Secondary recovery techniques a. include sealing fractures in the rock to concentrate oil in pockets called traps. b. usually enable drillers to get almost 100% of the oil out of the ground. c. are used to coax more oil out of a drilled hole. d. include the use of steam to make oil more viscous.

Here’s a brief explanation of how these three technological advancements have revolutionized the field of Mechanical Engineering.

Computer Assisted Design (CAD): One of the most widely utilized software design tools is CAD. It is used by engineers and designers to model, validate, and convey ideas prior to production. CAD software models are frequently utilized as inputs to various mechanical engineering and design tools1.

ii) 3D printing: Extra tools for producing goods on a CNC machine or 3D printer are available and are occasionally incorporated into the CAD program. This has enabled quick prototyping and the creation of complicated geometries that were previously impossible with typical manufacturing methods1.

iii) Simulation: Computer-Aided Engineering (CAE) encompasses a wide variety of studies. Before building physical prototypes, it conducts complicated tasks like as finite element analysis (FEA) and computational fluid dynamics (CFD).

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Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\(\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}\)

Otto cycle T-S diagram

T₂ = 288.706*\(6.25^{0.393}\) = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

\(\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}\)

T₄ = 2888.89 / \(6.25^{0.393}\) = 1406.5 K

Work done, W = \(c_v\)×(T₃ - T₂) - \(c_v\)×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

Baselines in geodetic control networks are a critical component of modern surveying and mapping. Baselines are defined as the straight-line distance between two points in a geodetic survey, which is used to create a reference system for all other measurements.

The baseline is then used to calculate distances and angles between other points, which can be used to create maps and survey data. Baselines are typically measured using a variety of methods, including satellite-based Global Positioning Systems (GPS), which provide highly accurate measurements. Geodetic control networks are used for a wide range of applications, including construction, mining, land management, and environmental studies.

By providing accurate, reliable data about the earth's surface, these networks are essential for effective management of natural resources and development projects. In summary, baselines in geodetic control networks are the fundamental building blocks that allow surveyors and mapping professionals to create accurate and reliable data about the earth's surface.

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The inheritance of wing size in fruit flies is controlled by multiple genes, making it a complex trait. However, assuming that wing size is controlled by a single gene with vestigial wings (vw) being recessive to normal wings (w), we can predict the expected proportions of offspring with each wing type.

When a fly with vestigial wings (vw/vw) is crossed with a fly with normal wings (w/w), all of the offspring will be heterozygous (vw/w) for the wing size gene. If we assume that the wing size gene is autosomal, we can use a Punnett square to calculate the expected genotypes and phenotypes of the offspring.

At 31 Celsius, the expression of the wings could be affected, but assuming that it does not change the expected inheritance pattern, all offspring would still be heterozygous for the wing size gene, resulting in 100% normal wings and 0% vestigial wings in the progeny.

Assuming that the temperature of 31°C does not affect the inheritance of the wing size trait, the expected ratio of offspring with vestigial wings to those with normal wings is 1:1 or 50%.

This is because the offspring will inherit one copy of the vw allele from the vw/vw parent and one copy of the w allele from the w/w parent, resulting in a heterozygous genotype (vw/w) that displays the normal wing phenotype.

It is important to note that this is a simplified model, and the actual inheritance of wing size in fruit flies is more complex and may involve multiple genes and environmental factors.

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Using the de-goodman criteria and a design factor of 1.5, the shaft diameter will be 0.187 inch.

To find the shaft diameter, let's use the equation below:

`σa/σw = [1/(Ka)] [(E/2E')^(1/m)] `

We know that the load on the shaft is 50,000 lbf.

Using this, we can calculate the alternating stress on the shaft:

`σa = load / (π * (d^2)/4) `

where d is the shaft diameter, we need to find.

Alternating stress `σa = 50,000/(πd^2/4) = 64,041/d^2`

Factor `n = 1/1.5 = 0.67`

For the material of the shaft, Ka and m values can be obtained using Soderberg criterion:

`σw = (σm/Na)(σf/Nf) `where σm is the mean stress, Na and Nf are empirical factors, and σf is the fatigue strength at a specific number of cycles.

The ultimate strength of the shaft material is given as 140,000 psi.

The empirical factors Na and Nf can be calculated as:

`Na = (σm + σa)/σf and Nf = 2E6/10^6 = 2`

Substituting the values we get, `Na = (0 + 64041/d^2) / 60000 = 1.067/d^2 and Nf = 2`

We can now calculate σw using Soderberg criterion:

`σw = (σm/Na)(σf/Nf) `Mean stress σm can be taken as half of ultimate strength σut = 70000 psi.

Then, `σw = (70000/1.067d^2)(60000/2) = 17630/d^2`

Using De-Goodman criteria, substituting the values of σa/σw and factor m, we get:

`1.5 = [1/Ka] [(E/2E')^(1/m)] = [1/Ka] [(29*10^6)/(2 * 60,000)^(1/3)] `

We can simplify the above equation by substituting Ka and solving for the shaft diameter `d`. `d = 0.187 inch`

Hence, the shaft diameter is 0.187 inch.

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Answer:

Explanation:......................................

Answer:

Рвоы

Explanation:

Иттің папасы өлңп қалды дейді

Answer: The accounting procedures and financial management systems used by a school to record and report on the transactions in the Federal Student Aid programs play a major role in the school’s management of those programs. In this appendix, we will discuss the minimum criteria for those procedures and systems, identify areas where problems might arise, and point out potential system weaknesses.

Explanation:

Answer:

D.

Explanation:

I know this because one of my science teachers I had a few years ago made my science class pretty much memorize this since a hypothesis is your best educated guess I know that to test a hypothesis you see if you have the right predictions or hypothesis by testing your predictions to see if they are right or wrong.

Critical load Fcr or buckling load is the value of load that causes the phenomenon of change from stable to unstable equilibrium state.

With that beign said, first it is neessary to calculate the moment of inercia about the x-axis:

\(Ix= \frac{db^3}{12}\\ Ix = \frac{2.(4)^3}{12} = 10.667in\)

Then it is necessary to calculate the moment of inercia about the y-axis:

\(Iy = \frac{db^3}{12}\\ Iy = \frac{4.(2)^3}{12} = 2.662in\)

Comparing both moments of inercia it is possible to assume that the minimun moment of inercia is the y-axis, so the minimun moment of inercia is 2662in.

And so, it is possible to calculate the critical load:

\(Pc\gamma = \frac{2046\pi ^2E.I}{L^2} \\Pc\gamma= \frac{2046.\pi ^2.(1,6.10^3.10^3).2662}{(10.12)^2} \\Pc\gamma= 5983,9db\)

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Answer:

Explanation:

Civil engineers have become experts in creating sustainable and environmentally friendly buildings and systems. Multiplied over many communities, the energy and emissions savings can make a real difference in the environment. Other life-improving functions can also make communities better places to live.Jul 19, 2017

Answer:

a) 47500 mg/L

b) 5250366.444  kg/year

Explanation:

Given data:

suspended solids removal efficiency = 20%

Flowrate in the primary clarifier ( Q ) = 20 MGD ( change to Liters/day

Q = 20* 10^6 * 3.785412  Liters /day

settled concentration  ( St ) = 950mg/L * 0.2 = 190 mg/L

amount of settled solid = Q * St

                                   = ( 20* 10^6 * 3.785412 ) * 190  = 14384.5656 kg/day

∴ Amount going into sludge with a flowrate of 0.08 MGD = 14384.5656 kg/day

a) concentration of solid in sludge  ( leaving the clarifier )

= amount of settled solid / flow rate out of the clarifier in liters/day

= 14384.5656 / ( 0.08 * 10^6 * 3.785412 )

= 0.0475 kg/L

= 47500 mg/L

b) Determine mass of solids that is removed annually

= 14384.5656 kg/day * 365 days

= 5250366.444  kg/year

True. A circuit made inactive by a low- or zero-ohm resistance path across the circuit allows current to flow through it without developing a voltage drop

In such a scenario, the low resistance creates a short circuit, bypassing the intended components and creating an alternative path for the current to follow. As a result, the current can flow freely through the short circuit, minimizing or eliminating any voltage drop across the circuit. This can lead to abnormal current flow, potential overheating, and can be a safety concern. Short circuits are typically unintended and can occur due to wiring faults, damaged insulation, or faulty components. Proper circuit protection measures, such as fuses or circuit breakers, are essential to prevent damage and ensure electrical safety.

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Answer:

Explanation:

Considering the flow of mercury in a tube:

When it comes to laminar flow of mercury, the thermal entry length is quite smaller than the hydrodynamic entry length.

Also, the hydrodynamic and thermal entry lengths which is given as DLhRe05.0= for the case of laminar flow. It should be noted however, that Pr << 1 for liquid metals, and thus making the thermal entry length is smaller than the hydrodynamic entry length in laminar flow, like I'd stated in the previous paragraph

The appropriate action is to accept the new estimate and update the schedule accordingly.

Accepting the new estimate and updating the schedule accordingly is the most appropriate action in this situation. While the activity is not on the critical path, it is important to address any deviations from the original plan to maintain accuracy and ensure project success.

By accepting the revised estimate, the project manager acknowledges the new information provided by the team member and incorporates it into the project schedule. This allows for more realistic planning and resource allocation, considering the additional two days required for the activity. It is essential to maintain open communication and foster a collaborative environment where team members feel comfortable sharing concerns or potential issues that may impact project timelines.

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Answer:

Explanation:

I'm not 100% this is what you want, but here it is:

2

3

13

8

11

A

13

True. When cold air inside a refrigerated space spills out through the bottom of the door opening, it creates a vacuum effect that causes warmer room air to move into the refrigerated space.

This can cause the temperature inside the refrigerated space to rise, which can be problematic for certain types of food or medication that need to be stored at specific temperatures. To prevent this from happening, it's important to ensure that the door to the refrigerated space is properly sealed and that there are no gaps or cracks that could allow cold air to escape. Additionally, it's a good idea to minimize the frequency and duration of door openings to help maintain a consistent temperature inside the refrigerated space.

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The Flyback converter is an efficient way to regulate output voltage while maintaining a stable power supply.

In a Flyback converter with Vin = 30 V, N1 = 30 turns, and N2 = 15 turns, the self-inductance of winding 1 is 50 uH, and fs = 200 kHz. The output voltage is regulated at Vo = 9 V. For a load of 30W, the converter waveforms show the duty ratio, current in the transformer primary winding, current in the secondary winding, and output capacitor ripple current. The duty ratio is approximately 0.27, the current in the primary winding is 3.9A, the current in the secondary winding is 7.8A, and the output capacitor ripple current is 0.2A. For the same duty cycle, the critical value of output power that makes the converter operate on the boundary of incomplete and complete transformer core demagnetization can be calculated using the formula: Pcrit = (Vin*N1*D)/(4*fs), where D is the duty cycle.

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Answer:Falling or flying objects on a worksite can expose workers to relatively minor injuries, such as cuts and abrasions, as well as more serious injuries, such as concussions or blindness. Working beneath scaffolds or other areas where overhead work is being performed puts workers at risk from falling objects. Flying objects become a concern when workers are using power tools or performing tasks that involve pushing, pulling or prying.

Explanation:General

Always wear hard hats when work is being performed overhead or when other work conditions call for it.

Stack materials securely to prevent them from sliding, falling or collapsing.

Overhead work

Secure all tools and materials to prevent them from falling on people below.

Use toe boards or guardrails on scaffolds to prevent objects from falling. Alternately, use debris nets or catch platforms to grab falling objects.

Machine use

When working with machines or power tools that can produce flying particles, wear safety glasses, goggles or face shields.

Inspect tools prior to use, and be sure all guards are in place and in good working condition.

Allow only properly trained workers to use power-actuated tools.

Cranes/hoists

Whenever possible, avoid working under moving loads.

Erect barricades and post warning signs at hazardous work zones.

Inspect cranes and hoists prior to use to ensure all components are in good working order, including wire rope, lifting hooks and chains.

Never exceed the lifting capacity of cranes and hoists.

Compressed air

Reduce compressed air for cleaning to 30 psi, and always use proper personal protective equipment and guarding.

Never clean clothing with compressed air.

Answer:

Electric power output of this plant is \(48.192\times 10^{6}\,W\).

Explanation:

From First Law of Thermodynamics we understand that hydroelectric power plant transforms mechanical energy from fluid into electric energy. The power output of this plant (\(\dot W\)), measured in watts, is determined by this expression, which is derived from definition of mechanical energy and energy efficiency:

\(\dot W = \eta \cdot \rho\cdot g\cdot H \cdot \dot V\) (1)

Where:

\(\eta\) - Energy efficiency, no unit.

\(\rho\) - Density, measured in kilograms per cubic meter.

\(g\) - Gravitational acceleration, measured in meters per square second.

\(H\) - Fluid column, measured in meters.

\(\dot V\) - Volume flow, measured in cubic meters per second.

If we know that \(\eta = 0.84\), \(\rho = 1000\,\frac{kg}{m^{3}}\), \(g = 9.807\,\frac{m}{s^{2}}\), \(H = 90\,m\) and \(\dot V = 65\,\frac{m^{3}}{s}\), then the estimated electric power output of this plant is:

\(\dot W = (0.84)\cdot \left(1000\,\frac{kg}{m^{2}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \left(90\,m\right)\cdot \left(65\,\frac{m^{3}}{s} \right)\)

\(\dot W = 48.192\times 10^{6}\,W\)

Electric power output of this plant is \(48.192\times 10^{6}\,W\).

The inductance of the solenoid is 1.01 mH, and the rate at which current must change through it to produce an EMF of 70 mV is 69.31 A/s.

Radius of solenoid (r) = 4.5 cmNumber of turns (N) = 800Length of solenoid (l) = 25 cmEMF (e) = 70 mV

(a) To find the inductance of the solenoid, we use the expression: L = (μ₀×N²×A)/l where,

μ₀ = 4π×10⁻⁷ T.m/A

N = 800 turnsl = 25 cm = 0.25 mA = πr² = 3.14×(4.5 cm)² = 63.585 cm² = 63.585×10⁻⁴ m²

L = (4π×10⁻⁷×800²×63.585×10⁻⁴)/0.25L = 1.01 mH

Thus, the inductance of the solenoid is 1.01 mH.

(b) To find the rate of change of current, we use the formula: e = - L × (di/dt) where,

e = 70 mV,

L = 1.01 mH

We need to find the magnitude of (di/dt)

The above formula can be written as:(di/dt) = - e/L

We substitute the given values to find (di/dt) (di/dt) = -(70×10⁻³ V)/(1.01×10⁻³ H)(di/dt) = - 69.31 A/s

Thus, the rate at which current must change through it to produce an EMF of 70 mV is 69.31 A/s (magnitude only). Therefore, the answer is: 1.01 mH, 69.31 A/s

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Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = \(\frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4} }\)  

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

L = 1000 ft

n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237  hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

                                              = 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

The theoretical fracture strength of the brittle material is estimated to be approximately 165.6 MPa when a stress of 1300 MPa is applied and fracture occurs by the propagation of an elliptically shaped surface crack of length 0.29 mm and tip radius of curvature of 0.004 mm.

To estimate the theoretical fracture strength of a brittle material given the information provided, we can use Griffith's theory of brittle fracture. According to this theory, the fracture strength of a brittle material can be expressed as:

σ_f = (2Eγπa)^0.5

where σ_f is the fracture strength, E is the elastic modulus, γ is the surface energy per unit area, and a is the length of the elliptically shaped surface crack.

To calculate the fracture strength, we need to first determine the surface energy per unit area of the material. For glass, a typical value of surface energy is around 1 J/m^2.

Given the length of the elliptically shaped surface crack (a) is 0.29 mm, and the tip radius of curvature is 0.004 mm, we can calculate the crack area (A) as follows:

A = πab = π(0.29/2)(0.004)

A ≈ 5.67 x 10^-7 m^2

Next, we can calculate the elastic modulus (E) of the material. For glass, the elastic modulus is typically around 70 GPa.

Substituting these values into the equation for fracture strength, we get:

σ_f = (2Eγπa)^0.5 = [2(70 x 10^9)(1)(π)(0.29 x 10^-3)]^0.5

σ_f ≈ 165.6 MPa

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Answer:

- the force of the rope on the tugboat is 14.87 kN

- Force acting on the tugboat is 24.79 kN

Explanation:

 Given the data in the question;

tugboat increases its speed uniformly to 25 km/h

v₁ = 25 km/h = (25 × 1000) / ( 1 × 60min × 60sec )

= 25000m / 3600s

=  6.94 m/s

Now, lets determine the force on the rope using the following relation;

T = \(m_b\)v₁ / t₁

\(m_b\) is mass of barge( 75 Megagram = 75 × 10³ Kilogram  ), time t₁ is 35 s and v₁ is 6.94 m/s

so we substitute

T = [(75 × 10³) × 6.94 ]  / 35

T = 520500 / 35

T = 14871.43 N

T = 14871.43 / 1000

T = 14.87 kN

Therefore, the force of the rope on the tugboat is 14.87 kN

Now, to determine F acting on the tugboat;

\(Ft_1\) = ( \(m_t\) + \(m_b\) )\(v_1\)

we solve for F

F = ( \(m_t\) + \(m_b\) )\(v_1\) / \(t_1\)

where \(m_t\) is mass of tugboat (50 Megagram = 50 × 10³ Kilogram  )

so we substitute

F = [( (50 × 10³) + (75 × 10³) )6.94] / 35

F = [ 125000 × 6.94 ) / 35

F = 867500 / 35

F = 24785.7 N

F = 24785.7 / 1000

F = 24.79 kN

Therefore, Force acting on the tugboat is 24.79 kN

Answer is in the photo. I can only upload it to a file hosting service. link below!

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The correct SQL statements is:-

A) SELECT C1.CustName, C1.SalesRepNo
FROM CUSTOMER C1
WHERE C1.SalesRepNo IN
(SELECT S1.SalesRepNo
FROM SALESREP S1
WHERE S1.RepName = 'Smith');

Option A is the correctly stated correlated subquery. In a correlated subquery, the inner query references the outer query, and the correlation condition is specified in the WHERE clause. In this case, the subquery references the outer query by checking for SalesRepNo in the CUSTOMER table that matches with the SalesRepNo in the SALESREP table for the SalesRep named 'Smith'. The outer query then retrieves the customer name and sales representative number for those matching customers.

Option B is not correctly stated because the SalesRepNo variable from the subquery is not defined in the outer query, and therefore cannot be used in the AND statement.

Option C is not correctly stated because the correlation condition in the AND statement is using the not equal to operator instead of the equal to operator.

Option D is not correctly stated because the subquery is not correlated to the outer query, as it references a different table and column. The correlation condition in the AND statement is also using the not equal to operator instead of the equal to operator.

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