PROCESS DYNAMIC+CONTROL-EBOOK>I<
4th Edition
ISBN: 2819480255712
Author: Seborg
Publisher: INTER WILE
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 1, Problem 1.6E
Interpretation Introduction
(a)
Interpretation:
The ways by which body temperature is regulated by the body using feedback control should be explained.
Concept introduction:
The feedback control system observed in two types such as:
- Positive control system: The output values and setpoints are added into this type of system.
- Negative control system: In this system, the output values and setpoints are subtracted.
The positive control system is less stable than a negative control system.
Interpretation Introduction
(b)
Interpretation:
The feedback control system for the regulation of another physiological variable should be discussed.
Concept introduction:
The feedback control system observed in two types such as
- Positive control system: The output values and setpoints are added into this type of system.
- Negative control system: In this system, the output values and setpoints are subtracted.
The positive control system is less stable than the Negative control system.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Q1: Consider the following transfer function
G(s)
5e-s
15s +1
1. What is the study state gain
2. What is the time constant
3. What is the value of the output at the end if the input is a unit step
4. What is the output value if the input is an impulse function with amplitude equals
to 3, at t=7
5. When the output will be 3.5 if the input is a unit step
give me solution math not explin
give me solution math not explin
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemical-engineering and related others by exploring similar questions and additional content below.Similar questions
- give me solution math not explinarrow_forwardgive me solution math not explinarrow_forwardExample (6): 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 O Transcribed Image Text: Example (7): Determine thearrow_forward
- 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/kgarrow_forwardExample(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 5:49arrow_forwardالعنوان 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 ostrarrow_forward
- Example 8: 900 Kg dry solid per hour is dried in a counter current continues dryer from 0.4 to 0.04 Kg H20/Kg wet solid moisture content. The wet solid enters the dryer at 25 °C and leaves at 55 °C. Fresh air at 25 °C and 0.01Kg vapor/Kg dry air is mixed with a part of the moist air leaving the dryer and heated to a temperature of 130 °C in a finned air heater and enters the dryer with 0.025 Kg/Kg alry air. Air leaving the dryer at 85 °C and have a humidity 0.055 Kg vaper/Kg dry air. At equilibrium the wet solid weight is 908 Kg solid per hour. *=0.0088 Calculate:- Heat loss from the dryer and the rate of fresh air. Take the specific heat of the solid and moisture are 980 and 4.18J/Kg.K respectively, A. =2500 KJ/Kg. Humid heat at 0.01 Kg vap/Kg dry=1.0238 KJ/Kg. "C. Humid heat at 0.055 Kg/Kg 1.1084 KJ/Kg. "C 5:42 Oarrow_forwardQ1: From the Figure below for (=0.2 find the following 1. Rise Time 2. Time of oscillation 3. Overshoot value 4. Maximum value 5. When 1.2 which case will be? 1.6 1.4 1.2 12 1.0 |=0.8- 0.6 0.4 0.8 0.2- 0.6 0.4 0.2 1.2 = 1.0 0 2 4 6 8 10 10 t/Tarrow_forwardPlease, I need solution in detailsarrow_forward
- please, I need solution in detailsarrow_forwardplease, I need solution in detailsarrow_forwardA system, in a closed container, consists of an unknown number of components and three phases. You are told that the system is fully defined by giving you only one mole fraction! What is the number components that is present? 3 1 2 The question is ill-posed.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Introduction to Chemical Engineering Thermodynami...Chemical EngineeringISBN:9781259696527Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark SwihartPublisher:McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind...Chemical EngineeringISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY
Elements of Chemical Reaction Engineering (5th Ed...Chemical EngineeringISBN:9780133887518Author:H. Scott FoglerPublisher:Prentice Hall
Industrial Plastics: Theory and ApplicationsChemical EngineeringISBN:9781285061238Author:Lokensgard, ErikPublisher:Delmar Cengage Learning
Unit Operations of Chemical EngineeringChemical EngineeringISBN:9780072848236Author:Warren McCabe, Julian C. Smith, Peter HarriottPublisher:McGraw-Hill Companies, The
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
Elementary Principles of Chemical Processes, Bind...
Chemical Engineering
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Elements of Chemical Reaction Engineering (5th Ed...
Chemical Engineering
ISBN:9780133887518
Author:H. Scott Fogler
Publisher:Prentice Hall
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:9781285061238
Author:Lokensgard, Erik
Publisher:Delmar Cengage Learning
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:9780072848236
Author:Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:McGraw-Hill Companies, The