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JulijaS [17]
2 years ago
13

A wind turbine designer is considering installing a horizontal axis wind turbine at a location in Michigan. To increase the powe

r extracted from wind, the designer is debating between increasing the blade radius by 10% versus increasing the height from 40 m to 60 m, which would increase the average wind speed from 6 mph to 6.5 mph. Considering the only the impact on power extraction, the designer should go with the height increase.
a. True
b. False
Engineering
1 answer:
bogdanovich [222]2 years ago
3 0

Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P =  n * 1/2 *<em> p</em> *Av^3

where: n = turbine efficiency

           <em>p = air density </em>

<em>            </em>A = πd^2 / 4

           v =  speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

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Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

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C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

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x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

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Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

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\mathbf{C_{10} = 137.611 \ kN}

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Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizon
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The complete Question is:

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

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What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?

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Explanation:

Find the given attachments for complete explanation

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