The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable horizontal tray is set 1m below the surface of the water, a) determine the percent removal of particles having a settling velocity of 1m/h. b) Could the removal efficiency of the clarifier to be improved by moving the tray? If so, where should the tray be located and what would be the maximum removal efficiency? c) What effect would moving the tray have if the particle settling velocity were equal to 0.5m/h?

Answer:

the speed of your poop

Explanation:

Answer:

a) 570 kWh of electricity will be saved

b) the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

[tex]P_{incandescent}[/tex] = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find  the total power consumption by the fluorescent lamp

[tex]P_{fluorescent}[/tex] = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

[tex]P_{saved}[/tex] = [tex]P_{incandescent}[/tex]  - [tex]P_{fluorescent}[/tex]

[tex]P_{saved}[/tex] = 750 - 180

[tex]P_{saved}[/tex]  = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved =  570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved =  570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ /  million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c)  If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

Explanation:

Dry air doesn't contain water vapor .

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

Answer:

procedure attached below

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Explanation:

Given data:

Minimum tensile strength = 865 MPa

Ductility = 10%EL

Desired Final diameter = 6.0 mm

20% cold worked 7.94 mm diameter 1040 steel stock

Describe the procedure you would follow to obtain this material.

assuming 1040 steel experiences cracking at 40%CW

attached below is a detailed procedure of obtaining the material

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Answer:

N0

Explanation:

It does not seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm

Given data :

Diameter ( d ) = 10 mm

length ( l ) = 150 mm

elasticity ( ∈ ) = 100 GPa

longitudinal strain ( б ) 200 MPa

Poisson ratio ( μ )  ( assumed ) =0.3

Assumption : deformation totally elastic

attached below is the detailed solution to why it is not reasonable .

The Sd value = 0.08 > the calculated Sd value ( 6*10^-3 ) hence it is not reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen

Answer:

True

Explanation:

Those are the exact uses of a resistor

Answer:

attached below

Explanation:

Attached below is  a detailed solution to the question above

Step 1 : determine the Reynolds number using the characteristics of Air at 45°c

Step 2 : calculate the Nusselt's number

Step 3 : determine heat transfer coefficient

Step 4 : calculate heat transfer ratio and thermal resistance

Repeat steps 1 - 4 for each value of diameter from 0.05 to 0.5 m

attached below is a detailed solution

Answer:

required flow rate is 75.44 gal/min  

Explanation:

Given the data in the question;

Power developed = 250 psi = 1.724 × 10⁶ Pa

hydraulic power W = 11 hp = 11 × 746 = 8206 Watt

now, Applying the formula for pump power

W = pgQμ

where p is density of fluid, Q is flow rate, μ is heat and W is power developed;

W = pgQμ

W = pgμ × Q  

W = P × Q -------- let this be equ 1

so we substitute in our values;

8.2027 kW = 1.724 × 10⁶ Pa × Q

Q = 8206 / 1.724 × 10⁶

Q = 4.75986 × 10⁻³ m³/sec

We know that, 1 cubic meter per seconds = 15850.3 US liquid gallon per minute, so

Q = 4.75986 × 10⁻³  × 15850.3 gallon/min

Q = 75.44 gal/min    

Therefore, required flow rate is 75.44 gal/min    

Answer:

B. The Hall-Petch Relation

Explanation:

The Hall-Petch relation indicates that by reducing the grain size the strength of a material is increased up to the theoretical strength of the material however when the material grain size is reduced below 20 nm the material is more susceptible to creep deformation and displays an "inverse" Hall-Petch Relation as the Hall-Petch relation then has a negative slope (k value)

The Hall-Petch relation can be presented as follows;

[tex]\sigma_y[/tex] = [tex]\sigma_0[/tex] + k·(1/√d)

Where;

[tex]\sigma_y[/tex] = The strength

σ₀ = The friction stress

d = The grain size

k = The strengthening coefficient

The model equation for the reverse Hall-Petch effect is presented here as follows;

[tex]\sigma_y[/tex] = 10.253 - 10.111·(1/√d)

Answer:

a. The time required for the tank to empty halfway is presented as follows;

[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]

b. The time it takes for the tank to empty the remaining half is presented as follows;

[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]

The total time 't', is presented as follows;

[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

[tex]\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}[/tex]

[tex]\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}[/tex]

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

[tex]dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh[/tex]

The time for the tank to drop halfway is given as follows;

[tex]\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]

[tex]t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]

[tex]t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]

[tex]t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]The time required for the tank to empty halfway, t₁, is given as follows;

[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

[tex]\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]

[tex]t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)[/tex]

[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]

The total time, t, to empty the tank is given as follows;

[tex]t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}[/tex]

[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]

Answer:

the entrained fluid flowrate is 150 liters/s

Explanation:

Given the data in the question;

we determine the flow rate of water though the jet by using the following expression;

Q₂ = A₂ × V₂

where Q₂  is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )

so we substitute

Q₂ = 0.01 m² × 30 m/s

Q₂ = 0.3 m³/s

Next we determine the flow rate of water through the pump by using the following expression

Q₃ = A₃ × V₃

where Q₃  is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )

so we substitute

Q₃ = 0.075 m² × 6 m/s

Q₃ = 0.45 m³/s

so to calculate the flow pumping rate of water into the water jet pump, we use the expression;

Q₁ + Q₂ = Q₃

we substitute

Q₁ + 0.3 m³/s = 0.45 m³/s

Q₁ = 0.45 m³/s - 0.3 m³/s

Q₁ = 0.15 m³/s

we know that 1 m³/s = 1000 Liter/second

so

Q₁ = 0.15 × 1000 Liter/seconds

Q₁ = 150 liters/s

Therefore, the entrained fluid flowrate is 150 liters/s

Answer:

i dont know th answer can u help ?

Explanation:

Answer:

is a mathematical representation of something three-dimensional.

Explanation:The typical base of a the model is a 3D mesh; the structural build consists of polygons.

Answer:

a) 24.07 m

b)  4 m

c)  14 number of drops

d) p = number of passes

e)  Dcd = 2.27

0.69 m

Explanation:

Given data:

Depth  ( D )= 7.6 m below ground surface

dynamic compaction ( w )  = 15-ton , diameter of tamper = 2.0 m , thickness = 1.4 m

Determine :

A) drop height ( H )

  D = n √wH

 therefore H = 361 / 15 = 24.07 m

where : D = 7.6 m ,  n = 0.4 , w = 15

B) Drop spacing

drop spacing = average of ( 1.5 to 2.5 )  * diameter  of tamper

                       = 2 * 2.0m =  4 m

C) number of drops

since the applied energy for fine grained soils and day fills range from 250 - 350 kj/m^2  the number of drops can be calculated using the relation below

AE = [tex]\frac{NWHP}{SPACING ^2}[/tex]

w = 15, H = 24.07 , Np = ?  , AE = 300 kj/m^2

∴ Np = 4800 / 361.05  = 13.3

the number of drops at one pass =  14

D) number of passes

p = number of passes

E) estimated crater depth and settlement

crater depth ( Dcd ) = 0.028 [tex]N_{d} ^{0.55} \sqrt{wtIt}[/tex]

Nd = 14 ,  wt = 15, It = 24.07

therefore : Dcd = 2.27

estimate settlement is within 3 to 5% therefore the improved settlement

= 2.27 * 0.04 * 7.6 = 0.69 m

Answer:

B. schematic design

Explanation:

This correct for Plato/edmentum

Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.

A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.

The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.

The schematic design consists of a rough sketch with markings and measurements.

Therefore, the correct option is B. schematic design.

To learn more about schematic design, refer to the link:

https://brainly.com/question/14959467

#SPJ5

Answer:

represnt

Explanation: