How do you get v in part c? The solution here says it\'s from table 7E but that\

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How do you get v in part c? The solution here says it's from table 7E but that's skipping some step because the table doesn't give that info, it only gives vf and vg Thermo nobles in other lable TTE Properties of saturated Refrigerant 134a a iquid por Pressure Table Specific Volume Internal Energy Btu/lb Temp. Sat, Sat. Liquid Vape Liquid vapor Liquid vapor Liquid Evap Vapor 53.48 0.01113 9153 93,79 0,000 29,71 0.01143 8.3508 3,74 3,73 00068 0.22 14.25 0,01164 4.3581 289 89.30 2.91 9445 97.37 0.0171 022 2.9747 7,36 91.40 7,40 92,27 99,66 0.22 2.48 0.0118 2.266 10,84 93,00 0.89 90.50 101.39 022 30 15.38 0.01209 1.5408 16,24 95.40 16.31 8765 103,96 0,0364 29.04 1.1692 20.57 85,31 105.88 00452 0.21 0.01232 20,48 0.2 40 27 001 252 0.9422 24.02 98.71 24,14 83.29 07,43 0,0523 0.2 0,01270 0.7887 27.10 99.96 27.24 81.48 108,72 0.0584 0.2 58.35 0,01286 0.6778 29,85 01.05 3001 79.82 109.83 0.0638 65.93 0.01302 0.5938 32.33 02.02 32.53 78.28 1081 111,68 0.0729 02 76,84 34.84 102.89 0.01317 0.5278 72.83 112,46 0.0768 0.2 75.47 36.99 103.68 36.75 04747 0.01332 79.17 0.2 0.0839 113,82 72.91 105.06 40,61 100.56 0.01386 0.3358 106.25 44,43 14,95 0.0902 0.2 160 109.56 0.01412 0.2916 47.23 107.28 47.65 68.26 115,91 0.0958 02 108,18 66,10 116.74 0.1009 0.2 117.74 0.01438 6401 117 A4 0.1057 200 0.01463 0.2288 52.90 108.98 53 44 61.96 118.05 0,1101 0.2 25.28 56.09 132.27 0.01489 0.2056 55.48 109,68 59.96 0.1142 138.79 0.01515 0.1861 57.93 110.30 58.61 57.97 118.99 0.1181 110,84 61.02 260 144.92 0.01541 0.1695 150.70 0.01568 0.1550 62.53 111.31 63.34 56.00 119.35 0.1219 64.71 65.59 119,62 300 156.17 0.01596 0.1424 70.97 49.03 120 00 350 168.72 0.01671 0.1166 69.88 112.45 76.11 43.80 119.91 9.95 0.01758 0.0965 74.81 112.77 450 190.12 0.01 79,63 112.60 81.18 38.08 119.26 199.38 0.0657 111.76 86.39 31 44 117.83 0.1338 0.1417 0.1493 0.1570

Explanation / Answer

When you compare the temperature of refrigerant given that is 100 F with that of the saturation temperature of the refrigerant at 100lb/in^2 which is 79.17F

It can be inferred that the refrigerant is in saturated condition. So v= vg at 100lb/in^2 or v=0.4747ft3/lb

Mass flow rate m = Area of Flow * Velocity of flow/specific volume

M= pi*1^2/(4*144)*30/0.4747

M= 0.3447 lb/sec

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