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3 years ago

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A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

e inboard channels set on a diameter of 18m. for a flow of 12900m^3/d, calculate the overflow rate, detention time,and weir loading.

1 answer:

jasenka [17]3 years ago

6 0

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

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Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

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3 years ago

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2 years ago

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Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

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Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$

$\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}$

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$\theta_2=90^o-\theta_1$

Given Vo=37 m/s and R=70 m

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$\theta_1=15^o$

And

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