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Chemical Engineering Interview Questions

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Question 70 - According to Adolf Eugen Fick (1829 - 1901) : rate of diffusion v increases with less wall thickness t, increased area A and decreased molecular weight of a fluid M. The diffusion constant D decreased with increasing M. (a) By assuming v, t, dP, A, M and D changes proportionally of each other, find the equation of v as a function of t, dP, A and D. (b) The ratio of self diffusion constant D, at T = 273 K and P = 0.1 MPa, for gases B and C are 1.604 : 0.155. If only 2 gases exist in such a system : hydrogen and nitrogen, find the type of gas for B and C with reference to their molecular weights M. (c) By using the equation of kinetic energy 0.5 MV = constant where V = square of v, find the ratio of V for B and V for C, or V(B) / V(C), as a function of M(B) and M(C), where M(B) is molecular weight of B and M(C) the molecular weight of C : Graham's Law of Diffusion.

Question 71 - (a) The byte is the basic building block of computer data used in chemical engineering process simulation where 16 bits make a word, 4 bits make a nibble, 32 bits make a quad word and 8 bits make a byte. Then how many nibbles are there in a megabytes? (b) In computer data items, let : 1 bit - counts from 0 to 1, 8 bits - counts from 0 to 255, 16 bits - counts from 0 to A. What is the value of A? (c) In a binary system of 4 bits, if 1100 = 12, 1101 = 13, 1110 = 14, 1111 = 15, B = 16, then guess the value of B. (d) By using any form of tools, find the exact value of 2 power 64 or 2^64.

Question 72 - (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calory, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?

Question 73 - (a) A dryer reduces the moisture content of 100 kg of a potato product from 80 % to 10 % moisture. Find the mass of the water removed in such drying process. (b) During the drying process, the air is cooled from 80°C to 71°C in passing through the dryer. If the latent heat of vaporization corresponding to a saturation temperature of 71°C is 2331 kJ / kg for water, find the heat energy required to evaporate the water only. (c) Assume potato enters at 24°C, which is also the ambient air temperature, and leaves at the same temperature as the exit air. The specific heat of potato is 3.43 kJ / (kg °C). Find the minimum heat energy required to raise the temperature of the potatoes. (d) 250 kg of steam at 70 kPa gauge is used to heat 49,800 cubic metre of air to 80°C, and the air is cooled to 71°C in passing through the dryer. If the latent heat of steam at 70 kPa gauge is 2283 kJ / kg, find the heat energy in steam. (e) Calculate the efficiency of the dryer based heat input and output, in drying air. Use the formula (Ti - To)/(Ti - Ta) where Ti is the inlet (high) air temperature into the dryer, To is the outlet air temperature from the dryer, and Ta is the ambient air temperature.

Question 75 - The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18. The fraction of polymer chain X of 3 different monomers a, b and c in a polymer are Xa = 0.5, Xb = 0.3 and Xc = 0.2. (i) Calculate number average molecular weight by using the formula Ma Xa + Mb Xb + Mc Xc. (ii) Calculate weight average molecular weight by using the formula (Ma Xa Ma + Mb Xb Mb + Mc Xc Mc) / (Ma Xa + Mb Xb + Mc Xc). (iii) Calculate the polydispersity by using the answer in (ii) divided by answer in (i). (iv) If the molecular weight of repeat unit is 12, calculate the degree of polymerization by using the formula (Ma Xa + Mb Xb + Mc Xc) / (molecular weight of repeat unit).

Question 76 - Let C% be the fractional crystallinity, Rs = density of sample, Ra = density of amorphous form and Rc = density of crystalline form. In a polymer, these unknowns could be related by the equation C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra). (a) Find the equation of Rc as a function of C%, Rs and Ra. (b) Two samples of a polymer, C and D exist. For sample C, C% = 0.513 when Rs = 2.215 unit. For sample D, C% = 0.742 when Rs = 2.144 unit. Both samples C and D have the same values of Ra and Rc. Find the values of Ra and Rc in 6 decimal places.

Question 77 - The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18, with their respective quantities in N units having the ratio of Na : Nb : Nc = 2 : 3 : 5. (a) Find the numerical average molecular weight of the polymer by using the formula (Ma Na + Mb Nb + Mc Nc) / (Na + Nb + Nc). (b) Find the weighted average molecular weight of the polymer by using the formula (Ma Na Ma + Mb Nb Mb + Mc Nc Mc) / (Ma Na + Mb Nb + Mc Nc). (c) Calculate the polydispersity Q by using the answer in (b) divided by answer in (a). (d) Find the volumetric average molecular weight of the polymer by using the formula (Ma Na Ma Ma + Mb Nb Mb Mb + Mc Nc Mc Mc) / (Ma Na Ma + Mb Nb Mb + Mc Nc Mc). (e) Estimate the polydispersity Q by using the answer in (d) divided by answer in (b).

Hallow everybody, we want to know in a NPk fertilizer process plant (Bulk Blender) we are in doubt if use square or rectangular hopper jus to avoid the vault effect durin the flow of of raw material like Phosphorus, Potasium and zeolite..Please any advise will be appresated Isdray

2327Question 78 - Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, 0.93 % argon, 0.039% carbon dioxide, and small amounts of other gases by volume. Fact 2 : Volume occupied is directly proportional to the number of moles for ideal gases at constant temperature and pressure. Fact 3 : 12.5 moles of pure oxygen is required to completely burn 1 mole of pure octane. Fact 4 : Air–fuel ratio (AFR) is the mass ratio of dry air to fuel present in a combustion process such as in an internal combustion engine or industrial furnace. Fact 5 : Molecular weight of oxygen gas is 31.998 g / mole and molecular weight of nitrogen gas is 28.014 g / mole. (a) Find the molar ratio of nitrogen and oxygen, or (moles of nitrogen) / (moles of oxygen) in dry air, by assuming ideal features of nitrogen and oxygen gases. (b) How many moles of nitrogen are available if dry air is used to completely burn the 1 mole pure octane? (c) Find the mass of fuel of 1 mole of octane with molecular weight of 114.232 g / mole. (d) Find the mass of dry air with 12.5 moles of pure oxygen by assuming only oxygen and nitrogen gases exist in the air. (e) Find the air-fuel ratio (AFR) when octane is used as fuel. (f) Find the fuel-air ratio (FAR) when octane is used as fuel.

CHEMICAL MATERIAL BALANCE – EXAMPLE 2.2 : Three hundred gallons of a mixture containing 75.0 wt % ethanol and 25 wt % water (mixture specific gravity = 0.877) and a quantity of a 40.0 wt % ethanol - 60 wt % water mixture (specific gravity = 0.952) are blended to produce a mixture containing 60.0 wt % ethanol. The specific gravity of a substance is the ratio of density of a substance compared to the density of water. The symbol of weight percent is wt %. (a) Estimate the specific gravity of the 60 % mixture by assuming that y = mx c where y is wt % ethanol, x is mixture specific gravity. Values for m and c are constants. (b) Determine the required volume of the 40 % mixture.

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.1 : Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt % methanol and the second contains 70.0 wt % methanol. If 200 kg of the first mixture is combined with 150 kg of the second, what are the mass and composition of the product? The symbol of weight percent is wt %.

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.3 : A 1.5 weight % aqueous salt solution is concentrated to 4 weight % in a single-effect evaporator. The feed rate to the evaporator is F = 7500 kg / h and the feed is at 85 degree Celsius. The evaporator operates at 1 bar. By assuming that only pure solvent of water exists in the form of vapor from the feed, calculate the flow rate of such vapor V.

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.4 : A mixture consists of benzene (B), toluene (T) and xylene (X). At a temperature of 353 K, the data of vapor pressures : B : 754.12, T : 289.71, X : 91.19. Unit is mm Hg. The pressure P is 0.5 atm. The value of k for each substance is k = (vapor pressure) / P. (a) Calculate k for B, T and X. Let L / V = 0.65. (b) By using the equation V = F / [ (L / V) + 1 ], find the value of V when F = 100, then what is the value of L?

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.5 : In a non-dilute absorber, the inlet gas stream consists of 8 mol % carbon dioxide in nitrogen. By contact with room temperature water at atmospheric pressure, 65 % of the carbon dioxide from a gas stream has been removed. (a) Find the mole ratio of carbon dioxide and nitrogen gases at inlet and outlet gas streams. (b) The Henry's Law provides y = 1640 x for carbon dioxide in water. Find the mole ratio when x = 0.0000427. Mole ratio is y / (1 - y) for y.

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