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3 years ago

8

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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A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulu

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Answer:

The correct answer to the following question will be "1.23 mm".

Explanation:

The given values are:

Average normal stress,

$\sigma=200 \ MPa$

Elastic module,

$E = 77 \ GPa$

Length,

$L = 570 \ mm$

To find the deformation, firstly we have to find the equation:

⇒ $\delta=\Sigma\frac{N_{i}L_{i}}{E \ A_{i}}$

⇒ $=\frac{P(\frac{L}{H})}{E(bt)} +\frac{P(\frac{L}{2})}{E (bt)(\frac{2}{3})}+\frac{P(\frac{L}{H})}{Ebt}$

On taking " $\frac{PL}{Ebt}$ " as common, we get

⇒ $=\frac{\frac{PL}{Ebt}}{[\frac{1}{4}+\frac{3}{4}+\frac{1}{4}]}$

⇒ $=\frac{5PL}{HEbt}$

Now,

The stress at the middle will be:

⇒ $\sigma=\frac{P}{A}$

⇒ $=\frac{P}{(\frac{2}{3})bt}$

⇒ $=\frac{3P}{2bt}$

⇒ $\frac{P}{bt} =\frac{2 \sigma}{3}$

Hence,

⇒ $\delta=\frac{5 \sigma \ L}{6E}$

On putting the estimated values, we get

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3 years ago

What is the lowest Temperature in degrees C?, In degrees K? in degrees F? in degrees R

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Answer:

-273.16 °C

-459.677 °F

0 °K

0 °R

Explanation:

The lowest temperature is the absolute zero.

Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.

Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.

Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.

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3 years ago

How can the use of local materials improve the standard of living of Filipinos?

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Choose a quality one, and don't use it as necessary

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2 years ago

A 50kg block of nickel at 90°C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25°C. Determine the fol

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Answer:

a).Control volume

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Explanation:

a). It is a control volume system because mass is varying in the system.

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$m_{nickel}\times c_{nickel}(T_{2}-T_{nickel})=(volume_{water}\times density_{water})\times c_{water}(T_{2}-T_{water})$

$50\times 502.416\times (T_{2}-90)=(0.5\times 1000)\times 4.187\times (T_{2}-25)$

$25120.8\times (T_{2}-90)=2093.5\times (T_{2}-25)$

$T_{2}$ = 87.91°C

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3 years ago

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A 55-μF capacitor has energy ω (t) = 10 cos2 377t J and consider a positive v(t). Determine the current through the capacitor.

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Given :

Capacitor , C = 55 μF .

Energy is given by :

$\omega(t)=10cos^2 (377t)\ J$ .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

$\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)$

Now , current i is given by :

$i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)$

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .

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3 years ago