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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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7 0
2 years ago
A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a
vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60    .................1

put here value

15 ×10^{3} = 2π×1750×T / 60

so

T = 81.84 Nm

and

torsion = T / Z                        ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³     .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

so

torsion = σy / factor of safety

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so d = 0.0133 m

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3 0
3 years ago
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3 0
3 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
Anettt [7]

Answer:

Q=4.98\times 10^{-3}\ m^3/s

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

For Copper tube is 3/4 standard type K drawn tube

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

\mu =0.00164\ Pa.s

Pressure difference given as

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where

L is length of tube

μ is dynamic viscosity

Q is volume flow rate

d is inner diameter of tube

ΔP is pressure drop

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s

7 0
3 years ago
Which of the following reduces friction in an engine A)wear B)drag C)motor oil D)defractionation
kobusy [5.1K]

It is motor oil, as oil is used to reduce friction

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3 years ago
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