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Firlakuza [10]
3 years ago
7

A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving thr

ough the middle of a wind farm when an idea snapped into her head. She thought to herself, what if she mounted a small windmill generator on the roof of her Prius and wired it into the battery? She thought, with a little head wind to get her started she could just drive off under wind power! The faster she drove, the more power she would make, and the faster she could go… What’s wrong with her idea? a. As a side question, Priuses make use of regenerative braking. Why is it called regenerative braking? Why don’t cars advertise regenerative accelerators?
Engineering
1 answer:
Eva8 [605]3 years ago
6 0

Answer:

Explanation:

It wouldn't work because the wind energy she would be collecting would actually come from the car engine.

The relative wind velocity observed from a moving vehicle is the sum of the actual wind velocity and the velovity of the vehicle.

u' = u + v

While running a car will generate a rather high wind velocity, and increase the power generated by a wind turbine, the turbine would only be able to convert part of the wind energy into electricity while adding a lot of drag. In the end, it would generate less energy that what the drag casuses the car to waste to move the turbine.

Regenerative braking uses an electric generator connected to the wheel axle to recover part of the kinetic energy eliminated when one brakes the vehicle. Normal brakes dissipate this energy as heat, a regenerative brake uses it to recharge a batttery. Note that is is a fraction of the energy that is recovered, not all of it.

A "regenerative accelerator" makes no sense. Braking is taking kinetic energy out of the vehicle, while accelerating is adding kinetic energy to it. Cars accelerate using the power from their engines.

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\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}

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Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}

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\ Q_{in} =6 \cdot 583 \text { tons }

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