1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
yKpoI14uk [10]
3 years ago
10

What process is used to remove collodal and dissolved organic matter in waste water ​

Engineering
1 answer:
Juli2301 [7.4K]3 years ago
3 0

Answer:

Aerobic biological treatment process

Explanation:

Aerobic biological treatment process in which micro-organisms, in the presence of oxygen, metabolize organic waste matter in the water, thereby producing more micro-organisms and inorganic waste matter like CO₂, NH₃ and H₂O.

You might be interested in
The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to
skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 kN/m^{2}

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = 0.02\ m^{3}/s

(Since, 1 litre = 10^{-3} m^{3})

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, H_{loss} = 45 cm = 0.45\ m

Height, h_{roof} = 2.5\ m

Now,

The velocity in the bore is given by:

v = \frac{Q}{\pi (\frac{d}{2})^{2}}

v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s

Now, using Bernoulli's eqn:

\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k                  (1)

The velocity head is given by:

\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553

Now, by using energy conservation on the surface of water on the roof and that in the tank :

\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}

\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45

P_{tank} = 5.5\times \rho \times g

P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}

4 0
3 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0
3 years ago
A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. W
irakobra [83]

Answer:

P_p = 27000 psf

Explanation:

given,

height of the retaining wall = h = 12 ft

internal angle of friction (∅)= 30°

unit weight = 125 pcf

Rankine passive earth pressure = ?

k_p is the coefficient of passive earth pressure

k_p = \dfrac{1 + sin\phi}{1 - sin\phi}

k_p = \dfrac{1 + sin30^0}{1 - sin30^0}

k_p = 3

Passive earth pressure

P_p = \dfrac{1}{2}k_p \gamma H^2

P_p = \dfrac{1}{2}\times 3\times 125 \times 12^2

      P_p = 27000 psf

Rankine passive earth pressure on the wall is equal to P_p = 27000 psf

7 0
3 years ago
Technician A says that a lack of lubrication on the back of the disc brake pads can cause brake noise. Technician B says that pa
sweet-ann [11.9K]

Answer:

Technician A and B both are correct.

Explanation:

According to Technician A, we can stop noise by applying lubricant (Break friction).

According to Technician A, we can stop noise by set-up right position for break pad.

7 0
3 years ago
A large plate is at rest in water at 15?C. The plate is suddenly translated parallel to itself, at 1.5 m/s. The resulting fluid
ratelena [41]

Answer:

Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s

   

Explanation:

du/u² = dt = ∫du/2.3183 = ∫dt

0.4313 u = t + c

(a) t = 0, u= 15m/s, c = 0.647

u = t+c/0.4313 = t + 0.647/0.4313

(a) when t= 1   u = 1+ 0.647/0.4313 = 3.8187m/s

(b) when t= 10   u = 10 + 0.647/0.4313 = 24.0858m/s

(c)when t= 1000  u = 1000 + 0.647/0.4313 = 3220.071m/s

5 0
3 years ago
Other questions:
  • Circular fins are employed around the cylinder of a lawn mower engine to dissipate heat. The fins are made of aluminum are 0.3-m
    10·1 answer
  • In which type of restaurant do customers order and pay before eating?
    5·1 answer
  • How to get on your screen on 2k20 in every mode
    15·2 answers
  • Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3-kW electric burner. If 60 percen
    9·1 answer
  • Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a
    13·1 answer
  • A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
    9·1 answer
  • Why water parameters of Buriganga river vary between wet and dry seasons?<br> Explain.
    7·1 answer
  • I wuv little space :)
    8·1 answer
  • There do u know it.............
    15·1 answer
  • _____ can be defined as the rate at which work is done or the amount of work done based on a period of time. (2 Points) voltage
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!