Float Calculation
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Figure 1: Free Body Diagram of Lever Arm
Abstract
In the fall semester of 2009, the Non-linear Chemical Doser team developed a Mathcad file to help plant operators choose a float given a non-linear dosing system.
For our prototype, we calculated that our float needs to weigh 4.95 lbs with a diameter of 6 inches and 4 inches of thickness. The calculations for these numbers will be enumerated below.
Method
The float design parameters can be determined using a moment balance around the pivot of the lever arm. As seen in Figure 1 above, the major forces acting on the lever arm are the center of masses of the lever arm on either side of the pivot, the weight of the dosing slide, the alum dosing tube, and the components of the float. A moment balance was performed in order to determine what mass of float would be required in order to balance the lever arm at the lever arm angle corresponding to maximum plant flow. The forces due to the masses of the lever arm to the left and right of the lever arm cancel out. All that remains then acting on the lever arm is the force due to the alum tube, the weight of the sliding scale, and the tension caused by the float. The weight for the float is then changed until the moment about the pivot point becomes zero. The weight found makes the lever arm perfectly balanced at a maximum dosing rate of 100 mg/l at the maximum angle the doser arm will experience (max flow rate).
Figure 2: Lever arm/float orientation
Figure 2 shows the orientation the doser arm experiences through min/max plant flow rate and how the float level changes accordingly.
The formula for the moment balance can be seen below.
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\large
$$
\sum {Moments_
Unknown macro: {pivot}
= 0;}
$$
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\large
$$
T(
Unknown macro: {L over 2}
)\cos (\alpha ) = W_
Unknown macro: {alumtube}
(
Unknown macro: {L over 4}
)\cos (\alpha ) + W_
Unknown macro: {slide}
(
- {{L_
Unknown macro: {slide}
} \over 2})\cos (\alpha )
$$
Where:
L = Length of the lever arm
Alpha = The angle the lever arm is with the horizontal
T = Tension Force in the string
Walumtube = The force caused by the weight alum in the dosing tube
Wslide = The force caused by the weight of the slide
The cosine terms cancel out of the equation and we can then substitute in the expression for the expression relating our float characteristics to the Tension force. This equation is shown below:
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\large
$$
T = F_b - W_
Unknown macro: {float}
$$
The force Archimedes formula for the calculation of buoyancy can be seen below
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\large
$$
F_b = \gamma V
$$
Where:
Fb = The force of buoyancy [Newtons]
\gamma = Unit weight of water [1000 Kg/m^3]
V = Volume of displaced water
After plugging the formula for the tension in the rope into our moment balance equation we can solve for our weight of the float required to cause our moments around the pivot point to be equal to 0. This approximation can be shown below.
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\large
$$
0 = - W_
Unknown macro: {float}
+ {{W_
Unknown macro: {alumtube}
} \over 2} + W_
Unknown macro: {slide}
+ {{\gamma _
Unknown macro: {H_2 O}
\pi D^2 H} \over 4}
$$
The solution for the weight of the float was determined to be .
Possible sources of Error
An assumption was made here which may cause some error in our calculation; in order to calculate the tension force for the line connecting the lever arm and the float, we had to make an assumption for the amount of displaced water caused by the float. It was assumed that the float was positioned halfway up in the water. This assumption for the position of the float allowed for the force of buoyancy and thus the tension force to be calculated. However, as the sliding arm is moved up and down the lever arm, this will change the amount of tension applied to the float, causing the float to not be halfway up in the water.
Form
Results/Discussion
The scale portion of the lever where the dual scale and the slider and the doser tube are connected is heavy enough that there will always be tension on the float. As the water level in the entrance tank drops from the maximum plant flow, the float will drop as well. As the float drops it maintains tension in the rope and drops the left side of the lever, raising the right side. The right side therefor has less head between the constant head tank and the dosing orifice, causing less flow of alum. Maintaining tension on the rope is essential for the accurate dosing of alum into the entrance tank. This tension should be maintained since the right side of our lever arm is heavier than the left side of our lever arm due to the added components. As the float drops, the tension force will cause the left side of the lever arm to drop, but the weight of the components on the right will not allow there to be "slack" in the tension line. Steps should be taken to ensure that the turbulence in the entrance tank is minimized so that there will be no additional forces on the float, which could cause errors in the alum dosage.
Conclusion and Future Work
We require a float with a weight of 4.95 lbs, diameter of 6 inches and a 4 inch height for our lever arm prototype.
Further work will have to be done analyzing the errors in dosing caused be the assumptions made in the calculation of the float size. The extent to which these errors will cause an improper amount of alum to be dosed is not fully known yet.
Upon construction of the lever arm, we will utilize a makeshift float in which we can adjust the weight of the float easily to verify our calculations here. This will be done before we order the final float for the dosing system.
Bibliography
- CEE4540 Flow Control Measurement Notes at https://confluence.cornell.edu/display/cee4540/Syllabus
Deliverables
- Final Float Design Parameters and Calculations
- Float protype for March 2010 EPA Competition