Process Dynamics and Control, 4e
Process Dynamics and Control, 4e
4th Edition
ISBN: 9781119285915
Author: Seborg
Publisher: WILEY
Question
Book Icon
Chapter 6, Problem 6.2E
Interpretation Introduction

(a)

Interpretation:

The given transfer function should be written in a standard constant form of time or gain.

4(s+2)(0.5s+1)(2s+1)e-5s.

Concept introduction:

A transfer function model characterizes the dynamic relationship of two process variables:

  1. A dependent variable (or output variable).
  2. An independent variable (or input variable).

Transfer function models are only directly applicable to processes that exhibit linear dynamic behavior, such as a process that can be modeled by a linear ODE. If the process is nonlinear, a transfer function can provide an approximate linear model.

Interpretation Introduction

(b)

Interpretation:

The value of gain, poles and zeroes of standard transfer function should be determined.

Concept introduction:

  • Gain: At steady state, the proportional value which shows the relationship between the magnitude of input and output signal, is known as Gain of the Transfer Function.
  • Poles: The values of (s) at which the denominator of the transfer function is equal to zero, is known as Poles of Transfer function.
  • Zeroes: The values of (s) at which the Numerator of the transfer function is equal to zero, is known as Zeroes of Transfer Function.

Interpretation Introduction

(c)

Interpretation:

The values of gain, poles and zeroes if the time delay function is replaced by (11) should be determined.

Concept introduction:

The Pade approximation for a time delay is a ratio of two polynomials in (s) with coefficients selected to match the terms of a truncated Taylor series expansion of (e-θs).

  • Gain: At steady state, the proportional value which shows the relationship between the magnitude of input and output signal, is known as Gain of the Transfer Function.
  • Poles: The values of (s) at which the denominator of the transfer function is equal to zero, is known as Poles of Transfer function.
  • Zeroes: The values of (s) at which the Numerator of the transfer function is equal to zero, is known as Zeroes of Transfer Function.

Blurred answer
Students have asked these similar questions
14.9. A forward feed double-effect vertical evaporator, with equal heating areas in each effect, is fed with 5 kg/s of a liquor of specific heat capacity of 4.18 kJ/kg K. and with no boiling point rise, so that 50 per cent of the feed liquor is evaporated. The overall heat transfer coefficient in the second effect is 75 per cent of that in the first effect. Steam is fed at 395 K and the boiling point in the second effect is 373 K. The feed is heated by an external heater to the boiling point in the first effect. It is decided to bleed off 0.25 kg/s of vapour from the vapour line to the second effect for use in another process. If the feed is still heated to the boiling point of the first effect by external means, what will be the change in steam consumption of the evaporator unit? For the purpose of calculation, the latent heat of the vapours and of the steam may both be taken as 2230 kJ/kg
Example(3): It is desired to design a double effect evaporator for concentrating a certain caustic soda solution from 12.5wt% to 40wt%. The feed at 50°C enters the first evaporator at a rate of 2500kg/h. Steam at atmospheric pressure is being used for the said purpose. The second effect is operated under 600mmHg vacuum. If the overall heat transfer coefficients of the two stages are 1952 and 1220kcal/ m2.h.°C. respectively, determine the heat transfer area of each effect. The BPR will be considered and present for the both effect 5:49
العنوان ose only Q Example (7): Determine the heating surface area 개 required for the production of 2.5kg/s of 50wt% NaOH solution from 15 wt% NaOH feed solution which entering at 100 oC to a single effect evaporator. The steam is available as saturated at 451.5K and the boiling point rise (boiling point evaluation) of 50wt% solution is 35K. the overall heat transfer coefficient is 2000 w/m²K. The pressure in the vapor space of the evaporator at atmospheric pressure. The solution has a specific heat of 4.18kJ/ kg.K. The enthalpy of vaporization under these condition is 2257kJ/kg Example (6): 5:48 An evaporator is concentrating F kg/h at 311K of a 20wt% solution of NaOH to 50wt %. The saturated steam used for heating is at 399.3K. The pressure in the vapor space of the evaporator is 13.3 KPa abs. The 5:48 1 J ۲/۱ ostr
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introduction to Chemical Engineering Thermodynami...
Chemical Engineering
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:McGraw-Hill Education
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemical Engineering
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Text book image
Elements of Chemical Reaction Engineering (5th Ed...
Chemical Engineering
ISBN:9780133887518
Author:H. Scott Fogler
Publisher:Prentice Hall
Text book image
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:9781119285915
Author:Seborg
Publisher:WILEY
Text book image
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:9781285061238
Author:Lokensgard, Erik
Publisher:Delmar Cengage Learning
Text book image
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:9780072848236
Author:Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:McGraw-Hill Companies, The