Question

A stream of air at 600.0 °C and 835 torr with a dew point of 30.0 °C flowing at a rate of 1915 L/s is to be cooled in a spray cooler. A fine mist of liquid water at 15.0 °C is sprayed into the hot air at a rate of 120.0 g/s and evaporates completely. The cooled air emerges at 1 atm. Physical Property Tables *Material Balances For the flow diagram below, use two material balances and three constraints to determine all unknowns except for the outlet temperature 73. g O (1)/s n2 (mol H20(1)/s) 160 1995/s na (mol/s) X1 (mol H20 (v) /mol) (1-x) (mol DAímol) 800. 0 toB5 Tdp = °80.0 ng (mol/s) xz (mol H2O(v)/mol) (1-x3) (mol DÁ/mol) T3 (°C), 1 atm C -1 mol/s -2 mol/s D D *3 mol/s = 4 mol H2O(v)/mol *3= D *5 mol H2O(v)/mol

Answer 1

(a) Draw and label the diagram as follows. C.813 - Determine the molar flow rates of dry air and water in the air stream (mol 5) = molar flow rate of dry air in the air stream (mol/s) = molar flow rate of H O in the air stream 1915 L 832 mm Hg| mol-K 14+*.= 5 873 K 162 36 L-mm Ha - 29.27 mol's (30°C) + P 31.824 mmHg 835 mmHg = 0.0381 mol Homol ait Solve the equations = 28.15 mol dry it's 11 =1.115 mol H,O'S Using the molar flow rates you just found, derive the expressions for the enthalpies of dry air and water at the outlet air temperature. Then substitute them into the energy balance to calculate the final temperature of the emerging air stream, assuming that the process is adiabatic by solving the following integral References: H, OIL. 25°C), Aisiv. 25°C) Substance (mol's dry air 28 15 Rulmal 1437 Aklot) LGT 2815 HOCv) 1.115 H00) 985 0 Use the energy balance with the enthalpies you found and use a spreadsheet to solve the following polynomial equation AH-0- -I-H. AH = (28.15).C), dT +(7.1) ) dI + +KG) -28.15)(14.37)-(1.00)(CC). dT+#+CC)... dT) = The final temperature of the emerging air stream which is assumed to be an adiabatic process is found to be T=142°C

The rate (kW) that beat is transferred from the hot air feed stream in the spray cooler will be calculated using the following equation. Use the final temperature you calculated above. Oun AH =* (C.T = (28.15) ) T-1.001 ), LT Q =-298 kW The heat will vaporize the entering liquid water and bring it to the final temperature of 142 °C. (c) Use the results of parts (a) and (b) to explain this process. Heat gets transferred from the air to the cold water mist when the cold water contacts the hot air. As a result, the temperature of the gas lowers as well as increasing the temperature of the cooling water

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