This set of Separation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Batch Crsytallization”.
1. What type of size distribution occurs in batch crystallization?
a) Trimodal
b) Monomodal
c) Bimodal
d) No distribution occurs
View Answer
Explanation: Formation of tiny crystals by a secondary nucleation occurs simultaneously with growth of added seeds. The nuclei formed also grow alongside. This ultimately leads to a bimodal or even multimodal size distribution of the final product.
2. Which of the following is not a common method in reducing the rate of nucleation?
a) To control supersaturation
b) To keep low level of turbulence of mixing
c) To keep the slurry density low
d) To keep high level of turbulence of mixing
View Answer
Explanation: A narrow crystal size distribution can be achieved if there is little or no nucleation, so that the final product is the result of growth of the seeds only. The common methods of reducing the rate of nucleation (a) to control the supersaturation level, (b) to keep a low level of turbulence or mixing, (c) to keep the slurry density low.
3. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 400 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.06lit/min
c) 0.07lit/min
d) 0.08lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.05lit/min.
4. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 800 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.06lit/min
c) 0.07lit/min
d) 0.1lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.1lit/min.
5. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 600 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.0625lit/min
c) 0.07lit/min
d) 0.08lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.0625lit/min.
6. Find the initial rate of evaporation if
Mass of seed=Ms= 0.924 kg,
Solubility C = 400 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.10lit/min
c) 0.15lit/min
d) 0.08lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.15lit/min.
7. Find the initial rate of evaporation if
Mass of seed=Ms= 3.08 kg,
Solubility C = 400 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.5lit/min
c) 0.7lit/min
d) 0.8lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.5lit/min.
8. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 200 kg/m3, G/Ls= 3.75 x 10-4
a) 0.0225lit/min
b) 0.06566lit/min
c) 0.0527lit/min
d) 0.5408lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.0225lit/min.
9. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 80 kg/m3, G/Ls= 3.75 x 10-4
a) 0.01lit/min
b) 0.02lit/min
c) 0.03lit/min
d) 0.04lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.01lit/min.
10. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 1600 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.15lit/min
c) 0.20lit/min
d) 0.25lit/min
View Answer
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.20lit/min.
Sanfoundry Global Education & Learning Series – Separation Processes.
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