Answer / kangchuentat

CHEMICAL MATERIAL BALANCE - ANSWER 2.3 : Mass balance of salt : Mass of salt in feed = Mass of salt in concentrated liquid. (0.015) (7500 kg / h) = 0.04L where L = flowrate of concentrated liquid = (0.015) (7500) / (0.04) = 2812.5 kg / h. Let F = V + L as overall mass balance, then V = F - L = 7500 - 2812.5 = 4687.5 kg / h.

Explain what are some common precipitating agents used to remove metals from aqueous waste streams?

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Question 80 - Liquid octane has a density of 703 kilograms per cubic metre and molar mass of 114.23 grams per mole. Its specific heat capacity is 255.68 J / (mol K). (a) Find the energy in J needed to increase the temperature of 1 cubic metre of octane for 1 Kelvin. (b) At 20 degree Celsius, the solubility of liquid octane in water is 0.007 mg / L as stated in a handbook. For a mixture of 1 L of liquid octane and 1 L of water, prove by calculations that liquid octane is almost insoluble in water.

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0 Answers   HPCL,

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MASS TRANSFER - EXAMPLE 4.1 : A concentric, counter-current heat exchanger is used to cool lubricating oil. Water is used as the coolant. The mass flow rate of oil into the heat exchanger is 0.1 kg / s = FO. For oil, the inlet temperature TIO = 100 degree Celsius and the outlet temperature TOO = 55 degree Celsius. For water, the inlet temperature TIW = 35 degree Celsius and the outlet temperature TOW = 42 degree Celsius. What is the mass flow rate of water in kg / s, FW needed to maintain these operating conditions? Constant for heat capacity of oil is CO = 2131 J /(kg K) and for water is CW = 4178 J /(kg K). Use the equation (FO)(CO)(TIO ?TOO) = (FW)(CW)(TOW ?TIW).

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why sphericle shaped vessele used for storage of petrol?

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Is there any way to repair a valve that is passing leaking internally without taking our process offline?

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Hi, Please give me chemical engineering paper for IOCL exam for entire written exam, GD and personel interview model questions to my email id. (chemistnathan@rediffmail.com) Rgds, Ragu

0 Answers   HAL, IOCL,

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In engineering economy, the future value of first year is FV = PV (1 i). For second year it is FV = PV (1 i) (1 i). For third year it is FV = PV (1 i) (1 i)(1 i) where FV = future value, PV = present value, i = interest rate per period, n = the number of compounding periods. By induction, what is the future value of $1000 for 5 years at the interest rate of 6%?

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how to calculate overall heat transfer coefficient in insulated pipeline? Please give a answer with formula

2 Answers   NPCIL,

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where can i get previous question papers of GATE in chemical engineering

7 Answers   CPCL,

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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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POLYMER ENGINEERING - QUESTION 24.2 : 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.

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