Select Page
Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

## Machine Design MCQ Set 1

1. Stages in the extractor are _______________
a) Equilibrium stages
b) Theoretical stages
c) Equilibrium or theoretical stages
d) None of the mentioned

Answer: c [Reason:] The extractor stages are maintained in equilibrium to get the effect mass transfer.

2. Extraction includes mixing alone.
a) True
b) False

Answer: b [Reason:] Extraction needs mixing and separation.

3. Every stage of extractor act as mixer and settler in extraction process.
a) True
b) False

Answer: a [Reason:] Every extracting process needs mixing and separating unit.

4. If the pure solvent is used find the value of “ys” by analysing the below single stage extractor.

Where, F,S are feed and solvent
a) 0
b) 1
c) 0-1
d) None of the mentioned

Answer: a [Reason:] Pure solvent in the sense no solute present initially so ys= 0.

5. Find the rate of extract in mol/hr if

Rate(mol/hr)Composition
Feed1000.5
Solvent500
Residue250.25

a) 100
b) 125
c) 150
d) 50

Answer: b [Reason:] Feed + solvent = Extract + residue 100+50-25 = 125 mol/hr.

6. Is the extraction is possible if a gas solvent is used for extracting solute from liquid mixture.
a) True
b) False

Answer: b [Reason:] The extraction is possible only if liquid-liquid equilibrium occurs. If gas solvent is used such equilibrium cannot be achieve.

7. Find the process.

a) Single stage batch process
b) Single stage continuous process
c) Multistage continuous process
d) None of the mentioned

Answer: b [Reason:] Here stage used is one and there is a continuous input and output so the process is a single stage continuous process.

8. Find the sum of R and E if the sum of F and S is 80 kmol.

a) 20 kmol
b) 40 kmol
c) 60 kmol
d) 80 kmol

Answer: d [Reason:] M= F+M= R+S =80.

9. Find the pure solvent rate if the feed ( F= 100 kmol/hr) composition is 0.65 and mixture solute composition is 0.5.
a) 10 kmol/hr
b) 20 kmol/hr
c) 30 kmol/hr
d) 40 kmol/hr

Answer: c [Reason:] S/F = 0.65- 0.5/0.5 -0 = 0.3 = 100 x 0.3 = 30 kmol/hr.

10. Find the mixture composition of feed and pure solvent if ratio of solvent to feed rate is 0.5 and the feed composition is 0.5.
a) 0.5
b) 0.25
c) 0.33
d) 0.4

Answer: c [Reason:] S/F= 0.5-x/x-0 0.5x+x = 0.5 then x= 0.33.

## Machine Design MCQ Set 2

1. For extraction, the distribution co-efficient should be greater that 1.
a) True
b) False

Answer: a [Reason:] If the distribution co-efficient is greater than 1 the solvent requirement for the extraction operation will be low.

2. The viscosity should be high for the solvents using in extraction operation.
a) True
b) False

Answer: b [Reason:] If the viscosity is high, the handling will be really tough so for better handling the viscosity should be low.

3. The Azeotropic composition of the solute makes the extraction process economical.
a) True
b) False

Answer: b [Reason:] For the Azeotropic composition, additional solvent is required to alter the azeotropes formations.

4. Find the separation factor if the ratio of weight fraction of solute in extract to raffinate is 0.75. Also given the ratio of dilutant in raffinate to extract is 0.5.
a) 1
b) 1.25
c) 1.5
d) 1.75

Answer: c [Reason:] Distribution co-efficient = the ratio of weight fraction of solute in extract to raffinate * the ratio of dilutant in raffinate to extract =0.75/0.5 = 1.5

5. The separation is possible if the ratio of weight fraction of solute in extract to raffinate is 0.15. Also given the ratio of dilutant in raffinate to extract is 0.5.
a) True
b) False

Answer: b [Reason:] Distribution co-efficient = the ratio of weight fraction of solute in extract to raffinate * the ratio of dilutant in raffinate to extract = 0.15/0.5 = 0.3 < 1 (Separation is not possible)

6. In absorption operation, the solubility of gas should be high on solvent.
a) True
b) False

Answer: a [Reason:] If the gas solubility is more in solvent, the amount of solvent requirement is less.

7. The solvent used in the extraction should be more volatile.
a) True
b) False

Answer: a [Reason:] After obtaining the distillate, the extract has to be separated from solvent by means of distillation. Generally for distillation more volatile compound needs less heat to evaporate.

8. By analysing this extraction equilibrium triangle, find the solvent requirement range.

a) Less solvent
b) More solvent
c) No solvent
d) None of the mentioned

Answer: b [Reason:] In this pair configuration, we need more solvent because the capacity to extract solute by the solvent is less.

9. For getting rapid absorption rate, low pressure drop and more heat transfer we need?
a) Low viscous solvent
b) More viscous solvent
c) No viscous solvent
d) None of the mentioned

Answer: a [Reason:] Low viscous solvents are easier to handle and provide us low pressure drop etc.

10. In extract the usage of solvent should be less.
a) True
b) False

Answer: a [Reason:] If less solvent is used we can reduce the heat cost for separation by distillation in extract phase.

## Machine Design MCQ Set 3

1. What is the value of diffusivity for ammonia?
a) 0.26
b) 0.27
c) 0.28
d) 0.29

Answer: c [Reason:] Ammonia is a colorless gas and it boils at a pressure of one atmosphere. It is easily liquefied due to the strong hydrogen bonding. Its unit is cm2/s.

2. Which one is having highest value of diffusivity?
a) Hydrogen
b) Oxygen
c) Water
d) Methanol

Answer: a [Reason:] Thermal diffusivity of hydrogen is 0.410 while that of oxygen, water and methanol are 0.206, 0.259 and .0159 respectively. Its unit is cm 2/s.

3. What is the value of diffusivity for hydrogen?
a) 0.210
b) 0.310
c) 0.410
d) 0.510

Answer: c [Reason:] It is highly flammable and will burn in air at a very wide range of concentrations. It reacts with every oxidizing agent. Its unit is cm 2/s.

4. Which one is having lowest value of diffusivity?
a) Acetic acid
b) Oxygen
c) Benzene
d) Toluene

Answer: d [Reason:] Thermal diffusivity of toluene is 0.084 while that of acetic acid, oxygen and benzene are 0.133, 0.206 and 0.081 respectively. Its unit is cm 2/s.

5. What is the value of diffusivity for water?
a) 0.233
b) 0.256
c) 0.276
d) 0.298

Answer: b [Reason:] It is transparent in the visible electromagnetic spectrum. It has high specific heat capacity and latent heat of vaporisation. Its unit is cm 2/s.

6. Which one is having highest value of diffusivity?
a) Water
b) Oxygen
c) Hydrogen
d) Carbon dioxide

Answer: a [Reason:] Thermal diffusivity of water is 0.256 while that of oxygen, hydrogen and carbon dioxide are 0.206, 0.210 and 0.164 respectively. Its unit is cm2/s.

7. What is the value of diffusivity for ethyl alcohol?
a) 0.117
b) 0.118
c) 0.119
d) 0.120

Answer: c [Reason:] There are extensive hydrogen bonding between its molecules. Its unit is cm2/s.

8. Which one is having lowest value of diffusivity?
a) Glucose
b) Urea
c) Ethanol
d) Oxygen

Answer: a [Reason:] Thermal diffusivity of glucose is 0.60 while that of urea, oxygen and ethanol are 8.06, 1.80 and 1.00 respectively. Its unit is cm2 /s.

9. What is the value of diffusivity for benzene?
a) 0.055
b) 0.066
c) 0.077
d) 0.088

Answer: d [Reason:] It is having higher diffusivity so they diffuse faster into each other. Its unit is cm2/s.

10. Which one is having highest value of diffusivity?
a) Ammonia
b) Urea
c) Glucose
d) Ethanol

Answer: b [Reason:] Thermal diffusivity of urea is 8.06 while that of ammonia, glucose and ethanol are 1.76, 0.60 and 1.00 respectively. Its unit is cm 2/s.

## Machine Design MCQ Set 4

1. Assume the following thermos-physical properties of air
D = 0.256 * 10 -4 m2/s
µ = 1.86 * 10 -5 kg/m s
c p = 1.005 k J/kg degree Celsius
p r = 0.701
p =1.165 kg/m3
Find the value of Schmidt number
a) 0.423
b) 0.523
c) 0.623
d) 0.723

Answer: c [Reason:] Sc = µ/p D.

2. Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of air
D = 0.256 * 10 -4 m2/s
µ = 1.86 * 10 -5 kg/m s
c p = 1.005 k J/kg degree Celsius
p r = 0.701
p =1.165 kg/m3
a) 0.1076 m/s
b) 0.2076 m/s
c) 0.3076 m/s
d) 0.4076 m/s

Answer: a [Reason:] h m = 0.0296 (Re) -0.2 (V)/ (Sc) 0.667.

3. Consider the above problem, find out the value of Reynolds number
a) 14.09 * 10 8
b) 14.09 * 10 7
c) 14.09 * 10 6
d) 14.09 * 10 5

Answer: d [Reason:] Re = V l p/ µ.

4. Assume the following thermos-physical properties of air
D =0.82 * 10 -6 m2/s
v = 15.5 * 10 -6 m2/s
Find the value of Schmidt number
a) 2.89
b) 1.89
c) 3.89
d) 0.89

Answer: b [Reason:] Schmidt number = v/D = 1.89.

5. Air at 1 atm and 25 degree Celsius temperature, flows inside a 3 cm diameter tube with a velocity of 5.25 m/s. Determine mass transfer coefficient for iodine transfer from the air stream to the weak surface. Assume the following thermos-physical properties of air
D =0.82 * 10 -6 m2/s
v = 15.5 * 10 -6 m2/s
a) 0.0176 m/s
b) 1.0176 m/s
c) 2.0176 m/s
d) 3.0176 m/s

Answer: a [Reason:] h m = (Sherwood number) D/d =0.0176 m/s.

6. Consider the above problem, find the value of Sherwood number
a) 84.44
b) 74.44
c) 64.44
d) 54.44

Answer: c [Reason:] Sherwood number = -0.023 (Re) 0.83 (Sc) 0.44 = 64.44.

7. The empirical correlation for local mass transfer coefficient for laminar boundary layer flow past a flat plate is given by
a) Sh X = 0.332 (Re) 0.5 (Sc) 0.23
b) Sh X = 0.332 (Re) 0.5 (Sc) 0.33
c) Sh X = 0.332 (Re) 0.5 (Sc) 0.43
d) Sh X = 0.332 (Re) 0.5 (Sc) 0.53

Answer: b [Reason:] Sh = h ml/D = .332 (Re) 0.5 (Sc) 0.33.

8. The empirical correlation for local mass transfer coefficient for turbulent boundary layer flow past a flat plate is given by
a) Sh X = 0.332 (Re) 0.5 (Sc) 0.33
b) Sh X = 0.332 (Re) 0.6 (Sc) 0.33
c) Sh X = 0.0298 (Re) 0.7 (Sc) 0.33
d) Sh X = 0.0298 (Re) 0.8 (Sc) 0.33

Answer: d [Reason:] Sh = h ml/D.

9. The expression for average mass transfer coefficient is
a) Sh = 0.664 (Re) 0.5 (Sc) 0.33
b) Sh = 0.664 (Re) 0.6 (Sc) 0.33
c) Sh = 0.332 (Re) 0.7 (Sc) 0.33
d) Sh = 0.332 (Re) 0.8 (Sc) 0.33

Answer: a [Reason:] Sh = h ml/D = 0.664 (Re) 0.5 (Sc) 0.33.

10. Sherwood number is a function of
a) Lewis number and Reynolds number
b) Prandtl number and Lewis number
c) Reynolds number and Schmidt number
d) Schmidt number and Lewis number

Answer: c [Reason:] It is a function of Reynolds number and Schmidt number.

## Machine Design MCQ Set 5

1. Spray tower is a _______ process.
a) Co-current
b) Counter current
c) Continuous
d) Batch

Answer: b [Reason:] Here, the gas is passed through bottom and the liquid is sprayed at the top.

2. Find the rate of non-diffusing solute, if the mole fraction of the gas phase is 0.65 and the diffusing rate is 70 moles/hr.
a) 2 4.5 moles/hr
b) 200 moles/hr
c) 18.18 moles/hr
d) 37.7 moles/hr

Answer: a [Reason:] Gs= G(1-y) =70 x(1-0.65) =24.5 moles/hr

3. In an absorber, lower denser materials are sent through the bottom and the higher denser materials are sent through top
a) True
b) False

Answer: a [Reason:] For reducing the pumping cost and to create sufficient pressure drop.

4. Stripper= (1/X)
Find X?
a) Condenser
b) Reboiler
c) Absorber
d) Sprayer

Answer: c [Reason:] Stripper is liquid to gas transfer but absorber is gas to liquid transfer.

5. Find the representation

Where, Y- gas phase concentration in mole ratio
X-liquid phase concentration in mole ratio
a) Stripper
b) Absorber
c) Condenser
d) Semi-Stripper

Answer: a [Reason:] For stripper, the solute gets transfer from liquid to gas so the operating line is below equilibrium line.

6. Find the slope of the operating line

Here, the concentrations are represented in mole ratio. The value of (Y1-Y2)/(X1-X2) =2. Find the slope of the operating line.
a) 1
b) 2
c) 3
d) Infinity

Answer: b [Reason:] For a counter current process, Y1- Y2 = Ls/Gs(X1-X2) Ls/Gs= slope of the operating line = 2

7. Have a look on the below diagram

Here, DE and DP are the operating line. Also Y and X are the concentrations in mole ratio for gas and liquid phase.
It’s clearly a representation of an absorber. For the operating line DE minimum liquid is required
a) True
b) False

Answer: b [Reason:] DP line requires minimum liquid because the composition is increased to X2.

8. If the operating line touches the equilibrium line, the separation is possible in an absorber.
a) True
b) False

Answer: b [Reason:] At that point, the slope of operating line will be equals to the slope of equilibrium line. So the Absorption factor becomes equal to one, the separation is not possible.

9. For the Absorber design the plotting with mole ratio helps to find the
a) Slope of operating line
b) Slope of equilibrium curve
c) Minimum number of trays
d) Maximum number of trays