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Crazy boy [7]
3 years ago
5

Describe the steps involved in accomplishing a life-cycle cost analysis (LCCA). I have listed the steps. just describe. a)-Opera

tional Performance. b)-Salvage externalities c)- Value vs. Risk d)- Initial expenditure e)- Maintenance implications
Engineering
1 answer:
Tasya [4]3 years ago
8 0

Explanation:

a). <u>Operational Performance</u>

     It is defined as the parameter that describes the percentage of accuracy of performing an operation or carrying out an activity.

b).  <u>Salvage externalities</u>

   Salvage in Life cycle cost analysis is a process of estimating the value of the remaining assets in the organisation,.

c). <u>Value Vs Risk</u>

   When we take risk in doing any activity we know the value of accomplising the activity. So value relates directly with risk. When the value of a certain task is high, the risk involve in it is also high.

d). <u>Initial expenditure</u>

   Initial expenditure is nothing but the cost involve in starting a particular acitvity or task at the starting phase.

e). <u>Maintenance implications</u>

   It lays emphasis in maintaining the cost of every possible parameters that are involve in the activity. It includes labour, machines, positions, energy, facilities, etc.

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A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of
OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0
2 years ago
Should aircraft wings have infinite stiffness?
Colt1911 [192]

Answer:

No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.

3 0
3 years ago
Read 2 more answers
If the slotted arm rotates counterclockwise with a constant angular velocity of thetadot = 2rad/s, determine the magnitudes of t
astraxan [27]

Answer:

Magnitude of velocity=10.67 m/s

Magnitude of acceleration=24.62 ft/s^{2}

Explanation:

The solution of the problem is given in the attachments

3 0
3 years ago
Kam
jolli1 [7]
I think D i’m not sure
7 0
3 years ago
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

so from idea gas equation

\frac{P1*V1}{T1} = \frac{P2*V2}{T2}   ................1

so {P2} = \frac{P1*T2}{T1}

{P2} = \frac{2*10^5*350}{300}

P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

so gauge pressure is 133 kPa

4 0
3 years ago
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