Control De Proceso

PRESENTACIÓN ASIGNADA
SECCIÓN 3-6 & EJERCICIO 3-4

CURSO DE
CONTROL DE PROCESOS

UNIVERSIDAD DE CARTAGENA

FACULTAD DE INGENIERÍAS

PROGRAMA DE INGENIERÍA QUÍMICA

CARTAGENA DE INDIAS D. T. Y C., 23 DE MARZO DE 2012

FIRST ORDER – DYNAMIC SYSTEMS

SECTION 3-6 – GAS PROCESS EXAMPLE

Consider the gas vessel shown in Fig. 1. A fan blows air into a tank, and from the tank theair flows out through a valve. For purposes of this example, let us suppose that the air flow delivered by the fan is given by

fit=0.16mi(t)

Where

fit=gas flow in scfmin, scf is cubic feet at standard conditions (60°F- 1 atm)

mit=signal to fan, %

Figure 1

The flow through the valve is expressed by

fot=0.00506mo(t)p(t)pt-p1(t)

Where

fot=gas flow, scf/min

mot=signal tovalve, %

pt=pressure in tank, psia

p1t=downstream pressure from valve, psia

The volume of the tank is 20 ft3, and it can be assumed that the process occurs isothermally at 60°F. The initial steady-state conditions are

fis=fos=8 scfm ps=40 psia p1s=1 atm mis=mos=50%

We want to develop the mathematical model, transfer functions, and block diagram that relate thepressure in the tank, p(t), to changes in the signal to the fan, mi(t), in the signal to the valve, mo(t) and in the downstream pressure, p1(t).

We must first develop the mathematical model for this process. An unsteady-state mole balance around the control volume, defined as the fan, tank, and outlet valve, provides the starting relation. That is

Rate of accumulation of moles in control volume=Rate of moles into control volume-Rate of moles out of control volume

Or, in equation form:

dn(t)dt=ρfi(t)-ρfo(t)

Where

ρ=molar density of gas at standard conditions, 0.00263 lbmoles/scf

nt=moles of gas in tank, lbmoles

Also we know that the ideal gas law is:

ptV=ntRT

All the equations above are the mathematical model of this process. Here we have p1s=14.69 psia,T=519.67°R and R=10.73 psia ft3lbmol °R.

Now we can write the mathematical model in another way, like:

VRTdp(t)dt=ρ0.16mit-ρ0.00506motptpt-p1t [1]

As we see, the second term on the right side of the equation [1] is non-linear, so we proceed to linearize it as follows:

fo(t)≈fos+∂fo(t)∂mo(t)smo(t)-mos+∂fo(t)∂p(t)sp(t)-ps+∂fo(t)∂p1(t)sp1(t)-p1s

C1=∂fo(t)∂mo(t)s=0.00506psps-p1sC1=0.005064040-14.69=0.161 psia

C2=∂fo(t)∂p(t)s=0.00506mos2ps-p1s2psps-p1s

C2=0.00506*50*2*40-14.6924040-14.69=0.2596 %

C3=∂fo(t)∂p1(t)s=0.00506mos-ps2psps-p1s

C3=0.00506*50*-4024040-14.69=-0.159 %

fos=0.00506mospsps-p1s

Replacing this in equation [1] we have:

VRTdp(t)dt=ρ0.16mit-ρ0.00506mospsps-p1s-ρC1mo(t)-mos-ρC2p(t)-ps-ρC3p1(t)-p1s [2]

Equation [1] in steady-state is:VRTdpsdt=ρ0.16mis-ρ0.00506mospsps-p1s [3]

Now [2]-[3] is:

VRTdp(t)-psdt=ρ0.16mit-mis-ρC1mo(t)-mos-ρC2p(t)-ps-ρC3p1(t)-p1s [4]

Applying the deviation variables in equation [4] pt=pt-ps, mit=mit-mis , mot=mo(t)-mos and p1t=p1(t)-p1s we obtain

VRTdptdt=ρ0.16mit-ρC1mot-ρC2pt-ρC3p1t [5]

Equation [5] can be written as:

VRTdptdt+ρC2pt=ρ0.16mit-ρC1mot-ρC3p1tVRTρC2dptdt+pt=ρ0.16ρC2mit-ρC1ρC2mot-ρC3ρC2p1t

VRTρC2dptdt+pt=0.16C2mit-C1C2mot-C3C2p1t [6]

Here we have:

τ=VRTρC2 K1=0.16C2 K2=C1C2 K3=C3C2

τ=2010.73*519.69*0.00263*0.2596 K1=0.160.2596 K2=0.1610.2596 K3=-0.1590.2596

τ=5.25 min K1=0.616 psia% K2=0.62 psia% K3=-0.612

Then, the equation [6] is:

τdptdt+pt=K1mit-K2mot-K3p1t [7]Applying the Laplace transform of the equation [7] we get:

Lτdptdt+Lpt=LK1mit-LK2mot-LK3p1t

τsps-p0+ps=K1mis-K2mos-K3p1s

τsps+ps=K1mis-K2mos-K3p1s

psτs+1=K1mis-K2mos-K3p1s

ps=K1τs+1mis-K2τs+1mos-K3τs+1p1s

The transfer functions that we can obtain here are:

psmis=K1τs+1

psmos=-K2τs+1

psp1s=-K3τs+1

The block diagram of this process is:

EXERCISE 3-4

Consider the...