Loose Leaf For Introduction To Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259878084
Author: Smith Termodinamica En Ingenieria Quimica, J.m.; Van Ness, Hendrick C; Abbott, Michael; Swihart, Mark
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Question
Chapter 2, Problem 2.12P
(a)
Interpretation Introduction
Interpretation:
To determine the change in internal energy △Ut in kJ.
Concept Introduction:
Internal Energy - The energy which is present within the system itself. It eliminates kinetic energy causes due to motion of the system and the potential energy of the system.
Formulae for the internal energy
(b)
Interpretation Introduction
Interpretation:
To determine change in elevation △z, in meters
Concept Introduction:
Potential energy of any object is the energy which causes due to position of the object.
Potential Energy is given by
△Z= change in elevation
g= acceleration due to gravity
And
(c)
Interpretation Introduction
Interpretation:
To determine change in velocity v in m/s
Concept Introduction:
The energy that comes through the movement of an object is kinetic energy.
Kinetic Energy is given by
Where
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Example (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 the
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
Chapter 2 Solutions
Loose Leaf For Introduction To Chemical Engineering Thermodynamics
Ch. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10P
Ch. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26PCh. 2 - Prob. 2.27PCh. 2 - Prob. 2.28PCh. 2 - Prob. 2.29PCh. 2 - Prob. 2.30PCh. 2 - Prob. 2.31PCh. 2 - In the following take Cv=20.8 and Cp=29.1Jmol1C1...Ch. 2 - Prob. 2.33PCh. 2 - Prob. 2.34PCh. 2 - Prob. 2.35PCh. 2 - Prob. 2.36PCh. 2 - Prob. 2.37PCh. 2 - Prob. 2.38PCh. 2 - Prob. 2.39PCh. 2 - Prob. 2.40PCh. 2 - Prob. 2.41PCh. 2 - Prob. 2.42PCh. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Prob. 2.45PCh. 2 - Prob. 2.46PCh. 2 - Prob. 2.47PCh. 2 - Prob. 2.48P
Knowledge Booster
Similar questions
- 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 ostrarrow_forwardExample 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_forward
- A 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_forwardA mixture of 2 components in 2 phases are present. You are given the temperature and mole fraction. How many additional variables can be specified before the system is completely determined? none 2 the system is overspecified 1 3arrow_forwardAt a Pressure of 600 mm Hg, match the substance with the boiling temperature. 54.69°C 1. n-Pentane 49.34°C 2. n-Hexane 3. Acetone 29.32°C く 61.40°C 4. Chloroformarrow_forward
- A mixture of oil and gas flows through a horizontal pipe with an inside diameter of 150 mm. The respective volumetric flow rates for the oil and gas are 0.015 and 0.29 m³s-1. Determine the gas void frac- tion and the average velocities of the oil and gas. The friction factor may be assumed to be 0.0045. The gas has a density of 2.4 kgm³ and viscosity of 1 x 10-5 Nsm-2. The oil has a density of 810 kgm³ and density of 0.82 Nsm². Answer: 0.79, 20.8 ms-1, 4 ms-1arrow_forward4. An experimental test rig is used to examine two-phase flow regimes in horizontal pipelines. A particular experiment involved uses air and water at a temperature of 25°C, which flow through a horizontal glass tube with an internal diameter of 25.4 mm and a length of 40 m. Water is admitted at a controlled rate of 0.026 kgs at one end and air at a rate of 5 x 104 kgs in the same direction. The density of water is 1000 kgm³, and the density of air is 1.2 kgm3. Determine the mass flow rate, the mean density, gas void fraction, and the superficial velocities of the air and water. Answer: 0.02605 kgs 1, 61.1 kgm³, 0.94, 0.822 ms-1, 0.051 ms-1arrow_forward1. Determine the range of mean density of a mixture of air in a 50:50 oil-water liquid phase across a range of gas void fractions. The den- sity of oil is 900 kgm³, water is 1000 kgm³, and gas is 10 kgm³. 2. Describe, with the use of sketches, the various flow regimes that can exist in a vertical pipe carrying two-phase flow (liquid and gas).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