EBK INTRODUCTION TO CHEMICAL ENGINEERIN
8th Edition
ISBN: 9781259878091
Author: SMITH
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 9, Problem 9.20P
Interpretation Introduction
Interpretation:
Discuss the pros and cons of the proposal on home refrigerator in which position of condenser has changed.
Concept Introduction:
In order to find pros and cons of the proposal, we need to think over the factors like installation cost, electricity cost, surrounding effects after the installation, and for that we need to calculate its performance or efficiency. The coefficient of performance
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give me solution math not explin
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
Chapter 9 Solutions
EBK INTRODUCTION TO CHEMICAL ENGINEERIN
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Similar questions
- 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 5:49arrow_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_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_forward
- 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/Tarrow_forwardPlease, I need solution in detailsarrow_forwardplease, I need solution in detailsarrow_forward
- please, 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_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_forward
- At 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_forwardA 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_forward
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