Answer / kangchuentat

CHEMICAL MATERIAL BALANCE – ANSWER 2.2 : (a) Let X = specific gravity of the 60 % mixture. By interpolation for a straight line with constant m as gradient, then (75 - 40) / (0.877 - 0.952) = (60 - 40) / (X - 0.952). Then 35 / (-0.075) = -466.67 = 20 / (X - 0.952). X = -0.0429 + 0.952 = 0.9091. (b) Let V = the required volume of the 40 % mixture. Mass of ethanol in 75 % mixture + Mass of ethanol in 40 % mixture = Mass of ethanol in 60 % mixture. Then, 300 (0.877) (0.75) + V (0.952) (0.4) = (V + 300) (0.9091) (0.6). Then 197.325 + 0.3808 V = 0.5455 V + 163.638. V = 204.53 gallons.

FOOD ENGINEERING - QUESTION 23.2 : (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.

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How to caliculate the boil up rate, dia and height in both packed and tray distillation columns and any referance books for the design.

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Should slurry pipes be sloped during horizontal runs?

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what is Vx-EHT(RED) complex chemical compound.

1 Answers   Excel,

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hi im appering for the written test of vizag steel plant for MT-2009.plz send me the pattern of placement paper plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.8 : Cash flow agreement where the sign + or - of the amounts are start negative and at a point switch to positive and stay positive is referred to has an investment. The cash flow could be denoted by the symbol of [a, b, c ...]. Consider the following two cases in a biochemical factory : (a) You pay $1 today. You receive $1.05, one year from now. (b) You pay $3 today. You receive $5, one year from today. You pay $2, two years from today. Determine if cases (a) and (b) are an investment or a business transaction with cash flow symbol. Find the interest rate r in the cash flow in both cases (a) and (b).

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How to size the cooling tower sump / basin for any given circulation rate? What is the guideline?

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what is mean by chemical engineering and its scope in industry

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Question 59 – There are 5 simultaneous equations with 5 unknowns as follow : 16 V – 4 X – Z = 36, 4 X – 4 W – Y + Z = 3, 8 V + 4 W – X + Y = 10, X – W = 1, V + W = 0. Find the values of V, W, X, Y and Z accurately within 15 minutes. State the computer program that you use to get the answers.

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In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).

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Question 94 - The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.

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I want previous year question paper of ONGC for chemical engg.

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