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abruzzese [7]
3 years ago
15

You are considering building a residential wind power system to produce 6,000 kWh of electricity each year. The installed cost o

f the system is $1.50/Winstalled. The system is estimated to have a capacity factor of 22% and a 10 year life. The interest rate is 8% with an operations and maintenance cost of $0.04/kWhgenerated. What is the leveled cost of electricity for the system ($/kWh)?
Engineering
1 answer:
alisha [4.7K]3 years ago
8 0

Answer:

leveled cost of electricity LCOE is 0.1159/kwh

Explanation:

given data:

installation cost of system = $1.50/winstalled

capacity factor = 22%

life = 10 year

rate of interest 8%

operation and maintenance cost = $0.04 / kwh generated

capcaitence recovery factorCRF = \frac{i(i+1)^n}{(i+1)^n -1}

i= rate of interest

n = annuity period

CRF = \frac{0.08(1+0.08)^10}}{(1+0.08)^{10} -1} = 0.14903

LCOE = \frac{cost\ of \installation \times CRF}{hours/ available \times capacity\ factor} + variable\ o&M\ cost

=  \frac{1.50\times 1000/kw \times 0.14903}{8760\times 0.22} + 0.04/kwh

          = 0.1159/kwh + 0.04 /kwh

          = 0.1559 /kwh

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balu736 [363]

Answer:

2543 k

Explanation:

This problem can be resolved by applying the first law of thermodynamics

<u>Determine the adiabatic flame temperature</u> when the furnace is operating at a mass air-fuel ratio of 16 for air preheated to 600 K

attached below is a detailed solution

cp = 1200

8 0
3 years ago
Technician A says that the paper test could detect a burned valve. Technician B says that a grayish white stain on the engine co
Bumek [7]

Answer:

Both Technician A and B are correct.

Explanation: Both are correct, but keep note that different color coolants leave different color  stains.

6 0
3 years ago
Answer back to question for la ,lot points
Lerok [7]

Answer:

yes

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5 0
3 years ago
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

6 0
3 years ago
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat
zlopas [31]

Answer: 10.631\times 10^3\ N/m^2

Explanation:

Given

Discharge is Q=12.5\ L

Diameter of pipe d=150\ mm

Distance between two ends of pipe L=800\ m

friction factor f=0.008

Average velocity is given by

\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s

Pressure difference is given by

\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow  \Delta P=10.631\ kPa

8 0
3 years ago
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