PROCESS DESIGN - EXAMPLE 21.2 : The names of the flow streams could be represented by : H1 for first hot stream, H2 for second hot stream, C1 for first cold stream, C2 for second cold stream. Data of supply temperature Ts in degree Celsius : 150 for H1, 170 for H2, 30 for C1, 30 for C2. Data of target temperature Tt in degree Celsius : 50 for H1, 169 for H2, 150 for C1, 40 for C2. Data of heat capacity Cp in kW / degree Celsius : 3 for H1, 360 for H2, 3 for C1, 30 for C2. (a) Find the enthalpy changes, dH for all streams of flow H1, H2, C1 and C2 in the unit of kW. Take note of the formula dH = (Cp) (Tt - Ts). (b) Match the hot streams H1 and H2 with the suitable cold streams C1 and C2 to achieve the maximum energy efficiency.
PROCESS DESIGN - ANSWER 21.2 : (a) dH for streams H1 : 3 (50 - 150) = -300 kW. H2 : 360 (169 - 170) = -360 kW. C1 : 3 (150 - 30) = 360 kW. C2 : 30 (40 - 30) = 300 kW. (b) dH (H1 + C2) = 0 kW; dH (H2 + C1) = 0 kW where all heats from hot utilities are supplied to cold utilities. Then H1 is matched to C2 and H2 is matched to C1. 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 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.
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