Hi gokulkrish2!
The short answer: 32.2 is a unit conversion associated with British/English/American units. Some people carry around gc but I don't. The way I look at it is simply 1 = [lbf sec^2 / (32.2 lbm ft)] (easy to remember since F=MA => 1=F/MA and since we know that a force on one lbf occurs when a mass of 1 lbm is subjected to acceleration of gravity which is 32.2 ft/sec^2).
The long answer is to repeat the full derivation from the beginning, this time accounting explicitly for the units. Here it goes:
Let T(w) = accelerating torque = Telec – Tmech
The rotational equivalent of F = m*a is:
T(w) = J * dw/dt
Rearrange:
dw/dt = T(w)/J
dt = [J/T(w)] dw
t = Int [J/(T(w)] dw [equation 1]
Define the right hand side of equation 1 as "Integral 1"
Integral 1 = Int [J/(T(w)] dw
We know speed N = w/(2*pi)
dN = dw/(2*pi)
Substitute dw = (2*pi)*dN
Integral 1 = Int [J/(T(w)] dw = 2 * Pi * Int [J/(T(w)] dN
Multiply by 1 = [lbf *sec^2 / (32.2 lbm *ft)]
Integral 1 = 2 * Pi * Int [J/(T(w)] dN * lbf *sec^2 / (32.2 lbm *ft)
Multiply by ft/ft
Integral 1 = 2 * Pi * Int [J/(T(w)] dN * ft-lbf *sec^2 / (32.2 lbm *ft^2)
Rearrange:
Integral 1 = 2 * Pi * Int [{/J/<lbm-ft^2>} (T/<ft-lbf>) )] dN * sec^2 / 32.2
multiply by minute/<60*sec>
Integral 1 = 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] dN * minute * sec / (32.2 *60)
Substitute minute = 1/min^-1
Integral 1 = 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} * sec / (32.2 *60)
Put Integral 1 back into equation 1:
t= 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} * sec / (32.2 *60)
Divide both sides by sec
{t/sec}= 2 * Pi * Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} / (32.2 *60)
Collect Constants at the end
{t/sec}= Int [{/J/lbm-ft^2} (T/ft-lbf) )] d{N/min^-1} *[2*Pi]/ (32.2 * 60)
Now we only need recognize that t/sec is a unitless quantity we would call "time in seconds". J/lbm-ft^2 is a unitless quantity we would call "rotating inertia in lbm-ft^2". T/ft-lbf is a unitless quantity we can call "Torque in ft-lbf". N/min^-1 is a unitless quantity we can call speed in rpm.
Based on the above we can rewrite in a more compact form:
t = Int [J/T)] dN * (2*pi) / (32.2 * 60)
t = Int [J/T)] dN * / 307 [equation 2]
where t in seconds, J in lbm-ft^2, T in lbf-ft, N in rpm.
Now look at some small interval from speed N1 to N2 over which we will perform numerical integration. From here to the end we will ASSUME the torque changes linearly with speed over each interval as given by:
T = T1 + (N-N1)/(N2-N1) * (T2-T1)
Let delta = N2-N1 be the width of the interval
T = T1 + (N-N1)/ delta * (T2-T1)
Call this linear function f(N) = T1 + (N-N1)/ delta * (T2-T1)
Let dt12 be the time to get from N1 to N2
dt12 = J / 307 * Int ( 1/[f(N) ), N=N1..N2 from equation 2.
Evaluate integral by chain rule:
dt12 = J/307 * [ln(f(N) / df/dN] evaluated at N2 minus the same thing evaluated at N1 (provided that df/dN is not itself a function of N)
df/dN = (T2-T1)/delta (we now confimred that df/dN is not a function of N)
dt12 = J*delta/(307*(T2-T1)) * [ln(f(N)] evaluated at N2 minus the same thing evaluated at N1
But we know f(N) evaluated at N2 and N1... it is T2 and T1!
dt12 = J*delta/(307*(T2-T1)) * [ln(T2) -ln(T1)]
Simplify by noting ln(T2)-ln(T1) is ln(T2/T1)
dt12 = J*delta/(307*(T2-T1)) * [ln(T2/T1)] [equation 3]
where
dt12 - time to accelerate through speed range (in seconds)
J - rotating inertia (in lbm-ft^2)
T1, T2 = Torque at beginning and end of speed range (in lbf-ft)
delta = width of the speed range (in rpm)
Should we prefer to avoid British units, instead of having unit factor the factor 307 =(32.2 * 60)/(2*pi) in the denominator, we would not need the 32.2 and instead have a factor (60)/(2*pi) = 9.55 in the denominator.
dt12 = J*delta/(9.55*(T2-T1)) * [ln(T2/T1)]
where
dt12 - time to accelerate through speed range (in seconds)
J - rotating inertia (in kg-m^2)
T1, T2 = Torque at beginning and end of speed range (in N-m)
delta = width of the speed range (in rpm)
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
