Re: How do you calculate need horsepower increase for RPM increase?

Jim,

Well, I told my wife about this problem and her response was "Keep It 
Simple, Stupid", so I will try my best.  I pulled out my trusty design book 
(Mechanical Engineering Design by Shigley & Mischke, 5th edition. pg. 670) 
and looked at the formula for power transmission in a flat belt.  This 
seemed like a decent model for your conveyor problem.  The simplest model 
states that power P = (F1 - F2)V.
Where:
P = power transmitted by the belt (in Watts)
F1 = tension in drive side belt (in Newtons)
F2 = tension in return side belt (in Newtons)
V = belt velocity (in meters/second)

To convert to horsepower the equation simply becomes Hp = (F1 - F2)V / 
33,000.  This corresponds to the common equation for power of P = F*V.

Ignoring a number of things that I will get into later, let's assume that 
the horsepower output of our 5 Hp motor is 5 Hp.  And you gave us belt 
speed at 150 feet/min. which is .762 m/s.  So we can solve for our tensions 
with 5 = (F1-F2)*.762 / 33,000 (I'm being careless and not writing units to 
make readability better but it is always a good idea to write your units in 
the equation to ensure you are not adding apples and oranges).  Anyway, 
solving for (F1-F2) = 216,536 N .
Now that we have a figure for that, we can plug it back in to the original 
equation and solve for an unknown V.  This is Hp = 216,536*V / 33,000 = 
6.56*V where V is again in m/s.  Using 1 m/s = 196.85 ft/min we can write 
the equation as 

Hp = .0333*V 

Where:
Hp = Horsepower
V = belt speed in ft/min

Therefore, for test, .0333*150 = 4.995 which is very close to our original 
5 Hp.  For a 10% increase, V=165 ft/min and therefore Hp = .0333*165 = 5.5 
Hp.  For 20% V= 180 and Hp = 6 and for 30% Hp = 6.5.  Since the 
relationship is linear, it makes sense that for every 10% increase in 
speed, horsepower increases by 0.5.

Now that is the simple explanation and it may be good enough for what you 
are doing, but it may not be right, and here is why. I don't know the 
application and I don't know the characteristics of your motor.  I assume 
it is a three phase induction motor running at 240V of 480V.  What I don't 
know is the current.  If I had the current and the voltage we could solve 
for the REAL power being used to drive the belt.  With a few more 
characteristics, we could have better idea of the torque characteristics of 
the motor as well.
The torque curve for an induction motor looks like the following curve.  I 
copied this from the Virginia Tech website that had an AC motor section 
that was over MY head.  But the visual is important to understand.



Now first off this is a picture of a typical curve, not necessarily for 
your motor.  The important parts are the starting torque, max torque, and 
one that isn't shown, rated torque.  Slip refers to (essentially) rotor 
speed, or RPM.  So starting torque is when RPM is zero.  This is usually 
greater than the rated torque so that the motor can spool up to its 
operating speed.  (If the initial load is greater than the starting torque, 
the motor would never move and burn up.)  The max torque is usually at a 
slower RPM than the operating RPM.  This is so that a motor can accommodate 
larger loads by slowing down instead of stalling.  The part that isn't 
shown is the rated load.  This is sometimes (but not always) the output 
stated on the motor.  This is located on the far right part of the curve 
somewhere below the starting torque and the max speed at T=0.  The last 
section is the synchronous motor speed when 
torque equals zero.  This is the max speed of the motor with no external 
loads (the furthest right of the curve).  This is pretty much the speed of 
the motor if you just plugged it in and ran it with no load.  So, by 
looking at this curve and without knowing the characteristics of the motor 
or the current it uses, we don't know where on this curve we are.  
Somewhere faster than max torque and slower than max speed.  What this 
means is that for a given torque load, a motor has an RPM associated with 
that load and will happily turn at that speed regardless of if it is ½ Hp 
or 4.5 Hp being used.  So without knowing details, it is possible (in fact 
probable) that a 5 Hp motor with different curve characteristics would 
drive the belt faster.  However, if you use the same type of motor with the 
same speed rating and higher power, the above equations should be valid for 
an estimate.  Hope this helps and wasn't too confusing.  If you need 
additional help you can always write me at bradk@jymis.com.  Good luck and 
feel free to ask me for additional assistance.
BK