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2 years ago

How are project deliverables determined?

Engineering

2 answers:

Greeley [361]2 years ago

7 0

Answer:

The essence including its problem is listed throughout the clarification section following.

Explanation:

Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.

For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:

It should be within the development of the work.
The interested parties-external or internal-must consent to the above. This is perhaps the product of hard effort.

So that the above seems to be the right answer.

Alex17521 [72]2 years ago

7 0

It is determined by:

For any output to be classified as a “deliverable” within a project, it has to meet a few criteria: It must be within the scope of the project. Stakeholders external or internal must agree to it. It must be the result of deliberate work.

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A 230 V shunt motor has a nominal armature current of 60 A. If the armature resistance is 0.15 ohm, calculate the following: a.

hjlf

Answer:

a) $E_{b} = 221 V$

b) P = 13,800 W

c) $P_{mech} = 13260 W$

di) $I_{initial} = 1533.33 A$

dii) $R_{start} = 1.85 ohms$

Explanation:

Voltage, Vt = 230 V

Armature current, $I_{a} = 60 A$

Armature Resistance, $R_{a} = 0.15 ohms$

a) The back emf is calculated as follows:

$E_{b} = V_{t} - I_{a} R_{a} \\E_{b} = 230 - (60 * 0.15)\\E_{b} = 230 - 9\\E_{b} = 221 V$

b) The power supplied to the armature (W)

$P =V_{t} I_{a}$

P = 230 * 60

P = 13,800 W

c) Mechanical power developed by the motor

$P_{mech}$ = Power supplied to the armature - Power lost in the armature

Power lost in the armature, $P_{a} = I_{a} ^{2} R_{a}$

$P_{a} = 60^{2} *0.15\\P_{a} = 540 W$

$P_{mech} = 13800 - 540\\P_{mech} = 13260 W$

d)( i)The initial starting current if the motor is directly connected across the 230 V line

At starting, there is no back emf, $E_{b} = 0$

$V_{t} = I_{initial} R_{a}$

$230 = I_{initial} * 0.15\\ I_{initial} = 230/0.15\\ I_{initial} = 1533.33 A$

ii) Value of the starting resistor needed to limit the initial current to 115 A

$V_{t} = I_{initial} (R_{a} + R_{start})$

$230= 115 (0.15 + R_{start})\\230/115 = 0.15 + R_{start}\\2 - 0.15 = R_{start}\\ R_{start} = 1.85 ohms$

5 0

3 years ago

You must yield the right-of-way to all of the following EXCEPT:

Elena-2011 [213]

Answer:

You must yield the right-of-way to a police car, fire engine, ambulance, or other emergency vehicle that uses a siren and flashing lights. Pull as close to the right of the road as possible and stop until the emergency vehicle(s) has passed. However, don't stop in an intersection because that would cause a wreck.

Explanation:

please have a good day. and also mark as brainllest bye

[o]-[o]

\___/

7 0

2 years ago

A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted

Pavlova-9 [17]

Answer:

a) The minimum acceptable value is 387.5 HV using Vickers hardness test.

b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.

Explanation:

To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.

According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):

$\mbox{Tensile strength}=HV*3.2\\\\HV = \frac{\mbox{Tensile strength}}{3.2} =\frac{1240}{3.2}=387.5$

According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:

$\mbox{Tensile strength}=HRC*31.5\\\\HRC = \frac{\mbox{Tensile strength}}{31.5} =\frac{1240}{31.5}=39.4$

8 0

2 years ago

Problem 4.041 SI Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at -26oC with a vo

Rom4ik [11]

Answer:

0.0297M^3/s

W=68.48kW

Explanation:

Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

state 1

X=quality=1

T=-26C

density 1=α1=5.27kg/m^3

entalpy1=h1=234.7KJ/kg

state 2

T2=70

P2=8bar=800kPa

density 2=α2=31.91kg/m^3

entalpy2=h2=306.9KJ/kg

Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.

m1=m2

(Q1)(α1)=(Q2)(α2)

$\frac{(Q1)(\alpha 1) }{\alpha 2} =Q2\\Q2=\frac{(0.18)(5.27) }{31.91} =0.0297M^3/s$

the volumetric flow rate at the exit is 0.0297M^3/s

To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation

W=m(h2-h1)

m=Qα

W=(0.18)(5.27)(306.9-234.7)

W=68.48kW

the compressor power is 68.48kW

4 0

2 years ago

Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the w

soldi70 [24.7K]

Answer:

Step 1

Given

Diameter of circular grill, D = 0.3m

Distance between the coal bricks and the steaks, L = 0.2m

Temperatures of the hot coal bricks, T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

4 0

2 years ago