determine pressure drop 806040-Pr= 1.251.62아1.5.Repmax = 10²1.42105101036.1.2|2.0104101.010221050.40.6 0,8 I2.0.80.60.42.50.2Sp = Sr3.50.14 6 810! 2 4 68102 2 4 6 810³ 2 4 6 8104 2 4 68 105 24 6 8106 2Rep,max(b) Staggered arrangementFriction factor, f Required informationAir 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 at0°C and 1 atm with a mean velocity of 5 m/s. The tubes are arranged staggered with longitudinal and transverse pitches ofSI = ST = 1.5 cm. There are 25 rows in the flow direction with 15 tubes in each row. Assume the mean temperature as-5°C and the properties of air at this temperature and 1 atm are (Table A-15) k= 0.02326 W/m-K, Cp= 1.006 kJ/kg-K, p=1.316 kg/m3, Pr = 0.7375, µ = 1.705 × 10-5 kg/m-s, and Prs = Pra Ts =-20°C = 0.7408. Also, the density of air at the inlettemperature of 0°C (for use in the mass flow rate calculation at the inlet) is pj = 1.292 kg/m³.NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.0°C1 atmV m/s0.8 mAirSr= 1.5 cm0.8 cmS, = 1.5 cm/Refrigerant, –20°C

Name: determine pressure drop 806040-Pr= 1.251.62아1.5.Repmax = 10²1.42105101
Uploaded: 2024-03-20T06:56:55.000Z
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Description: determine pressure drop 806040-Pr= 1.251.62아1.5.Repmax = 10²1.42105101036.1.2|2.0104101.010221050.40.6 0,8 I2.0.80.60.42.50.2Sp = Sr3.50.14 6 810! 2 4 68102 2 4 6 810³ 2 4 6 8104 2 4 68 105 24 6 8106 2Rep,max(b) Staggered arrangementFriction factor,