Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.6P
Interpretation Introduction
Interpretation:
The partial molar pressure and partial molar temperature has to be explained. The result should be expressed in relation to the T and P of the mixture.
Concept introduction:
In order to understand partial molar temperature or pressure, it is important to understand the concept of partial molar properties. In a solution, more than two components are present which are mixed homogeneously and thus, the properties of the solutions are non-additive. The partial molar properties are the set of properties such as Vi, Hi, Ui,Si etc. which are possessed by one mole of “i” component in the solution at particular temperature and pressure.
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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
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Q1: 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/T
Please, I need solution in details
Chapter 10 Solutions
Introduction to Chemical Engineering Thermodynamics
Ch. 10 - Prob. 10.1PCh. 10 - Prob. 10.2PCh. 10 - Prob. 10.3PCh. 10 - What is the ideal work for the separation of an...Ch. 10 - Prob. 10.5PCh. 10 - Prob. 10.6PCh. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Prob. 10.10P
Ch. 10 - Prob. 10.11PCh. 10 - Prob. 10.12PCh. 10 - Prob. 10.13PCh. 10 - For a particular binary liquid solution at...Ch. 10 - Prob. 10.15PCh. 10 - From the following compressibility-factor data for...Ch. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - Estimate the fugacity of one of the following...Ch. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Rationalize the following expression, valid at...Ch. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - The following expressions have been proposed for...Ch. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Prob. 10.40PCh. 10 - Prob. 10.41PCh. 10 - Prob. 10.42PCh. 10 - Prob. 10.43PCh. 10 - Prob. 10.44P
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