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3 years ago

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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)?

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 factor $CRF = \frac{i(i+1)^n}{(i+1)^n -1}$

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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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

The mathematical relation between stress to strain on the plastic region of deformation is

$\sigma _T =K\epsilon^n_T$

Where K is a constant

n is known as the strain hardening exponent

This constant K can be obtained as follows

$K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

$K = \frac{345}{(0.02)^{0.22}}$

$= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

$\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

$=0.0443$

From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

$l_i = l_o e^{\epsilon_T}$

Where

$l_i$ is the instantaneous length

$l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

$l_i = 470 * e^{0.0443}$

$=491.28mm$

We can also obtain the elongated length mathematically as follows

$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

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