17-06-2020

GP-CPC – PGP 01

GATE Questions from Previous Papers

2011-2020

Prof.G.Prabhakar

Chem Engg, SVU

GP-CPC-PGP

2020 - A feed stream containing pure species L flows into a reactor, where L is partly converted to M as shown in the figure.

The mass flow rate of the recycle stream is 20% of that of the product stream. The overall conversion of L (based on mass units) in the process is 30%. Assuming steady state operation , the one- pass conversion of L (based on mass units) through the reactor is(a) 34.2%(b) 30% (c) 26.3% (d) 23.8%

Fraction Conversion = Converted / Fed

Overall Conversion is based on Net quantities. Once through Conversion is based on Gross quantities

If 1 kg of Fresh feed stream enters, 1 kg of product stream comes out, out of which 30 % is converted and the balance

70 % is unconverted L.

Recycle Stream = 20 % of the product stream = 0.2 kg

L in the Recycle Stream = L in the Product Stream = 0.7 ( mass fraction) = L in the Reactor Discharge ( as no separation indicated)

Pure Species L is taken as feed stream and is mixed with recycle stream

Mass of L in the Gross in ( Fresh + Recycle) = 1 (1) + 0.2(0.7) = 1.14 kg (Fed to the Reactor)

Mass of L in the Gross Out ( Product + Recycle) = 1.2 ( 0.7) = 0.84 kg

Mass of L converted in the reaction = 1.14 – 0.84 = 0.3 kg (Converted)

Once through Conversion = 0.3 / 1.14 = 0.263 or 26.3 %

Use of Recycle - It improves the performance of the event. Here, the conversion is improved.

GP-CPC-PGP

2020 - An irreversible gas phase reaction 2P → 4Q + R is conducted in an isothermal and isobaric batch reactor. Assume

ideal gas behaviour. The feed is an equimolar mixture of the reactant P and an inert gas. After complete conversion of P, the

fractional change in volume is (round off to 2 decimal places).

Ideal Gas Law P Vt = n R T

Isothermal ( Constant T) Isobaric ( Constant P) So, V is proportional to n

2 moles of reactants, on complete conversion, yield 5 moles of products ( 4 moles of Q and 1 mole of R )

1 mole of feed . 0.5 moles of P and 0.5 moles of inert

O.5 moles of P give rise to 0.5 * 5/2 = 1.25 moles of Q & R

Final moles = 0.5 (inert) + 1.25 ( Q & R) = 1.75

Fractional change in moles = (1.75 - 1) / 1 = 0.75

Fractional change in volume = (1.75 - 1) / 1 = 0.75

Fraction Change in volume is calculated in a similar way, in case of variable volume reacting systems ,

assuming linear variation

GP-CPC-PGP

2020- A tank initially contains a gas mixture with 21 mol % oxygen and 79 mol % nitrogen. Pure nitrogen enters the tank, and a gas mixture of nitrogen and oxygen exits the tank. The molar flow rate of both the inlet and exit streams is 8 mol/s .

In addition, use the following data and assumptions

• Assume the tank contents to be well mixed

• Assume ideal gas behaviour.

• The temperature and pressure inside the tank are held constant.

• Molar density of the gas mixture in the tank is constant at 40 mol/ m3

If the volume of the tank is 20 m3 , then the time (in seconds) required for oxygen content in the tank to decrease to 1 mol % is

1. 100.45 2. 304.45 3. 10 4. 3.445

Unsteady State Mass Balance, No reaction

Total moles at any time in the tank = 40 mol / m3 * 20 m3 = 800 moles

Change in Accumulation = Inflow – Outflow

Concentration of Oxygen changes wrt time, So is that of nitrogen.

Oxygen balance : d ( 800 C) = 8 (0) - 8 ( C)

dt

Well mixed tank. Therefore, leaving stream composition is the same as the concentration of the tank contents.

d ( 800 C) / 8C = - dt

At t = 0, C = 0.21 At t = t , C = 0.01

On integration, 100 ln (0.0.1/ 0.21) = t t = 304.45 sec

GP-CPC-PGP

2019 - 100 kg of feed containing 50 wt% of a solute C is contacted with 80 kg of a solvent containing 0.5 wt% of C in a

mixer-settler unit. From this operation, the resultant extract and raffinate phases contain 40 wt% and 20 wt% of C ,

respectively. If E and R donate the mass of the extract and raffinate phases, respectively, the ratio E/R is

a) 1/4 (b) 2/3 (c) 1/2 (d) 1

Mass Balance without Reaction , Extraction, All components affected, Simultaneous Equations

3 balance equations possible.

Overall : F + S = M = E + R 100 + 80 = E + R

Solute C : F Xf + S Ys = E YE + R Xr 100 ( 0.5) + 80 ( 0.005) = E (0.4) + R ( 0.2)

100 ( 0.5) + 80 ( 0.005) = E (0.4) + (180 - E) ( 0.2)

50 + 0.4 = (0.4 – 0.2) E + 36

E = 14.4 / 0.2 = 72

R = 180 – 72 = 108

E / R = 72 / 108 = 2 / 3

GP-CPC-PGP

2018 - The reactant (M) is converted into product(N) in the presence of catalyst in a fixed bed reactor.

All the flow rates (F,G,H,P and R) in mol/s, and mole fraction of reactant(M) in these streams (XF,XG,XH, and

XR) are shown in figure

Fraction Conversion = Converted / Fed Overall fraction conversion is based on net quantities

Moles M in , fresh feed = F *XF

Moles M out, net product = P * XF

Moles M converted + F * XF - P * XP

Overalll fraction conversion = (F * XF) - (P * XP)

( F *XF)

GP-CPC-PGP

2018 - The ammonia (NH3 ) oxidation process occur over a catalyst as 4NH3+ 5O2 →6H2O +4N0 . Air is

supplied such that oxygen(O2) is 20% in excess of that required for complete conversion of NH3. The

mole fraction of O2 in the inlet gas mixture (NH3 + AIR) is ________

Reaction – Stoichiometry - Excess

Basis : 1 kmol ammonia

Oxygen theoretically required = 1 * 5/4 = 1.25 kmol

Oxygen supplied in 20 % excess and as air

Oxygen supplied = 1.25 * 1.2 = 1.5 kmol

Nitrogen in, with oxygen = 1.5 * 79 /21 = 5..643 kmol

Total moles of mixture, ( NH3 + Air) = 1 + 1.5 + 5.643 = 8.143 kmol

Mole fraction of Oxygen = Moles of Oxygen / Moles total

= 1.5 / 8.143 = 0.184

GP-CPC-PGP

2017 - An aqueous salt solution enters a crystallizer operating at steady state at 25o C. The feed temperature is 90o C and

the salt concentration in the feed is 40 weight %. The salt crystallizer as a pentahydrate. The crystals and the mother liquor leave

the crystallizer. The molecular weight of the anhydrous salt is 135. The solubility of the salt at 25oC is 20 weight%.

The feed flow rate required for a production rate of 100 kg/s of the hydrated salt, rounded to the nearest integer, is __kg/s.

Mass Balance without reaction - All components affected - Simultaneous equations

Feed Solution F : Weight percent of salt = 40 Weight fraction , XF = 0.4

Crystals C : Salt. 5 H20 Weight percent on of salt , XC = 135 / (135 + 90) = 0.6

Mother Liquor M : Will always be saturated, after crystallization

Solubility = 20 wt % Weight fraction of salt, XM = 0.2

Balance Equations

Basis : 100 kg / s of hydrated salt

Overall : F = C + M F = 100 + M

Salt : F *XF = C * XC + M * XM F (0.4) = 100 ( 0.6) + (F - C) ( 0.2)

0.2 F = 60 - 100 (0.2) = 40

F = 40 / 0.2 = 200 kg/s

GP-CPC-PGP

2017- Reaction A→B is carried out in a reactor operating at a steady state and 1 mol/s of pure A at 425° c enters the reactor. The

outlet stream leaves the reactor at 325° c. The heat input to the reactor is 17KW. The heat of the reaction at the reference

temperature of 25°c is 30Kj/mol. The specific heat capacities (in KJ/mol.k) of A and B are 0.1 and 0.15 respectively. The molar

flowrates of B leaving the reactor, rounded to two decimal places, is ____ mol/s.

Energy Balance - Incomplete reaction Basis : 1 mol/s of pure A

Let x be the moles of A converted /s

Energy in, with entering stream = 1 mol /s * 0.1 kJ /mol-k * (425-25) = 40 kJ /s

Heat input to the reactor = 17 kJ /s

Total Energy in = 40 + 17 = 57 kJ/s

Energy out , with leaving material = [x mol of B /s * 0.15 + (1-x) mol of A /s * 0.1] (325-25) = 30 - 0.05x kJ/s

Energy effect of reaction = 30 * x kJ/s Endothermic – Output term

Total Energy out = 30 – 0.05x + 30x = 30-29.95 x kJ/s

Equating 30 – 29.95 x = 57

x = 27 / 29.95 = 0.90 moles of A are converted

Molar flow rate of B = 0.9015 mol /s = 0.90 mol /s

GP-CPC-PGP

2016 - A liquid mixture of ethanol and water is flowing as inlet

stream P into a stream splitter. It is split into two streams, Q and R,

as shown in the figure below.

Q

P

R

The flowrate of P, containing 30 mass% of ethanol, is 100 kg/h.

What is the least number of additional specification(s) required to

determine the mass flowrates and compositions (mass%) of the two

exit streams?

1. 0 2. 1 3. 2 4. 3

Stream splitter merely splits a flow in to two or more parts and does not cause any change in the composition.

Composition of Q and R are the same as that of P = 30 mass %Mass flow rate of either P or Q is needed to find the flowrate. Ans : 1

GP-CPC-PGP

2016 - A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K the vapor pressures of pure

benzene and pure toluene are 92 kPa and 35 kPa, respectively. The mixture follows Raoult’s law. The equilibrium

vapor phase mole fraction (rounded off to the second decimal place) of benzene in contact with this liquid mixture

at 350 K is _______.

Raoult’s Law - Binary System

Benzene : Partial Pressure = 0.2 * 92 = 18.4 kPa

Toluene : Partial Pressure = 0.8 * 35 = 28 kPa

Total Pressure = 18.4 + 28 = 46.4 kPa

Vapor mole fraction of benzene = 18.4 / 46.4 = 0.3965 = 0.40

GP-CPC-PGP

2016 - A catalytic reforming plant produces hydrogen and benzene from cyclohexane by de-hydro aromatization. In

order to increase the production of hydrogen, the owner plans to change the process to steam reforming of the same

feedstock that produces hydrogen and carbon dioxide. Stoichiometrically, what is the maximum ratio of pure hydrogen

produced in the proposed process to that in the existing process?

(A) 1 (B) 2 (C) 5 (D) 6

Stoichiometry, Maximum ( complete conversion)

Existing : C6H12 C6H6 + 3 H2 3 mol of hydrogen per mole of cyclohexane

Proposed : C6H12 + 12 H2O 6 CO2 + 18 H2 18 mol of hydrogen per mole of cyclohexane

Ratio of hydrogen produced, proposed to existing = 18 / 3 = 6

GP-CPC-PGP

2015.. The schematic diagram of a steady state process is shown below. The fresh feed (F) to the reactor consists of 96

mol% reactant A and 4 mol% inert I. The stoichiometry of the reaction is A → C. A part of the reactor effluent is recycled.

The molar flow rate of the recycle stream is 0.3 F. The product stream P contains 50 mol% C. The percentage conversion of

A in the reactor based on A entering the reactor at point 1 in the figure (up to one decimal place) is _____.

Recycle - One pass Conversion – based on Gross quantities - % Conversion = 100 * Converted / Fed

No change in the number of moles as per the stoichiometry. 1 mol leaves as stream P ( 50 % C, 46 % A & 4 % inert)

Basis : 1 mol of fresh feed - 96 % A and 4 % inert ; 0.3 F = 0.3 mol join as recycle

A entering as fresh feed = 0.96 moles ; A in the recycle stream = 0..3 * 0.46 = 0.138 moles

A in the Gross Feed ( point 1) = 0.96 + 0.138 = 1.098 moles

A in the discharge from the reactor = 1.3 * 0.46 = 0.598 moles

A converted = A in – A out = 1.098 - 0.598 = 0.5 mol

Percentage A converted = 100 * 0.5 / 1. 098 = 45. 5 % Note overall at (100 * 0.5 / 0.96 = 52) is higher

GP-CPC-PGP

2014 - Two elemental reactions (A and B) are reacting to form a liquid(C) in a steady state process as per the

reaction A + B → C. . The single-pass conversion of the reaction is only 20% and hence recycle is used. The product is

separated completely in pure form. The fresh feed has 49 mol% of A and B each along with 2 mol% impurities. The maximum

allowable impurities in the recycle stream is 20 mol%. The amount of purge stream (in moles) per 100 moles of the fresh feed

is___

Purge Stream – Mass Balance

100 mol gross feed 20 moles impurities 80 mol of (A +B)

20 % of (A + B) converted

Recycle Stream after separation of C = 64 moles of (A+R) and 20 moles of impurities

Mole fraction of Impurities = 20 / 84 = 0.2381 Mole fraction of (A+B) = 64 / 84 = 0.7619

Around Recycle point, overall: F + R = 100

(A+B) balance: F * 0.98 + R * 0.7619 = 80

F = 80 – 76.19 = 17.47 moles

0.98 - .0.7619

Impurities entering through the fresh feed = Impurities leaving through the fresh feed

17.47 * (2/98) = P ( 0.2381) P = 1.497

For 100 moles of fresh feed, 1.497 * 100 / 17.47 = 8.57 mol

GP-CPC-PGP

2014 - Two elemental reactions (A and B) are reacting to form a liquid(C) in a steady state process as per the

reaction A + B → C. . The single-pass conversion of the reaction is only 20% and hence recycle is used. The product is

separated completely in pure form. The fresh feed has 49 mol% of A and B each along with 2 mol% impurities. The maximum

allowable impurities in the recycle stream is 20 mol%. The amount of purge stream (in moles) per 100 moles of the fresh feed

is___

Purge Stream – Mass Balance

100 mol gross feed 20 moles impurities 80 mol of (A +B)

20 % of (A + B) converted

Recycle Stream after separation of C = 64 moles of (A+R) and 20 moles of impurities

Mole fraction of Impurities = 20 / 84 = 0.2381 Mole fraction of (A+B) = 64 / 84 = 0.7619

Around Recycle point, overall: F + R = 100

(A+B) balance: F * 0.98 + R * 0.7619 = 80

F = 80 – 76.19 = 17.47 moles of ( A + B)

0.98 - .0.7619

Impurities entering through the fresh feed = Impurities leaving through the purge stream

17.47 * (2/98) = P ( 0.2381) P = 1.497

For 100 moles of fresh feed, 1.497 * 100 / 17.47 = 8.57 mol

GP-CPC-PGP

2014 - Carbon monoxide (CO ) is burnt in the presence of 200% excess pure oxygen and the flame temperature achieved

is 2298 K. The inlet streams are at 25°C. The standard heat of formation (at 25°C) of CO and CO2 are (-) 110 kJ / mol and

(– ) 390 KJ / mol , respectively. The heat capacities (in J / mol K) of the components are

Cp O2 = 25 + (14 * 10-3 T), CP, CO2 = 25 + (42 * 10-3 T)

where T is in K . The heat loss (in kJ) per mole of CO burnt is ………..

Energy Balance – Heat Loss - CO + 0.5 O2 = CO2 Assume Complete combustion

Basis : 1 mol Carbon monoxide,

0.5 mol oxygen theoretically required. Oxygen actually supplied, @ 200 % Excess = 0.5 * 300/100 = 1.5 mol

After the reaction, 1 mol CO2 ( from 1 mol CO) and 1 mol oxygen ( excess ) remain.

Energy in, with entering stream ( both CO and O2) = 0 ( inlet temperature = 25 O C)

Energy in , reaction effect = 1 mol * ( 390 – 110) kJ / mol = 280 kJ ( exothermic, input entry)

Heat in , total = 0 + 280 = 280 kJ

TMEAN = (2298 + 298) / 2 = 1298

Energy Out , CO2 : 1 mol * [25 + (0.042 * 1298)] J/mol-K * ( 2298-298) K = 159032 J = 159.032 kJ

O2 : 1 mol * [25 + (0.014 * 1298)] J/mol-K * ( 2298-298) K = 86344 J = 86.344 kJ

Heat loss is an output term. It is the difference between Total heat in and heat out( excepting heat loss)

Heat loss = 280 – (159.032 + 86.344) = 34.624 kJ per mole of CO burnt

GP-CPC-PGP

2013 - Calculate the heat required (in kJ, upto 1 digit after the decimal point) to raise the temperature of 1 mole of a solid

material from 100 O C to 1000 O C. The specific heat (Cp) of the material ( in J/mol-K) is expressed as Cp = 20 + 0.005 T ,

Where T is in K. Assume no phase change.

Thermophysics – Sensible Energy

Q = m Cp ∆T

Cp is given as a function of T. Hence, mean is to be calculated

T1 = 100 OC = 373 K & T2 = 1000 OC = 1273 K , TMEAN = ( 373 + 1273) /2 = 823 K

Heat required = 1 mol * (20 + 0.005 * 823) J/mol-K * (1273- 373) K = 19846.64 J = 19.8 kJ

GP-CPC-PGP

2013 - A reverse osmosis unit treats feed water (F) containing fluoride and its output consists of a permeate stream and a

reject stream (R). Let CF, CP, and CR denote the fluoride concentrations in the feed, permeate, and reject streams,

respectively. Under steady state conditions, the volumetric flow rate of the reject is 60 % of the volumetric flow rate of

the inlet stream, and CF = 2 mg/L and CP = 0.1 mg/L.

The value of CR in mg/L, up to one digit after the decimal point, is

Mass Balance with out reaction – All components affected - Simultaneous equations

Assume constant density. Mass / density = volume. Justified in writing volume ( instead of Mass)

Overall : Feed F = Permeate P + Reject R F = P + 0.6 F P = 0.4 F

Fluoride : F * CF = P * CP + R * CR

1 * 2 = 0.4 * 0.1 + 0.6 CR CR = 3.267 = 3.27 mg/L

GP-CPC-PGP

2013 – Common Data Question - A reverse osmosis unit treats feed water (F) containing fluoride and its output consists of a

permeate stream and a reject stream (R). Let CF, CP, and CR denote the fluoride concentrations in the feed, permeate, and

reject streams, respectively. Under steady state conditions, the volumetric flow rate of the reject is 60 % of the

volumetric flow rate of the inlet stream, and CF = 2 mg/L and CP = 0.1 mg/L.

A fraction f of the feed is bypassed and mixed with the permeate to obtain treated water having a fluoride

concentration of 1 mg/L. Here, also the flow rate of the reject stream is 60% of the flow rate entering the reverse osmosis

unit (after the bypass). The value of f (up to 2 digits after the decimal points) is ______.

Bypass Stream - Mass Balance

Let 1 be the volume of feed. f bypassed

(1 - f) enters the treatment cell - 0.6 (1 – f) is reject and 0.4 (1-f) is permeate , having 0.1 mg/L Fluoride

Bypass stream will have 2 mg / L of fluoride

Treated water is obtained by mixing permeate and bypass stream

Overall balance around mixing point : f + 0.4 ( 1- f ) = 0.6 f + 0.4 ( treated water)

Fluoride Balance around mixing point : f * 2 mg/L + 0.4 (1 - f) * 0.1 mg/L = (0.6 f + 0.4) * 1 mg/L

(2 – 0.04 – 0.6) f = 0.4 – 0.04 = 0.36 f = 0.36 / 1.36 = 0.26

GP-CPC-PGP

2012 - The reaction A (LIQ) + B(GAS) → C (LIQ) + D (GAS) is carried out in a reactor

followed by a separator as shown below :

Notation: Molar flow rate of fresh B is FFB

Molar flow rate of A is FA

Molar flow rate of recycle gas is FRG

Molar fraction of B in recycle as is YRB

Molar flow rate of purge gas is FPG ; Molar flow rate of C is FC

Here, FFB = 2 mol /s FA = 1 mol/s FB/FA = 5 and A is completely converted.

GP-CPC-PGP

Here, FFB = 2 mol /s FA = 1 mol/s FB/FA = 5 and A is completely converted.

If YRB = 0.3, the ratio of recycle gas to purge gas (FRG / FPG ) is (A) 2 (B) 5 (C) 7 (D) 10

Mass Balance – Recycle – Purge Assume solubility of D in C = 0, C does not vaporize.

FA = 1 mol /s, FB = 5 FA = 5 mol/s = FFB + FRG, FRG = 5 – 2 = 3 mol/s A completely converted

In Converted / formed Out A 1 -1 0B 2 + 3*0.3 = 2.9 -1 1 + 0.9 = 1.9 Distributed between liquid (C) & gas (D) phasesC 0 +1 1 Liquid Phase - A componentD 3 * 0.7 = 2.1 +1 3.1 Gas Phase - A component

As far as B is concerned, 1.9 / (1.9 + 3.1) = 0.38 > 0.3 ( data value ). Hence, B is distributed

Mol of gas phase leaving the separator = 3.1 / 0.7 = 4.43 mol /s

FRG = 3 FPG= 1.43 FRG / FPG = 3 / 1.43 = 2.09

4.43 * 0.3 = 1.329 moles of B will be in gas phase . (1.9 – 1.329=) 0.571 moles B with C

GP-CPC-PGP

2011 - Ammonia is synthesized at 200 bar and 773K by the reaction N2 +3H2 = 2NH3. The yield of ammonia is 0.45 mol/mol

of fresh feed. Flow sheet for the process (along with available composition) is shown in the figure..

The single pass conversion for H2 in the reactor is 20%. The amount of H2 lost in the purge as a percentage of H2 in fresh feed is (a) 10 (b) 20 (c) 45 (d) 55

Mass Balance – Reaction - Purge

Basis :1 k mol of fresh feed.

Recycle stream and purge stream will have the same composition

Purge – inert balance : Inert in fresh feed = Inert in Purge1 * 0.01 = P (0.1) P = 0.1 mol

Hydrogen lost - Percent of Purge / Fresh Feed = (0.1 * 0.75 / 1 * 0.75 ) * 100 = 10 %

GP-CPC-PGP

2011 - The following combustion reactions occur when methane is burnt.

CH4 + 2O2 = CO2 + 2 H2O

2CH4 + 3 O2 = 2 CO + 4 H2O equivalent to CH4 + 1.5 O2 = CO + 2 H2O

20% excess air is supplied to the combustor. The conversion of methane is 80% and the molar ratio of C0 to C02 in the

flue gas is 1:3 . Assume air to have 80 mol% N2 and rest O2. The O2 consumed as a PERCENTAGE of O2 entering the

combustor is (a) 20 (b) 62.5 (c)80 (d) 83.3

Reaction - Stoichiometry – Excess - Partial Conversion

Basis : 1 mol of CH4

Theoretical Oxygen ( for complete conversion ) = 2 mol

Oxygen supplied @ 20 % excess = 2 * 1.2 = 2.4 mol

Methane converted = 1 * 0.8 = 0.8 mol Part to CO2 and the balance to CO .

CO to CO2 ratio in the flue gas is 1 :3. If 4 units are converted, 1 unit is converted to CO, 3 to CO2

CH4 converted to CO = 0.8 * 1 / 4 = 0.2 mol Oxygen consumed = 0.2 * 1.5 = 0. 3 mol

CH4 converted to CO2 = 0.8 * 3 / 4 = 0.6 mol Oxygen consumed = 0.6 * 2 = 1. 2 mol

Total Oxygen used = 0.3 + 1.2 = 1.5 mol Oxygen Fed = 2.4 mol

Percent - Oxygen consumed / Oxygen fed = 100 * 1.5 / 2.4 = 62.5 %

GP-CPC-PGP

Good day…..