You are viewing an old version of this page. View the current version.

Compare with Current View Page History

« Previous Version 85 Next »

Float Calculation

Figure 1: Lever arm/float orientation

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. This float will ensure that the angle of the lever arm will be kept at the proper position to ensure the accurate dosing of alum as the plant flow rate is varied. For our prototype, we calculated that our float needs to have a diameter of 15.25 cm and 30.5 cm of height. The mass of the float will depend on the dosage that we want to calibrate our float on. 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 2 below, 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, the tube connecting the lever arm with the rapid mix tube, and the components of the float. In this case, we are going to balance the lever, or calibrate it, with the dosage set to maximum dosage(100 mg/L). This means that the slider is pushed all the way to the far end of the scale. 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 tubes, 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. With the alum tube completely filled with solution, 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: Free Body Diagram of Lever Arm

The formula for the moment balance can be seen below.

Unknown macro: {latex}

\large $$ \sum {Moments_

Unknown macro: {pivot}

= 0;} $$

Unknown macro: {latex}

\large
$$
T(

Unknown macro: {L over 2}

)\cos (\alpha ) = W_

Unknown macro: {alumtube1}

(

Unknown macro: {L over 4}

)\cos (\alpha ) + (W_

Unknown macro: {slide}

+ W_

Unknown macro: {alumtube2}

)(

- {{L_

Unknown macro: {slide}

} \over 2})\cos (\alpha )
$$

Unknown macro: {latex}

\large
$$
F_

Unknown macro: {Buoy}

= \gamma _

Unknown macro: {H2O}

\times \pi \times {\textstyle{{D_

Unknown macro: {Float}

^2 } \over 4}} \times \left( {{\textstyle{{D_

Unknown macro: {Length}

} \over 2}} - Y} \right)
$$

Where:
L = Length of the lever arm

Unknown macro: {latex}

$$ \alpha $$

= The angle the lever arm is with the horizontal
T = Tension Force in the string

Unknown macro: {latex}

$$ W_

Unknown macro: {alumtube}

$$

= The force caused by the weight of alum in the dosing tube

Unknown macro: {latex}

$$ W_

Unknown macro: {slide}

$$

= The force caused by the weight of the slide

Unknown macro: {latex}

$$ W_

Unknown macro: {alumtube2}

$$

= The force caused by the weight of alum in the tube which extends into the rapid mix unit.

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:

Unknown macro: {latex}

\large $$ T = W_

Unknown macro: {float}

-F_b $$

Archimedes principle, which predicts the upward force on the float cause by the displacement of water, can be seen below.

Unknown macro: {latex}

\large $$ F_b = \gamma V $$

Where:

Unknown macro: {latex}

$$ F_

Unknown macro: {b}

$$

= The force of buoyancy (Newtons)

Unknown macro: {latex}

$$ \gamma $$

= Unit weight of water (1000 Kg/m^3)
V = Volume of displaced water

After substituting 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.

Unknown macro: {latex}

\large

$$
0 = - W_

Unknown macro: {float}

+ {{W_

Unknown macro: {alumtube1}

} \over 2} + W_

Unknown macro: {slide}

(1 - {{L_

} \over L}) + W_

Unknown macro: {alumtube2}

(1 - {{L_

Unknown macro: {slide}

} \over L}) + {{\gamma _

Unknown macro: {H_2 O}

\pi D^2 H} \over 4}
$$

The solution for the mass of the float for calibration at maximum dosage was determined to be 2.849 kg. This solution was obtained after inputing a reasonable value for the height and diameter of the float. The float's height was set to 0.305 m, which is reasonably tall enough to provide stability. The diameter was set at a high value of 15.2cm in order to reduce the amount of vertical movement of the float. We can also apply the above equations to derive the mass of the float needed to calibrate the lever arm at dosages different from the maximum. The values of calibration mass versus dosage are shown below:

Table 1: Float Calibration Mass

Dosage(mg/L)

Mass(kg)

20

2.783

30

2.786

40

2.791

50

2.797

60

2.805

70

2.814

80

2.824

90

2.836

100

2.849

Click here for the error analysis of the float calculation.

  • No labels