For the hydraulic system shown here the source pressure P1 i

For the hydraulic system shown here the source pressure P_1 is constant and the oil flow rate q due to the opening of x is is proportional to x as well as the pressure difference of (P_1 - P_2) with constant coefficients of C_1 and C_2. Mass and cross sectional area of the piston are M and A. Oil leak flow rate can be assumed to be q_L = L(P1 - P2) where L is a constant value. Considering displacements of x and y are input and output of the system. Determine a differential equation which relates the output to the input in terms of constants C_1, C_2, L, M, A, c, and k. Assuming the following values: C_1 = 1 m^2/s, C_2 = 0.01m^4s/kg, L = 0.02 m^4s/kg, M = 5 kg, A - 0.01 m^2, c = 49.99 Ns/m, and k = 12, 625 N/m. Also, consider the input as a ramp function x (t) = t. Determine the function y (t) using time domain solution, and plot x(t) and y(t) for the range of [0 to 1] Determine the function y (t) using Laplace transform, and plot x(t) and y(t) for the range of [0 to 1] Determine the function y (t) using state space analysis, and plot x(t) and y(t) for the range of [0 to 1]

Solution

solution:

1)here given that when gate are poen ramp input x(t) it will flood system and creates pressure over piston and make to move but at same time spring system pull it back ward and due to damper it finally damp to steady state,if we assume damping is not provided by oil in chamber

2) as given flow rate is proportinal distant x and pressure difference ,hence

q1=c1x

q2=C2(P1-P2)

and leakage is

ql=L(P1-P2)

hence resultant flow rate is

Qr=q1+q2-ql

Qr=C1x+(c2-L)(P1-P2)

where fluid velocity is given by

Qr=A*velocity of output

Qr=A*y\'

3) let consider x as input and y as output then for given system at piston has external force as pressure force at piston

and at piston equation of motion as

my\'\'+cy\'+ky\'=Fp

where Fp=pressure force=(P1-P2)A

where from flow rate we have area as

my\'\'+cy\'+ky=(P1-P2)(Qr/y\' )

3) final differential equation of motion as

my\'\'*y\'+c(y\' )^2+ky*y\'=(p1-P2)*(c1*x+(C2-L)(P1-p2))

on putting value we get

5y\'\'*y\'+49.99(y\' )^2+12.625y*y\'=(p1-P2)*(x-.01(P1-p2))

it\'s solution is given by

yt=yc+yp

4) above equation is differential equation of above system in wrt time and it can be converted in laplace transform\'s s domain by following way

y\'\'(t)=s2y(s)-sy(0)-y\'(0)=s2y(s)

y\'(t)=sy(s)-y(0)=sy(s)

so finally above equation in laplace form as

(5s3+49.99s2+12.625s)y^2(s)=(p1-P2)*(x(s)-.01(P1-p2))

5) input vary linearly between 0 to 1 as x(t)=t

t=0s,x=0 m

t=1s,x=1m