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In a simple way the pressure will decrease because there's less air above you, so there's less mass pushing you.

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You can calculate air velocity as v = f/A where f=volumetric air flow and A = cross-sectional area of the air passage. Also, you have m=p*f, where m=mass flow rate and p=air density. Here 1 it says that "By conservation of energy, the energy consumed in rotating the fan is the same as the energy required to deliver the air: Pfan ∝ υ"

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There have been attempts to use ionized air to reduce drag on aircraft It's a generator that sends a beam of microwaves upstream into the Mach 6 flow, ripping apart the gas ahead of the model so that it is flying through a plasma--a boiling mix of positive ions and electrons--rather than ordinary gas. The experiment, at NASA's Langley ...

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From an engineering perspective, there are many different fan designs including axial and centrifugal configurations, along with various blade designs including forward curved, backward curved, and radial. There is no formula to calculate the required fan speed, but if a specific fan configuration is known then fan similarity laws can be used to calculate ...

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Using a dimensional analysis indeed can help you find dimensionless numbers as The Dark Side suggested Floris might help with, but beyond that there is no direct analytical method, no closed form solutions that can relate fan speed, shaft torque, flow rate and delta pressure. The issue is that at best the flow behavior is two dimensional, but more likely ...

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Wind chill is really due to two things: 1) colder air moves across the surface of your skin, replacing the air you heated with your body: this in essence takes away the blanket of warm air you keep making for yourself. 2) As your body loses moisture through evaporation, there is a humidity gradient of stagnant vapor around your body. The higher the ...

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Consider the total energy of the particle $$E=\frac{mv^2}{2}+mgh$$ Then (assuming $k>0$): $$\dot{E}=mv\dot{v}+mgv=mv[-g-mkv+g]=-m^2kv<0$$ So when the particle is thrown up and returns to a given height it has less energy than when it was first there. Since the potential energies are the same the speed has fallen. That is it comes down slower than ...

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You are making this rather hard for yourself. You correctly solved for the velocity, which is of the form $$v(t) = c_1 e^{-\alpha t} - \frac{g}{a}$$ where $a = mk$ and $c_1$ is found from the initial conditions. Integrating this expression should just give you $$x(t) = -\frac{c_1}{a} e^{-at} - \frac{gt}{a}$$ I think that because you ended up splitting ...

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