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3 years ago

Wastewater flows into a _________ once it is released into a floor drain.

Engineering

1 answer:

rodikova [14]3 years ago

6 0

Answer:

A) Sump pit

Explanation:

A wastewater typically refers to a body of water that has contaminated through human use in homes, offices, schools, businesses etc. Wastewater are meant to be disposed in accordance with the local regulations and standards because they are unhygienic for human consumption or use.

Generally, many homes use a floor drain in their bathrooms and toilets to remove wastewater in order to mitigate stagnation and to improve hygiene. A floor drain can be defined as a material installed on floors for the continuous removal of any stagnant wastewater in buildings. Wastewater flows into a sump pit once it is released into a floor drain through the use of a pipe such as a polyvinyl chloride (PVC) pipe, which directly connects the floor drain to the sump pit. The wastewater can the be removed from the sump pit when it is filled up through the use of a pump.

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Alla [95]

Answer:

how are we supposed to help?

4 0

3 years ago

Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain

fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

$N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })$

Where

$D_{AB}$ = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

$P_{A1}$ = Pressure at first boundary

$P_{A2}$ = Pressure at the destination boundary

T = System temperature

$P_{T}$ = System pressure

Where $P_{T}$ = 101.3 kPa $P_{A2} =0$ , $P_{A1} =y_{A}$ , $P_{T} =$ 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R = $\frac{kJ}{(kmol)(K)} ,$ T = 298 K and $D_{AB}$ = 1.18 $\frac{cm^{2} }{s}$ = 1.8×10⁻⁵ $\frac{m^{2} }{s}$

$N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65})$ = 5.153×10⁻⁴ $\frac{kmol}{m^{2}s }$

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0

3 years ago

WILL MARK BRAINLEST PLEASE HELP

TEA [102]

I put

People who pursue a career in the creative imaging fields have qualities like a good imagination, creativity, open minds, good with ideas, and handling situations. If you enter that field, you need imagination to create things and an open mind to be open to all creations. You need good ideas to make good thing that will work.

please don't plagiarise tho, re-word it.

6 0

3 years ago

The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the

gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, $v^{2}=\frac {ks^{2}}{m}$ hence $v=s\sqrt {\frac {k}{m}}$ where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

k is given as 200 N/m and mass is 0.75 Kg

Substituting the given values

$v=0.1 m\sqrt \frac {200 N/m}{0.75 Kg}=1.632993162 m/s\approx 1.633 m/s$

Therefore, <u>the speed is 1.633 m/s</u>

The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as $9.81 m/s^{2}$

Therefore, $F_y=0.75\times 9.81=7.3575 N\approx 7.36 N$

The sum of forces in normal direction is given by $Ma_n=Ks$ therefore

$Ma_n=200*0.1=20 N$

Therefore, <u>normal force on the rod is 20 N</u>

5 0

3 years ago

2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11

ahrayia [7]

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² = 1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² = 9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ = (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵) × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )

0 - 0.6014925 + Rₓ = 0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R $_y$ = m'( V₂sin(60°) - 0.5 )

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R $_y$ = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R $_y$ = 0.092728( 0.305187 )

⇒ 1.5484 + R $_y$ = 0.028299

R $_y$ = 0.028299 - 1.5484

R $_y$ = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R $_y$ ² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

7 0

2 years ago