Multiple choice question for engineering
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
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
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
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
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.
7. The total mass and number of atom of each element these two quantities is constant during any Chemical and Biochemical reactions.
8. Reactant which controls amount of products is _________
a) Deficient reactant
b) Non- Deficient reactant
c) Limiting reactant
d) Excess reactant
9. There is always a limiting reactant.
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
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
2. In heterogeneous reaction, if a reaction proceeds slowly even in the presence of adequate substrate then mass transfer will be?
b) Very slow
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
4. Factors affecting speed of reaction does not include
b) Particle size
c) Change in mass
5. Which of the following is not an example of bulk catalyst?
a) Silica-alumina catalyst
b) Molybdenum oxide supported on alumina
d) Iron doped with alumina and potassium oxide
6. A catalyst provides an alternative pathway with a lower activation energy.
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
8. In which type of reactor, the perfect mixing can occur?
9. The overall conversion of reactants can be increased in spite of increasing 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
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
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
13. The rate of a chemical reaction can be expressed in Molarity per seconds?
14. Factors affecting speed of reaction include ________
a) Volume of gas
b) Change in mass
d) Temperature change
15. Concentration of the reactant is _____ proportional to the number of collisions?
d) Not proportional
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
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
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?
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)
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
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%
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______
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
2) The range or spread
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
(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?
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˚
11. You will read the meniscus reading as you measure the level at the horizontal center or inside part of the meniscus. (True/False)
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
2. What is the unit of oxygen solubility “C*AL”?
3. The addition of ions and sugars added to the fermentation increases the oxygen solubility?
4. The solubility of oxygen in water is temperature and pressure dependent?
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
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
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
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
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
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
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?
12. Does yeast need oxygen in fermentation process?
13. Which type of homebrewer is best for 8 ppm of dissolved oxygen in solution?
a) Siphon sprays
c) Splashing and shaking
d) Pure air through a stone with an aquarium pump
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
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
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%
(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
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.
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
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
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
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
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
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
[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
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.