ −2{ˤ=8. ˖˖$% =˖=˖#$ −2−2{1˦ˮ−2{ˤ =11.66˦ˮ−2{1˦ˮ−2{0. ˦ˮ−2{1˦ˮ −2{.55˦ˮ=. ˦ˮ =.XYXY
Now that the wheel dimensions are known, we need to calculate how many paddle whe els per arm. To accomplish this, we take the total width of the basin and subtract out the minimum wheel spacing between wheels and the minimum spacing between the walls. Note: the spacing between wheels will be a function of the number of wheels themselves, but the wall spacing will remain constant. Once we have this number, we will round down to the nearest whole integer. We cannot round up because this would mean we have exceeded our basin width.
+2ͼ {1˦ˮ + {−1{2.5˦ˮ ˲=90˦ˮ={10˦ˮ∴;=
We must now check our answer to make sure this number of wheels will work. First, attempt to hold the wall clearances constant at 1ft and increase wheel spacing, being sure not to cross the 36 in. clearance limit.
+2{1˦ˮ + {7−1{˳ ˲=90˦ˮ= {7{10˦ˮ∴=XY
c) First, we must check that the total paddle blade areas are between 15 and 20% of the total cross-sectional area of each compartment. We start by calculating out the total cross-sectional area of the blades in each compartment:
˓ = ×ˤ ×6×7={10˦ˮ{0.5˦ˮ{6{7=210˦ˮ$ $ ˓ 210˦ˮ ˯˩˥ˤ= ˲×ˤ ×100= 90˦ˮ×14.17˦ˮ ×100=16.5 $ 54187.5˦ˮ % ∀ ˬ−J 50 $ ' ˜# =˙ 3 =2.73×10 ˦ˮ$ FJ˥F Ӟ 3 ӟ= XY−ͼ $ 54187.5˦ˮ % ˬ−J 20 ' ˜$ =2.73×10 ˦ˮ$ FJ˥F Ӟ 3 ӟ= XY−ͼ $ 54187.5˦ˮ % XY−ͼ ˬ−J 10 ' ˜% =2.73×10 ˦ˮ$ FJ˥F Ӟ 3 ӟ=
We now compare this to the area of each compartment that the blades cover (width times depth):
This falls firmly between the bounds of 15-20%. Therefore, the following equation can be used in solving for the power imparted on each shaft:
We simply repeat this step for compartments 2 and 3, changing the G values to match:
To obtain the total energy required, we simply sum these: Total power imparted = 1479 ft-lb/s
˰ = {J ˖˥˰F0.75 → ˰# = {J{{11.6˦ˮ{0.75 =27.33{J ˰˰$% ={J{{8. 6 ˦ˮ{0. 7 5=20. 2 6{J = {J{{5.6˦ˮ{0.75 =13.19{J
d) To start, we will find the velocity of each blade relative to the water:
We repeat this step for the other blades at their respective diameters:
Now that we have these relative velocities, we can use this in the power equation supplied to each wheel individually (take the total power in each compartment and divide by the number of wheels per compartment) to solve for the revolutions per second. This can be converted into and rpm value. This rpm value will be the upper rpm bound for the 1:4 variable speed driver. So, to get the complete bounds on this driver, we simply take the maximum rpm value and divide by four. We start with the power supplied per wheel in each compartment:
˕˭Iˮ˭˥ˮ 1:˜ = ˜7 = 1233˦ˮ∙7 ˬ/J =176.14˦ˮ∙ˬ/J ˕˭Iˮ˭˥ˮ 2:˜ = ˜7 = 197˦ˮ∙ˬ/J =28.14˦ˮ∙ˬ/J 7 ˕˭Iˮ˭˥ˮ 3:˜ = ˜7 = 49˦ˮ∙ˬ/J 7 =7.00˦ˮ∙ˬ/J ˜ =0.97˕ ˓{˰#% +˰$% +˰%%
We now equate these individual powers to the overall power equations:
So, plugging in numbers for each value, we can solve for the rps in each component: Compartment 1
176.14˦ˮ∙ ˬJ ={0.97{2{1.5{10˦ˮ×0.5˦ˮ{{27.33J% +{20.26J% +{13.19J% ∴J=0. 0 73× 160˭˩J˥˯ˮ˥ =4.38˭ ˔ˮˮ˭ ˔˯ˤ: ˭4 = 4.38˭ 4 =1.10˭ Answer: 1.10rpm to 4.38rpm
% +{20.26J% +{13.19J% {10˦ˮ×0. 28.14˦ˮ∙ˬ/J={0.9607{2{1. 5 5 ˦ˮ{{27. 3 3J ∴J=0. 0 40× 1 ˭˩J˥˯ˮ˥ =2.4˭ ˔ˮˮ˭ ˔˯ˤ: ˭4 = 2.4˭ 4 =0.6˭ Compartment 2
Answer: 0.60rpm to 2.40rpm
% +{20.26J% +{13.19J% {10˦ˮ×0. 7˦ˮ∙ˬ/J={0.9607{2{1. 5 5 ˦ˮ{{27. 3 3J ∴J=0. 0 25× 1 ˭˩J˥˯ˮ˥ =1.50˭ ˔ˮˮ˭ ˔˯ˤ: ˭4 = 1.50˭ 4 =0.38˭ Compartment 3
Answer: 0.38rpm to 1.50rpm
Problem 8.5 Problem: A pneumatic flocculation basin is to be designed for a tertiary treatment plant having a flow of 5.0 MGD. The plant is to employ high-pH lime coagulation, and the pertinent data for -1 O the flocculation basin are as follows: detention time = 5min, G = 150s (at 50 F), length = 2 times width, depth = 9ft-10in, diffuser depth = 9ft-0in, and air flow = 4 cfm per diffuser. Determine: a) The basin dimensions if 1in increments are used. 3 b) The total air flow in ft /min. c) The number of diffusers. Approach: First, we’ll solve for the basin dimensions the same way as was solved for in the previous problem. Then, we can solve for the total air flow by using a relation between power Solution: a) First, we need to calculate the volume of the basin. We accomplish this through the following:
% 5×10 ˧Iˬ J 1 ˤI˳ 1 ℎ˯ 1 ˦ˮ ∀ = ˝× = 1 ˤI˳ × 24 ℎ˯J × 60 ˭˩ × 7.48 ˧Iˬ ×5 ˭˩˯ˮ˥J=2321˦ˮ% ∀ =2˱$ ∀ = ˱××ˤ =2˱ ˤ=9˦ˮ−10˩ {9.8333˦ˮ ∴ 9.833˦ˮ % ∀ 2321˦ˮ ˱= 2×9.8333 = 2×9.8333˦ˮ =10.86 ˦˥˥ˮ ≈10˦ˮ−10˩ =2˱=2×10.86 ˦˥˥ˮ=21.72 ˦˥˥ˮ ≈21˦ˮ−9˩
We now take this volume and equate it to the following:
Now, solving for length:
Summary: Basin Dimensions: 10ft-10in x 21ft-9in x 9ft-10in b) To begin, we’ll solve for the power:
'×{10.8333˦ˮ ×21.75˦ˮ ×9.8333˦ˮ ˜=˙$∀={150J#$ ×{2. 7 3×10 =1423.2 ˦ˮ−ˬ J˥ ˜=˕#˙ logℎ+$ $F
Now, we use this power in the following equation to solve for air flow:
Where:
h = height of diffuser c2 = 34 (British Units) Rearranging and solving:
C1 = 81.5 (British Units)
% ˜ 1423. 2 ˦ ˮ ˙ = ˕# logӘℎ+$ә = 81.5×log9˦ˮ+34F =171.2 ˭˩ $ 34 c) For this, we simply divide the total air flow provided by the amount required per diffuser (4 cfm):
% ˦ ˮ 171. 2 ˖˩˦˦˯J˥J= 4 ˦ˮ% ˥ ˤ˩˭˩˦˦˯J˥ =42.8 ˤ˩˦˦˯J˥J →43 ˤ˩˦˦˯J˥J ˭˩
