Elements of Chemical Reaction Engineering (5th Edition) (Prentice Hall International Series in the Physical and Chemical Engineering Sciences)
5th Edition
ISBN: 9780133887518
Author: H. Scott Fogler
Publisher: Prentice Hall
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Chapter 18, Problem 18.7P
A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m3/s. The reaction is a first-order irreversible isomerization. The reactor is 3 m long, with a cross-sectional area of 25 dm2. After being built, a pulse tracer test on the reactor gave the following data: tm = 10 s and σ2 = 65 s2. What conversion can be expected in the real reactor?
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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
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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 18 Solutions
Elements of Chemical Reaction Engineering (5th Edition) (Prentice Hall International Series in the Physical and Chemical Engineering Sciences)
Ch. 18 - The second-order liquid-phase reaction AB+C is to...Ch. 18 - The elementary liquid-phase reaction...Ch. 18 - Prob. 18.5PCh. 18 - Use the data in Example 16-2 to make the following...Ch. 18 - A tubular reactor has been sized to obtain 98%...Ch. 18 - Prob. 18.11PCh. 18 - Prob. 18.13PCh. 18 - Prob. 18.14PCh. 18 - Prob. 18.15PCh. 18 - Consider the following system in Figure P18-16C...
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- 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
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