ENGINEERING COLLEGE

An activated sludge plant is being designed to handle a feed rate of 0.438 m3 /sec. The influent BOD concentration is 150 mg/L and the cell concentration (MLVSS) is 2,200 mg/L. If you wish to operate the plant with a food-to-microorganism ratio of 0.20 day-1 , what volume of aeration tank should you use? Please give your answer in m3 .

Answers

Answer 1

Answer:

Answer:

Volume of aeration tank = 1.29 x 10^4 m³

Explanation:

Food/Micro- organism Ratio = 0.2/day

Feed Rate (Q) = 0.438 m³/s

Influent BOD = 150 mg/L

MLVSS = 2200 mg/L

The above mentioned parameters are related by the equation

F/M = QS₀/VX

where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get

V = 0.4380 x 150/0.2 x 2200

V = 0.1493 (m³/s) x day

V = 0.1493 x 24 x 60 x 60

V = 1.29 x 10^4 m³

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The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)and is placed in the clamp when the temperature is T1 = 15°C. If the two 304-stainless-steel (E =193GPa, a = 17x10^-6/°C) carriage bolts of the clamp each have a diameter of 10mm, and they holdthe cylinder snug with a negligible force against the rigid jaws, determine the temperature at whichthe average normal stress in either the magnesium or steel becomes 12 MPa.
Aluminum alloys are made by adding other elements to aluminum to im prove its properties, such as hardness or tensile strength. The following table shows the composition of ve commonly used alloys, which are known by their alloy numbers (2024, 6061, and so on) [Kutz, 1999]. Obtain a matrix algorithm to compute the amounts of raw materials needed to pro duce a given amount of each alloy. Use MATLAB to determine how much raw material of each type is needed to produce 1000 tons of each alloy. Composition of aluminum alloys Alloy %Cu %Mg %Mn %Si %Zn 2024 6061 7005 0 7075 356.0 0 0.6 0 0 0 0 4.5 5.6 0 0.6 0 0 2.5 0.3 0
What is the function of deaerator in thermal power plant?
Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12 psia with a quality of 1.0, through an outlet duct 1 ft in diameter. Calculate the turbine power output

Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/kg, respectively. Use these values to estimate the total energy required to melt 100 kg of ice at 0 °C and boil off 40 kg of water at 100 °C. a) 239,028 kJ b) 95,250 kJ c) 185,500 kJ d) 362,628 kJ e) 123,600 kJ

Answers

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .

Sensible heat for water Q

$Q=mC_p\Delta T$

For water

$C_p=4.178\ KJ/Kg.K$

Q=4.178 x 40 x 100 KJ

Q=16,712 KJ

So total heat

Total heat =100 x 333.23+16,712 + 60 x 2257 KJ

Total heat =185,455 KJ

Approx Total heat = 185,500 KJ

So the answer C is correct.

What is a SAFETY CHECK

Answers

Answer:

Safety check is defined as rounding to make sure that the patients and the milieu (patients living quarters) is secured and free of harmful items that can be used to hurt someone.

Determine the degrees of superheat of steam at 101.325 kPa and 170°C. a. 50°C b. 70°C c. 60°C d. 80°C

Answers

Answer:

b) 70°C

Explanation:

Given that super heat temperature at 101.325 KPa= 170°C.

We know that saturation temperature at 101.325 KPa is 100°C.

Degree of super heat =Super heat temperature - saturation temperature ( at constant pressure)

Now by putting the values

Degree of super heat=170-100

Degree of super heat=70°C.

So our option b is right.

When circuit switching is used, what is the maximum number of circuit-switched users that can be supported? Explain your answer in your own words.

Answers

Answer & Explanation:

When circuit switching is used, maximum 15 circuit switched user can be supported. Because for each of the user is allocated 10 MBPS bandwidth, and given link capacity is 150 MBPS.

(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinusodal force given by p(t) = P sin ωt. All positions are measured from equilibrium. Use m_1=1.5 kg, m_2=2 kg, k=7 N/m, b=3.2 (N∙s)/m, P=15 N, =12 rad/sec. Hint: first create the state space model for the system. Then use SS2TF to make the two transfer functions and then the two Bode plots (include with submission). Use the plots to find the steady-state equations.

Answers

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Convert the angles of a triangle to radians.Part A31∘43′53′′, 90∘32′11′′, 57∘43′56′′Express your answers, separated by commas, to six significant figures.nothingrad, rad, radRequest AnswerPart B94∘22′19′′, 40∘54′53′′, 44∘42′48′′Express your answers, separated by commas, to six significant figures.

Answers

Answer:

Explanation:

To convert to radians

A31∘43′53′′, 90∘32′11′′, 57∘43′56′′

using DMS approach ; 1degree = 60minutes = 3600 seconds

1° = 60' = 3600"

And degree to radian = multiply by π/180

A) 31∘43′53′′ = 31degree + 43minutes + 53 seconds

= 31 degree + 43minutes + 53/60

= 31 degree + 43.88minutes

= 31 degree + 43.88/60 = 31.73 degree x π/180 = 0.5534radians

FOR 90∘32′11′′ = 90 degree + 32minutes + 11seconds

= 90degree + 32minutes + 11/60

= 90 degree + 32.183minutes

= 90 degree + 32.183/60 = 90.54degree x π/180

= 1.580radians

FOR 57∘43′56′′ = 57degree + 43minutes+ 56seconds

= 57degree + 43minutes + 56/60

= 57 degree + 43.93minutes

= 57degree + 43.93/60 = 57.73degree X π/180

= 1.00radians

PART B

FOR 94∘22′19′′ = 94degree + 22minutes + 19seconds

= 94degree + 22minutes + 19/60

= 94degree + 22.32minutes

= 94degree + 22.32/60

= 94.37degree X π/180 = 1.65radians

FOR 40∘54′53′′ = 40degree + 54minutes + 53seconds

= 40 degree + 54minutes + 53/60

= 40 degree + 54.88minutes = 40 degree + 54.88/60

= 40.91degree X π/180 = 0.714radians

FOR 44∘42′48′′ = 44degree + 42minutes + 48seconds

= 44degree + 42.8minutes

= 44.71degree X π/180 = 0.780radians

Answer:

0.176270π rad, 0.502980π rad, 0.320735π rad

0.524289π rad, 0.227304π rad, 0.248407π rad

Explanation:

We know that,

1° = 60' 180° = π

1 ' = 1°/60 1° = π/180

a. 31°43'53''

Step 1

53'' = 53 * 1/60

= 53'/60

Step 2

43'53''

= 43'+53'/60

= (2580+43)/60

= 2623'/60

-------- Convert to degrees

= 2623/60 * 1/60

= 2623/3600

Step 3

31°43'53''

= 31+ 2623/3600

= (111600 + 2623)/3600

= 114223°/3600

Now, we convert to radians

= 114223/3600 * π/180°

= 0.176270π rad

90°32'11''

Step 1.

11' = 11 * 1/60

= 11/60

Step 2

32'11'

= 32 + 11/60

= 1931/60

-------- Convert to degrees

= 1931/60 * 1/60

= 1931/3600

Step 3

90°31'11''

= 90 + 1931/3600

= 325931°/3600

Now we convert to radians

= 325931°/3600 * π/180°

= 0.502980π rad

57°43'56''

Step 1

56' = 56 * 1/60

= 56/60

= 14/15

Step 2

43'56''

= 43 + 14/15

= 659/15

Now we convert to degrees

= 659/15 * 1/60

= 659°/900

Step 3

57°43'56''

= 57 + 659/900

= 51959/900

Now we convert to radians

= 51959°/900 * π/180°

= 0.320735π rad

94∘22′19′′

Step 1

19'' = 19/60

Step 2

22'19''

= 22 + 19/60

= 1339/60

Now we convert to degrees

= 1339/60 * 1/60

= 1339°/3600

Step 3

94°22'19"

= 94 + 1339/3600

= 339739°/3600

Now we convert to radians

= 339739°/3600 * π/180

= 0.524289π rad

40∘54′53′′

Step 1

53" = 53/60

Step 2

54'53"

= 54'+ 53/60

= 3293/60

Now we convert to degrees

= 3293/60 * 1/60

= 3293/3600

Step 3

40°54'53"

= 40 + 3293/3600

= 147293/3600

Now we convert to radians

= 147293/3600 * π/180

= 0.227304π rad

44∘42′48′

Step 1

48' = 48/69

= 4/5

Step 2

42'48"

= 42 + 4/5

=214/5

Nowz we convert to degrees

= 214/5 * 1/60

= 107/150

Step 3

44°42'48"

= 44 + 107/150

= 6707/150

Now we convert to radians

= 6707/150 * π/180

= 0.248407π rad

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