An equimolar liquid mixture of n-pentane and n-hexane at 80C and 5.00 atm is fed into a

Chapter 8, Problem 8.61

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An equimolar liquid mixture of n-pentane and n-hexane at 80C and 5.00 atm is fed into a flash evaporator at a rate of 100.0 mol/s. When the feed is exposed to the reduced pressure in the evaporator, a substantial amount is vaporized. The temperature in the tank is maintained at 65C by adding heat. The vapor and liquid phases, which are in equilibrium with each other, are separated and discharged as separate streams. The liquid product stream contains 41.0 mole% pentane. A flowchart and an inlet outlet enthalpy table for the process are given below. 100 mol/s, liquid Liquid product 65C, P0(atm) Vapor product 65C, P0(atm) FLASH EVAPORATOR P = pentane H = hexane 50 mole% P 50 mole% H 80C, 5.0 atm yP[mol P(v)/mol] xP = 0.41 mol P(l)/mol Q(kW) References: P(l, 65C), H(l, 65C) Substance n_in H in n_ out H out P(l) n_ a H a n_ c H c P(v) n_ d H d H(l) n_ b H b n_ e H e H(v) n_f H f (a) Using Raoults law for vaporliquid equilibrium calculations, calculate (i) the system pressure, P0(atm), (ii) the mole fraction of pentane in the vapor product, yP, (iii) the volumetric flow rate of the vapor product, V_(L/s), and (iv) the fractional vaporization of pentane, f(mol vaporized/mol fed). (b) Determine values for all the n_s and H^s in the enthalpy table and calculate the required rate of heat addition to the evaporator, Q_(kW). (c) How would each of the variables calculated in Parts (a) and (b) change if the evaporator temperature were increased (increase, decrease, no change, cannot tell)? Explain your reasoning.