Answer / kangchuentat

CHEMICAL ENERGY BALANCE - ANSWER 11.5 : Let x(1) = x(2) = 0.5 for 50 : 50 mole fraction mixture, then P = x(1) p(1) g(1) + x(2) p(2) g(2) or 1.08 = 0.5 (0.82) exp [ (0.5) A (0.5) ] + 0.5 (1.93) exp [ (0.5) A (0.5) ]. Then exp [ (0.5) A (0.5) ] = 1.08 / [ 0.5 (0.82 + 1.93) ] = 0.785, A = - 0.97. P for 30 : 70 mixture is P = x(1) p(1) g(1) + x(2) p(2) g(2) = (0.3) (0.82) exp [ (-0.97) (0.7) (0.7) ] + (0.7) (1.93) exp [ (-0.97) (0.3) (0.3) ] = 1.39 bar (proven) where g(1) = exp [ A x(2) x(2) ], g(2) = exp [ A x(1) x(1) ]. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question 35 – 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?

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DIFFERENTIAL EQUATIONS - EXAMPLE 20.1 : By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?

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ENGINEERING MECHANIC - EXAMPLE 15.2 : A cantilever beam of length L = 3 m carries a uniformly distributed load (UDL) of W = 20 N / m throughout the length. Calculate the bending moment, BM of the beam near the fixed end. What is the shear force, SF at this point? Let SF = -WL, BM = -LWL/2.

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CHEMICAL MATERIAL BALANCE – EXAMPLE 2.6 : 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.

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Question 69 - A well delivers 225 US-gallons per minute of water to a chemical plant during normal system operation. (a) Calculate its flowrate in the unit of mega US-gallon per day or MGD. (b) The following formula is written next to the chlorine feed point : (chlorine feed rate, lb / day) = (flowrate, MGD) X (dose, mg / L) x (8.34). If this formula is correct, then what should the chlorine feed rate to be in pounds per day (lb / day) if the desired dose is 2 mg / L. (c) Prove by calculations that the constant 8.34 in the formula next to the chlorine feed point is correct. Let 1 US-gallon = 3.78541 L and 1 mg = 0.0000022046 pound.

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