Lecture02_10thJan2009

download Lecture02_10thJan2009

of 8

Transcript of Lecture02_10thJan2009

  • 8/6/2019 Lecture02_10thJan2009

    1/8

    Lecture: 02

    Basic Terms in

    Instrumentation and Control

  • 8/6/2019 Lecture02_10thJan2009

    2/8

    AN INTRODUCTORY EXAMPLE

    The System

    Steady-state Design

    Process Control The Unsteady State

    Feedback Control

    Transient Responses Integral Control

    Thermocouple

    Block Diagram

  • 8/6/2019 Lecture02_10thJan2009

    3/8

    The System

    A liquid stream at temp. Ti is available at a constant flowrate ofw in units of mass per time.

    It is desired to heat this stream to a higher temp. TR.

    The fluid flows into a well-agitated tank equipped with aheating device.

    It is assumed that the agitation is sufficient to ensurethat all fluid in the tank will be at the same temp., T.

    Heated fluid is removed from the bottom of the tank at

    the flow rate w as the product of this heating process. Under these conditions, the mass of fluid retained in the

    tank remains constant in time, and the temp. of theeffluent fluid is the same as that the fluid in the tank.

    For a satisfactory design this temp. must be TR.

  • 8/6/2019 Lecture02_10thJan2009

    4/8

    The System

    T

    w, Ti

    w, T

    Heater

    Agitator

    Figure-1:Agitated heating tank.

  • 8/6/2019 Lecture02_10thJan2009

    5/8

    Steady-state Design

    Definition: A process is said to be at steady state when none ofthe variables are changing with time.

    At the desired steady state, an energy balance around theheating process may be written as follows:

    qs = w C (Ts Tis) (1)

    qs the heat input to the tank and subscript s is added toindicate a steady-state design value. For a satisfactory design, the steady-state temp. of the effluent

    stream Ts must be equal TR. Henceqs = w C (TR Tis) (2)

    Problem: However, it is clear from the physical situation that, ifthe heater is set to deliver only the constant input qs, then if

    process conditions change, the tank temperature will alsochange from TR. A typical process condition that may change is the inlet temp.

    Ti. Solution: An obvious solution to the problem is to design the

    heater so that its energy input may be varied as required tomaintain T at or near TR.

  • 8/6/2019 Lecture02_10thJan2009

    6/8

    Process Control

    Need: It is necessary to decide how much the heat inputq is to be changed from qs to correct any deviations ofTfrom TR.

    Solution-1:

    To hire a process operator, who would be responsible forcontrolling the heating process.

    The operator would observe the temp. in the tank,presumably with a measuring instrument (thermocouple andthermometer), and compare this temp. with TR. If T wereless than TR, he would increase the heat input and viceversa.

    As he became experienced at this task, he would learn justhow much to change q for each situation.

    Solution-2: This task can be easily and less expensively performed by a

    machine.

    The use of machines for this and similar puroposes is knownas AUTOMATIC PROCESS CONTROL.

  • 8/6/2019 Lecture02_10thJan2009

    7/8

    The Unsteady State

    If a machine is to be used to control the process, it isnecessary to decide in advance precisely what changesare to be made in the heat input q for every possiblesituation that might occur.

    We can not rely on the judgment of the machine as wecould on that of the operator.

    Machines do not think; they simply perform a predeterminedtask in a predetermined manner.

    To be able to make these control decisions in advance, we

    must know how the tank temp. T changes in response tochanges in Ti and q.

    This necessitates writing the unsteady-state, or transient,energy balance for the process.

    The input and output terms in this balance are the same

    as those used in steady-state balance, [Eq.(1)].

  • 8/6/2019 Lecture02_10thJan2009

    8/8

    The Unsteady State

    In addition, there is a transient accumulation of energy inthe tank, which may be written:

    Accumulation = V C dT/dt [energy units/ time]

    Accumulation = Input Output

    V C dT/dt = q [w C (T Ti)]

    V C dT/dt = w C (Ti T) + q ..(3)

    Eq.(1) is the steady-state solution of Eq. (3), obtaining by

    setting the derivative to zero.