agitator hw3 solution

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

width. The velocity gradient is to be 900 s-1

(at 50O 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 x = length and width of basin

G = velocity gradient d = depth of basin

Q = flow rate P = Power input

µ = absolute viscosity of water D = diameter of impeller

n = impeller velocity ∀ = Volume of basin

W = Power imparted � = water density

Solution:

a) First, we need to calculate the volume of the basin. We accomplish this through the following:

∀ = � × � = 4 × 10�� 1 � � × 1 � �24 ℎ�� × 1 ℎ��60 �� × 1 ��60 �� × 1 ��7.48 � � × 30 ��

= 185.68��

We now take this volume and equate it to the following: ∀ = %& × � � = 1.25% ∴ ∀ = 1.25%�

% = ( ∀1.25) = (185.68��1.25) = 5.3 �� ≈ 5�� − 4��

Now, solving for depth: � = 1.25% = 1.25 × 5.3 �� = 6.675 �� ≈ 6�� − 8��

Summary:

Basin Dimensions: 5ft-4in x 5ft-4in x 6ft-8in

b) First, we need to calculate the power imparted:

, = -& × . = /900�123& × /2.73 × 10143 = 22.113 �� − �5�� − ��

Now, we can solve for the total power:

6 = , × ∀= 22.113 �� − �5�� − �� × /5.3373& × /6.6773 = 4193 �� − �5��

Lastly, we convert to horsepower:

6 = 4193 �� − �5�� × ��550 �� − �5 × ℎ8 = 7.62ℎ8

c) For this, we simply need the following equation:

� = / 69:;4�32/� = / 4193 �� − �5��5.75 × /2.667��34 × 1.93632/� = 1.41 ��=�� × 60 ��1 �� = 84.5�8�

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

mean velocity gradient of 26.7 s-1

(at 500F), 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

10s-1

. 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 x = width of basin

G = velocity gradient d = depth of basin

Q = flow rate µ = absolute viscosity of water

d’ = diameter of paddle wheel v = blade velocity relative to water

∀ = Volume of basin P = Power imparted

L = Length of Basin GT = GT value

Solution:

a) To begin, we need to check that the GT value will fall in the appropriate range (50,000-

100,000):

-> = - × � = 26.7�� × 45�� × 60 ��1 �� = 72,090

@A, AAA ≤ CD, AEA ≤ FAA, AAA → HI! Next, we compute the volume of the entire basin:

∀ = �� = 13.0 K-;24 ℎ�� × 1 ℎ��60 �� × 45�� × 1 ��7.48 � � = 5.43 × 10L ��

We already know the width of the basin, so we can divide that out to obtain the profile area of

the basin:

� × M = ∀% = 5.43 × 10L ��90 �� = 603.46��&

To obtain that each of profiles of the compartments is a square, we assume that per compartment,

length = depth. Therefore: � × 3� = 603.46��& → � = 14�� − 2�� MNON = 42�� − 6��

Final basin dimensions:

L = 42’6” x = 90’ d = 14’2” ∀ = 54187.5 ft3

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.

;2 = � − 2/1��3 − 2 P12 �7Q = 14.17�� − 2/1��3 − 2 P12 × 0.5��Q = 11.67�� ≈ FF. RST

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.

;& = ;2 − 2/1��3 − 2/�73 = 11.6�� − 2/1��3 − 2/. 5��3 = U. RST ;� = ;& − 2/1��3 − 2/�73 = 8.6�� − 2/1��3 − 2/0.5��3 = @. RST

Now that the wheel dimensions are known, we need to calculate how many paddle wheels 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. % = 90�� = �/10��3 + 2/1��3 + /� − 13/2.5��3 ∴ W = C XYZZ[\

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. % = 90�� = /73/10��3 + 2/1��3 + /7 − 13/�3 ∴ ] = ^ST 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: _ = M7 × �7 × 6 × 7 = /10��3/0.5��3/63/73 = 210��&

We now compare this to the area of each compartment that the blades cover (width times depth):

% ��8�� = _abcdef% × � × 100% = 210��&90�� × 14.17�� × 100% = 16.5%

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:

62 = .-& ∀3 = P2.73 × 1014 �5 − ��& Q P 50��Q& g54187.5��3 h = FD^^ ST − [i\

We simply repeat this step for compartments 2 and 3, changing the G values to match:

6& = P2.73 × 1014 �5 − ��& Q P 20��Q& g54187.5��3 h = FEC ST − [i\

6� = P2.73 × 1014 �5 − ��& Q P 10��Q& g54187.5��3 h = jE ST − [i\

To obtain the total energy required, we simply sum these:

Total power imparted = 1479 ft-lb/s

d) To start, we will find the velocity of each blade relative to the water:

=k = /�8�3 Pl;k��= Q 0.75 → =2 = /�8�3/l3/11.6��3/0.753 = 27.33/�8�3

We repeat this step for the other blades at their respective diameters: =& = /�8�3/l3/8.6��3/0.753 = 20.26/�8�3 =� = /�8�3/l3/5.6��3/0.753 = 13.19/�8�3

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:

m��8 �� 1: 6opeeb = 67 = 1233�� ∙ �5/�7 = 176.14�� ∙ �5/�

m��8 �� 2: 6opeeb = 67 = 197�� ∙ �5/�7 = 28.14�� ∙ �5/�

m��8 �� 3: 6opeeb = 67 = 49�� ∙ �5/�7 = 7.00�� ∙ �5/�

We now equate these individual powers to the overall power equations: 6k = 0.97mr_/=2� + =&� + =��3

So, plugging in numbers for each value, we can solve for the rps in each component:

Compartment 1

176.14�� ∙ �5� = /0.973/23/1.53/10�� × 0.5��3//27.33�8�3� + /20.26�8�3� + /13.19�8�3�3

∴ �8� = 0.073 × 60 ��1 �� = 4.38�8� s�� s��: �8�tcu4 = 4.38�8�4 = 1.10�8�

Answer: 1.10rpm to 4.38rpm

Compartment 2 28.14�� ∙ �5/� = /0.973/23/1.53/10�� × 0.5��3//27.33�8�3� + /20.26�8�3� + /13.19�8�3�3

∴ �8� = 0.040 × 60 ��1 �� = 2.4�8� s�� s��: �8�tcu4 = 2.4�8�4 = 0.6�8�


Compartment 3 7�� ∙ �5/� = /0.973/23/1.53/10�� × 0.5��3//27.33�8�3� + /20.26�8�3� + /13.19�8�3�3

∴ �8� = 0.025 × 60 ��1 �� = 1.50�8� s�� s��: �8�tcu4 = 1.50�8�4 = 0.38�8�


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

the flocculation basin are as follows: detention time = 5min, G = 150s-1

(at 50O 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.

b) The total air flow in ft3/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�� 1 � � × 1 � �24 ℎ�� × 1 ℎ��60 �� × 1 ��7.48 � � × 5 �� = 2321��

We now take this volume and equate it to the following:

∀ = v × M × � M = 2v � = 9�� − 10�� /9.8333��3 ∴ ∀9.833�� = 2v&

v = ( ∀2 × 9.8333 = ( 2321��2 × 9.8333�� = 10.86 �� ≈ 10�� − 10��

Now, solving for length: M = 2v = 2 × 10.86 �� = 21.72 �� ≈ 21�� − 9��

Summary:

Basin Dimensions: 10ft-10in x 21ft-9in x 9ft-10in

b) To begin, we’ll solve for the power: 6 = -&.∀= /150�123& × /2.73 × 10143 × /10.8333�� × 21.75�� × 9.8333��3= 1423.2 �� − �5��

Now, we use this power in the following equation to solve for air flow:

6 = m2-c log Pℎ + �&�& Q

Where:

h = height of diffuser C1 = 81.5 (British Units)

c2 = 34 (British Units)

Rearranging and solving:

-c = 6m2 log zℎ + �&�& { = 1423.2

81.5 × log P9�� + 3434 Q = 171.2 ��

c) For this, we simply divide the total air flow provided by the amount required per diffuser (4

cfm):

;�� = 171.2 ��4 �� 8�� = 42.8 �� → 43 ��

agitator hw3 solution

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