Horsepower vs Pulley Ratio

There is also a loss of power at the pulley depending on pulley and belt type.

Flat belts create less power loss but are hard to get tracking right.

V-Belts have more resistance against the pulley and create more loss.

To answer your question, the ratio between pulleys is directly related to the power differential. 1":1" remains the same, 1":2" doubles the power but cuts the speed in half. (I was taught that the ratio is read DRIVE to DRIVEN)

Please correct me if I am wrong.

I really like your transformer analogy! Never thought of it that way... GA!

It apears that your pulley ratio is correct, if 1 is for fan and 0.57 is for motor because 20 hp must be induction motor, and it will rotate slightly lesser rpm than 1800 at 60 Hz supply.

Changing pulley ratio will only change the speed of fan and the formula for Power at changed speed can be obtained by simple one of the fan laws:

N1/N2 = (P1/P2)**3

Your flow rate will reduce directly with speed, discharge pressure by square of speed and power by cube of speed symultaniously.

Several good answers relative to power, torque and speed.

BUT- your question was "what does the decrease in pulley pitch have on the output HP on the fan side?" The answer goes back to the fan- FAN horsepower is based on mass flow vs. pressure in a defined time (like all HP calculations). Fan HP is computed (CFM x Density of Air x pressure in Inches W.C.) / (6356 x fan efficiency (%)).

NOTE- 6356 is a constant for this application, W.C, refers to "water column"- the height of a column of water that the fan "total" pressure could support. Total pressure is outlet pressure minus inlet pressure- if inlet pressure is negative, its absolute value is added to output pressure.

Your fan is moving 5,000 CFM of air against some pressure while spinning at 3131 RPM. Assuming a fan efficiency of, say, 75% (typical for this type of fan, but directly available from the fan curve), this application would have a Total pressure of about 16.97 inches W.C. If you disconnected the fan from the duct, the air flow would increase and the HP would likely decrease (the fan curve would give actual values).

As indicated earlier, an increase in fan speed would increase air volume directly, increase fan air pressure by the square of the speed change ratio and increase fan HP (and corresponding motor output HP) by the cube of the speed change ratio.

The "decrease" in pulley pitch DOES NOT HAVE ANY DIRECT IMPACT on fan HP- you could use a 1:1 ratio pulley and a VFD to control motor speed and, so long as the fan was spinning at 3131 RPM, everything else would be the same (assuming that the pulley friction was also be essentially equal).

Given a specific circumstance, the formula above will ALWAYS work- relative to FAN horsepower required. The drive losses are added to the fan power for Total Power required by the device (0utoput of the motor).

I think I am on topic to remark that I agree that the amount of 'torque transferring ability' is lost by changes in pulley pitch ratios is negligible compared to 'torque transferred', but at some point (which you will probably never reach anyway) the gripping ability of the belt would be lost and slippage would occur.

I am taking the notion to the limit, and not wanting to burden everybody with needless "what-ifs" (as is often done).