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PROBLEM 6.18 Air at 90°C and 1.00 atm (absolute) contains 10.0 mole% water. A continuous stream of this air enters a compressor-condenser, in which the temperature is lowered to 15.6°C and the pressure is raised to 3.00 atm. The air leaving the condenser is then heated isobarically to 100°C. Calculate the fraction of water that is condensed from the air, the relative humidity of the air at 100°C, and the ratio m' outlet air @ 100°C /m' feed air @ 90°C. Solution Basis: 1 mol feed. Since the problem statement asks us to calculate the ratio of the volumes of feed air and exit air, we will label both volumes on the flowchart. DA = dry air Compressor Heater Vi (m³) 1 mol 0.100 mol H2O(v)/mol 0.900 mol DA/mol 90°C, 1 atm V2 (m³) n2 (mol) Пz (mol) y2 (mol H20(v)/mol) (1-y2) (mol DA/mol) 15.6°C, 3 atm V2 (1-y2) 100 C, 3 atm n3 [mol H2O(1)] 15.6°C, 3 atm DEGREE-OF-FREEDOM ANALYSIS ON COMPRESSOR UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION + 4 unknowns (V1, n1, n2, y2) 2 balances 1 gas law at inlet 1 DOF Problem is unsolvable Since we have more unknowns than equations, unless we can come up with another relationship among the compressor variables we're stuck. (If you do the DOF analyses for the overall system and the heater you'll run into the same problem-try it.) Fortunately, there is another relationship. Can you state what it is and justify your claim? (Hint: What do you know about the two streams leaving the compressor?) (6.18-2) The solution strategy is straightforward.

PROBLEM 6.18 Air at 90°C and 1.00 atm (absolute) contains 10.0 mole% water. A continuous stream of this air enters a compressor-condenser, in which the temperature is lowered to 15.6°C and the pressure is raised to 3.00 atm. The air leaving the condenser is then heated isobarically to 100°C. Calculate the fraction of water that is condensed from the air, the relative humidity of the air at 100°C, and the ratio m' outlet air @ 100°C /m' feed air @ 90°C. Solution Basis: 1 mol feed. Since the problem statement asks us to calculate the ratio of the volumes of feed air and exit air, we will label both volumes on the flowchart. DA = dry air Compressor Heater Vi (m³) 1 mol 0.100 mol H2O(v)/mol 0.900 mol DA/mol 90°C, 1 atm V2 (m³) n2 (mol) Пz (mol) y2 (mol H20(v)/mol) (1-y2) (mol DA/mol) 15.6°C, 3 atm V2 (1-y2) 100 C, 3 atm n3 [mol H2O(1)] 15.6°C, 3 atm DEGREE-OF-FREEDOM ANALYSIS ON COMPRESSOR UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION + 4 unknowns (V1, n1, n2, y2) 2 balances 1 gas law at inlet 1 DOF Problem is unsolvable Since we have more unknowns than equations, unless we can come up with another relationship among the compressor variables we're stuck. (If you do the DOF analyses for the overall system and the heater you'll run into the same problem-try it.) Fortunately, there is another relationship. Can you state what it is and justify your claim? (Hint: What do you know about the two streams leaving the compressor?) (6.18-2) The solution strategy is straightforward.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Date: 1 of 1 PROBLEM 6.18 Air at 90°C and 1.00 atm (absolute) contains 10.0 mole% water. A continuous stream of this air enters a compressor-condenser, in which the temperature is lowered to 15.6°C and the pressure is raised to 3.00 atm. The air leaving the condenser is then heated isobarically to 100°C. Calculate the fraction of water that is condensed from the air, the relative humidity of the air at 100°C, and the ratio m outlet air @ 100°C /m feed air @ 90°C. Solution Basis: 1 mol feed. Since the problem statement asks us to calculate the ratio of the volumes of feed air and exit air, we will label both volumes on the flowchart. DA = dry air Compressor Heater V1 (m³) V2 (m³) П2 (mol) n2 (mol) V2 (mol H20(v)/mol) (1-y2) (mol DA/mol) 15.6°C, 3 a 1 mol 0.100 mol H2O(v)/mol 0.900 mol DA/mol 90°C, 1 atm V2 (1-y2) 100 C, 3 atm N3 [mol H2O(1)] 15.6°C, 3 atm DEGREE-OF-FREEDOM ANALYSIS ON COMPRESSOR UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION 4 unknowns (V1, n1, n2, y2) 2 balances 1 gas law at inlet 1 DOF Problem is unsolvable Since we have more unknowns than equations, unless we can come up with another relationship among the compressor variables we're stuck. (If you do the DOF analyses for the overall system and the heater you'll run into the same problem-try it.) Fortunately, there is another relationship. Can you state what it is and justify your claim? (Hint: What do you know about the two streams leaving the compressor?) (6.18-2) The solution strategy is straightforward.
Transcribed Image Text:Date: 1 of 1 PROBLEM 6.18 Air at 90°C and 1.00 atm (absolute) contains 10.0 mole% water. A continuous stream of this air enters a compressor-condenser, in which the temperature is lowered to 15.6°C and the pressure is raised to 3.00 atm. The air leaving the condenser is then heated isobarically to 100°C. Calculate the fraction of water that is condensed from the air, the relative humidity of the air at 100°C, and the ratio m outlet air @ 100°C /m feed air @ 90°C. Solution Basis: 1 mol feed. Since the problem statement asks us to calculate the ratio of the volumes of feed air and exit air, we will label both volumes on the flowchart. DA = dry air Compressor Heater V1 (m³) V2 (m³) П2 (mol) n2 (mol) V2 (mol H20(v)/mol) (1-y2) (mol DA/mol) 15.6°C, 3 a 1 mol 0.100 mol H2O(v)/mol 0.900 mol DA/mol 90°C, 1 atm V2 (1-y2) 100 C, 3 atm N3 [mol H2O(1)] 15.6°C, 3 atm DEGREE-OF-FREEDOM ANALYSIS ON COMPRESSOR UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION 4 unknowns (V1, n1, n2, y2) 2 balances 1 gas law at inlet 1 DOF Problem is unsolvable Since we have more unknowns than equations, unless we can come up with another relationship among the compressor variables we're stuck. (If you do the DOF analyses for the overall system and the heater you'll run into the same problem-try it.) Fortunately, there is another relationship. Can you state what it is and justify your claim? (Hint: What do you know about the two streams leaving the compressor?) (6.18-2) The solution strategy is straightforward.
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