Multiple choice question for engineering
1. When the boundary layer is present in the shell mass transfer then?
a) CAs = CAb
b) CAs ≠ CAb
c) CAs < CAb
d) CAs > CAb
Answer: d [Reason:] The term CAs is the concentration of substrate A at the external surface of the catalyst. This term made its way into the analysis in the boundary conditions used for solution of the shell mass balance.
Reduction in substrate concentration from CAb in the bulk liquid to CAs at the catalyst surface occurs across the boundary layer surrounding the solid. In the absence of the boundary layer, CAS = CAb, which is easily measured. When the boundary layer is present, CAs takes some value less than CAb.
2. What is the dimension of liquid – phase mass-transfer coefficient “kS”?
Answer: a [Reason:] Rate of mass transfer across a liquid boundary layer is represented by the following equation:
NA = kSa (CAb — CAS)
where NA is the rate of mass transfer, ks is the liquid-phase mass-transfer coefficient with dimensions LT-1, and a is the external surface area of the catalyst. If NA is expressed per volume of catalyst with units of, for example, kg mol s-1 m-3, to be consistent, a must also be expressed on a catalyst-volume basis with units of, for example, m2 m-3 or m-1.
3. If Ω << 1, then _________
a) CAS = CAb
b) CAS < CAb
c) CAS> CAb
d) CAS = CAb = 0
Answer: a [Reason:] Ω is an observable modulus for external mass transfer. If Ω << 1, CAS = CAb and external mass-transfer effects are insignificant. Otherwise, CAS < CAb and external mass-transfer effects are significant.
4. Removing the boundary layer, CAS would decrease.
Answer: b [Reason:] Removing the boundary layer would increase the value of CAS, thus establishing a greater driving force for internal mass-transfer and reducing the likelihood of CA falling to zero inside the particle.
5. Mass transfer resistance leads to an increase in the concentration of reactants at the catalyst surface in bulk fluid.
Answer: b [Reason:] In absence of mass transfer there is no difference in concentration of reactant at bulk and near the catalyst surface. However, presence of significant mass transfer resistance results in a decrease in the concentration of reactants at the catalyst surface compared to that in the bulk fluid. Consequently, the observed rate is less than the intrinsic rate evaluated at bulk fluid reactant concentration.
6. Denitrifying bacteria are immobilised in gel beads and used in a stirred reactor for removal of nitrate from groundwater. At a nitrate concentration of 3 g m-3, the conversion rate is 0.011 g s-1 m-3 catalyst. The effective diffusivity of nitrate in the gel is 1.5×10-9 m2 s-1, the beads are 6 mm in diameter, and the liquid-solid mass-transfer coefficient is 10-5 m s-1. Km for the immobilised bacteria is approximately 25 g m-3. What is the value of an observable modulus for external mass transfer, Ω?
7. Refer to Q6 and calculate the value of CAs.
a) 1.9 g m-3
b) 0.9 g m-3
c) 2.9 g m-3
d) 3.9 g m-3
Explanation: CAs/CAb = 1 – Ω = 0.63
CAs = 0.63 CAb = 0.63 (3 gm-3) = 1.9 g m-3.
8. If the reaction is endothermic, the temperature at the catalyst surface will be less.
Answer: a [Reason:] If the reaction is endothermic, the temperature at the catalyst surface will be less than the bulk fluid temperature and the observed rate will be less than that determined at the bulk fluid temperature.
9. Rate at the surface will be higher due to heat transfer limitations.
Answer: a [Reason:] If the reaction is endothermic, the temperature at the catalyst surface will be less than the bulk fluid temperature and the observed rate will be less than that determined at the bulk fluid temperature. If the reaction is exothermic, the temperature of catalyst surface will be more than the bulk fluid temperature. Therefore, the observed rate will be higher or lower than corresponding bulk fluid conditions, depending on both heat transfer and mass transfer effects. Rate at the surface will be higher due to heat transfer limitations and lower due to mass transfer limitations. Depending on the relative magnitude, the overall rate will be increased or decreased.
10. If a species appears in the numerator of the rate law, it is probably a product.
Answer: b [Reason:] If a species appears in the numerator of the rate law, it is probably a reactant and if a species appears in the denominator of the rate law, it is probably adsorbed in the surface.
11. Molar flux consist of ______________
a) Molar flux = partial flux + diffusion
b) Total flux = bulk motion + diffusion
c) Molar flux = solid motion + flow rate
d) Total flux = partial flux + flow rate
Answer: b [Reason:] Molar flux consists of two parts:
– Bulk motion of the fluid, BA
– Molecular diffusion flux relative to the bulk motion of the fluid produced by a concentration gradient, JA
– WA = BA + JA (total flux = bulk motion + diffusion).
12. What is the unit of heat transfer coefficient?
a) Watt m2/ k
b) Watt m2 k
c) Watt/ m2 k
d) Watt/ m2k2
Answer: c [Reason:] Heat flux form a bulk fluid at T∞ to the surface of a spherical particle of diameter dp at TS: Newtons law of cooling:
q = h (T∞ – TS)
Where, q (Heat flux) = Watt/ m2 h (Heat transfer coefficient) = Watt/ m2K.
13. What is the unit of mass transfer coefficient?
Answer: b [Reason:] Molar flux form a bulk fluid at concentration CAb to the surface of a spherical particle of diameter dp at concentration CAs:
WA = kc (CAb-CAs)
Where, WA (Molar flux) = moles/m2s
kC (Mass transfer coefficient) = m/s.
14. The ratio of momentum diffusivity (kinematic viscosity) and mass diffusivity is referred as:
a) Nusselt number
b) Schimdt number
c) Sherwood number
d) Reynolds number
Answer: b [Reason:] Schmidt number (Sc) is a dimensionless number defined as the ratio of momentum diffusivity (kinematic viscosity) and mass diffusivity, and is used to characterize fluid flows in which there are simultaneous momentum and mass diffusion convection processes.
15. The ratio of inertial forces to viscous forces is called as ________
a) Nusselt number
b) Schimdt number
c) Sherwood number
d) Reynolds number
Answer: d [Reason:] The Reynolds number is the ratio of inertial forces to viscous forces and is a convenient parameter for predicting if a flow condition will be laminar or turbulent. It can be interpreted that when the viscous forces are dominant (slow flow, low Re) they are sufficient enough to keep all the fluid particles in line, and then the flow is laminar. Even very low Re indicates viscous creeping motion, where inertia effects are negligible. When the inertial forces dominate over the viscous forces (when the fluid is flowing faster and Re is larger) then the flow is turbulent.
1. In the following Vand’s equation what does “φ”, represents?
μ = μL (1+ 2.5 φ + 7.25 φ2)
a) Mass fraction of solids
b) Mass fraction of liquids
c) Volume fraction of solids
d) Volume fraction of liquids
Answer: c [Reason:] The viscosity of a suspension of spheres in Newtonian liquid can be predicted using the Vand equation, where μL is the viscosity of the suspending liquid and φ is the volume fraction of solids.This equation has been found to hold for yeast and spore suspensions up to 14 vol% solids.
2. The filamentous cells will become flexible under which of the following condition?
a) Increased osmotic pressure and low turgor pressure
b) Decreased osmotic pressure and high turgor pressure
c) Increased osmotic pressure and high turgor pressure
d) Decreased osmotic pressure and low turgor pressure
Answer: a [Reason:] Osmotic pressure of the culture medium affects cell turgor pressure. This in turn affects the hyphal flexibility of filamentous cells; increased osmotic pressure gives a lower turgor pressure making the hyphae more flexible. Improved hyphal flexibility reduces broth viscosity, and can also have a marked effect on yield stress.
3. The temperature is directly proportional to viscosity.
Answer: b [Reason:] A fluid’s viscosity strongly depends on its temperature. Along with the shear rate, temperature really is the dominating influence. The higher the temperature is, the lower a substance’s viscosity is. Consequently, decreasing temperature causes an increase in viscosity. The relationship between temperature and viscosity is inversely proportional for all substances. A change in temperature always affects the viscosity – it depends on the substance just how much it is influenced by a temperature change. For some fluids a decrease of 1°C already causes a 10 % increase in viscosity.
4. The pressure is directly proportional to viscosity?
Answer: a [Reason:] Viscosity increases with increasing pressure because the amount of free volume in the internal structure decreases due to compression. Consequently, the molecules can move less freely and the internal friction forces increase. The result is an increased flow resistance.
5. Convert 5 cP to Pa.s and estimate the correct answer.
Answer: d [Reason:] 1 centipoise [cP] = 0.001 pascal second [Pa•s]
5 centipoise = 0.005 pascal second.
6. If the particle size is small in fluid than viscosity will ___________
c) Remain the same
d) Slightly increase
Answer: b [Reason:] The size of the particles of a substance will greatly affect its viscosity. Small particles can move more easily past each other and can therefore flow faster, meaning they have a lower viscosity. Large particles would mean a higher viscosity.
7. The force of attraction between the particles is directly proportional to viscosity.
Answer: a [Reason:] Particles of the same substance have an attractive force on one another. Some substances have a strong attraction while some substances have a weaker attraction. The stronger the attraction of particles, the higher the viscosity.
8. Conversion of 5 Pa-s into Poise will be ____________
Answer: b [Reason:] Ten poise equal one pascal second [Pa s] making the centipoise [cP] and millipascal second [mPa s] identical.
1 Pa-s = 10 P
5 Pa-s = 50 P.
9. When a gas is heated its viscosity will ___________
c) Remain same
d) Slightly increase
Answer: a [Reason:] A gas’s viscosity increases when heated (more energy causing more movement + therefore more friction because particles are hitting each other). Decreases when cooled.
10. A fluid is flowing between two layers. Calculate the shearing force if the shear velocity is 0.25 m/s and has length 2 m and dynamic viscosity is 2Ns/m2.
a) 0.2 N/m2
b) 0.4 N/m2
c) 0.5 N/m2
d) 0.6 N/m2
Answer: c [Reason:] Given: Shear velocity u = 0.25 m/s,
shear stress y = 0.125/s
dynamic viscosity μ = 2 Ns/m2
The shearing stress is given by,
F = μ A u/y
F = 2 Ns/m2 × 0.125 /s
F = 0.5 N/m2 Therefore, the shearing stress is 0.5 N/m2.
11. Calculate the density of fluid having an absolute viscosity of 0.89 Ns/m2 and kinematic viscosity of 2 m2/s.
a) 0.445 kg/m3
b) 0.440 kg/m3
c) 0.544 kg/m3
d) 0.550 kg/m3
Answer: a [Reason:] Known: Absolute viscosity μ = 0.89 Ns/m2
, kinematic viscosity ν = 2 m3
The density is given by,
Therefore, the density of fluid is 0.445kg/m3.
1. A concentrated solution of the limiting substrate is added in fed-batch at a rate?
a) Greater than the solution of the limiting substrate and the same medium used to establish the batch culture
b) Less than the solution of the limiting substrate and the same medium used to establish the batch culture
c) Equal to the solution of the limiting substrate and to the same medium used to establish the batch culture
d) Negligible to the solution of the limiting substrate and to the same medium used to establish the batch culture
Answer: b [Reason:] A concentrated solution of the limiting substrate is added in fed-batch at a rate less than the solution of the limiting substrate and the same medium used to establish the batch culture, resulting in an increase in volume.
2. Which of the following Fed-batch system is described as a fixed volume?
a) The same medium used to establish the batch culture
b) A solution of the limiting substrate
c) A concentrated solution of the limiting substrate
d) A very concentrated solution of the limiting substrate
Answer: d [Reason:] A very concentrated solution of the limiting substrate added at a rate less than the same medium used to establish the batch culture, a solution of the limiting substrate and a concentrated solution of the limiting substrate, resulting in an insignificant increase in volume.
3. What do you mean by “Quasi steady state”?
a) Cell concentration remains virtually constant
b) Cell concentration is virtually variable
c) Total biomass remains constant with time
d) Total biomass decreases with time
Answer: a [Reason:] The total biomass in the culture (X) increases with time, cell concentration (x) remains virtually constant, that is dx/dt ≅ 0 and therefore μ ≅ D. This situation is termed as a quasi steady state.
4. Quasi – steady state in fed batch is when?
a) Growth rate remains constant
b) Dilution remains constant
c) Growth rate changes variably
d) μ remains constant
Answer: c [Reason:] The major difference between the steady state of a chemostat and the quasi steady state of a fed batch culture is that μ is constant in the chemostat but decreases in fed-batch. However, in the genuine steady state of a chemostat, dilution rate and growth rate are constant whereas in a fed-batch quasi steady state they change over the time of fermentation.
5. In a fixed-volume fed- batch culture μ declines when?
a) Biomass increases
b) Biomass decreases
c) Biomass remains constant
d) Biomass is equal to zero
Answer: a [Reason:] The μ declines when the limiting substrate concentration remains virtually constant, biomass increases and the concentration of the non-limiting nutrients declines.
6. Dilution refilling is done with______
a) Non- sterile water
b) Sterile water
Answer: b [Reason:] Dilution would be achieved by withdrawing culture and refilling to the original level with sterile water or medium not containing the feed substrate.
7. What will be the condition when pH will be high?
a) Glucose is low
b) Glucose is equal with biomass concentration
c) Glucose is high with biomass concentration
d) Excess of glucose
Answer: a [Reason:] Glucose starvation may result in the organic nitrogen in the medium being used as a carbon source, resulting in high pH and inadequate biomass formation.
8. Fed-batch culture is superior to conventional batch culture.
Answer: a [Reason:] The advantage of the fed-batch culture is that one can control concentration of fed-substrate in the culture liquid at arbitrarily desired levels (in many cases, at low levels). Generally speaking, fed-batch culture is superior to conventional batch culture when controlling concentrations of a nutrient (or nutrients) affect the yield or productivity of the desired metabolite.
9. Fed-batch culture is not used for substrate inhibition.
Answer: b [Reason:] Nutrients such as methanol, ethanol, acetic acid, and aromatic compounds inhibit the growth of microorganisms even at relatively low concentrations. By adding such substrates properly lag-time can be shortened and the inhibition of the cell growth markedly reduced.
10. In fed-batch culture, the feed solution is ________
a) Less concentrated
b) Highly concentrated
c) Highly diluted
Answer: b [Reason:] The fed-batch strategy is typically used in bio-industrial processes to reach a high cell density in the bioreactor. Mostly the feed solution is highly concentrated to avoid dilution of the bioreactor.
11. Organisms in which phase are adapting to the new environment?
a) Lag phase
b) Death phase
c) Exponential phase
d) Stationary phase
Answer: a [Reason:] During batch culture, a typical bacterial growth curve shows five distinct phases of growth: lag phase, the delay before the start of exponential growth; exponential phase, where cell division proceeds at a constant rate; stationary phase, when conditions become unfavorable for growth and bacteria stop replicating; death phase, when cells lose viability; and, finally, long-term stationary phase, which can extend for years. It has been assumed that lag phase allows the adaptation required for bacterial cells to begin to exploit new environmental conditions. This process could include the repair of macromolecular damage that accumulated during stationary phase and the synthesis of cellular components necessary for growth.
12. Which type of media is used for fungi cultivation?
a) Non nutrient agar
b) Sabouraud’s dextrose agar
c) MacConkey’s agar
Answer: b [Reason:] Sabouraud Agar or Sabouraud Dextrose Agar or SDA is a type of agar growth medium containing peptones. It is used to cultivate dermatophytes and other types of fungi, and can also grow filamentous bacteria such as Nocardia. It has utility for research and clinical care.
13. Conversion of substrate to amino acids is called __________
Answer: b [Reason:] Proteolysis is the breakdown of proteins into smaller polypeptides or amino acids. Uncatalysed, the hydrolysis of peptide bonds is extremely slow, taking hundreds of years. Proteolysis is typically catalysed by cellular enzymes called proteases, but may also occur by intra-molecular digestion. Low pH or high temperatures can also cause proteolysis non-enzymatically.
14. Conversion of substrate to ammonia and organic acid is by _______
Answer: a [Reason:] Nitrification is the biological oxidation of ammonia or ammonium to nitrite followed by the oxidation of the nitrite to nitrate. The transformation of ammonia to nitrite is usually the rate limiting step of nitrification. Nitrification is an important step in the nitrogen cycle in soil. Nitrification is an aerobic process performed by small groups of autotrophic bacteria and archaea.
15. In the fed-batch method the critical elements of the nutrient solution are added in which amount of concentrations at the beginning of the fermentation?
c) Very large
Answer: a [Reason:] Fed-batch cultivation can provide the solution to substrate inhibition problem by slow feeding of nutrients to the bioreactor; however it can still not address the severe inhibition problem due to accumulating high product concentrations. The optimal design of fed-batch cultivation has to take in to account several factors in to consideration for example time to start the fresh nutrient feed (in the end or when the culture is exponentially growing) what should be the substrate concentration in the feed and its rate of addition and when to finish the nutrient feeding so that the highest concentration of product is produced and no unconverted substrate when the reactor is full.
1. What does the term Positive pressure means?
a) The contaminants of the room should not flow out into the surrounding
b) There should be the balance between the room and surrounding of air and contaminants
c) The pressure makes room isolated from the surrounding and contaminants
d) The air pressure is lower than the surrounding
Answer: c [Reason:] Positive pressure refers to pressure that exceeds the surrounding pressure of any room, chamber or confined space. Positive pressure is maintained in a closed zone to ensure no outside contaminated gaseous or liquid substance can get into that protected zone.
2. For which phase filter aid is used to filter?
a) Gas and liquid
b) Solid and Liquid
Answer: b [Reason:] Filtration is the separation of solids from liquids by forcing the liquid to flow through a porous medium and depositing the solids on the medium. A filter aid (finely divided material added to the liquid to be filtered) helps control flow and solids removal.
Precoat filtration is a type of mechanical filtration that can be used to clarify liquids. Filter aids such as diatomaceous earth, perlite and cellulose enables a liquid to pass through while retaining the haze-causing particles.
3. “Filter aid can also be added to the fermentation broth to increase the porosity of the cake as it forms”, what does this statement means?
a) To increase the pH of the fermentation broth
b) To decrease the filtration rate
c) To decrease the pI of the fermentation broth
d) To increase the filtration rate
Answer: d [Reason:] The additive should provide a thin layer of solids having high porosity (0.85 to 0.90) over the filter medium’s external surface. As Suspension particles will ideally form a layered cake over the filter aid cake layer. The high porosity of the filter aid layer will ensure a high filtration rate. Porosity is not determined by pore size alone. High porosity is still possible with small size pores.
And at high pH and low pI the proteins will not denature and not coagulate and therefore more aggregates will not be formed so there will be less filtration rate.
4. Is plate frame filter a batch filter?
Answer: a [Reason:] It is a batch filter so in between batches time wastage is considerable. Batch discharge after each cycle requires detailed consideration of ways of receiving and storing cake, or of converting it to a continuous stream for delivery to an ultimate disposal method.
5. A rotary drum filter with 30% submergence is used to filter concentrated aqueous slurry of calcium carbonate containing 15lb of solids per cubic foot of water. The pressure–drop is 2011Hg. If the filter cake contains 50% mixture (wet basis), calculate the filter area required to filter 10 gal/min of slurry. The filter cycle time is 5 min. The specific cake resistance depends on the pressure–drop as:
α (ft/lb) = α ̥(Δp)0.26, where α = 2.90×1010
Viscosity of the filtrate = 6.72×10-4 lb/ft-s. Density of calcium carbonate = 168.8 lb/ft3.Neglect filter medium resistance.
a) ≅ 88 ft2
b) ≅ 86 ft2
c) ≅ 82 ft2
d) ≅ 90 ft2
Answer: c [Reason:] The expression for the production rate of cake.
6. A city is to install rapid sand filters downstream of the clarifiers. The design loading rate is selected to be 160 m3/ (m2.d). The design capacity of the water works is 0.35 m3/s. The maximum surface per filter is limited to 50 m2. Design the number and size of filters.
a)189 m2, 4.78
b) 189 m2, 3.78
c) 167 m2,2.66
d) 167 m2,2.78
7. Refer to Q6, and find out the normal filtration rate?
a) 154.3 m3/(m2.d)
b) 154.6 m3/(m2.d)
c) 155.6 m3/(m2.d)
d) 155.3 m3/(m2.d)
Answer: a [Reason:] Select four filters .
The surface area for each filter is, a = 189 m2
/4 = 47.25 m2
We can use 7 m×7 m or 6 m×8 m or 5.9 m×8 m (exact)
If a 7 m×7 m filter is installed, the normal filtration rate is
V = Q/A = (0.35 m3/s×86,400s/d)/(4×7 m×7 m) = 154.3 m3/(m2.d)
8. Can Microfiltration be used to remove viruses?
Answer: b [Reason:] Microfiltration cannot be used to remove viruses as a microfiltration filter has a pore size around 0.1 micron, so when water undergoes microfiltration, many microorganisms are removed, but viruses remain in the water. Ultra filtration would remove these larger particles, and may remove some viruses.
9. What type of filter does not come under the Membrane filters?
a) Ultra filtration
c) Precoat (filter aid) filtration
Answer: c [Reason:] A filter cake is formed by the substances that are retained on a filter. The filter cake grows in the course of filtration, becomes “thicker” as particulate matter is being retained. With increasing layer thickness the flow resistance of the filter cake increases whereas membrane filters do not use any filter aid.
10. What do you mean by the term”Backwashing”?
a) Excluding intermittent use of compressed air during the process
b) Is a form of destructive maintenance
c) Washing at the backside of the filter
d) including intermittent use of compressed air during the process.
Answer: d [Reason:] In terms of water treatment, including water purification and sewage treatment, backwashing refers to pumping water backwards through the filters media, sometimes including intermittent use of compressed air during the process. Backwashing is a form of preventive maintenance so that the filter media can be reused.
11. Which of the following do not comes under continuous filtration?
a) Vacuum rotary drum filter
b) Rubber belt filter
c) Belt filter cake discharge
d) Disc filter
Answer: d [Reason:] Disc filter, this type of filter is fed continuously for a certain amount of time. Then feed is stopped and next batch steps follow, like cake wash, cake blow and cake discharge.
1. Which of the following fluid flow can form “eddies”?
a) Streamline flow
b) Laminar flow
c) Turbulent flow
d) Non-turbulent flow
Answer: c [Reason:] The slower the flow the more closely the streamlines represent actual motion. Slow fluid flow is therefore called streamline or laminar flow. In fast motion, fluid particles frequently cross and recross the streamlines. This motion is called turbulent flow and is characterized by formation of eddies.
2. What do you mean by “u” in the following equation of Reynolds number and for what type of conduit is it applicable?
a) Fluid viscosity, Stirred vessel
b) Average linear velocity, Pipe
c) Average linear velocity, Stirred vessel
d) Fluid viscosity, Pipe
Answer: b [Reason:] A parameter used to characterise fluid flow is the Reynolds number. For full flow in pipes with circular cross-section, Reynolds number Re is defined as:
where D is pipe diameter, u is average linear velocity of the fluid, ρ is fluid density, and μ is fluid viscosity. For stirred vessels there is another definition of Reynolds number:
where Rei is the impeller Reynolds number, Ni is stirrer speed, Di is impeller diameter, ρ is fluid density and μ is fluid viscosity.
3. In smooth pipes, Laminar flow is encountered at what value of Reynolds number ?
a) More than 2100
c) More than 4000
d) Less than 2100
Answer: d [Reason:] One of the most significant outcomes of Reynolds’ experiments is that there is a critical Reynolds number which marks the upper boundary for laminar flow in pipes. In smooth pipes, laminar flow is encountered at Reynolds numbers less than 2100. Under normal conditions, flow is turbulent at Re above about 4000. Between 2100 and 4000 is the transition region where flow may be either laminar or turbulent depending on conditions at the entrance of the pipe and other variables.
4. What do you mean by incompressible flow?
a) Temperature is constant
b) Density is constant
c) Pressure is constant
d) Velocity is constant
Answer: b [Reason:] In fluid mechanics or more generally continuum mechanics, incompressible flow(isochoric flow) refers to a flow in which the material density is constant within a fluid parcel an infinitesimal volume that moves with the flow velocity.
5. Is water incompressible?
Answer: b [Reason:] An important concept in fluid mechanics is that liquids, like water and oil, cannot be compressed much when you push down on them in an enclosed container.So, liquids are considered to be incompressible fluids.
6. A garden hose of diameter 2 cm is used to fill a 20 liter bucket. If it takes 1 minute to fill the bucket, what is the speed at which water enters the hose?
a) 106.1 cm/s
b) 106 cm/s
c) 106.5 cm/s
d) 106.8 cm/s
Answer: a [Reason:] The cross-sectional area of the hose will be given by:
A1= π r2 = π (2 cm/2)2 = π cm2
To find the velocity, v1, we use
7. Refer to Q6 and estimate a practical joker pinches the open end of the hose down to a diameter of 5 mm, and sprays his neighbor with it. What is the speed at which water comes out of the hose?
a) 1690 cm/s
b) 1660 cm/s
c) 1698 cm/s
d) 1668 cm/s
Answer: c [Reason:] The flow rate (A1v1) of the water approaching the constriction must be equal to the flow rate leaving the hose (A2v2). This gives:
8. Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What must be the diameter of the nozzle in order that the water emerge at 16 m/s?
a) 0.884 cm
b) 0.891 cm
c) 0.881 cm
d) 0.894 cm
Answer: d [Reason:] The area is proportional to the square of diameter, so:
v1d12 = v2d22
9.Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What is the rate of flow rate of flow in m3/min?
a) 0.0754 m3/min
b) 0.0750 m3/min
c) 0.0764 m3/min
d) 0.0760 m3/min
Answer: a [Reason:] R = v1
10. Water ideally flows through a pipe of radius 6 centimeters at a rate of 5 meters per second. The pipe then narrows to a radius of 2 centimeters. What is the new velocity of the water?
a) 50 m/s
b) 55 m/s
c) 60 m/s
d) 65 m/s
Answer: b [Reason:] First convert everything to basic metric units.
(6cm)((1m)/(100cm)) = .06m
(2cm)((1m)/(100cm)) = .02m
Now find the cross-sectional area in both sections of the pipe.
A = π(r)²
π(.06)² ≈ 0.011m²
π(.02)² ≈ 0.001m²
Next, use the equation of continuity with the areas you have obtained and the given velocity.
A₁v₁ = A₂v₂
(0.011m²)(5m/s) = (.001m²)v₂
0.055m³/s = (.001m²)v₂
55m/s = v₂.
11. Venturi relation is one of the applications of ___________
a) Light equation
b) Bernoulli’s equation
c) Speed equation
d) Equation of continuity
Answer: b [Reason:] The flow speed of a fluid can be measured using a device such as a Venturi meter or an orifice plate, which can be placed into a pipeline to reduce the diameter of the flow. For a horizontal device, the continuity equation shows that for an incompressible fluid, the reduction in diameter will cause an increase in the fluid flow speed. Subsequently, Bernoulli’s principle then shows that there must be a decrease in the pressure in the reduced diameter region. This phenomenon is known as the Venturi effect.
12. Simplified equation of continuity is represented as _____________
a) A1V2 = A2V2
b) A1V1 = A1V2
c) A2V1 = A1V1
d) A1V1 = A2V2
Answer: d [Reason:] Continuity uses the conservation of matter to describe the relationship between the velocities of a fluid in different sections of a system. The simple observation that the volume flow rate, Av, must be the same throughout a system provides a relationship between the velocity of the fluid through a pipe and the cross-sectional area. Continuity works in tandem with Bernoulli’s principle in the design and construction of systems of irrigation, plumbing, etc.
13. Relative density of mercury is _____________
Answer: a [Reason:] Relative density of a substance is the ratio of its density to that of water.ie. So relative density of mercury is 13.6 means that density of mercury is 13.6 times the density of water.
14. The unit of pressure one bar is ___________
a) 1 Pascal
b) 1 kilo Pascal
c) 100 kPascal
d) 1000 kPascal
Answer: c [Reason:] A bar (b) is a metric measurement unit of pressure. One bar is equivalent to ten newtons (N) per square centimeter (cm²).
1 bar = 100,000 Pascal = 100,000 N/m² = 100,000 N / (100*100cm²) = 10 N/cm².
15. Reynolds number signifies the ratio of ____________
a) inertia forces to gravity forces
b) buoyant forces to inertia forces
c) inertial forces to viscous forces
d) gravity forces top viscous forces
Answer: c [Reason:] The Reynolds number is the ratio of inertial forces to viscous forces within a fluid which is subjected to relative internal movement due to different fluid velocities, in which is known as a boundary layer in the case of a bounding surface such as the interior of a pipe.