APJ ABDUL KALAM TECHNOLOGICAL UNIVERSITY Previous Years Question Paper & Answer

Course : B.Tech

Semester : SEMESTER 3

Year : 2020

Term : SEPTEMBER

Scheme : 2015 Full Time

Course Code : CH 201

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bottom product contains 96% toluene. The stream entering the condenser from
the top of the column is 8000 kg/hr. Find the reflux ratio.
A hot solution of Ba(NO3)2 from an evaporator contains 30.6 kg of Ba(NO3)2 per
100 kg of water and goes to a crystallizer where the solution is cooled and
Ba(NO3)2 crystallizes. On cooling 10% of the original water present evaporates.
For a feed solution of 100 kg, calculate the yield of crystals and water evaporated
if the solution is cooled to 290 K. Solubility of Ba(NO3), is 8.6 kg per 100 kg
total water at 290 K.
Ammonia is recovered from a gas mixture containing 25% (volume) CO» and
75% (volume) NH3 by scrubbing with water. Assuming that CO> is insoluble in
water, determine the percent of ammonia in the entering gas that is absorbed if
the gas leaving the scrubber analyses 35% 1411.
An air- water vapour sample at 1 atm has a dry bulb temperature of 328 K and is
10% saturated with water vapour. Using psychrometric chart, determine (i) the
absolute humidity, (ii) partial pressure of water vapour, (iii) absolute saturation
humidity at 328 K, (iv) the vapour pressure of water at 328 K, (v) percent relative
saturation and (vi) dew point of the system.
PART C
Answer any two full questions, each carries 20 marks.

Pure propane is burnt in an excess of air to give the following analysis of
combustion products in volume percent: (02 — 5%, CO — 3.5%, 110 — 11.4%,
ഠാ — 7%, No - 73.1%. Calculate the percentage excess air and find the Orsat
analysis.
State and explain Hess’s law of constant heat summation.
Coal is burnt to a gas of the following composition by mole: CO2— 9.2, CO — 1.5,
O, — 7.3, 14, — 82%. Compute the enthalpy difference for this gas between the
bottom and the top of the stack if the temperature at the bottom is 550 K and at
the top is 200 K.

Cp of 02 is 8.448 + 5.757 x 10° T— 21.59 x 10 12 + 3 x 10191 J/mol K

Cp of CO is 6.865 + 0.8024 x 10° T — 0.736 x 107 12 J/mol K

Cp ௦11, is 6.895 + 0.7624 x 103 T —0.7 x 107 T* J/mol K

Cp of Op is 7.104 + 0.7851 x 10° ¶ _ 0.5528 x 107 T? J/mol K
Explain recycle, bypass and purge operations with examples.

CO and hydrogen reacts to give methanol. The conversion of CO entering the

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