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Multiple choice question for engineering

Set 1

1. For a faithful reproduction of the input signal linearity over the required range of signal amplitudes must be satisfied by the individual parts of the system?
a) True
b) False

View Answer

Answer: a [Reason:] For a faithful reproduction of the input signal, three basic conditions must be satisfied by the individual parts of the system. These requirements are linearity over the required range of signal amplitudes and an adequate passband for the frequencies involved without producing any phase shift between the input and recorded signal.

2. CRO stands for _______
a) Common Ray Oscilloscope
b) Cathode Ray Oscilloscope
c) Cathode Ray Oscillator
d) Common Ray Oscillator

View Answer

Answer: b [Reason:] Cathode ray oscilloscopes are widely used for the display of waveforms encountered in the medical field. These waveforms can be recorded from the CRO screen by running a photographic film through a recording camera fixed in front of the screen. Recorders are either of the single channel type or of the type which record several channels simultaneously.

3. Which of the following is not a passive transducer?
a) Strain gauge
b) Ultrasonic transducer
c) IR sensor
d) Doppler effect transducer

View Answer

Answer: a [Reason:] Strain gauge transducer is an active transducer. Its working principle is based on change in resistance. Ultrasonic transducers need power to operate. So do the IR sensors and doppler effect transducers.

4. _____________ refers to the degree of repeatability of a measurant.
a) accuracy
b) precision
c) resolution
d) sensitivity.

View Answer

Answer: b [Reason:] Precision refers to the degree of repeatability of a measurant. Accuracy describes the algebraic difference between the indicated value and the true or theoretical value of the measurand. Resolution is the ability of the transducer or sensor to see small differences in reading.

5. Accuracy is ______
a) ability of the transducer or sensor to see small differences in reading
b) ability of the transducer or sensor to see small differences in reading
c) algebraic difference between the indicated value and the true or theoretical value of the measurand
d) total operating range of the transducer

View Answer

Answer: c [Reason:] Accuracy describes the algebraic difference between the indicated value and the true or theoretical value of the measurand. Resolution is the ability of the transducer or sensor to see small differences in reading. Precision refers to the degree of repeatability of a measurant.

6. ____________ filter amplifies signals below a certain frequency.
a) band stop filter
b) high pass filter
c) band pass filter
d) low pass filter

View Answer

Answer: d [Reason:] Low pass filter amplifies signals below a certain frequency. Band pass filter amplifies frequencies with in a certain band. Band stop filter amplifies all the frequencies except those in a certain band. High pass filter amplifies signal above a certain frequency.

7. Which of the following filter amplifies frequencies with a certain band?
a) band pass filter
b) band stop filter
c) low pass filter
d) high pass filter

View Answer

Answer: a [Reason:] Band pass filter amplifies frequencies with in a certain band. Band stop filter amplifies all the frequencies except those in a certain band. High pass filter amplifies signal above a certain frequency. Low pass filter amplifies signals below a certain frequency.

8. ________________ amplifies all the frequencies except those in a certain band.
a) high pass filter
b) low pass filter
c) band pass filter
d) band stop filter

View Answer

Answer: d [Reason:] Band stop filter amplifies all the frequencies except those in a certain band. High pass filter amplifies signal above a certain frequency. Low pass filter amplifies signals below a certain frequency. Band pass filter amplifies frequencies with in a certain band.

9. AAMI stands for ____________________
a) American Association of Medical Instrumentation
b) Association for the Advancement of Medical Instrumentation
c) Association of American Medical Instrumentation
d) American Association of Measurement Instruments

View Answer

Answer: b [Reason:] Association for the Advancement of Medical Instrumentation, USA is an association for propelling the advancement, and protected and powerful utilization of restorative innovation established in 1965 by Robert D. Corridor, Jr. what’s more, Robert J. Allen, President and Vice President individually of Tech/Reps, Inc. AAMI is an intentional association, and in spite of the fact that its suggested practices and gauges once in a while reverberate other social insurance rules, consistence with these models isn’t really required by administrative associations that review medicinal services offices.

10. A wavelet transform is almost always implemented as a bank of filters that decompose a signal into multiple signal bands.
a) True
b) False

View Answer

Answer: a [Reason:] It is true. A wavelet transform is almost always implemented as a bank of filters that decompose a signal into multiple signal bands. Wavelets are a relatively new signal processing method. . Thus, one of the biggest advantages of using the wavelet transform is that signal features can be easily extracted. In many cases, a wavelet transform outperforms the conventional FFT when it comes to feature extraction and noise reduction. predictable, and software simulations can exactly reflect product performance.

Set 2

1. WHat is an arrhythmia monitor ?
a) Patient monitoring system
b) Sophisticated alarm system
c) Sophisticated monitoring system
d) ECG interpretation system

View Answer

Answer: b [Reason:] An arrhythmia monitor is basically a sophisticated alarm system.It is not an ECG interpretation system. It constantly scans ECG rhythm patterns and issues alarms to events that may be premonitory or life threatening.

2. Which task is performed after the Ventricular fibrillation detection in automated arrhythmia monitoring system ?
a) Noise detection
b) Beat labeling
c) Atrial fibrillation detection
d) Rhythm definition

View Answer

Answer: d [Reason:] In automated arrhythmia monitoring system, Rhythm definition is performed after the Ventricular fibrillation detection. Rhythm definition is also performed after the beat labeling and atrial fibrillation detection in automated arrhythmia monitoring and analysis system.

3. In arrhythmia monitoring system, it gives alarm light signals whenever the prematured or widened ectopic beats exist up to the rate of __________
a) 6/min to 10/min
b) 6/min to 12/min
c) 6/min or 10/min
d) 6/min or 12/min

View Answer

Answer: d [Reason:] In arrhythmia monitoring instrument, , it gives alarm light signals whenever the prematured or widened ectopic beat exist up to the rate of 6/min or 12/min.It is one of the operating sequence of the arrhythmia monitoring instrument.

4. ECG signal is amplified and filtered with 0.05-100 Hz for diagnostic purposes and 1-40 Hz for monitoring purposes in signal conditioning.
a) True
b) False

View Answer

Answer: a [Reason:] It is True. ECG signal is amplified and filtered with 0.05-100 Hz for diagnostic purposes and 1-40 Hz for monitoring purposes in signal conditioning.

5. _________ resolution analog-to-digital converter is used in digitization of ECG signal in signal conditioning.
a) 16 bit
b) 12 bit
c) 32 bit
d) 64 bit

View Answer

Answer: b [Reason:] In signal conditioning, ECG signal is amplified, filtered and digitized using an 8- or 12- bit analog-to-digital converter with a typical sampling rate of 250 Hz.

6. A disturbance in the heart’s normal rhythmic contraction is called ____________
a) Heart stroke
b) Cardiac arrest
c) Arrhythmias
d) Premature contraction

View Answer

Answer: c [Reason:] Any disturbance in the heart’s normal rhythmic contraction is called an arrhythmias or cardiac dysrhythmia. In this arrhythmias heart can’t beat in regular rhythm. In arrhythmia heart-rate will be higher than normal rate or will be less than the normal rate.

7. The steep, large amplitude variation of the QRS complex is the obvious characteristics to use and this is the function of the Q wave detector.
a) True
b) False

View Answer

Answer: b [Reason:] It is False. Arrhythmia monitors require reliable R wave detectors as a prerequisite for subsequent analysis. The steep, large amplitude variation of the QRS complex is the obvious characteristics to use and this is the function of the R wave detector.

8. Which of the following two elements are removed by the detection filter in the process of the ECG waveform?
a) Low frequency noise, motion noise
b) Muscle artifact, motion noise
c) Baseline wander, motion noise
d) Baseline wander, muscle artifact

View Answer

Answer: a [Reason:] The detection filter removes low frequency noise (baseline wander) and muscle artifact. The ECG waveform is processed by two digital filters: a detection filter and a classification filter. P waves and T waves are diminished.

9. The number of steps involved in detection of QRS complex are ______
a) One step
b) Two steps
c) Three steps
d) Four steps

View Answer

Answer: b [Reason:] The ECG is first preprocessed to enhance the QRS complex while suppressing noise, artifact and non-QRS portions of the ECG. QRS detection is now almost universally performed digitally in a two-step process.The output of the preprocessor stage is subjected to a decision rule that confirms detection of QRS if the processor output exceeds a threshold.

10. Which of the following is based on analyzing the shape of the QRS complexes and separating beats into groups or clusters?
a) Morphology characterization
b) Noise detection
c) Beat labeling
d) Timing classification

View Answer

Answer: a [Reason:] Morphology characterization is based on analyzing the shape of the QRS complexes and separating beats into groups or clusters of similar morphology.Most algorithms for real time arrhythmia analysis maintain no more than 10-20 clusters at a time , order to limit the amount of computation needed to assign a QRS complex to a cluster.

11. What is the condition in which the R-R interval is declared premature?
a) If it is less than 75% of the predicted interval
b) If it is greater than 75% of the predicted interval
c) If it is less than 85% of the predicted interval
d) If it is greater than 85% of the predicted interval

View Answer

Answer: c [Reason:] In timing classification, the observed R-R interval is compared to an estimate of the expected R-R interval. An R-R interval will be declared premature if it is less than 85% of the predicted interval. Similarly, an R-R interval is long if it is greater than 110% of the predicted value.

12. ___________ is the final stage in arrhythmia analysis.
a) Summary statistics
b) Alarms
c) Rhythm labeling
d) Beat labeling

View Answer

Answer: c [Reason:] It is based on defined sequences of QRS complexes. Rhythm labeling is the final stage in arrhythmia analysis. The analysis systems are heavily oriented towards detecting ventricular arrhythmias, particularly single PVCs.

13. ______________ techniques are used in a new algorithm proposed by Jen and Hwang to obtain the long term ECG signal feature and extract the meaningful information hiding in the QRS complex.
a) Cepstrum time warping and Dynamic coefficient
b) Cepstrum coefficient and Dynamic time warping
c) QRS detection and Dynamic coefficient
d) QRS detection and Cepstrum time warping

View Answer

Answer: b [Reason:] This algorithm may also be used for arrhythmia detection by simply checking the difference of R-R wave intervals through signal feature extraction comparison for a certain period of time. Jen and Hwang proposed a new algorithm using cepstrum coefficient and the dynamic time warping techniques to obtain the long term ECG signal feature and extract the meaningful information hiding in the QRS complex.

14. Which of the following is used to detect Ventricular Fibrillation?
a) Shape of the QRS complexes
b) Frequency domain analysis
c) Timing sequence of QRS complexes
d) Difference of the R-R interval

View Answer

Answer: b [Reason:] Ventricular fibrillation is usually detected by frequency domain analysis. It can be distinguished from noise by appropriately designing band-pass filters. The system is characterized as a narrow-band, low frequency signal with energy concentrated in a band around 5-6 Hz.

15.____________ is the sampling rate of analog-to-digital converter in digitizing of ECG signal in signal conditioning.
a) 200-215 Hz
b) 215 Hz
c) 40-100 Hz
d) 250 Hz

View Answer

Answer: d [Reason:] In signal conditioning, ECG signal is amplified, filtered (0.05-100 Hz for diagnostic purposes, 1-40 Hz for monitoring purposes) and digitized using an 8- or 12-bit analog-to-digital converter with a typical sampling rate of 250 Hz.

Set 3

1. Which type of forces stretch and distort the bubbles?
a) Shear forces
b) Strain forces
c) Surface tension
d) Frictional forces

View Answer

Answer: a [Reason:] Shear forces in turbulent eddies stretch and distort the bubbles and break them into smaller sizes; at the same time, surface tension at the gas-liquid interface tends to restore the bubbles to their spherical shape. In the case of solid material such as cell flocs or aggregates, shear forces in turbulent flow are resisted by the mechanical strength of the particles.

2. Viscosity is inversely proportional to the size of eddies?
a) True
b) False

View Answer

Answer: b [Reason:] If the viscosity of the fluid is increased, the size of the smallest eddies also increases. Increasing the fluid viscosity should, therefore, reduce shear damage in bioreactors.

3. Larger foam is produced in bioreactors with increased headspace and small workspace?
a) True
b) False

View Answer

Answer: a [Reason:] A bioreactor is divided in a working volume and a headspace volume. The working volume is the fraction of the total volume taken up by the medium, microbes, and gas bubbles. The remaining volume is called the headspace. Typically, the working volume will be 70-80% of the total fermenter volume. This value will however depend on the rate of foam formation during the reactor. If the medium or the fermentation has a tendency to foam, then a larger headspace and smaller working volume will need to be used.

4.” Instrument air compressor” should be used generally as a compressor?
a) True
b) False

View Answer

Answer: b [Reason:] Note that it is very important that an “instrument air” compressor is not used. Instrument air is typically generated at higher pressures but is aspirated with oil. Instrument air compressors are used for pneumatic control. Air compressors used for large scale bioreactors typically produce air at 250 kPa. The air should be dry and oil free so as to not block the inlet air filter or contaminate the medium.

5. What is the function of Pluronic f-68?
a) To increase the foaming
b) To decrease the foaming
c) Increase cell attachment
d) To change the composition of the cells

View Answer

Answer: b [Reason:] Pluronic F68 is used in cell culture as a stabilizer of cell membranes protecting from membrane shearing and additionally acts as an anti-foaming agent. It is a non-ionic surfactant used to control shear forces in suspension cultures. It can also be used to reduce foaming in stirred cultures and reduce cell attachment to glass. It is provided at a concentration of 10% and effective at a working concentration of 0.1%.

6. Microcarrier beads 120 μm in diameter are used to culture recombinant CHO cells for production of growth hormone. It is proposed to use a 6-cm turbine impeller to mix the culture in a 3.5-1itre stirred tank. Air and carbon dioxide are supplied by flow through the reactor headspace. The microcarrier suspension has a density of approximately 1010 kg m-3 and a viscosity of 1.3 x 10-3 Pa s. Estimate the kinematic viscosity.
a) 1.30 × 10-6 m2s-1
b) 1.29 × 10-6 m2s-1
c) 1.50 × 10-6 m2s-1
d) 1.49 × 10-6 m2s-1

View Answer

Answer: b [Reason:] Damage due to eddies is avoided if the Ko ̈1mogorov scale remains greater than 2/3–1/2 the diameter of the beads. Let us determine the stirrer speed required to create eddies with size, λ = 2/3 (120 μm) = 80 μm = 8 x 10-5 rn. The stirrer power producing eddies of this dimension can be estimated: bioprocess-engineering-assessment-questions-answers-q6

7. Refer to Q6, and calculate the power dissipated per unit mass of fluid.
a) 0.050 m2 s-3
b) 0.042 m2 s-3
c) 0.040 22 s-3
d) 0.052 m2 s-3

View Answer

Answer: d bioprocess-engineering-questions-answers-role-shear-stirred-fermenters -q7

8. Referring to Q6 and Q7 and, calculate the stirrer power.
a) 1.13×10-2 W
b) 1.10×10-2 W
c) 1.20×10-2 W
d) 1.23×10-2 W

View Answer

Answer: a [Reason:] The stirrer power P is equal to ε multiplied by ρDi3:

P= (0.052m2s-3) (1010kgm-3) (6x 10-2m)3 P= 1.13×10-2kgm2s -3 = 1.13x 10-2W.

9. Referring to Q6, 7 and Q8, and calculate the stirrer speed. (Given: Np’ = 5)
a) 80.5 rpm
b) 85.0 rpm
c) 85.5 rpm
d) 80.0 rpm

View Answer

Answer: c bioprocess-engineering-questions-answers-role-shear-stirred-fermenters -q9

10. What is the function of microcarrier beads?
a) To give the cells the shape of beads
b) It provides non-buoyancy condition
c) It helps in the lysis of cells
d) It provides protection and surface area

View Answer

Answer: d [Reason:] The microcarrier beads are used to increase the number of adherent cells per flask and are either dextran or glass based, they come in a range of densities and sizes. The beads are buoyant and therefore can be used with spinner culture flasks. The surface area available for cell growth on these beads is huge. It is a support matrix allowing for the growth of adherent cells in bioreactors.

Set 4

1. What is the process of making biomass energy?
a) Oxidation
b) Combustion
c) Reduction
d) Vaporization

View Answer

Answer: b [Reason:] It is burning (combustion). The combustion process generates heat that’s transformed into energy. In the combustion process biomass is burned and converted into energy.

2. A solid-oxide fuel cell is fed with carbon monoxide and reacts with air to produce CO2. This reaction will produce 2 electrons which are used to power an electric circuit external to the fuel cell. The reaction equation is shown below:

2CO(g) + O2(g) -> 2CO2(g) ΔHr° = -565.96 kJ/mol

This reaction does not occur for other types of fuel cells which use a catalyst, such as polymerelectrolyte membrane or phosphoric-acid fuel cells. The presence of carbon monoxide on the anode side of these types of fuel cells will cause catalyst poisoning, reducing the efficiency and voltage of the fuel cell.

Determine the rate of enthalpy change for a carbon dioxide production rate of 208 mol/hr.

The extent of the reaction occurring in the fuel cell can be obtained by the following equation:

bioprocess-engineering-interview-questions-answers-experienced-q2
where:

bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q2-1
a) -16.35 kW
b) -15.35 kW
c) -16.45 kW
d) -15.45 kW

View Answer

Answer: a

Explanation: bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q3

Now the rate of change in the enthalpy for the oxidation of carbon monoxide can be calculated as follows: ΔH ̇ = ξ ̇ ΔHr° Entering the known quantities for the extent of reaction and the enthalpy of reaction into this equation yields:

ΔH ̇ = (104 mol/hr) (-565.96 kJ/mol) . (1hr/3600 s) = -16.35 kJ/s ΔH ̇ = -16.35 kW.

3. Refer to Q2 and estimate that the synthesis gas obtained from a coal gasification process can be used for producing methanol, which is used as fuel in direct-methanol fuel cells. Determine the rate of production of methanol if the reaction shown below is releasing 21.6 kW of energy.

CO(g) + 2H2(g) -> CH3OH(l) ΔHr° = -128.08 kJ/mol
a) 455.6 mol/hr
b) 457.2 mol/hr
c) 450.9 mol/hr
d) 454.5 mol/hr

View Answer

Answer: b

Explanation: Solving for ξ ̇ and substituting the corresponding quantities into this equation yields:

bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q3-1

The rate in enthalpy change was considered to be negative since the problem is stating that the reaction is releasing energy (exothermic reaction).

Now we can enter the calculated extent of reaction into the equation previously solved for the molar production rate of methanol, to get: bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q3-3

4. In this problem we wish to develop the combustion equation and determine the air-fuel ratio for the complete combustion of n-Butane (C4H10) with theoretical air.
bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q4

a) 15.4 kg-air/kg-fuel
b) 12.5 kg-air/kg-fuel
c) 12.4 kg-air/kg-fuel
d) 15.5 kg-air/kg-fuel

View Answer

Answer: d [Reason:] i) Theoretical air :

bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q4-1

5. Refer to Q4, and calculate 50% excess air.
a) 22.2 kg-air/kg-fuel
b) 23.2 kg-air/kg-fuel
c) 20.2 kg-air/kg-fuel
d) 20.3 kg-air/kg-fuel

View Answer

Answer: b [Reason:] 50% Excess air ( 150% Theoretical air): bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q5

6. In this problem Propane (C3H8) is burned with 61% excess air, which enters a combustion chamber at 25°C. Assuming complete combustion and a total pressure of 1 atm (101.32 kPa), determine the air-fuel ratio [kg-air/kg-fuel].

61% Excess air (161% Theoretical air):
C3H8 + (1.61) z (O2 + 3.76 N2) => 3 (CO2) + 4(H2O) + (0.61)z (O2) + (1.61) (3.76) z (N2 )

Equating coefficients, z= 3+2 =5 (oxygen component balance), thus:

C3H8 + 8.05 (O2 + 3.76 N2) => 3 (CO2) + 4(H2O) + 3.05 (O2) + 30.27 (N2)

bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q6

a) 26.3 kg-air/kg-fuel
b) 25.5 kg-air/kg-fuel
c) 25.3 kg-air/kg-fuel
d) 26.5 kg-air/kg-fuel

View Answer

Answer: c [Reason:] bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q6-1

7. Refer to Q6 and calculate the percentage of carbon dioxide by volume in the products.
a) 7.0%
b) 6.0%
c) 7.4%
d) 6.4%

View Answer

Answer: c [Reason:] NCO2 = 3, Ntotal = 3+4+3.05+ 30.27 = 40.32 bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q7

8. Refer to Q6 and Q7 and calculate the dew point temperature of the products.
a) 45.8°C
b) 40.5°C
c) 30.5°C
d) 35.8°C

View Answer

Answer: a [Reason:] bioprocess-engineering-questions-answers-heat-reaction-processes-biomass-production-q8

Tdew-point = T = 45.8°C.

9. “Thermal conversion of organic matter with an oxidant (normally oxygen) to produce primarily carbon dioxide and water”, which term is used for this process?
a) Oxidation
b) Pyrolysis
c) Combustion
d) Gasification

View Answer

Answer: c [Reason:] Combustion is defined as thermal conversion of organic matter with an oxidant (normally oxygen) to produce primarily carbon dioxide and water. The oxidant is in stoichiometric excess, i.e., complete oxidation.

10.” Thermal conversion (destruction) of organics in the absence of oxygen”, which term is used for this process?
a) Reduction
b) Pyrolysis
c) Combustion
d) Gasification

View Answer

Answer: b [Reason:] Pyrolysis is defined as thermal conversion (destruction) of organics in the absence of oxygen. In the biomass community, this commonly refers to lower temperature thermal processes producing liquids as the primary product. Possibility of chemical and food byproducts.

Set 5

1. Which of the following represents the figure corresponding to the stream?
a) Recycle system
b) By-pass system
c) Purge system
d) Recover stream

View Answer

Answer: b bioprocess-engineering-interview-questions-answers-freshers-q1 [Reason:] A by-pass is one where a portion of the inlet to a process unit is split from the feed and instead of entering the process is combined with the outlet from that process. This practice is far less common than recycle, but may be used if your ultimate goal is a material with properties” in-between” the untreated reactant and the process outlet product.

2. Consider the following labeled flowchart for a simple chemical process based on reaction A->B and predict the overall and single-pass conversion of the process?
a) 100%, 70%
b) 100%, 50%
c) 100%, 75%
d) 100%, 55%

View Answer

Answer: c [Reason:] Overall conversion: Based on the streams that enter and leave the overall process. Overall conversion = (moles of reactants in fresh feed-moles in the output of the overall process)/(moles of reactant in fresh feed)

bioprocess-engineering-questions-answers-material-balances-recycle-pass-purge-systems -q2

Therefore, the overall conversion of A is from equation: Overall conversion = ((75 mol A/min)in –(0 mol/min)out )/((75 mol A/min)out) ×100% =100% Single-pass conversion: Based on streams that enter and leave the reactor. Overall conversion= (moles of the reactants fed into the reactor-moles that exiting the reactor)/(moles of reactant fed into reactor) Therefore, the single-pass conversion from equation is: ((100 mol A/min)in-(25 mol A/min)out)/((100 mol A/min)in) × 100% = 75%.

3. What is the combined feed ratio from the following?
a) Quantity of mixed feed stream to the quantity of fresh feed stream
b) Quantity of recycle stream to the quantity of fresh feed stream
c) Quantity of fresh feed stream to the quantity of recycle stream
d) Quantity of fresh feed stream to the quantity of mixed feed stream

View Answer

Answer: a [Reason:] Combined feed ratio is the ratio of the quantity of mixed feed stream to the quantity of fresh feed stream. Combined feed ratio = M/F.

4. A single effect evaporator is fed with 10000 kg / h of weak liquor containing 15 % caustic by weight and is concentrated to get thick liquor containing 40 % by weight caustic. Calculate:
(i) kg / h of water evaporated and
(ii) kg / h of thick liquor
a) 4560 kg/h, 3720 kg/h
b) 3460 kg/h, 7680 kg/h
c) 4350 kg/h, 6732 kg/h
d) 6250 kg/h, 3750 kg/h

View Answer

Answer: d [Reason:] 10000 kg / h of weak liquor. Let x be the kg / h thick liquor obtained and y be the kg / h water evaporated.

Overall Material balance: Total mass Input = Total mass output kg / h weak liquor = kg / h water evaporated + kg / h thick liquor 10000 = x + y Material Balance of NaOH: NaOH in the liquid stream = NaOH in output stream NaOH in the weak liquor = NaOH in thick liquor

0.15 × 10000= 0.40x x = 3750 kg/hr Hence, y= 6250 kg/hr Water evaporated = 6250 kg/h Thick liquor obtained = 3750 kg/h

5. A solution of potassium dichromate in water contains 15% w/w Kr2Cr2O7. Calculate the amount of Kr2Cr2O7 that can be produced from 1500 kg of solution if 700 kg of water is evaporated and remaining solution is cooled to 293K.
Data: Solubility of Kr2Cr2O7 at 293 K is 115 kg per 1000 kg of water.
a) 158.875 kg
b) 156.678 kg
c) 145.478 kg
d) 148.875 kg

View Answer

Answer: a [Reason:] 1500 kg of solution. Kr2Cr2O7 content of solution = 0 15×1500 = 225 kg Water in 15 % solution = 1500 – 225 = 1275 kg M.B of water: Water in final solution = 1275 – 700 = 575 kg Solubility of Kr2Cr2O7 at 293 K = 115 kg / 1000 kg water Amount of Kr2Cr2O7 at 293 K = 115/1000 × 575 = 66.125 kg M.B of Kr2Cr2O7: [Kr2Cr2O7 in feed solution] = [Kr2Cr2O7 in solution at 293k] + [Kr2Cr2O7 produced as crystals] Amount of Kr2Cr2O7crystals produced = 225 – 66.125= 158.875 kg

6. Which of the following is not the application of recycle system?
a) Increased reactant conversion
b) Decreased reactant conversion
c) Continuous catalyst regeneration
d) Circulation of the working fluid

View Answer

Answer: b [Reason:] The most common application of recycle for systems involving chemical reaction is the recycle of reactants, an application that is used to increase the overall conversion in a reactor and not the decreased reactant conversion.

7. The fresh feed to an ammonia synthesis reactor contains nitrogen, hydrogen and 2.0 mole per cent inerts. The molar ratio of H2:N2 is 3:1. The product stream consists of pure ammonia. Since conversion in the reactor is only 15%, a recycle stream is used and in order to avoid build-up of inerts, a purge stream is withdrawn. The rate of purge stream is adjusted to keep inert concentration in the recycle stream at 8 mole per cent. For a fresh feed rate of 100 moles/hr. Note that recycle stream contains only nitrogen, hydrogen and inerts. The N2:H2 ratio of 1:3 is maintained in every process stream. Calculate the moles of ammonia produced.
a) 38.90 moles/hr
b) 28.90 moles/hr
c) 37.50 moles/hr
d) 27.50 moles/hr

View Answer

Answer: c bioprocess-engineering-questions-answers-material-balances-recycle-pass-purge-systems -q7 [Reason:] The concentration of inerts in the purge stream is 8 mole percent. The amount of inerts must be 2 moles/hr in order to prevent accumulation of inerts. Therefore the flow rate of purge stream is 25 moles/hr.

The flow rate of nitrogen and hydrogen in the purge stream is 23 moles/hr (5.75 nitrogen, 17.25 hydrogen). Therefore moles of ammonia produced: (24.50 – 5.75)x2= 37.50 moles/hr of ammonia.

8. Refer to Q7, and calculate the moles of nitrogen entering the reactor and in the recycle stream?
a) 125 moles/hr, 100.50 moles
b) 135 moles/hr, 50 moles
c) 125 moles/hr, 50 moles
d) 185 moles/hr, 100.50 moles

View Answer

Answer: a [Reason:] Moles of nitrogen entering the reactor= (37.50/2)/0.15 = 125 moles/hr Therefore there are 125 – 24.50= 100.50 moles of nitrogen in the recycle stream.

9. Refer to Q7 and Q8, and calculate the number of moles, moles of inerts and moles of hydrogen in the recycle stream?
a) 437 moles/hr,35 moles/hr, 301.5 moles/hr
b) 237 moles/hr, 30 moles/hr, 200 moles/hr
c) 567 moles/hr, 35 moles/hr, 205 moles/hr
d) 347 moles/hr, 30 moles/hr, 500 moles/hr

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Answer: a [Reason:] Total number of moles in the recycle stream = 100.50×4/0.92 = 437 moles/hr Moles of inerts in the recycle stream= 437- 100.50×4=35 moles/hr Moles of hydrogen in the recycle stream = 100.50×3 = 301.50 moles/hr.

10. What do you mean by the splitting point?
a) The two streams split with different composition
b) The two streams split with equal composition
c) Assuming it’s not a reactor and there’s only 2 streams
d) Assuming it’s not a reactor and there’s only 1 stream

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Answer: b [Reason:] Splitting Point: 6 variables – 2 mass balances – 1 knowing compositions are the same – 1 splitting ratio = 2 DOF (Degree of freedom). The splitting point is special because when a stream is split, it generally is split into two streams with equal composition.