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2 years ago

Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe

Engineering

1 answer:

lys-0071 [83]2 years ago

4 0

Answer:

Dalton

The part of the close-out report that would describe how he would plan and manage projects in the future is:

summary of project management effectiveness

Explanation:

The Project Close-out Report is a project management document, which identifies the variances from the baseline plans. These variances are specified in terms of project performance, project cost, and schedule. The project close-out report records the completion of the project and the subsequent handover of project deliverables to others. The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.

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An electrical utility delivers 6.25E10 kWh of power to its customers in a year. What is the average power required during the ye

Sindrei [870]

Answer:

The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh

However, the average power P, required for a year, t = ? Kw

Expressing their relationship, we will have

U = P x t

Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)

t = 8760 hours

P = $\frac{62500000000}{8760}$

P = 7134.7Kw

Hence, the average power required during the year is 7,135Kw

Now to calculate the energy used by the power plant in a year (in quads)?

Recall, Efficiency, η = Power Output/Power Input (100)

so, we have

η = P₀/P₁, given

0.45 = $\frac{7134.7Kw}{P₁}$

P₁ = 15,855Kw

the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules

So to convert this to quads, Note;

1 quads of energy = 10¹⁵ Joules

The total energy used is 0.000000139 quads

Now to find the cubic feet of natural gas required to generate this power?

Note: 0.29Kwh of Power generated = 1 cubic feet of natural gas used

Since, the power plant generated = 62500000000Kwh

The cubic feet of natural gas used = $\frac{62500000000}{0.29}$

Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.

8 0

2 years ago

A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensil

Gemiola [76]

The radius of the specimen is 60 mm

<u>Explanation:</u>

Given-

Length, L = 60 mm

Elongated length, l = 10.8 mm

Load, F = 50,000 N

radius, r = ?

We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this

elongation using Equation:

ε = Δl / l₀

ε = 10.8 / 60

ε = 0.18

We know,

σ = F / A

Where A = πr²

According to the stress-strain curve of brass alloy,

σ = 440 MPa

Thus,

$sigma = 50,000 / \pi (r)^2\\\\440 X 10^6 = \frac{50,000}{3.14 X (r)^2}\\\\r = 0.06m\\r = 60mm\\\\\\$

Therefore, the radius of the specimen is 60 mm

3 0

2 years ago

mechanically cleaned wastewater bar screen is constructed using 6.5-mm-wide bars spaced 5.0 cm apart center to center; the bars

Annette [7]

Answer: The head loss is 1.5 cm

Explanation: Equation needed to solve is: $h= (vs^2-v^2)/2gC^2$

where C is the friction coefficient. To determine the area of flow, we nee to realize that the area available for flow over an area of 1*1 m, Aₓ= 0.87 m², observed from the correlation that for every width of 5.0 cm, we lose 6.5 mm, roughly about 12.9 or 13%.

For a fresh clean screen, the velocity will tend to increase by a factor of (1/0.87).

The velocities through the screen equals to (0.4/0.87) = 0.46 m/s and (0.8/0.87) = 0.92 m/s.

Apply the above mentioned equation:

$h= (0.92^2-0.8^2)/2*0.84^2*9.81 \\\\=0.01499\\=0.015 m = 0.015*100 = 1.50 cm$

6 0

2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

bagirrra123 [75]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.

Explanation:

Given:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

To calculate:

The magnitude of applied stress in the direction of [101] and [011].

Formula:

zcr=σ cosФ cosλ

Solution:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.

6 0

2 years ago

A shopaholic has to buy a pair of jeans , a pair of shoes l,a skirt and a top with budgeted dollar.Given the quantity of each pr

fredd [130]

Answer:

you might be facing some difficulty in observing this point of division between your question and me

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2 years ago