Chemical Engineering

Chemical Engineering

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Sunday, 22 April 2018

ONGC jobs for Graduate Trainees Across India. Last Date to apply: 03 May 2018

April 22, 2018 0
ONGC jobs for Graduate Trainees Across India. Last Date to apply: 03 May 2018


ONGC jobs for Graduate Trainees Across India. Last Date to apply: 03 May 2018

Advt. No. 3/2018 (R&P)Recruitment of Graduate Trainees at E-1 level in Engineering & Geo-Sciences posts through GATE-2018ONGC recruiting  1032 Vacancies for the post of Graduate Trainees 

AEE(Cementing) - Mechanical/11 Post
Mechanical Engineering (ME)
Qualification : Graduate Degree in Mechanical Engineering with minimum 60% marks

AEE(Cementing) - Petroleum/01 Post
Petroleum Engineering (PE)
Qualification : Graduate Degree in Petroleum Engineering with minimum 60% marks

AEE(Civil)/27 Post
Civil Engineering (CE)
Qualification : Graduate Degree in Civil Engineering with minimum 60% marks

AEE (Drilling) – Mechanical/129 Post
Mechanical Engineering (ME)
Qualification : Graduate Degree in Mechanical Engineering with minimum 60% marks

AEE(Drilling) - Petroleum/08 Post
Petroleum Engineering (PE)
Qualification : Graduate Degree in Petroleum Engineering with minimum 60% marks

AEE (Electrical)/127 Post
Electrical Engineering (EE)
Qualification : Graduate Degree in Electrical Engineering with minimum 60% marks Should have Certificate of competency as Electrical Supervisor

AEE (Electronics)/30 Post
Electronics and Communication Engineering (EC)
Qualification : Graduate Degree in Electronics Engineering with minimum 60% marks
Qualification : Graduate Degree in Telecom Engineering with minimum 60% marks
Qualification : Graduate Degree in E&T Engineering with minimum 60% marks
Physics (PH)
Qualification : Post Graduate Degree in Physics with Electronics with minimum 60% marks

AEE (Instrumentation)/40 Post
Instrumentation Engineering (IN)
Qualification : Graduate Degree in Instrumentation Engineering with minimum 60% marks

AEE (Mechanical)/106 Post
Mechanical Engineering (ME)
Qualification : Graduate Degree in Mechanical Engineering with minimum 60% marks

AEE (Production)- Mechanical/76 Post
Mechanical Engineering (ME)
Qualification : Graduate Degree in Mechanical Engineering with minimum 60% marks

AEE(Production)- Petroleum/46 Post
Petroleum Engineering (PE)
Qualification : Graduate Degree in Petroleum Engineering / Applied Petroleum Engineering with minimum 60% marks

AEE(Production)- Chemical/101 Post
Chemical Engineering (CH)
Qualification : Graduate Degree in Chemical Engineering with minimum 60% marks

AEE (Reservoir)/18 Post
Geology & Geophysics (Part A and Section 2 of Part -B) (GG)
Qualification : Post Graduate Degree in Geophysics with minimum 60% marks (Must have Mathematics / Physics at B.Sc. level)
Geology & Geophysics (Part A and Section 1 of Part -B) (GG)
Qualification : Post Graduate Degree in Geology (Must have Mathematics / Physics at B.Sc. level) with minimum 60% marks
Chemistry (CY)
Qualification : Post Graduate Degree in Chemistry with minimum 60% marks (Must have Mathematics / Physics at B.Sc. level)
Mathematics (MA)
Qualification : Post Graduate Degree in Mathematics with minimum 60% marks (Must have Mathematics / Physics at B.Sc. level)
Physics (PH)
Qualification : Post Graduate Degree in Physics with minimum 60% marks (Must have Mathematics / Physics at B.Sc. level)
Petroleum Engineering (PE)
Qualification : Post Graduate Degree in Petroleum Technology with minimum 60% marks (Must have Mathematics / Physics at B.Sc. level)
Chemical Engineering (CH)
Qualification : Graduate Degree in Chemical Engineering with minimum 60% marks
Petroleum Engineering (PE)
Qualification : Graduate Degree in Petroleum Engineering / Applied Petroleum Engineering with minimum 60% marks

Chemist/73 Post
Chemistry (CY)
Qualification : Post Graduate Degree in Geology with minimum 60% marks
Qualification : M.Sc. or M. Tech in Petroleum Geoscience with minimum 60% marks
Qualification : M.Sc. or M. Tech in Petroleum Geology with minimum 60% marks
Qualification : M. Tech in Geological Technology with minimum 60% marks

Geophysicist (Surface)/41 Post
Geology & Geophysics (Part A and Section 2 of Part -B) (GG)
Qualification : Post Graduate Degree in Geophysics with minimum 60% marks
Qualification : M. Tech. in Geophysical Technology with minimum 60% marks
Physics (PH)
Qualification : Post Graduate Degree in Physics with Electronics with minimum 60% mark

Geophysicist (Wells)/26 Post
Geology & Geophysics (Part A and Section 2 of Part -B) (GG)
Qualification : Post Graduate Degree in Geophysics with minimum 60% marks
Qualification : M. Tech. in Geophysical Technology with minimum 60% marks
Physics (PH)
Qualification : Post Graduate Degree in Physics with Electronics with minimum 60% marks

Materials Management Officer/49 Post
Mechanical Engineering (ME) 
Qualification : Graduate Degree in Mechanical Engineering with minimum 60% marks
Electrical Engineering (EE) 
Qualification : Graduate Degree in Electrical Engineering with minimum 60% marks
Instrumentation Engineering (IN) 
Qualification : Graduate Degree in Instrumentation Engineering with minimum 60% mark
Petroleum Engineering (PE) 
Qualification : Graduate Degree in Petroleum Engineering/Applied Petroleum Engineering with minimum 60% marks
Chemical Engineering (CH) 
Qualification : Graduate Degree in Chemical Engineering with minimum 60% marks
Electronics and Communication Engineering (EC) 
Qualification : Civil Engineering (CE) : Graduate Degree in Civil Engineering with minimum 60% marks
Qualification : Graduate Degree in Electronics Engineering with minimum 60% marks
Qualification : Graduate Degree in E&T Engineering Engineering (EC) with minimum 60% marks
Qualification : Graduate Degree in Telecom Engineering with minimum 60% marks
Computer Science & Information Technology (CS)
Qualification : Graduate Degree in Computer Engineering with minimum 60% marks
Qualification : Graduate Engineering Degree in Information Technology with minimum 60% marks

Programming Officer/16 Post
Computer Science & Information Technology (CS)
Qualification : Graduate Degree in Computer Engineering with minimum 60% marks
Qualification : Graduate Engineering Degree in Information Technology with minimum 60% marks
Qualification : Post Graduate in Computer Applications (MCA) with minimum 60% marks
Qualification :Post Graduate in Computer Science with minimum 60% marks
‘B’ level diploma as defined by Dept., of Electronics, GOI.

Transport Officer/14 Post
Mechanical Engineering (ME)
Qualification : Graduate degree in Auto Engineering with minimum 60% marks
Qualification : Graduate degree in Mechanical Engineering with minimum 60% marks

Age Limit (with age relaxation) as on 01.01.2018 : 30-35 Yrs
Fee : 1. General & OBC Rs. 299.20 /- , 2. SC , ST & PWD No Fees payable
Selection Process : 
The selection process will comprise of the following: a) GATE – 2018 : The desirous and eligible candidates having the essential qualification for the advertised posts should have appeared in Graduate Aptitude Test in Engineering (GATE) 2018 as indicated above in para 1.1.Essential Qualification & GATE Subject with Code under Eligibility Criteria and declared qualified by the GATE-2018 authorities . b) The GATE-2018 Score in the subject mentioned against the posts of ONGC will be considered by ONGC for shortlisting the candidates for the further selection process of Personal Interview as per the criteria decided by the Management.
Weightage of GATE-2018 Score : 60 Marks
Interview : 15 Mark
Qualification 25 Marks : ( 20 Marks for Essential Qualification & 05 marks for in line PhD*)
Hiring Process : Written-test

How to apply

Important Dates : 
Start Date for receiving Online Applications :16/04/2018 (From 1400 Hrs.)
Last date for receipt of Online Applications :03/05/2018 (Up to 1730 Hrs.)
Period for uploading the scanned copies of certificates and downloading the Interview Call letters :10/05/2018 to 22/05/2018(Tentative)
Starting of Interviews : 28 May,2018 (Tentative


Apply Online

Monday, 2 April 2018

Zeolite filters out carcinogens from smoke flavoring

April 02, 2018 0
Zeolite filters out carcinogens from smoke flavoring


Zeolite filters out carcinogens from smoke flavoring : -


Besides removing harmful compounds, the filtering process also improves the smoke’s flavor

Infusing foods with smoke can create delicious flavors, but the process also can introduce small amounts of carcinogenic compounds into the food. At the American Chemical Society national meeting in New Orleans on Wednesday, University of Reading’s Jane K. Parker reported that filters made from a microporous aluminosilicate material called a zeolite can reduce the levels of carcinogenic polycyclic aromatic hydrocarbons (PAHs) in smoke by as much as 93% while retaining desirable flavors.
Parker’s team used a naturally occurring type of zeolite called clinoptilolite to filter smoke from oak wood before using it to prepare food. They found that the filter worked best when they activated the zeolite with a heat treatment and ground it into a fine powder to maximize the surface area in contact with the smoke. The zeolite reduced the levels of benzo[a]pyrene, a type of PAH that the International Agency for Research on Cancer classifies as a known human carcinogen, by 93%. Concentrations of several other PAHs also dropped, many by more than 70%, said Parker at a symposium organized by the Division of Agricultural & Food Chemistry.
Removing these compounds had no adverse effect on the smoke flavor, the researchers found. A panel of 12 expert tasters evaluated chicken brined in smoked water and cream cheese mixed with smoked tomato flakes, prepared with either filtered or unfiltered smoke. The tasters described the filtered smoke flavor as more balanced, with fewer bonfire- and diesel-like flavors than foods prepared with unfiltered smoke.
The researchers used mass spectrometry to analyze the compounds in the two types of smoke and found that the zeolite filter primarily removed higher molecular weight components, which Parker said may be the chemicals responsible for the harsher flavors.
Jonathan D. Beauchamp, a food scientist at Fraunhofer Institute for Process Engineering & Packaging, who co-organized the symposium, said he thinks the work could have a broad impact on the food industry, as a zeolite-filtered smoking process could be applied to a wide variety of smoked food products.

Wednesday, 28 March 2018

Heat of Sublimation

March 28, 2018 0
Heat of Sublimation
                                                              Heat of Sublimation


Calculate the enthalpy of sublimation of Iodine from the following data:
H2(g) + I2(s) à 2 HI(g)      DHR = 51.9 kJ
H2(g) + I2(g) à 2 HI(g)      DHR = -9.2 kJ
Calculations:
H2(g) + I2(s) à 2 HI(g) -- I      DHR = 51.9 kJ
H2(g) + I2(g) à 2 HI(g) -- II      DHR = -9.2 kJ
From Hess's law of constant heat of summation,
Subtracting equation II from I gives,
I2(s) - I2(g) = 0      DHR = 51.9 - (-9.2) = 61.1 kJ
i.e.,
I2(s) à I2(g)      DHR = 61.1 kJ

That is enthalpy of sublimation (transformation of iodine from solid phase to directly vapor phase) of iodine is 61.1 kJ

Solved Problem On Flue Gas Analysis & Orsat Analysis

March 28, 2018 0
Solved Problem On Flue Gas Analysis & Orsat Analysis
                                                                                 



    Flue Gas Analysis & Orsat Analysis

Natural gas containing 80% CH4, 15% C2H6 and 5% C3H8 is burnt with 50% excess air. Assuming that 90% of the carbon in the hydrocarbons are converted to CO2 and the rest to CO, determine
  1. Flue gas analysis
  2. Orsat analysis
Calculations:
Basis: 1 mole of natural gas
ComponentMoleAtoms of CAtoms of H
CH40.800.8 x 1 = 0.80.8 x 4 = 3.2
C2H60.150.15 x 2 = 0.30.15 x 6 = 0.9
C3H80.050.05 x 3 = 0.150.05 x 8 = 0.4
Total1.01.254.5
Reactions:
C + O2 à CO2 -- I
C + 1/2 O2 à CO -- II
H + 1/4 O2 à 1/2 H2O -- III
90% of Carbon is converted by reaction I, and 10% of carbon is converted by II.
Amount of CO2 produced = 1.25 x 0.9 = 1.125 mole
Amount of CO produced = 1.25 x 0.1 = 0.125 mole
Amount of H2O produced = 4.5 / 2 = 2.25 mole
Amount of O2 used by hydrocarbon = O2 used by reactions I, II and III.
= 1.125 + 0.125 x (1/2) + 4.5 x (1/4) = 2.3125 mole
Theoretical O2 needed = Oxygen for complete conversion of C to CO2 and H to H2O.
= 1.25 + 4.5 x (1/4) = 2.375 mole
Oxygen entering = 150% of theoretical = 1.5 x 2.375 = 3.5625 mole
Therefore, nitrogen entering = 3.5625 x 79/21 = 13.4018 mole = N2 in the flue gas
O2 in the flue gases = O2 entering - O2 used = 3.5625 - 2.3125 = 1.25 mole
Flue gas analysis:
ComponentMolesMole %
CO21.1256.20
CO0.1250.69
H2O2.2512.39
O21.256.89
N213.401873.83
Total18.1518100
Orsat analysis (Water free):
ComponentMolesMole %
CO21.1257.07
CO0.1250.79
O21.257.86
N213.401884.28
Total15.9018100


Balance of Individual Reactions:
Basis: 1 mole of Natural gas
CH40.80
C2H60.15
C3H80.05
Total1.0
CH4 + 2 O2 à CO2 + 2 H2O -- 1
CH4 + 3/2 O2 à CO + 2 H2O -- 2
C2H6 + 7/2 O2 à 2 CO2 + 3 H2O -- 3
C2H6 + 5/2 O2 à 2 CO + 3 H2O -- 4
C3H8 + 5 O2 à 3 CO2 + 4 H2O -- 5
C3H8 + 7/2 O2 à 3 CO + 4 H2O -- 6
In the above reactions CO2 is produced from reactions 1, 3 and 5.
Since 90% of Carbon is converted to CO2 and 10% to CO,
CO2 produced = (1 x 0.8 + 2 x 0.15 + 3 x 0.05) x 0.9 = 1.125 mole
Similarly CO is obtained from reactions 2, 4 and 6.
CO produced = (1 x 0.8 + 2 x 0.15 + 3 x 0.05) x 0.1 = 0.125 mole
H2O produced = (2 x 0.8 + 3 x 0.15 + 4 x 0.05) x 0.9 + (2 x 0.8 + 3 x 0.15 + 4 x 0.05) x 0.1 = 2.25 mole
O2 used up in these reactions = (2 x 0.8 + 3.5 x 0.15 + 5 x 0.05) x 0.9 + (1.5 x 0.8 + 2.5 x 0.15 + 3.5 x 0.05) x 0.1
= 2.3125 mole
Theoretical O2 needed = moles of O2 needed for Conversion of C to CO2 and H to H2O.
= 2 x 0.8 + 3.5 x 0.15 + 5 x 0.05 = 2.375 mole
O2 entering = 50 % excess = 150% of theoretical = 2.375 x 1.5 = 3.5625 mole
N2 entering along with O2 in the air = 3.5625 x 79/21 = 13.4018 mole (sine air is 21% O2 and 79% N2 by volume).
O2 in the flue gas = O2 entering - O2 used up = 3.5625 - 2.3125 = 1.25 mole
N2 in the flue gas = N2 entering = 13.4018 mole
Therefore, for 1 mole of Natural gas entering, the flue gas coming out are:
ComponentMoles
CO21.125
CO0.125
H2O2.25
O21.25
N213.4018
Total18.1518

On comparing with the data obtained from atomic balance and balance of individual reactions, it can be seen that the results are same from either method. But instead of lengthy calculations for individual reactions, we can very well make use of atomic balances.

Solved Problem On Recycle Reactor With Separators

March 28, 2018 0
Solved Problem On Recycle Reactor With Separators


Methanol vapor can be converted into formaldehyde by the following reaction scheme:
CH3OH + 0.5 O2 à HCHO + H2O
CH3OH à HCHO + H2
The fresh feed to the process was 0.5 kmol / hr of O2 and an excess methanol. All of the O2 reacts in the reactor. Formaldehyde and water are removed from the product stream first, after which H2 is removed from the recycled methanol. The recycle flow rate of methanol was 1 kmol/hr. The ratio of methanol reacting by decomposition to that by oxidation was 3. Draw the flow diagram and then calculate the per pass conversion of methanol in the reactor and the fresh feed rate of methanol.

Calculations:
The flow diagram for the above process is as given below:
From the problem statements,
  1. all of the O2 entering, reacts in the reactor and
  2. the ratio of methanol reacting by decomposition to that by oxidation = 3.
The number of moles taking parts in the reactions are given below:
CH3OH + 0.5 O2 à HCHO + H2O à 1
1              0.5              1         1
CH3OH à HCHO + H2 à 2
1              1          1
from reaction 1,
0.5 mole of O2 h 1 mole of CH3OH
Therefore, for complete conversion of oxygen, 0.5 mole/hr of O2 needs 1 mole/hr of CH3OH.
And from the ratio of methanol conversion, methanol reacting by decomposition = 3 times that by oxidation = 3 mole/hr of CH3OH.
Therefore,
Total methanol entering the reactor = Methanol reacted by oxidation + methanol reacted by decomposition + recycle methanol = 1 + 3 + 1 = 5 kmol/hr.
Conversion per pass = 100 x (Total methanol reacted /Total methanol entering the reactor)
= 100 x 4/5 = 80%

Fresh methanol rate = Total methanol entering the reactor - recycle methanol = 5 - 1 = 4 kmol/hr.

Solved Problem On Humidity

March 28, 2018 0
Solved Problem On Humidity
       
                      
                                  Humidity

Air at 30oC and 150 kPa in a closed container is compressed and cooled. It is found that the first droplet of water condenses at 200 kPa and 15oC. Calculate the percent relative humidity of the original air. The vapor pressures of water at 15oC and 30oC are 1.7051 kPa and 4.246 kPa respectively.
Calculations:
At 15oC, and 200 kPa water is at 100% humidity.
i.e., Percentage humidity or percentage absolute humidity = 100 x [pA(p - pS)]/[ pS(p - pA)] = 100
Therefore,
pA/(p - pA) = pS/(p - pS)
where pA = partial pressure of water vapor;
pS = vapor pressure of water vapor
p = total pressure of system
i.e., no of moles of water vapor per mole of dry air = 1.7051/(200 - 1.7051) = 0.0086
This ratio (moles of water vapor / mole of dry air) is not going to change for a closed system. Therefore, partial pressure of A at 30oC and 150 kPa is found as follows:
pA/(p - pA) = 0.0086
pA/(150 - pA) = 0.0086
pA + 0.0086 pA = 1.29
pA = 1.29/1.0086 = 1.279 kPa
Percentage relative humidity of the original air:

percentage relative humidity = 100 x (pA/pS) = 100 x 1.279/4.246 = 30.12%

Solved Problem On Recycle Reactor

March 28, 2018 0
Solved Problem On Recycle Reactor


                                                        Recycle Reactor

The reaction A à 2B + C takes place in a catalytic reactor (diagram is given below). The reactor effluent is sent to a separator. The overall conversion of A is 95%. The product stream from the separator consists of B, C and 0.5% of A entering the separator, while the recycle stream consists of the remainder of the unreacted A and 1% of B entering the separator. Calculate the
  1. single pass conversion of A in the reactor
  2. molar ratio of recycle to feed. 
 

Calculations:
Basis: 1 mole of pure A in the feed
From the reaction, A à 2B + C
1 mole of A produces 2 moles of B and 1 mole of C for complete conversion.
From the problem statement, overall conversion is 95%. Therefore, 1 mole of feed A will produce
2 x 0.95 mole of B ,
1 x 0.95 mole of C and the
unreacted A is 0.05 mole.
i.e.,
Product stream contains:
A : 0.05 mole
B : 1.9 mole
C : 0.95 mole
Material Balance for the compound B:
From the problem statement, B in the recycle stream = 1% of B entering the separator.
Therefore, B in the product stream = 99% of B entering the separator = 1.9 mole
Total B entering the separator = 1.9/0.99 = 1.919 mole
And, amount of B in the recycle stream = 1.919 x 0.01 = 0.019 mole.
Material Balance for the compound A:
From the problem statement, A in the product stream = 0.5% of A entering the separator.
i.e., A in the product stream = 0.5% of A entering the separator = 0.05 mole.
Therefore, amount of A entering the separator = 0.05/0.005 = 10 mole.
And, the amount of A in the recycle stream
= Amount of A entering the separator - Amount of A in the product stream
= 10 - 0.05 = 9.95 mole.
Amount of A entering the reactor = fresh A + recycle A = 1 + 9.95 = 10.95 mole.
Amount of A converted in the reactor = moles of A entering the reactor - moles of A entering the separator = 10.95 - 10 = 0.95 mole
Single pass conversion of A in the reactor
= 100 x Amount of A converted in the reactor / Amount of A entering the reactor
= 100 x (0.95/10.95) = 8.676%
Total moles of recycle stream = moles of A in recycle stream + moles of B in recycle stream
= 9.95 + 0.019 = 9.969 mole

Molar ratio of recycle to feed = 9.969/1 = 9.969