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maxonik [38]
3 years ago
15

How to connect the wind turbines to the grid? Give advantages and disadvantages to your answer.​

Engineering
1 answer:
taurus [48]3 years ago
7 0

Answer:

Electricity from the wind turbine generator travels to a transmission substation where it is converted into extremely high voltage, between 155,000 and 765,000 volts, for long distance transmission on the transmission grid. This grid comprises a series of power lines that connect the power sources to demand centers.The advantages of wind energy are more apparent than the disadvantages. The main advantages include an unlimited, free, renewable resource (the wind itself), economic value, maintenance cost, and placement of wind harvesting facilities.

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Which examples demonstrate tasks commonly performed in Maintenance/Operations jobs? Check all that apply.
Verdich [7]

1.Ross fixes a dishwasher for a homeowner.

3.Cassandra fixes holes in an old road.

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Select the correct answer.
boyakko [2]
I think the answer is b
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You are testing a new jet engine in a test cell at sea level conditions. You measure the mass flow through the engine and find i
bulgar [2K]

Answer:

43248 newtons.

Explanation:

Force = mass x accelerations and units of force are newtons which are given in the question.

here mass = 125 of air and 2.2 of fuel, total = 125+2.2=127.5kg/s  and the velocity of the exhaust is 340m/s.

force = 340m/s * 127.5kg/s = 43248 newtons technically this is wrong (observe units) but i will expalin how i have taken acceleration as a velocity here and mass/unit time as simply mass.

see force is mass times acceleration or deceleration, here our velocity is not changing therefore it is constant 340m/s but if it were to change and become 0 in one second then there would be -340m/s^2 (note the units ) of deceleration and there would be force associated with it and that force is what i have calculated here. similarly there would be mass in flow rate of mass per second, which is also in that one second of time.

let's calculate error.

error = (actual-calculated)/actual. = (43248-60000)/43248= -38.734% less is ofcourse greater than 2%.

So the load cell is not reading correct to within 2% and it should read 43248newtons.

5 0
3 years ago
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A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip
JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})

Above equation can be written as:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})

First Consider no wire i.e d/D=0

Above expression will become:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}

Part A: (d/D=0.1)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574

DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.574}{1}*100

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783

DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.783}{1}*100

DeltaV percent=21.7%

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Answer:

The city will experience high levels of rainfall.

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