Multiple choice question for engineering
1. The Macronutrients concentration needed is __________
a) Greater than 10-2 M
b) Less than 10-2 M
c) Greater than 10 -4 M
d) Less than 10-4 M
Answer: c [Reason:] Macronutrients are needed in concentrations larger than 10-4 M. Carbon, nitrogen, oxygen, hydrogen, sulfur, phosphorus, Mg2+, and K+ are major macronutrients.
2. Which of the following components is not a Micronutrient?
Answer: d [Reason:] Elements used in large quantities by the plant are termed macronutrients, which can be further defined as primary or secondary. The primary nutrients include nitrogen (N), phosphorus (P), and potassium (K). These elements contribute to plant nutrient content, function of plant enzymes and biochemical processes, and integrity of plant cells. Deficiency of these nutrients contributes to reduced plant growth, health, and yield; thus they are the three most important nutrients supplied by fertilizers. The secondary nutrients include calcium (Ca), magnesium (Mg), and sulfur (S).
3. Viruses contain lipids.
Answer: a [Reason:] Viruses with a lipoprotein sheath may contain 25% lipid. Composition as a dry weight % includes <1.
4. Facultative autotrophs can grow under heterotrophic condition.
Answer: a [Reason:] Facultative autotrophs normally grow under autotrophic conditions; however, they can grow under heterotrophic conditions in the absence of CO2 and inorganic energy sources.
5. Which of the following can grow in absence of CO2 or carbon sources?
d) Facultative autotrophs
Answer: d [Reason:] Facultative autotrophs normally grow under autotrophic conditions; however, they can grow under heterotrophic conditions in the absence of CO2 and inorganic energy sources.
6. Which of the following is the key element in the regulation of cell metabolism?
Answer: b [Reason:] The main role of potassium is to provide the ionic environment for metabolic processes in the cytosol, and as such functions as a regulator of various processes including growth regulation. Plants require potassium ions (K+) for protein synthesis and for the opening and closing of stomata, which is regulated by proton pumps to make surrounding guard cells either turgid or flaccid. Phosphorus constitutes about 3% of cell dry weight and is present in nucleic acids and in the cell wall of some gram-positive bacteria such as teichoic acids. Phosphorus is a key element in the regulation of cell metabolism.
7. Which of the following ion is required by the ribosomes?
Answer: b [Reason:] Magnesium, serves as a structural component and is involved as a cofactor in many enzymatic reactions. Magnesium is a component of the chlorophyll structure. Magnesium is required to maintain ribosome integrity. Ribosomes specifically require Mg2+ ions.
8. What do you mean by the term “Trace elements”?
a) Very small amount
b) Medium amount
c) High amount
d) Very high amount
Answer: a [Reason:] Trace elements are essential to microbial nutrition. Lack of essential trace elements increases the lag phase (the time from inoculation to active cell replication in batch culture) and may decrease the specific growth rate and the yield. A trace element is a chemical element whose concentration (or other measure of amount) is very low (a “trace amount”).
9. What is “EDTA”?
a) Magnesium calcium edentate
b) Nitrogen phosphorus edentate
c) Sodium calcium edentate
d) Magnesium phosphorus edentate
Answer: c [Reason:] A specific salt of EDTA, known as sodium calcium edetate, is used to bind metal ions in the practice of chelation therapy, e.g., for treating mercury and lead poisoning. It is used in a similar manner to remove excess iron from the body. EDTA is a chelating agent and a chelating agent is a substance whose molecules can form several bonds to a single metal ion. In other words, a chelating agent is a multidentate ligand.
10. Which of the following is not a defined media?
b) Magnesium chloride
c) Calcium chloride
d) Phenyl acetic acid
Answer: d [Reason:] Defined media contain specific amounts of pure chemical compounds with known chemical compositions. A medium containing glucose, (NH4)2SO4, KH2PO4, and MgCl2 is a defined medium. Complex media contain natural compounds whose chemical composition is not exactly known. Complex medium used in penicillin fermentation includes Phenyl acetic acid by continuous feed in a concentration of about 0.5-0.8% of total (g/liter).
11. Which device is used to measure the amount of energy stored in foods?
b) Adiabatic calorimeter
c) Bomb calorimeter
d) Heat flow calorimeter
Answer: c [Reason:] Complex organic food molecules such as sugars, fats, and proteins are rich sources of energy for cells because much of the energy used to form these molecules is literally stored within the chemical bonds that hold them together. Scientists can measure the amount of energy stored in foods using a device called a bomb calorimeter. With this technique, food is placed inside the calorimeter and heated until it burns. The excess heat released by the reaction is directly proportional to the amount of energy contained in the food.
12. Complex carbohydrates which make up cell wall in plants are?
Answer: c [Reason:] Cellulose is a very important polysaccharide because it is the most abundant organic compound on earth. Cellulose is a major component of tough cell walls that surround plant cells, and is what makes plant stems, leaves, and branches so strong.
13. Micronutrient which is important in transport of sugar, synthesis of enzymes and cell division is?
Answer: a [Reason:] Boron (B) is a micronutrient critical to the growth and health of all crops. It is a component of plant cell walls and reproductive structures. It is a mobile nutrient within the soil, meaning it is prone to movement within the soil. Because it is required in small amounts, it is important to deliver B as evenly as possible across the field. Boron plays a key role in a diverse range of plant functions including cell wall formation and stability, maintenance of structural and functional integrity of biological membranes, movement of sugar or energy into growing parts of plants, and pollination and seed set. Adequate B is also required for effective nitrogen fixation and nodulation in legume crops.
14. The minerals involved in water-splitting reaction during photosynthesis are ___________
a) Manganese and chloride
b) Magnesium and sulfur
c) Magnesium and potassium
d) Magnesium and phosphorus
Answer: a [Reason:] Water splitting is catalyzed by the oxygen-evolving complex (OEC) of protein complex photosystem II (PSII), producing dioxygen gas, protons and electrons. O (2) is released into the atmosphere, sustaining all aerobic life on earth; product protons are released into the thylakoid lumen, augmenting a proton concentration gradient across the membrane; and photo-energized electrons pass to the rest of the electron-transfer pathway. The OEC contains four manganese ions, one calcium ion and (almost certainly) a chloride ion.
15. Energy is measured in ________
Answer: c [Reason:] A calorie is a unit that is used to measure energy. The Calorie you see on a food package is actually a kilocalorie, or 1,000 calories. A Calorie (kcal) is the amount of energy needed to raise the temperature of 1 kilogram of water 1 degree Celsius.
1. “The density of liquids is practically dependent of pressure”?
Answer: b [Reason:] The density of liquids is practically independent of pressure; liquids are incompressible fluids. If the density of a fluid changes with pressure, the fluid is compressible. Gases are generally classed as compressible fluids.
2. What is the property of an ideal or perfect fluid?
a) Compressible and zero viscosity
b) Compressible and zero density
c) Incompressible and zero viscosity
d) Incompressible and zero density
Answer: c [Reason:] Viscosity is the property of fluids responsible for internal friction during flow. An ideal or perfect fluid is a hypothetical liquid or gas which is incompressible and has zero viscosity. The term inviscid applies to fluids with zero viscosity.
3. Fluids which undergo strain rates proportional to the applied shear stress are termed as?
a) Inviscid fluid
b) Newtonian fluid
c) Non- Newtonian fluid
d) Viscous fluid
Answer: b [Reason:] Newtonian fluids undergo strain rates proportional to the applied shear stress. It is defined to be a fluid whose shear stress is linearly proportional to the velocity gradient in the direction perpendicular to the plane of shear. This definition means regardless of the forces acting on a fluid, it continues to flow.
4. Which of the following is not an example of a Newtonian fluid?
c) Non- Drip Paints
Answer: c [Reason:] Stirring a non-Newtonian fluid can cause the viscosity to decrease, so the fluid appears “thinner” (this is seen in non-drip paints). There are many types of non-Newtonian fluids, as they are defined to be something that fails to obey a particular property – for example, most fluids with long molecular chains can react in a non-Newtonian manner.
5. Which of the following is not an example of a Non- Newtonian fluids?
Answer: b [Reason:] Water is a Newtonian fluid, because it continues to display fluid properties no matter how much it is stirred or mixed. A slightly less rigorous definition is that the drag of a small object being moved slowly through the fluid is proportional to the force applied to the object. (Compare friction). Important fluids, like water as well as most gases, behave – to good approximation – as a Newtonian fluid under normal conditions on Earth.
6. What do you mean by the term “Rheology”?
a) Study of materials with both solid and fluid characteristics
b) Study of materials with only solid characteristics
c) Study of materials with only fluid characteristics
d) Study of material with both fluid and gas characteristics
Answer: a [Reason:] It is the study of the flow of matter, primarily in a liquid state, but also as “soft solids” or solids under conditions in which they respond with plastic flow rather than deforming elastically in response to an applied force. It is a branch of physics which deals with the deformation and flow of materials, both solids and liquids.
7. What is the unit of viscosity of fluids in C.G.S?
Answer: c [Reason:] The unit of viscosity (.in C.G.S) is poise.
One poise = 0.1 N.s / m2,
µwater = 10-3 Ns / m2 µ air = 1.81 x 10-5 Ns / m2.
8. If 5 m3 of certain oil weighs 45 kN calculate the specific weight of the oil.
a) 10 kN/m3
b) 9 kN/m3
c) 5 kN/m3
d) 2 kN/m3
Answer: b [Reason:] Given data: Volume = 5 m3
Weight = 45 kN
9. Refer to Q8 and Calculate specific gravity of the oil. (Specific weight of water = 9.807 kN/m3)
Answer: c [Reason:] Given data: Volume = 5 m3
Weight = 45 kN
Specific weight of water = 9.807 kN/m3
10. A liquid has a mass density of 1550 kg/m3. Calculate its specific weight.
a) 1.50×102 N/m3
b) 1.52×104 N/m3
c) 1.54×104 N/m3
d) 1.50×104 N/m3
Answer: b [Reason:] Given data: Mass density = 1550 kg/m3
Specific weight = mass density × Acceleration due to gravity
= 1550 kg/m3 × 9.8 m/s2
= 1.52×104 N/m3.
11. Refer to Q10 and calculate specific gravity.
Answer: a [Reason:] Given data: Mass density = 1550 kg/m3
12. Refer to Q 10 and Q11, and calculate specific volume.
a) 6.51 × 10-5 m3/N
b) 6.78 × 10-5 m3/N
c) 6.45 × 10-5 m3/N
d) 6.57 × 10-5 m3/N
Answer: d [Reason:] Specific volume = 1/(Specific weight )
= 1/(1.52×104 N/m3 )
= 6.57 × 10-5 m3/N.
13. If the equation of a velocity profile over a plate is v = 5y2 + y (where v is the velocity in m/s) determine the shear stress at y =0. Given the viscosity of the liquid is 8.35 poise.
Answer: c [Reason:] Given Data: Velocity profile v = 5y2 + y;
μ = 8.35 poise
Substituting y=0 on the above equation, we get shear stress at respective depths.
τ = 0.835
14. Refer to Q13 and determine the shear stress at y =7.5cm.
a) 1.46 N/m3
b) 1.45 N/m3
c) 1.40 N/m3
d) 1.43 N/m3
Answer: a [Reason:] Given Data: Velocity profile v = 5y2 + y;
μ = 8.35 poise
Substituting y=7.5 on the above equation, we get shear stress at respective depths.
τ = 1.46 N/m3.
15. “Mach is the Dimensionless quantity”.
Answer: a [Reason:] In fluid dynamics, the Mach number (M or Ma) is a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound.
M is the Mach number,
u is the local flow velocity with respect to the boundaries (either internal, such as an object immersed in the flow, or external, like a channel), and
c is the speed of sound in the medium.
1. In the following equation, what is the unit of Dilution rate”D”? (Given: F = dm3h-1 and V = dm3)
D = F/V
Answer: b [Reason:] The flow of medium into the vessel is related to the volume of the vessel by the term dilution rate, D, defined as:
D = F/V
Where, F is the flow rate (dm3h-1)
V is the volume (dm3)
Thus, D is expressed in units h-1.
2. under continuous culture which of the following condition is applicable?
a) μmax < D
b) μmax > D
c) μmax = D
Answer: b [Reason:] The net change in cell concentration over a time period may be expressed as:
dx/dt = growth – output
dx/dt = μx – Dx
Under steady-state conditions, the cell concentration remains constant, thus dx/dt = 0 and:
μ = D
Under steady-state conditions the specific growth rate is controlled by the dilution rate. Under batch culture conditions an organism will grow at its maximum specific growth rate and, therefore, it is obvious that a continuous culture may be operated only at dilution rates below the maximum specific growth rate. Thus, within certain limits the dilution rate may be used to control the growth rate of the culture.
3. If one starts with 10,000 (104 ) cells in a culture that has a generation time of 2 h, how many cells will be in the culture after 4 and 48 h?
a) 4.0×104 cells, 1.7×1011 cells
b) 4.2×104 cells, 1.1×1011 cells
c) 4.6×104 cells, 1.5×1011 cells
d) 4.8×104 cells, 1.3×1011 cells
Answer: a [Reason:] Use the equation X= 2n
, where X0
is the initial number of cells, n is the number of generations, and X is the number of cells after n generations.
After 4 h, n=4 h/2 h per generation= 2 generations:
X=22 (104) = 4.0×104 cells
After 48 h, n= 24 generations:
X = 224 (104) = 1.7×1011 cells
This represents an increase of less than one order of magnitude for the 4-h culture, and seven orders of magnitude for the 48-h culture.
4. “The growth rate of the cells will be greater than the dilution rate”, Is this statement applicable for decrease in biomass production?
Answer: b [Reason:] The growth rate of the cells will be less than the dilution rate and they will be washed out of the vessel at a rate greater than they are being produced, resulting in a decrease in biomass concentration.
5. An alternative of chemostat is turbidostat?
Answer: a [Reason:] An alternative type of continuous culture to the chemostat is the turbidostat, where the concentration of cells in the culture is kept constant by controlling the flow of medium such that the turbidity of the culture is kept within certain, narrow limits.
6. What do you mean by wall growth in bioreactors?
a) Growth of cells in the wall
b) Consumption of paints by the cells from the coated walls
c) Biomass concentration is increased
d) Immobilized cells consumes substrate within reactors
Answer: d [Reason:] Wall growth is commonly encountered practical difficulty in which the organism adheres to the inner surfaces of the reactor resulting an increase in heterogeneity. The immobilized cells are not subject to removal from the vessel but will consume substrate resulting in the suspended biomass concentration being lower than predicted.
7. The yield factor is proportional to the dilution rate?
Answer: a [Reason:] This is attributed to the phenomenon of microorganisms utilizing a greater proportion of substrate for maintenance at low dilution rates. Effectively, the yield factor decreases at low dilution rates.
8. Which is the correct meaning of the internal feedback from the following?
a) Effluent stream is more concentrated than in the vessel
b) Effluent stream is less concentrated than in the vessel
c) Effluent stream is equally concentrated with the vessel
d) Effluent stream is not concentrated
Answer: b [Reason:] Internal feedback means that limiting the exit of biomass form the chemostat such that the biomass in the effluent stream is less concentrated than in the vessel.
9. Mixed cultures can be maintained using chemostat cultures in _________
a) Fed-Batch culture
b) Semi-Batch culture
c) Batch culture
d) Continuous culture
Answer: d [Reason:] Important feature of chemostats and other continuous culture systems is that they are well-mixed so that environmental conditions are homogenous or uniform and microorganisms are randomly dispersed and encounter each other randomly. Therefore, competition and other interactions in the chemostat are global, in contrast to biofilms.
10. Which of the following is not an advantage of continuous culture?
a) Can be used for different reactions every day
b) Little risk of infection or strain mutation
c) Long growth periods of subtrates/microbes
d) Eliminating the inherent down time for cleaning and sterilization
Answer: c [Reason:] Long growth periods not only increase the risk of contamination but also dictate that the bioreactor must be extremely reliable and consistent, incurring a potentially larger initial expenditure in higher quality equipment.
11. Which of the following is not a disadvantage of continuous culture?
a) Long growth periods of subtrates/microbes
b) Maintenance of mixed cultures
c) Requires feed-batch culturing
d) Viscosity of mixture for filamentous organisms
Answer: b [Reason:] Continuous culture provides a higher degree of control than a batch culture. Growth rates can be regulated and maintained for extended periods. By varying the dilution rate, biomass concentration can be controlled. Secondary metabolite production can be sustained simultaneously along with growth. In steady state continuous culture, mixed cultures can be maintained using chemostat cultures – unlike in a batch process where one organism usually outgrows another.
12. Continuous addition of sugars in “Fed-batch” fermentation is done to __________
a) Produce methane
b) Purify enzymes
c) Degrade sewage
d) Obtain antibiotics
Answer: b [Reason:] A fed-batch is a biotechnological batch process which is based on feeding of growth limiting nutrient substrate to a culture continuous addition of sugars in fed batch fermentation is done to purify enzymes.
13. Name the phase which is a period of adaptation of the cells to the new environment.
a) Lag phase
b) Log phase
c) Exponential phase
d) Stationary phase
Answer: a [Reason:] Lag time is defined as the initial period in the life of a bacterial population when cells are adjusting to a new environment before starting exponential growth. Many factors influence the duration of lag time, including inoculum size, the physiological history of the cells, and the precise physiochemical environment of both the original and the new growth medium.
14. In chemostat, constant cell concentration is maintained?
Answer: b [Reason:] In the chemostat, in the steady state adjusting the concentration of one substrate controls cell growth. In the turbidostat, cell growth is kept constant by using turbidity to monitor the biomass concentration and the rate of feed of nutrient solution is appropriately adjusted. In the chemostat, constant chemical environment is maintained, while in a turbidostat constant cell concentration is maintained.
15. Which of the following is used to grow bacterial cultures continuously?
c) Bacteria cannot be grown in continuous culture
Answer: b [Reason:] A chemostat (from chemical environment is static) is a bioreactor to which fresh medium(Bacteria) is continuously added, while culture liquid containing left over nutrients, metabolic end products and microorganisms are continuously removed at the same rate to keep the culture volume constant.
1. Convection does not occur in _______________
d) Molten Solid
Answer: a [Reason:] Convection cannot take place in most solids because neither bulk current flows nor significant diffusion of matter can take place. Diffusion of heat takes place in rigid solids, but that is called heat conduction. Convection, however, can take place in soft solids or mixtures where solid particles can move past each other.
2. The unit of mass transfer coefficient is ______________
Answer: b [Reason:] (mol/s)/(m2•mol/m3) = m/s
Note, the units will vary based upon which units the driving force is expressed in. The driving force shown here as ‘ΔcA’ is expressed in units of moles per unit of volume, but in some cases the driving force is represented by other measures of concentration with different units. For example, the driving force may be partial pressures when dealing with mass transfer in a gas phase and thus use units of pressure.
3. Rate of mass transfer is inversely proportional to the rate of reaction at solid surface?
Answer: b [Reason:] Estimation of the interfacial concentration CAi is more difficult; measuring compositions at phase boundaries is not easy experimentally. To overcome this problem, we must consider the processes in the system which are linked to mass transfer of A. Transport of A is linked to reaction at the surface of the solid, so that the value of CAi will depend on the rate of consumption of A at the interface. In practical terms, we can therefore calculate the rate of mass transfer of A only if we have information about the rate of reaction at the solid surface. Simultaneous reaction and mass transfer occurs in many bioprocesses.
4. In the two-phase aqueous systems, in ordinary situations the bulk phases are not in equilibrium?
Answer: a [Reason:] Normally, it can be assumed that there is negligible resistance to mass transfer at the actual interface, i.e. within distances corresponding to molecular free paths on either side of the phase boundary. This is equivalent to assuming that the phases are in equilibrium at the interface; therefore, CA1i and CA2i are equilibrium concentrations. As a result it is known that there are special situations, such as when there is adsorption of material at the interface, where the assumption is invalid. However, in ordinary situations, the evidence is that equilibrium does exist at the interface between phases. If the bulk liquids were in equilibrium, no net mass transfer would take place.
5. The distribution coefficient is accurate only if?
a) Solvents are miscible
b) Solutes are miscible
c) Solvents are immiscible
d) Occurrence of chemical reaction
Answer: c [Reason:] Equilibrium distribution of one solute between two phases is conveniently described in terms of the distribution law. At equilibrium, the ratio of solute concentrations in the two phases is given by the distribution coefficient or partition coefficient, m. The distribution law is accurate only if both solvents are immiscible and there is no chemical reaction.
6. In a wetted-wall tower, an air-H2S mixture is flowing by a film of water which is flowing as a thin film down a vertical plate. The H2S is being absorbed from the air to the water at a total pressure of 1.50 atm abs and 30oC. The value of kc of 9.567×10-4 m/s has been predicted for the gas-phase mass-transfer coefficient. At a given point the mole fraction of H2S in the liquid at the liquid-gas interface is 2.0×10-5 and pA of H2S in the gas is 0.05 atm. The Henry’s law equilibrium relation is pA(atm) = 609xA (mole fraction in liquid). Calculate the rate of absorption of H2S.
a) 1.480×10-3 kmol/m2.s
b) 1.486×10-3 kmol/m2.s
c) 1.485×10-3 kmol/m2.s
d) 1.487×10-3 kmol/m2.s
Answer: b [Reason:] The rate of absorption of H2
S per unit area of the thin film is given by:
The partial pressure of H2S in the gas phase at the interface is determined from Henry’s law and the mole fraction of H2S in the liquid at the liquid-gas interface.
pAi = 609xAi = 609×2.0×10-5 = 1.218×10-2 atm
7. What do you mean by Sherwood number?
a) Dimension mass transfer number
b) Dimensionless mass transfer number
c) Dimension momentum transfer number
d) Dimensionless momentum transfer number
Answer: b [Reason:] The Sherwood number (Sh) (also called the mass transfer Nusselt number) is a dimensionless number used in mass-transfer operation. It represents the ratio of the convective mass transfer to the rate of diffusive mass transport, and is named in honor of Thomas Kilgore Sherwood.
8. A stream of air at 100 kPa pressure and 300 K is flowing on the top surface of a thin flat sheet of solid naphthalene of length 0.2 m with a velocity of 20 m/sec. The other data are:
Mass diffusivity of naphthalene vapor in air = 6 * 10 –6 m2/sec
Kinematic viscosity of air = 1.5 * 10 –5 m2.sc
Concentration of naphthalene at the air-solid naphthalene interface = 1 * 10 – 5 kmol/m3
Calculate the overage mass transfer coefficient over the flat plate.
Note: For heat transfer over a flat plate, convective heat transfer coefficient for laminar flow can be calculated by the equation.
Nu = 0.664 ReL1/2 Pr1/3
you may use analogy between mass and heat transfer.
a) 0.014 m/sec
b) 0.015 m/sec
c) 0.016 m/sec
d) 0.013 m/sec
Answer: a [Reason:] Given: Correlation for heat transfer
Nu = 0.664 ReL1/2 Pr1/3
The analogous relation for mass transfer is:
Sh = 0.664 ReL1/2 Sc1/3
Sh = Sherwood number = kL/D AB
Re L = Reynolds number = Lυρ/µ
Sc = Schmidt number = µ / (ρ D AB)
k = overall mass transfer coefficient
L = length of sheet
D AB = diffusivity of A in B
υ = velocity of air
µ = viscosity of air
ρ = density of air,
and µ/ρ = kinematic viscosity of air.
Substituting for the known quantities in equation:
9. Refer to Q8 and, calculate the rate of loss of naphthalene from the surface per unit width.
a) 0.100 gmol/m.hr
b) 0.102 gmol/m.hr
c) 0.101 gmol/m.hr
d) 0.103 gmol/m.hr
Answer: c [Reason:] Rate of loss of naphthalene = k (C Ai – C A∞)
= 0.014 (1 * 10 –5 – 0) = 1.4024 * 10–7 kmol/m2 sec
Rate of loss per meter width = (1.4024 * 10 –7) (0.2) = 2.8048 * 10 –8 kmol/m.sec
= = 0.101 gmol/m.hr.
10. Psychrometry deals with between which type of phases?
Answer: c [Reason:] Psychrometry is concerned with the physical and thermodynamic properties of gas-vapor mixtures. Although the principles of psychrometry apply to any physical system consisting of gas-vapor mixtures, the most common system of interest is the mixture of water vapor and air, because of its application in heating, ventilating, and air-conditioning and meteorology. In human terms, our thermal comfort is in large part a consequence of not just the temperature of the surrounding air, but (because we cool ourselves via perspiration) the extent to which that air is saturated with water vapor.
11. Which type of columns are used for liquid dispersion in a continuous gas phase?
b) Bubble cap
c) Sieve – plate
d) Fluidized bed
Answer: a [Reason:] Packed Beds. Although packed bed columns are used most often for absorption, they are also used for the distillation of vapor-liquid mixtures. The packing provides a large surface area for vapor-liquid contact, which increases the column’s effectiveness.
12. Heat transfer by molecular collision in ____________________
Answer: b [Reason:] Flow of heat through currents within a fluid (liquid or gas). Convection is the displacement of volumes of a substance in a liquid or gaseous phase. When a mass of a fluid is heated up, for example when it is in contact with a warmer surface, its molecules are carried away and scattered causing that the mass of that fluid becomes less dense. For this reason, the warmed mass will be displaced vertically and/or horizontally, while the colder and denser mass of fluid goes down (the low-kinetic-energy molecules displace the molecules in high-kinetic-energy states). Through this process, the molecules of the hot fluid transfer heat continuously toward the volumes of the colder fluid.
13. Convection is faster than conduction?
Answer: a [Reason:] Forced air heating and air conditioning are examples of heating (or cooling) by convection. This is an effective way of bringing a hot (or cold) fluid to a different area. Convection transfers heat over a distance faster than conduction.
14. Which is the fastest mode of transfer of heat?
Answer: d [Reason:] Radiation is the fastest mode of transfer of heat, because radiation travels at the speed of light, which is very quick. The slowest mode of transfer of heat is conduction because it takes place from particle to particle. The fastest is radiation at the speed of light 3,00,000 km in one second.
15. Heat transfer takes place as per ______________
a) Zeroth law of thermodynamics
b) First law of thermodynamics
c) Second law of thermodynamics
d) Kirchoff’s law
Answer: c [Reason:] Second Law of Thermodynamics states that It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.
1. What do you mean by the term “Ammoniacal Nitrogen”?
a) A safe disposal
b) An unsafe disposal
c) A metal
d) A non-toxic substance
Answer: b [Reason:] Ammoniacal nitrogen (NH3-N), is a measure for the amount of ammonia, a toxic pollutant often found in landfill leachate and in waste products, such as sewage, liquid manure and other liquid organic waste products. Ammonia can directly poison humans and upset the equilibrium of water systems.
2. Disposal in sea and rivers are safe.
Answer: b [Reason:] Beyond the standards of disposal in water bodies is toxic. In addition, levels of ammoniacal nitrogen may be stipulated. There are as well, often stringent upper limits for toxic metals and chemicals which might kill the fauna and flora.
3. Lagoon is a part of sea water.
Answer: a [Reason:] Lagoon is an area of sea water separated from the sea by a reef (= a line of rocks and sand).
4. What is a Facultative pond?
a) Top layer is anaerobic
b) Bottom layer is aerobic
c) Aerobic throughout depth
d) Top layer is aerobic
Answer: d [Reason:] The most often used ponds in domestic wastewater treatment are the stabilization pond and facultative lagoon. The stabilization pond is designed to be aerobic throughout its depth and the facultative lagoon will be anaerobic at the bottom and aerobic at the top.
5. Deep water ponds are mechanically agitated.
Answer: a [Reason:] Achieving good agitation is an important part of this. Failure to properly agitate the manure will result in a continuous buildup of settled solids within the storage, resulting in less and less available storage as time goes by. Good agitation of the manure will re-suspend those settled solids and facilitate their removal from the storage ensuring we maintain that capacity we need. Additionally, agitation of the manure helps homogenize it and provide a more consistent nutrient content as it is applied.
6. Which of the following step is required for the spray irrigation?
a) Waste should be chlorinated
b) Waste should not be chlorinated
c) Land high above 38mm rainfall
d) Land with below 38mm rainfall
Answer: a [Reason:] The waste are initially chlorinated to lower the BOD and reduce unpleasant odors and then sprayed on to land until the equivalent of 38mm of rainfall is reached. The process is repeated at monthly intervals.
7. What is micro spray irrigation?
Answer: c [Reason:] Micro irrigation is defined as the frequent application of small quantities of water directly above and below the soil surface; usually as discrete drops, continuous drops or tiny streams through emitters placed along a water delivery line. Drip Irrigation. Simcha Blass, an Israeli hydraulic engineer, is credited with the discovery and introduction of modern drip irrigation in the early 1930’s. Drip irrigation (also known as micro-irrigation) became more common with the introduction of plastics in the 1950’s.
8. Injection wells are used for waste disposal apart from oil and gas production.
Answer: a [Reason:] An injection well is a device that places fluid deep underground into porous rock formations, such as sandstone or limestone, or into or below the shallow soil layer. The fluid may be water, wastewater, brine (salt water), or water mixed with chemicals.
9. What do you mean by “Leachate”?
a) A worm
b) A solid
c) A fluid
d) A gas
Answer: c [Reason:] A leachate is any liquid that, in the course of passing through matter, extracts soluble or suspended solids, or any other component of the material through which it has passed.
10. Landfill is produced by microbes.
Answer: a [Reason:] Landfill gas is a complex mix of different gases created by the action of microorganisms within a landfill. Landfill gas is approximately forty to sixty percent methane, with the remainder being mostly carbon dioxide.
11. Landfill gas (LFG) contains Nonmethane organic compounds (NMOCs).
Answer: a [Reason:] By volume, landfill gas typically contains 45% to 60% methane and 40% to 60% carbon dioxide. Landfill gas also includes small amounts of nitrogen, oxygen, ammonia, sulfides, hydrogen, carbon monoxide, and nonmethane organic compounds (NMOCs) such as trichloroethylene, benzene, and vinyl chloride.
12. Landfills do not decompose.
Answer: b [Reason:] The solid waste layer becomes laced with these strips of dirt. Landfills are not designed to break down waste, only to store it. But garbage in a landfill does decompose, albeit slowly and in a sealed, oxygen-free environment.
13. Incineration is a “Cold treatment” process.
Answer: b [Reason:] Incineration is a waste treatment process that involves the combustion of organic substances contained in waste materials. Incineration and other high-temperature waste treatment systems are described as “thermal treatment”. Incineration of waste materials converts the waste into ash, flue gas and heat. The ash is mostly formed by the inorganic constituents of the waste, and may take the form of solid lumps or particulates carried by the flue gas. The flue gases must be cleaned of gaseous and particulate pollutants before they are dispersed into the atmosphere. In some cases, the heat generated by incineration can be used to generate electric power.
14. What do you mean by the term “Flue gas”?
a) Dry CO2
b) Chimney gas
c) Non-combustible gas
d) Cold gas
Answer: b [Reason:] A flue is a duct, pipe, or opening in a chimney for conveying exhaust gases from a fireplace, furnace, water heater, boiler, or generator to the outdoors. Historically the term flue meant the chimney itself. Flue gas is the gas exiting to the atmosphere via a flue, which is a pipe or channel for conveying exhaust gases from a fireplace, oven, furnace, boiler or steam generator. Quite often, the flue gas refers to the combustion exhaust gas produced at power plants.
15. Flue gas does not need pre-treatment.
Answer: b [Reason:] Flue Gas Treatment. Waste incineration and many other industrial processes generate flue gases. These often contain pollutants such as sulfur oxides (SO2 + SO3), hydrochloric acid (HCl), hydrofluoric acid (HF) as well as heavy metals, dioxins and furans. So it needs to undergo pre-treatment process.