Consider the system shown at right Assume the resistances R1
Solution
solution:
1)as here are two tank are available hence this system have two variable to govern it output hence we have to write two transfer function one for first tank and second for second tank
2)here for first tank ,as per mass balance equation we have
Qm+Q1=Q1\'
Qm=input flow rate
Q1=volume change of tank
Q1\'=output flow
2)where from capacitance of tank we have
V=C1P1
P1=density*g*h1
we can write further as
dP1/dt=Q1/C1
wherer flow resistance in pi[e is
P1\'=R1*Q1\'
where by KPE
P1+P1\'=Pa
Pa=0=it is not atmospheric pressure but pressure difference across pipe length,here it is zero
P1=-P1\'
hence putting all in first equation we get
dP1/dt=(Q1\'-Qm)/C1
dP1/dt=(P1\'/R1-Qm)/C1
dP1/dt=(-P1/R1-Qm)/C1
2)whaterver flow Q1\' moving out from first tank is adding to second tank hence mass balance equation become
Q1\'+Q2=Q2\'
where capacitance relation is written as
dP2/dt=Q2/C2
dP2/dt=Q2\'-Q1\'/C2
where Q2\'=flow in output pipe suffer from resistance R2,hence
Q2\'=P2\'/R2
as first tank here again
P2\'=-P2
dP2/dt=(-P2/R2-Q1\')/C2
Q1\'=Q1+Qm
hence puting value we get that
dP2/dt=(-P2/R2-Q1-Qm)/C2
P2=density*g*h2
on putting in equation we get
dh2/dt=(-P2/R2-Q1-Qm)/density*g*C2
on differentiating we get that
P2/R2=constant as it is pressure loss
hence we get height in second tank as function of Qm only as follows
d^2h2/d^2t=(1/density*g*c2)Qm\'(t)
hence on applying laplace transform we get that
S^2*h2(s)=(1/density*g*C2)SQm(s)
hence transfer function becomes
h2(s)/Qm(s)=(1/S)(1/density*g*c2)

