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Question 6: Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8-cm-outer-diameter and 0.8-m-long tubes inside which the refrigerant is evaporating at -20°C. Air approaches the tube bank in the normal direction at 0°C and 1 atm with a mean velocity of 5 m/s. The tubes are arranged in-line with longitudinal and transverse pitches of SL = Sr = 1.5 cm. There are 25 rows in the flow direction with 15 tubes in each row. Determine the refrigeration capacity of this system. Evaluate the air properties at an assumed mean temperature of -5°C and 1 atm. Is this a good assumption? 5

Question 6: Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8-cm-outer-diameter and 0.8-m-long tubes inside which the refrigerant is evaporating at -20°C. Air approaches the tube bank in the normal direction at 0°C and 1 atm with a mean velocity of 5 m/s. The tubes are arranged in-line with longitudinal and transverse pitches of SL = Sr = 1.5 cm. There are 25 rows in the flow direction with 15 tubes in each row. Determine the refrigeration capacity of this system. Evaluate the air properties at an assumed mean temperature of -5°C and 1 atm. Is this a good assumption? 5

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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Question 6: Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8-cm-outer-diameter and 0.8-m-long tubes inside which the refrigerant is evaporating at -20°C. Air approaches the tube bank in the normal direction at 0°C and 1 atm with a mean velocity of 5 m/s. The tubes are arranged in-line with longitudinal and transverse pitches of SL = ST = 1.5 cm. There are 25 rows in the flow direction with 15 tubes in each row. Determine the refrigeration capacity of this system. Evaluate the air properties at an assumed mean temperature of -5°C and 1 atm. Is this a good assumption? 0.8 m Air O°C I atm 5 m/s S = 1.5 cm 0.8 cm S = 1.5 cm Refrigerant, -20°C Hint: The refrigeration capacity is T;=-20°C Q = hA,ATm = (179.5 W/m? .°C)(7.540 m²(11.85°C) = 16,040 W SL V=5 Answers: 16,040W; The arithmetic mean fluid temperature is (T;+ m/s Te)/2 = (0-13.7)/2 = -6.9°C, which is fairly close to the assumed value ST of -5°C. Therefore, there is no need to repeat calculations. Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the assumed mean temperature of -5°C (will be checked later) and 1 atm (Table A-15): k = 0.02326 W/m-K p = 1.317 kg/m3 C, =1.006 kJ/kg-K µ = 1.705x105 kg/m-s Pr = 0.7375 Pr; = Pre Ts- -20°c = 0.7408 Also, the density of air at the inlet temperature of 0°C (for use in the mass flow rate calculation at the inlet) is p, = 1.292 kg/m.
Transcribed Image Text:Question 6: Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8-cm-outer-diameter and 0.8-m-long tubes inside which the refrigerant is evaporating at -20°C. Air approaches the tube bank in the normal direction at 0°C and 1 atm with a mean velocity of 5 m/s. The tubes are arranged in-line with longitudinal and transverse pitches of SL = ST = 1.5 cm. There are 25 rows in the flow direction with 15 tubes in each row. Determine the refrigeration capacity of this system. Evaluate the air properties at an assumed mean temperature of -5°C and 1 atm. Is this a good assumption? 0.8 m Air O°C I atm 5 m/s S = 1.5 cm 0.8 cm S = 1.5 cm Refrigerant, -20°C Hint: The refrigeration capacity is T;=-20°C Q = hA,ATm = (179.5 W/m? .°C)(7.540 m²(11.85°C) = 16,040 W SL V=5 Answers: 16,040W; The arithmetic mean fluid temperature is (T;+ m/s Te)/2 = (0-13.7)/2 = -6.9°C, which is fairly close to the assumed value ST of -5°C. Therefore, there is no need to repeat calculations. Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the assumed mean temperature of -5°C (will be checked later) and 1 atm (Table A-15): k = 0.02326 W/m-K p = 1.317 kg/m3 C, =1.006 kJ/kg-K µ = 1.705x105 kg/m-s Pr = 0.7375 Pr; = Pre Ts- -20°c = 0.7408 Also, the density of air at the inlet temperature of 0°C (for use in the mass flow rate calculation at the inlet) is p, = 1.292 kg/m.
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