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2 years ago

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What is the approximate average power output of a well-designed modern turbine in Des Moines, Iowa with a 10 m2 swept area and 5

0 m hub height

1 answer:

viva [34]2 years ago

7 0

The approximate average power output is mathematically given as

P=1097.6w

<h3>What is the approximate average power output?</h3>

Question Parameters:

Iowa with a 10 m2 swept area and 50 m hub height

Assume 80% of the Betz limit, 80% conversion efficiency, and air density of 1.0 kg/m3. Wind speed is 7 m/s2

Generally, the equation for the average output power is mathematically given as

$P=0.5 \phi BAu^3*n\\$

Where

B= Benz coefficient

n=0.8

Therefore

P=0.5*1*0.8*10*7^3*0.8

P=1097.6w

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Complete Question

What is the approximate average power output of a well-designed modern turbine in Des Moines, Iowa with a 10 m2 swept area and 50 m hub height? Assume 80% of the Betz limit, 80% conversion efficiency, and air density of 1.0 kg/m3. Wind speed is 7 m/s2

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Which principle of the software engineering code of ethics has gilbert violated?.

Yuki888 [10]

Answer:

Judgement

Explanation:

Gilbert is required by the Judgement Principle to "disclose those conflicts of interest that cannot reasonably be avoided or escaped." Since Gilbert professionally believes that the software meets specifications, secures documents, and satisfies user requirements, it is not clear if he violated any principle. However, he could have informed his client of his interest in the software and also presented other software packages of different companies from which the client could make its independent choice.

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3 years ago

25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an

Gnesinka [82]

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

$N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)$

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is $0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}$ .

Total pressure P = 1 atm = 101.325 KPa

$P_{A_{1} }$ = 23.76 mm Hg

$P_{A_{1} }$ = $\frac{23.76}{760}$

$P_{A_{1} }$ = 0.03126 atm

$P_{A_{1} }$ = 3.167 K Pa

When air surrounded is dry air, then $PA_{2}$ = 0 mm Hg

R = 8.314 $\frac{K P a \cdot m^{3}}{mol \cdot k}$

$P_{B/M}$ = $\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}$

$P_{B_{1}}=P_{T}-P_{A1_{}}$

= 101.325 - 3.167

$P_{B_{1} }$ = 98.158 K Pa

$P_{B_{2}}=P_{T}-P_{A_{2}}$

$P_{B_{2} }$ = 101.325 - 0

$P_{B_{2} }$ = 101.325 K Pa

$P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}$

$P_{B/M}$ = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

$N_{A}$ = $\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }$

$N_{A}$ = 0.11077 × $10^{-6}$ mol/ $m^{2}$ .s

$N_{A}$ = 0.11077 × $10^{-6}$ × 18 × (60×60×24)

$N_{A}$ = 0.1723 $lb_{m}$ / $m^{2}$ .day

$\tilde{N}_{A}=N_{A} \times Area$

Area of individual pipe is

$A=\frac{\pi}{4}(0.0254)^{2}$

A = 0.00051 $m^{2}$

$\bar{N}_{\boldsymbol{A}}$ = 0.1723 × 0.00051

$\bar{N}_{\boldsymbol{A}}$ = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

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3 years ago

(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator

MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

$100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\$

Carrier frequency 100kHz

Upper side band

$100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz$

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

$100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz$

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

$100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz$

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

$100k, 99.9k, 99.8k\ \ and \ \99.6kHz$

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

$100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz$

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3 years ago

If pure oxygen is fed in excess by 25%, what would the fractional conversion of methane be for the final concentration of CO2 in

inessss [21]

Answer:the fractional conversion of methane is 12.5%

Explanation:The reaction represent the combustion of methane to produce Co2 and steam.

CH4 +2O2_CO2 + 2H2O

From gay lussac law of proportionality

1mol of CH4 requires 2mol of Oxygen to produce 1 mol of CO2 and 2mol of H2O

So from the combining ratio,25% of O2 will fractional produce 25×1/2% of CH4.

While 12.5% of CO2 and 25% of steam is also produced .so in essence 2.5% of CO2 was lost in the reaction.

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4 years ago

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Andreyy89

I don’t know but good luck

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3 years ago