Process Dynamics and Control, 4e
4th Edition
ISBN: 9781119285915
Author: Seborg
Publisher: WILEY
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Question
Chapter 1, Problem 1.5E
Interpretation Introduction
(a)
Interpretation:
The use of the feedback control system to accomplish the given objective is to be explained. Also, the sensor, actuator, disturbances, and appropriate control action taken is to be determined.
Concept introduction:
A control system is a system that regulates or monitors the behavior or working of other devices/systems using control loops. It can be used to control a single device or large industrial processes. There are two types of control system which can be used to control the processes. These are the feedback control system and the feedforward control system.
The feedback control system considers the output of the system and then compares it with the desired value to adjust the measured value to provide accurate results whereas in the feedforward control system, disturbances are taken into account beforehand and they are rectified to control it from deviating from the desired set point.
Interpretation Introduction
(b)
Interpretation:
The information utilized by the driver in addition to the distance between the two vehicles to avoid collision of the car with another car ahead of it is to be determined in accordance with the feedforward control
Concept introduction:
A control system is a system that regulates or monitors the behavior or working of other devices/systems using control loops. It can be used to control a single device or large industrial processes. There are two types of control system which can be used to control the processes. These are the feedback control system and the feedforward control system.
Feedback control systems consider the output of the system and then compares it with the desired value to adjust the measured value to provide accurate results whereas in the feedforward control system, disturbances are taken into account beforehand and they are rectified to control it from deviating from the desired set point.
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LATIHAN
8.5-4. Concentration of NaOH Solution in Triple-Effect Evaporator. A forced-circulation
triple-effect evaporator using forward feed is to be used to concentrate a 10 wt
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gage. The absolute pressure in the vapor space of the third effect is 6.76 kPa.
The feed rate is 13 608 kg/h. The heat-transfer coefficients are U₁ = 6246, U2
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surface area and steam consumption.
8.5-1. Boiling Points in a Triple-Effect Evaporator. A solution with a negligible boiling-
point rise is being evaporated in a triple-effect evaporator using saturated steam
at 121.1°C (394.3 K). The pressure in the vapor of the last effect is 25.6 kPa abs.
The heat-transfer coefficients are U₁ = 2840, U₂ = 1988, and U₁ = 1420
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The power generation unit in a plant uses a hot exhaust gas from another process to produce work. The gas enters at 10 bar and 350°C and exits at 1 bar and 40°C. The process produces a net amount of work equal to 4500 J/mol and it exchanges an unknown amount of heat with the surroundings. 1.1 Determine the amount of heat exchanged with the surroundings. Is this heat absorbed or rejected by the system? 1.2 Calculate the entropy change of the exhaust gas. 1.3 As a young and ambitious chemical engineer, you seek ways to improve the process. What is the maximum amount of work that you could extract from this system? Assume that the inlet and outlet conditions of the exhaust gas remain the same.
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Latihan mandiri
Reaktor fluidisasi menggunakan katalis padat dengan diameter
partikel 0,25 mm, rapat massa 1,50 g/ml, sperisitas 0,90. Pada
kondisi unggun diam, porositas 0,35, tinggi unggun 2 m. Gas
masuk dari bagian bawah reaktor pada suhu 600°C pada
viskositas 0,025 CP serta rapat massa 0,22 lb/cuft. Pada fluidisasi
minimum, porositas tercapai pada 0,45. Hitung
Hitung
a. Laju alir semu minimum (VM) gas masuk kolom fluidisasi !
b. Tinggi unggun jika Vo = 2 VM
c. Pressure drop pada kondisi Vo = 2,5 VM
<
1 m
= 3,28084 ft
1 g/ml
= 62,43 lbm/ft³
1 cp
gc
= 6,7197 × 10-4 lbm/ft.s
= 32,174 ft/s²
=
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