Claudel Olivier Process Analysis and Modeling October 17, 2001

Claudel Olivier Process Analysis and Modeling October 17, 2001

Claudel Olivier
Process Analysis and Modeling
October 17, 2001
Project 8
Given,
(R + mS ) x 0 − (R + mS ) x m = (Vr + Vs )
dè
dx m
Find what xm.
Rearranging to get all like terms to one side yields:
dx m
dè
=
Vr + mVs (R + mS ) x 0 − (R + mS ) x m
Here a substitution will be made before integration to simplify the solution
process. Let U=(R+mS)x0-(R+mS)xm. Therefore taking the derivative of U with
respect to xm yields,
dU = −(R + mS )dx m
− dU
dx m =
R + mS
Now the integration can take place.
dè
du
=
(Vr + mVs ) − (R + mS )U
dè
du
∫ (Vr + mVs ) = ∫ − (R + mS)U
è
ln U
=
(Vr + Vs ) − (R + mS )
è
ln [(R + mS) x 0 − ( R + mS ) x m ]
=
(Vr + Vs )
− (R + mS)
− è(R + mS )
= ln [(R + mS )( x 0 − x m ) ]
(Vr + Vs )
−è (R + mS )
e
(Vr +Vs )
= (R + mS )( x 0 − x m )
− è (R + mS )
Rx0 + Sy 0 − e
xm =
(Vr +Vs )
= (R + mS ) x m
− è ( R + mS )
(Vr +Vs )
Rx 0 + Sy 0 − e
( R + mS )