What is screen analysis and what are its applications in the chemical industry?
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Question 45 - According to Raoult’s law for ideal liquid, x (PSAT) = yP where x is mole fraction of component in liquid, y is mole fraction of component in vapor, P is overall pressure and PSAT is saturation pressure. A liquid with 60 mole % component 1 and 40 mole % component 2 is flashed to 1210 kPa. The saturation pressure for component 1 is ln (PSAT) = 15 - 3010 / (T + 250) and for component 2 is ln (PSAT) = 14 - 2700 / (T + 205) where PSAT is in kPa and T is in degree Celsius. By assuming the liquid is ideal, calculate (a) the fraction of the effluent that is liquid; (b) the compositions of the liquid and vapor phases. The outlet T is 150 degree Celsius.
how should i prepare for my placements.i am a student of chemical eng,final year,thapar university
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Question - Chemical Engineering Material - In crystal material, hexagonal crystal system could form 4-digit index in certain direction of solid. For [1(-1)0] direction in the hexagonal crystal systems of particular catalyst applied in fume removal of incinerator, what is the four-digit index for this direction? Hint : The transformation equations between the 3-digit [h’k’l’] and the 4-digit [hkil] indices are : h = (1/3) (2h’ – k’); i = - (h + k); k = (1/3) (2k’ – h’); l = l’. A. [(-1)100] B. [1(-1)00] C. [(-1)000] D. [00(-1)(-1)] E. [(-1)0(-1)0]
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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Water flows through a pipe with circular cross sectional area at the rate of V / t = 80 L / s where V is the volume and t is time. Let Av = 80 L / s where A is cross sectional area and v is velocity of fluid. For point 1, the radius of the pipe is 16 cm. For point 2, the radius of the pipe is 8 cm. Find (a) the velocity at point 1; (b) the velocity at point 2; (c) the pressure at point 2 by using Bernoulli’s equation where P Rgy 0.5 RV = constant. P is the pressure, R = density of fluid, V = square of fluid’s velocity, g = gravitational constant of 9.81 N / kg and y = 2 m = difference of height at 2 points. The pressure of point 1 is 180 kPa.
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