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# Multiple choice question for engineering

## Set 1

1. Consider the growth of a microorganism in batch culture. When the substrate concentration is high, the cell density doubles every 0.75 h, the observed substrate yield coefficient is 0.3 g DCW/g, and substrate consumption is allocated towards biosynthesis (60%), maintenance (10%), as well as product formation (30%). The product formation is strictly growth-associated. The batch reactor is inoculated with 0.01 g DCW/L and 10 g/L substrate. Estimate the maximum cell density (after lag phase).
a) 1.21 g DCW/ l
b) 1.41 g DCW/ l
c) 1.61 g DCW/ l
d) 1.81 g DCW/ l

Answer: d [Reason:] Xfromsubstrate = (10 g substrate/L ).(0.6).0.3 g DCW/ g Substrate = 1.8 g DCW/l

Xmax = Xo + Xfromsubstrate = 0.01 g DCW/l + 1.8 g DCW/l = 1.81 g DCW/ l.

2. Refer to Q1 and, Estimate the time (after lag phase) required achieving it.
a) 5.60 hr
b) 5.61 hr
c) 5.62 hr
d) 5.63 hr

3. Refer to Q1 and Q2, and determine the value of the maintenance coefficient (g substrate/g DCW-h).
a) 0.099 g substrate/g DCW-h
b) 0.089 g substrate/g DCW-h
c) 0.079 g substrate/g DCW-h
d) 0.069 g substrate/g DCW-h

Answer: a [Reason:] 10% of the substrate goes to maintaining the cells.

m = (10 g substrate/l.0.10)/(1.81 g DCW/L.5.63 hr)

= 0.099 g substrate/g DCW-h.

4. Steady state condition is maintained in ____________
a) Continuous process
b) Batch process
c) Fed- batch process
d) Semi-batch process

Answer: a [Reason:] In continuous process steady state condition is maintained, in which – The rate of increase of cell biomass with time is equal to zero. – The total number of cells are constant and – The total volume in the bioreactor are constant.

5. What is the unit of Maintenance coefficient “m”?
a) kg substrate (kg biomass) S-1
b) kg substrate (kg biomass) S
c) kg substrate (kg biomass)-1 S-1
d) kg substrate (kg biomass)-1 S

Answer: c [Reason:] A maintenance coefficient is used to describe the specific rate of substrate uptake for cellular maintenance, or

m = – [dS/dt]m /X

Unit of maintenance coefficient , m. is kg substrate (kg biomass)-1S-1.

6. “Ypx” is defined as mass or mole of product produced per unit mass or mole of substrate consumed.
a) True
b) False

Answer: b [Reason:] (Ypx) is defined as mass or mole of product produced per unit mass or mole of Biomass formed.

7. The total mass and number of atom of each element these two quantities is constant during any Chemical and Biochemical reactions.
a) True
b) False

Answer: a [Reason:] Total mass and Number of atom of each element These two quantities are constant during any Chemical and Biochemical reactions. Therefore the estimation of biomass can be done by using Stoichiometric calculations with the help of – Biomass Yield – Product Yield.

8. Reactant which controls amount of products is _________
a) Deficient reactant
b) Non- Deficient reactant
c) Limiting reactant
d) Excess reactant

Answer: c [Reason:] The same thing happens in chemical reactions: there is always a limiting reactant, which is a chemical element or substance that limits the amount of product made during a chemical reaction. Typically, there is also an excess reactant, or the amount of an element or substance left over after the reaction stops.

9. There is always a limiting reactant.
a) True
b) False

Answer: b [Reason:] If all the reagents are added in exactly the right mole ratio then all of each reagent is used up, so there is no excess or limiting reagent. In this case you can use either reagent moles to work out the amount of product formed. Then both reagents will be fully consumed so neither is limiting.

10. A homogenous material is defined as being ____________
a) An element
b) Any material with uniform composition
c) Synonymous with “solution”
d) Any material with non-uniform composition

Answer: b [Reason:] “Homogeneous material” means one material of uniform composition throughout or a material, consisting of a combination of materials, that cannot be disjointed or separated into different materials by mechanical actions such as unscrewing, cutting, crushing, grinding and abrasive processes.

## Set 2

1. Which of the following is not an example of convectiton?
a) After a car is turned on, the engine becomes hot.
b) Steaming cup of hot tea
c) Ice melting
d) Hot air balloon

Answer: a [Reason:] After a car is turned on, the engine becomes hot. The hood will become warm as heat is conducted from the engine to the hood. So it is an example of the process conduction.

2. In heterogeneous reaction, if a reaction proceeds slowly even in the presence of adequate substrate then mass transfer will be?
a) Slow
b) Very slow
c) Rapid
d) Negligible

Answer: c [Reason:] If a reaction proceeds slowly even in the presence of adequate substrate, it is likely that mass transfer will be rapid enough to meet the reaction demand. In this case, the observed rate would be determined more directly by the reaction process than mass transfer. Conversely, if the reaction tends to be very rapid, it is likely that mass transfer will be too slow to supply substrate at the rate required.

3. During a chemical reaction, the
a) mass of catalyst remains unchanged
b) physical state of catalysts remain unchanged
c) changes in chemical composition of catalysts may be observed
d) chemical composition of catalyst may change

Answer: a [Reason:] Catalyst is defined as a substance, which alters the rate of a chemical reaction, itself remaining chemically unchanged at the end of the reaction. The process is called Catalysis.

4. Factors affecting speed of reaction does not include
a) Concentration
b) Particle size
c) Change in mass
d) Pressure

Answer: c [Reason:] A catalyst is a species that speeds up a chemical reaction without being chemically changed upon completion of the reaction. In other words, the mass of a catalyst is the same before and after a reaction occurs.

5. Which of the following is not an example of bulk catalyst?
a) Silica-alumina catalyst
b) Molybdenum oxide supported on alumina
c) Iron-molybdate
d) Iron doped with alumina and potassium oxide

Answer: b [Reason:] The supported catalysts are the catalytically active materials which are dispersed over the high surface area support material. For example, hydrodesulphurization is carried out over molybdenum oxide supported on alumina.

6. A catalyst provides an alternative pathway with a lower activation energy.
a) True
b) False

Answer: a [Reason:] A catalyst provides an alternative pathway with a lower activation energy. Therefore, more particles have enough energy to overcome the new activation energy requirement.

7. What is the unit of the reaction rate?
a) mol dm-3s-1
b) mol dm s
c) mol dm-3 s
d) mol-1 dm s-1

Answer: a [Reason:] The Reaction Rate for a given chemical reaction is the measure of the change in concentration of the reactants or the change in concentration of the products per unit time. Its units are mol dm-3s-1.

8. In which type of reactor, the perfect mixing can occur?
a) Ideal
b) Non-ideal
c) Constant
d) Variable

Answer: a [Reason:] In ideal reactor, composition and temperature are spatially uniform (i.e. perfect mixing). Ideal (perfectly mixed) reactor: spatially uniform temp, conc, & reaction rate :

9. The overall conversion of reactants can be increased in spite of increasing of the residence time.
a) True
b) False

Answer: b [Reason:] In heterogeneous reactors external mass transport can be increased by a better mixing of the fluid. A higher flow rate also contributes to a better mixing. Under the better mixing a laminar film around the catalyst particle is thinner and more reactants can be transported to the surface. Consequently, the overall conversion of reactants can be increased in spite of lowering of the residence time.

10. If the temperature of a reaction is decreased, what effect will this have on the rate of reaction?
a) The rate of reaction will increase
b) The rate of reaction will decrease
c) The reaction will stop
d) No effect on reaction

Answer: b [Reason:] Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

11. In homogeneous catalysis_________
a) All reactants are in the same rate
b) The catalyst is in a different state to the reactants
c) The reactants and catalyst are in the same rate
d) All the products are in the same state

Answer: c [Reason:] Homogeneous catalysis refers to catalytic reactions where the catalyst is in the same phase as the reactants. Homogeneous catalysis applies to reactions in the gas phase and even in solids.

12. The minimum amount of energy needed to start a reaction is called the ______
a) Activation energy
b) Energy of reaction
c) Entropy of reaction
d) Reaction mechanism energy

Answer: a [Reason:] Activation energy is the minimum energy required for a reaction to occur. Catalysis is the increase in the rate of a chemical reaction by lowering its activation energy. Transition state is an intermediate state during a chemical reaction that has a higher energy than the reactants or the products.

13. The rate of a chemical reaction can be expressed in Molarity per seconds?
a) True
b) False

Answer: a [Reason:] There are many factors that can either slow or speed up the rate of a chemical reaction such as temperature, pressure, concentration, and catalysts. The Rate of a Chemical Reaction is always positive. It can be confusing since the Rate of Disappearance is negative, however when you think about it, a rate should never be negative since the rate is describing how fast the concentration changes with time. The units for the rate is Molarity per Seconds (M/s).

14. Factors affecting speed of reaction include ________
a) Volume of gas
b) Change in mass
c) Pressure
d) Temperature change

Answer: c [Reason:] Increasing the pressure of a reaction improves the likelihood reactants will interact with each other, thus increases the rate of the reaction. As you would expect, this factor is important for reactions involving gases, and not a significant factor with liquids and solids.

15. Concentration of the reactant is _____ proportional to the number of collisions?
a) Directly
b) Indirectly
c) Inversely
d) Not proportional

Answer: a [Reason:] Concentration of the reactants is directly proportional to the number of collisions and also directly proportional to the reaction rate. The more the collision, more will be the reaction rate and hence more will be the concentration of reactants.

## Set 3

1. What is the rounding off of a number?
a) To change a number to its nearest prime number
b) To change a number to its nearest odd number
c) To change a number to its nearest whole number
d) to change a number to its nearest even number

Answer: c [Reason:] To reduce successively the number of digits to the right of the decimal point of a mixed number by dropping the final digit and adding 1 to the next preceding digit if the dropped digit was 5 or greater, or leaving the preceding digit unchanged if the dropped digit was 4 or less.

2. Round off the number 425.68 to Tenths (1 decimal place), Hundredths (2 decimal place) and to Tens place?
a) 425.6, 425.68, 426
b) 425.7, 425.68, 430
c) 425.5, 426.78, 425
d) 425.7, 4257, 430

Answer: b [Reason:] In the Tenths( 1 decimal place), we will take 68 as a digit after the decimal and round off it to the multiple of ten as 70, In the hundredths( 2 decimal place), we will take both 68 as a number after the decimal because we have to take 2 decimal places and 6,8 both numbers are greater than 5, In the Tens place, we will round the number before the decimal place to the closest multiple of ten, i.e. 430.

3. A piece of iron rod was measured and found to be 120cms. But the actual value of the wood is 123cms. Find the relative error?
a) 3%
b) 2.5%
c) 2.43%
d) 3.43%

Answer: c [Reason:] Given: True value = 123cms Measured value = 120cms Firstly, let us find out absolute error. We know by subtracting true value with measured value we get absolute error = 123 – 120 = 3cms Therefore, absolute error = 3cms Now, let us find out the relative error = absolute error/True value x 100 = 3/123 x 100 = 2.43% Therefore, the relative error for the given values = 2.43%.

4. Which of the following is correct according to calculating relative error as a measure of Precision?
a) (Absolute error/ True value)*100
b) (Absolute error/True value)
c) (measurement being taken/Absolute error)*100
d) (Absolute error/measurement being taken)

Answer: d [Reason:] Relative error (RE) — when used as a measure of precision — is the ratio of the absolute error of a measurement to the measurement being taken. In other words, this type of error is relative to the size of the item being measured. RE is expressed as a percentage and has no units.

5. From the below data which one is precise?
Sample A: 15.38, 15.37, 15.36, 15.33, and 15.32.
Sample B: 32.56, 32.55, 32.48, 32.49 and 32.48.
a) Sample A
b) Sample B
c) Sample A & B
d) None of the mentioned

Answer: a [Reason:] Subtract the lowest data point from the highest: Sample A: 15.38 – 15.32 = 0.06 Sample B: 32.56 – 32.48 = 0.08

Sample A has the lowest range (.06) and so is the more precise.

6. If a student measured the mass of an object on a balance three times and obtained the values 14.568 g, 14.566 g, and 14.565 g, the range and percent of the average value between the measurements would be ____________
a) 0.005gm, 0.01%
b) 0.004gm,0.02%
c) 0.003gm,0.02%
d) 0.001gm,0.01%

Answer: c [Reason:] Range = (14.568 – 14.565) g = 0.003 g , This is a reasonable range; the uncertainty appears in the last significant figure of the measured value, as it should. The average value for the mass of the object is 14.566 g. Therefore: % of average = (0.003 g/14.566 g) x 100% = 0.02% the range as a percent of average (RPA) has only one significant figure in this case, because the numerator of the fraction has only one.

7. The accepted value for the density of gold metal is 19.31 g/cm3. If a student measured the mass and volume of a sample of gold, and obtained a value of 19.03 g/cm3, the percent difference would be______
a) -1.5%
b) 1.5%
c) -0.5%
d) 0.5%

Explanation: % difference =[ (19.03 – 19.31) g/19.31 g] x 100% = – 1.5% A percent difference can be positive or negative; the sign shows whether the experimental value is higher or lower than the actual or theoretical value. This distinction is not used in precision, since all values are experimental.

8. A group of students worked in separate teams to measure the length of an object. Here are their data: Find out:-
1) The average length
3) The precision of the measurement

a) 2.7, 0.2, 2.7±0.2
b) 2.8, 0.1, 2.8±0.1
c) 2.0, 0.1, 2.0±0.1
d) 2.0, 0.2, 2.0±0.2

Answer: a [Reason:] The average length is the mean or average , in which the summation of the total number is divided from the total number, i.e.

(2.65+2.60+2.77+2.80+2.75)cm/5 = 13.57cm/5= 2.714cm or 2.7cm

The range or spread is the subtraction from the highest value from the lowest value , i.e., Highest value =2.80cm Lowest value =2.60cm

Therefore, Range= 2.80cm-2.60cm = 0.2cm

The precision of the measurement can be shown as average± range cm, i.e., Average=2.7cm Range = 0.2cm

Therefore, Precision= 2.7±0.2cm.

9. You are given a cube of pure copper. You measure the sides of the cube to find the volume and weigh it to find its mass. When you calculate the density using your measurements, you get 8.78 grams/cm3. Coppers accepted density is 8.96 g/cm3. What is your percent error?
a) 1%
b) 0.02%
c) 0.01%
d) 2%

Answer: d [Reason:] Experimental value = 8.78 g/cm3 Accepted value = 8.96 g/cm3 Step 1: Subtract the accepted value from the experimental value. 8.78 g/cm3 – 8.96 g/cm3 = -0.18 g/cm3 Step 2: Take the absolute value of step 1 |-0.18 g/cm3| = 0.18 g/cm3 Step 3: Divide that answer by the accepted value, 0.18g/cm3 ÷ 8.96g/cm3= 0.02 Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage. 0.02 x 100 = 2 The percent error of your density calculation was 2%.

10. What type of surface you would observe in the case of mercury with glass?
a) Concave meniscus
b) Convex meniscus
c) Contact angle =180˚
d) Contact angle= 0˚

Answer: b [Reason:] A convex meniscus occurs when the molecules have a stronger attraction to each other than to the container and Mercury does not wet glass – the cohesive forces within the drops i.e., bond between the molecules of mercury are stronger than the adhesive forces between the drops and glass.

11. You will read the meniscus reading as you measure the level at the horizontal center or inside part of the meniscus. (True/False)
a) True
b) False

Answer: a [Reason:] Place the graduated cylinder on a flat surface and view the height of the liquid in the cylinder with your eyes directly level with the liquid. The liquid will tend to curve downward. This curve is called the meniscus. Always read the measurement at the bottom of the meniscus.

## Set 4

1. The partial pressure of oxygen at 1atm is ________________
a) 0.2000 atm
b) 0.2098 atm
c) 0.2099 atm
d) 0.2096 atm

Answer: c [Reason:] The mole fraction of oxygen in air is 0.2099, so the partial pressure of oxygen at 1 atm air pressure is 0.2099 atm.

2. What is the unit of oxygen solubility “C*AL”?
a) mgl-1
b) mg-1l-1
c) m-1g-1l-1
d) mgl

Answer: a [Reason:] Respiration is expressed in molar units related to biochemical stoichiometries. C*AL is oxygen solubility in units of mg l-1.

3. The addition of ions and sugars added to the fermentation increases the oxygen solubility?
a) True
b) False

Answer: b [Reason:] Oxygen solubility is decreased by the ions and sugars normally added to fermentation media. In a typical fermentation medium, oxygen solubility is between 5% and 25% lower than in water as a result of solute effects. In simple systems, with all the components being dissolved, C0 represents the oxygen solubility and linearly decreases with increasing solute concentrations. In complex solutions with multi-phase structure an increase in C0 can be detected. It suggests that C0 consists of two components — one being oxygen solubility, the other being determined by the amount of oxygen adsorbed on the interphase and bound by macromolecules. The presence of biomass leads to a decrease in C0.

4. The solubility of oxygen in water is temperature and pressure dependent?
a) True
b) False

Answer: a [Reason:] The solubility of oxygen in water is temperature and pressure dependent. About twice as much (14.6 mg•L -1) dissolves at 0 °C than at 20 °C (7.6 mg•L-1), this estimates that at high temperature oxygen solubility is low. Less oxygen dissolves at high elevations compared to low elevations because the atmospheric pressure is less and thus the partial pressure is lower, this estimates that at low pressure the oxygen solubility is also low.

5. “The amount of air dissolved in a fluid is proportional to the pressure in the system”, which law is applicable to this statement?
a) Raoult’s law
b) Fick’s law
c) Henry’s law
d) Newton’s law

Answer: c [Reason:] Solubility of air in water follows Henry’s Law – “the amount of air dissolved in a fluid is proportional to the pressure in the system” – and can be expressed as: c = pg / kH where, c = solubility of dissolved gas kH = proportionality constant depending on the nature of the gas and the solvent pg = partial pressure of gas (Pa, psi) The solubility of oxygen in water is higher than the solubility of nitrogen. Air dissolved in water contains approximately 35.6% oxygen compared to 21% in air.

6. Henry Law’s Constants at a system temperature of 25oC (77oF) of oxygen is 756.7 atm/ (mol/litre). Molar Weight of O2 is 31.9988 g/mol and partial fraction in air is ~ 0.21. Calculate the Oxygen dissolved in the Water at atmospheric pressure.
a) 0.0090 g/liter
b) 0.0089 g/liter
c) 0.0080 g/liter
d) 0.0099 g/liter

Answer: b [Reason:] Oxygen dissolved in the Water at atmospheric pressure can be calculated as: c = pg / kH co = (1 atm) 0.21 / (756.7 atm/(mol/litre)) (31.9988 g/mol) = 0.0089 g/liter.

7. Henry Law’s Constants at a system temperature of 25oC (77oF) of nitrogen is 1600 atm/(mol/litre).Molar weight of N2 is 28.0134 g/mol and partial fraction in air is ~ 0.79. Calculate the Nitrogen dissolved in the Water at atmospheric pressure.
a) 0.0138 g/liter
b) 0.0130 g/liter
c) 0.0132 g/liter
d) 0.0134 g/liter

Answer: a [Reason:] Nitrogen dissolved in the Water at atmospheric pressure can be calculated as: c = pg / kH cn = (1 atm) 0.79 / (1600 atm/(mol/litre)) (28.0134 g/mol) = 0.0138 g/liter.

8. Refer to Q6 and Q7, and calculate the air dissolved in water?
a) 0.0228 g/liter
b) 0.0223 g/liter
c) 0.0227 g/liter
d) 0.0222 g/liter

Answer: c [Reason:] Since air mainly consists of Nitrogen and Oxygen – the air dissolved in the water can be calculated as:

ca = (0.0089 g/litre) + (0.0138 g/litre) = 0.0227 g/liter.

9. The dissolved oxygen decreases when?
a) The temperature is increased
b) The pressure is increased
c) The salinity is decreased
d) The salinity is increased

Answer: d [Reason:] Dissolved oxygen decreases exponentially as salt levels increase. That is why, at the same pressure and temperature, saltwater holds about 20% less dissolved oxygen than freshwater.

10. The Henry’s law constant for O2 in water at 25°C is 1.27×10−3M/atm and the mole fraction of O2 in the atmosphere is 0.21. Calculate the solubility of O2 in water at 25°C at an atmospheric pressure of 1.00 atm.
Strategy: ▪ Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen.
▪ Use Henry’s law to calculate the solubility, expressed as the concentration of dissolved gas.
a) 2.5×10-4 M
b) 2.1×10-4 M
c) 2.3×10-4 M
d) 2.7×10-4M

Answer: d [Reason:] According to Dalton’s law, the partial pressure of O2 is proportional to the mole fraction of O2: PA=XAPt=(0.21)(1.00atm)=0.21atm

From Henry’s law, the concentration of dissolved oxygen under these conditions is:

CO2 = KpO2= (1.27×10−3M/atm) (0.21atm) = 2.7×10-4 M.

11. The value of Henry’s law constant increases with increasing temperature?
a) True
b) False

Answer: a [Reason:] The value of the Henry’s law constant is found to be temperature dependent. The value generally increases with increasing temperature. As a consequence, the solubility of gases generally decreases with increasing temperature. The decrease in solubility of gases with increasing temperature is an example of the operation of Le Chatelier’s Principle. The heat or enthalpy change of the dissolution reaction of most gases is negative, which is to say the reaction is exothermic. As a consequence, increasing the temperature leads to gas evolution.

12. Does yeast need oxygen in fermentation process?
a) True
b) False

Answer: a [Reason:] Oxygen is a critical additive in brewing. Oxygen is the only necessary nutrient not naturally found in wort. Adding adequate oxygen to wort requires a fundamental understanding of why yeast need oxygen, how much oxygen they need, and how to get oxygen into solution and the factors affecting solubility of oxygen. Yeast use oxygen for cell membrane synthesis. Without oxygen, cell growth will be extremely limited. Yeast can only produce sterols and certain unsaturated fatty acids necessary for cell growth in the presence of oxygen.

13. Which type of homebrewer is best for 8 ppm of dissolved oxygen in solution?
a) Siphon sprays
b) Whipping
c) Splashing and shaking
d) Pure air through a stone with an aquarium pump

Answer: c [Reason:] Homebrewers have several aeration/oxygenation methods available to them: siphon sprays, whipping, splashing, shaking, pumping air through a stone with an aquarium pump, and injecting pure oxygen through a sintered stone. Pumping compressed air through a stone is not an efficient way to provide adequate levels of DO. Traditional splashing and shaking, although laborious, is fairly efficient at dissolving up to 8 ppm oxygen. To increase levels of oxygen, the carboy headspace can be purged with pure oxygen prior to shaking.

14. The atmospheric pressure is 1.0 atm and Henry’s law constant for O2 is 1.66 x 10-6 M/mm Hg at 25 °C. Assume air contains 21% oxygen. Calculate the partial pressure of oxygen. (21% of air is oxygen and the mole fraction of O2 is 0.21).
a) 180 mm Hg
b) 130 mm Hg
c) 120 mm Hg
d) 160 mm Hg

Answer: d [Reason:] The partial pressure of oxygen is:

P (O2)= (1.0 atm) ((760 mm Hg)/(1 atm) ) (0.21) = 160 mm Hg.

15. Refer to Q14 and, Calculate the solubility of oxygen in units of grams of oxygen per liter of water.
a) 0.0080 g/L
b) 0.0082 g/L
c) 0.0083 g/L
d) 0.0085 g/L

Answer: d [Reason:] The solubility of oxygen in units of grams of oxygen per liter of water is:

## Set 5

1. Skim milk is prepared by the removal of some of the fat from whole milk. This skim milk is found to contain 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original milk contained 4.5% fat, Calculate its composition assuming that fat only was removed to make the skim milk and that there are no losses in processing.
a) Fat = 4.5%, Water= 86.5%, protein= 3.3%, carbohydrate= 4.9%, ash = 0.8%
b) Fat = 4.5%, Water= 83.5%, protein= 3.0%, carbohydrate= 4.5%, ash = 0.9%
c) Fat = 4.6%, Water= 80.5%, protein= 3.5%, carbohydrate= 4.0%, ash = 0.9%
d) Fat = 4.6%, Water= 81.5%, protein= 3.3%, carbohydrate= 4.5%, ash = 0.8%

Answer: a [Reason:] Basis: 100 kg of skim milk. This contains, therefore, 0.1kg of fat. Let the fat which was removed from it to make skim milk be x kg. Total original fat = (x+ 0.1) kg Total original mass = (100+ x) kg and as it is shown that the original fat content was 4.5% so,

(x+0.1)/ (100+x) = 0.045 where = x+0.1 = 0.045(100+x) x = 4.6 kg So the composition of the whole milk is then fat = 4.5%, water= 90.5/ 104.6 = 86.5%, protein = 3.5/ 104.6 =3.3%, carbohydrate= 5.1/ 104.6 = 4.9% and ash = 0.8%.

2. If 35,000kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream with 45% fat, what are the flow rates of the two output streams from a continuous centrifuge which accomplishes this separation?
Basis 1 hour’s flow of whole milk.
a) 5678 kg/hr
b) 5368 kg/hr
c) 2567 kg/hr
d) 2578 kg/hr

Answer: b [Reason:] Mass in: Total mass = 35000/6 = 5833 kg. Fat = 5833 x 0.04 = 233 kg. And so Water plus solids-not-fat = 5600 kg.

Mass out: Let the mass of cream be x kg then its total fat content is 0.45x. The mass of skim milk is (5833 – x) and its total fat content is 0.0045 (5833 – x).

Material balance on fat: Fat in = Fat out 5833 x 0.04 = 0.0045(5833 – x) + 0.45x. And so x = 465 kg. So that the flow of cream is 465 kg / hr and skim milk (5833 – 465) = 5368 kg/hr.

3. A textile dryer is found to consume 4 m3 /hr of natural gas with a calorific value of 800 kJ/mole. If the throughput of the dryer is 60 kg of wet cloth per hour, drying it from 55% moisture to 10% moisture, estimate the overall thermal efficiency of the dryer taking into account the latent heat of evaporation only.
Assuming the natural gas to be at standard temperature and pressure at which 1 mole occupies 22.4 liters and Latent heat of evaporation = 2257 kJ/K.
a) 40%
b) 50%
c) 58%
d) 48%

Answer: d [Reason:] 60 kg of wet cloth contains 60 x 0.55 kg water = 33 kg moisture and 60 x (1-0.55) = 27 kg bone dry cloth. As the final product contains 10% moisture, the moisture in the product is 27/9 = 3 kg

And so Moisture removed / hr = 33 – 3 = 30 kg/hr Latent heat of evaporation = 2257 kJ/K Heat necessary to supply = 30 x 2257 = 6.8 x 104 kJ/hr

Assuming the natural gas to be at standard temperature and pressure at which 1 mole occupies 22.4 litres.

Rate of flow of natural gas = 4 m3 /hr = (4 x 1000)/22.4 = 179 moles/hr Heat available from combustion = 179 x 800 = 14.3 x 104 kJ/hr Approximate thermal efficiency of dryer = heat needed / heat used = 6.8 x 104 / 14.3 x 104 = 48%.

4. It is desired to freeze 10,000 loaves of bread each weighing 0.75 kg from an initial room temperature of 18°C to a final temperature of -18°C. The bread-freezing operation is to be carried out in an air-blast freezing tunnel. It is found that the fan motors are rated at a total of 80 horsepower and measurements suggest that they are operating at around 90% of their rating, under which conditions their manufacturer’s data claims a motor efficiency of 86%. If 1 ton of refrigeration is 3.52 kW, estimate the maximum refrigeration load imposed by this freezing installation assuming that fans and motors are all within the freezing tunnel insulation and the heat-loss rate from the tunnel to the ambient air has been found to be 6.3 kW.
Extraction rate from freezing bread (maximum) = 104 kW
a) 46 tons of refrigeration
b) 40 tons of refrigeration
c) 56 tons of refrigeration
d) 50 tons of refrigeration

Answer: a [Reason:] Extraction rate from freezing bread (maximum) = 104 kW Fan rated horsepower = 80 Now 0.746 kW = 1 horsepower and the motor is operating at 90% of rating, And so (fan + motor) power = (80 x 0.9) x 0.746 = 53.7 kW With motors + fans in tunnel Heat load from fans + motors = 53.7 kW Heat load from ambient = 6.3 kW Total heat load = (104 + 53.7 + 6.3) kW = 164 kW = 46 tons of refrigeration.

5. Refer to Q4, and estimate the maximum refrigeration load imposed by this freezing installation assuming the fans but not their motors are in the tunnel.
a) 50.5 tons of refrigeration
b) 40.5 tons of refrigeration
c) 44.5 tons of refrigeration
d) 55.5 tons of refrigeration

Answer: c [Reason:] With motors outside, the motor inefficiency = (1- 0.86) does not impose a load on the refrigeration

Total heat load = (104 + [0.86 x 53.7] + 6.3) = 156 kW = 44.5 tons of refrigeration.

6. Water is pumped from a storage tank through a tube of 3.0 cm inside diameter at the rate of 0.001 m3 /s. What is the kinetic energy per kg water in the tube?
a) 2.00 J/kg
b) 1.00 J/kg
c) 5.00 J/kg
d) 0.05 J/kg

Answer: b [Reason:] Kinetic energy Ek = ½ mv2, tube dia. D = 3.0 cm, m = 1 kg. Cross-section area of tube A = ¼ πD2 = ¼ π (3.0/100 m)2 = 7.0686×10-4m2 Average velocity of water v = Q/A = 0.001 m3 /(7.0686×10-4m2 ) = 1.415 m/s KE per kg = Ek/m = ½ v2 = ½ (1.415 m/s)2 = 1.00 m2/s2 = 1.00 J/kg.

7. Water is pumped from a storage tank (tank 1) to another tank (2) which is 40 ft above tank. Calculate the potential energy increase with each lb of water pumped from tank 1 to tank2.
a) 119.544 J/kg
b) 120.678 J/kg
c) 122.500 J/kg
d) 190.600 J/kg

Answer: a [Reason:] PE increase per lb = Ep /m = gh = (9.806 m/s2 ) (40 m/3.2808) = 119.54 m2/s2 = 119.544 J/kg.

8. Refer to Q7, and Express the answer in btu/lb.
a) 0.0421 btu/lb
b) 0.0532 btu/lb
c) 0.0514 btu/lb
d) 0.0432 btu/lb

Answer: c [Reason:] PE increase per lb = Ep /m = gh = (9.806 m/s2 ) (40 m/3.2808) = 119.54 m2 /s2 = 119.544 J/kg = (119.544 btu/1055.6)/2.2046 lb = 0.0514 btu/lb.

9. Concentrated fermentation liquid containing 20% (w/w) gluconic acid from an evaporator has a flow rate of 2000 kg/h and a temperature 90 °C. It needs to be cooled to 6 °C in a heat exchanger with cooling water. The cooling water has a flow rate 2700 kg/h and an initial temperature 2 °C. If the cooling water leaves the heat exchanger at 50 °C, what is the rate of heat loss from gluconic acid solution to the surroundings? Assume the heat capacity of gluconic acid is 0.35 cal/g-°C-1.

a) 69390.84 kJ/h
b) 65780.56 kJ/h
c) 67890.67 kJ/h
d) 65432.10 kJ/h

Answer: a [Reason:] Heat lost by gluconic acid = heat gained by water

[2000(0.2)×(0.35×4.184×1000/1000)×(90-6) + 2000×(0.8)×(376.92-25.2)]+Q = 2700×(209.33-8.37)×611955.84+Q=542565

Q = -69390.84 kJ/h

Negative sign means heat loss by the system to surrounding. Hence, heat loss to the surroundings is 69390.84 kJ/h.

10. How much work can be obtained from an adiabatic, continuous-flow turbine, if steam at 60 bar and 500oC is used and the outlet stream is at 1 bar and 400oC?

a) -144 kJ/kg
b) -155 kJ/kg
c) -166 kJ/kg
d) -177 kJ/kg

Answer: a [Reason:] Basis: 1 kg steam For superheated steam:

Inlet: (60 bar, 500oC) h1 = 3422 kJ/kg Outlet: (1 bar, 400oC) h2 = 3278 kJ/kg min = mout = 1 kg

DECV + Sni(Ek + Ep + H)i = Q + WS (General form of the energy balance)

Now we make the following simplifying assumptions for this turbine:

▪ Adiabatic: Q = 0 ▪ The inlet and outlet lines are not tremendously different in height: Ep = 0 ▪ There is little difference in the velocity of the fluid in the inlet and outlet lines: Ek = 0 ▪ There are no accumulation terms (steady-state operation with steady flow): DECV = 0

The simplified form of the energy balance is therefore:

Hout – Hin = Ws = mouthout – minhin = m(hout – hin)

Substituting in the values for the specific enthalpies gives:

Ws = m (hout – hin) or Ws/m = (3278 – 3422) kJ/kg = -144 kJ/kg.