Correct formula for inertia constant 1

[Mello84]

Hi!
Using data from a load rejection test, I have to calculate the inertia constant of a turbine-generator set. I know that I have to calculate the slope of the shaft speed v/s time graph. From the swing curve; I have the following:

H=(Po/Sb )/(2(dn/dt))
where Po = initial loading of machine
Sb = base MVA
dn/dt = slope of shaft-speed response at instant of rejection (n is in per unit).

However; I found that there also is the following expression for H (for e.g. Pg 17 in

http://www.wecc.biz/library/rro/Shared Documents/synctest.pdf).

H=(Sb(dn/dt))/(2Po)

Is there any convention for H I am unaware of? What is the correct expression?

 

[7anoter4]

See: thread238-257084

 

[7anoter4]

I'll try to recalculate the H formula based on IEEE 399/1990.
H=KE/S
KE=1/2*I*w^2 kinetic energy [Eq. 87]
The drive unit [motor, turbine or else] develops a torque Ta [ since P=Ta*w]
w=2*pi()*rpm/60 [rev/sec]
P=Ta*(2*pi()*rpm/60) P[W] Ta in [Nm] ; Ta=1000*P*60/2/pi() P[kw]
Ta=9549.3*KW/RPM [Nm] or Ta=9549.3*10^3*MW/RPM [Nm]
If there isn't a counter torque from the generator this motor torque will produce acceleration:
Ta=I*[d2(alpha)/dt2] where alpha=rotating angle[rad] and d(alpha)/dt=w
Ta=I*(dw/dt) I inertia moment [kg*m^2] I=Ta/(dw/dt)
dw=2*pi()*dn/60
I=1000*Po*60/2/pi()/2/pi()*60/(dn/dt) I=91189*Po/(dn/dt) Po [kw] I [kg.m^2]
As I [the sum of all rotating part I connected together] is constant and also for a while Po is constant
KE will grow with the velocity from rpm steady state up to rpmmax.
KE=1/2*91189*Po/(dn/dt)*(2*pi()*nmax/60)^2
KE=1/2*10^6*Po/(dn/dt)*nmmax^2 [Po in MW]
H=KE/S=1/2*(Po/Srated)/(dn/dt)*rpmmax^2
Srated [VA]=10^6 Srated[MVA]