Blower Calculations

Height of air plane is required when running pressure. Content of atsmophere and/conditions ALSo play a vital role in pressure of of tunnel not just air moved by fans. Type of plane and its dynamics also play a role in tunnel testing, dropping a plane through a cloud is a great way to test if your a billionaire lol. Great mistakes come from not examining enough possibilities. Equating probability is an expensive industry.

It's an air pump with a centrifugal fan, not a airplane with a propeller....

http://en.wikipedia.org/wiki/Centrifugal_fan

http://www.scribd.com/doc/16648848/Fans-BlowersCalculation-of-Power

In your formula:1/2m*v^2= 75w, you don't say what m is, but working back to get the right answer and units, it's the mass flow, 0.07 kg/s (air density 1.2 kg/m³). But this formula amounts to multiplying velocity pressure (0.5*ρ*V²) by flow. The discrepancy in power is because the pressure rise seen by the blower is higher the velocity pressure, caused by resistance in the nozzle, fittings etc (as you suggest), and because blower efficiency < 100%.

To find the power, you would multiply flow x actual pressure rise. You could estimate pressure rise from the blower dimensions, but as you already know the power, easier to work back to find pressure rise, assuming blower efficiency. 650W is probably the rated electrical input power (not the motor shaft power to the impeller). I would guess overall efficiency no more than ~ 40%.

For your 2nd question, I think you're right, for a propeller in free air power is as your formula.

The power requirement would be given by the formula

P=yQH/1000 Watts where

P=Power consumed

y=specific gravity =pg where p=density and g=gravitational constant

Q=Discharge in m3/sec

H=Head developed by the blower.

This power has to be divided by motor efficiency to get input power to the motor.

That could confuse him.

Specific gravity is not ρ*g, for liquids it's density kg/m³/water density, 1000kg/m³ (and is dimensionless). For gas flow it's usually based on air =1.

Power = H*ρ*g*Q if h is in m fluid, ρ = density, kg/m³ (not SG). But for gases, you're unlikely to know H in m head of gas, to get that you would have to work back from pressure rise. And if everything in SI units it gives power in watt, don't need the 1000.

Simpler to take power (watt) = p*Q, p in Pa. If p is given in say mm water, or mbar, it's easy to convert. For gases, this formula is only correct for small pressure rise.

That gives fluid power, must divide by blower efficiency to blower shaft power, as well as motor efficiency to get input power to the motor.