Question

Problems 7-10 are based on the following description Two kilograms of water inside a piston-cylinder device undergoes the process from state 1 to 3, shown in the p-V diagram below. The initial conditions at state 1 are P1 - 150 kPa and V1 -0.002106 m. At the final state 3 are p = 200 kPa and V3 = 0.4 m? A p(kPa) 3 1 V(m") 4. The work in kJ from state 1 to 3 is closer to: 59.7 39.8 159.1 136.9 2013 4220 75.6

please help solve all questions

Answer 1

Cruin Pi= 150kpa 3 P2 = P = 1 so kPa P3= 200kpa. V = 0.002106m3 12 = 1₂=0.4m² V3 = U2= 0.4 m² ܘ ܢܟ m=214 work in kJ fren I to ③ Wi-z= W2-3= Constant uolun. wi- 2t W 2-3 Wr3=0 w,-3= Wi-2 - Pi (U2-U,.). - Iso (0.4-0.002 106) 972 W ,-3= 59.684 kJ ~ 59.7KI 88) spefic internal energy at s tat State V2 = 0.4m² 3 m=2kg P2 = Isokpa, 0.4 Uz 0.2 ke 0 1 2 m3 specific noueme noueme B2 = m

using at slate steam tabel. at p= 150 ko 0.00 1053 la nf = 46713 KS kg hgan ke Ug = 1.1593 hops 222 6.0 KS IT ve allaf= 466.97 KI kg ug kg مد 2519.2 ks as Y Z Uz LV2LUg { state is liquid- vapour states. U₂ = y + x(Ug-Uf) 0-2 = 0.00 1053 + x (1.1593 - 0.001053) X=0.1717 Specific internal energy 42 U2 = Uptx (Ug-U) 42= 466.97 +0.1917 (2519.2 - 466.9+) U2 = 819.47 KJ ig 881 As

99) heat transfer from 1 to ③. at state 0 P = 15okpa, V = 0.002106 m3 Specific notune Ui=v, 0.002106 =o.solos3 m? 2 53 i. State 0 is saturated liquid. 0.0010 53.m 3 = Uz = kg hah= 467.13 kJ atstat

State hze hot xlhas) het 467.13 +0.717x (22260) hz- 849.33 kJ ig heat mausfor 1-2 9= mch₂-hi) 9= 2x1849.33 - 467.13) = 764 kJ gg - AI

810) solr specific volume 3 = state & 200 kPa. V3 = 0.4m3 U. P2= 0:4 0.2 m3 lg. from steam take at P=200kpa & Uz = ho c 504.7 KI 0.0010605 m3 14 Ug = 0.88568 m3 kg 104.7 x = 2201.5 y <U3 <Vg : { it is sliquid -reporn states has= 2201.5KJ kg as U3= Uf = xL Vg-U6) 0.2 = 0.000605 + X (0.88568- 0.0010 60s) [x = 0-23488 hf +X (hfg) h3= 504.7+0.22 488 x 2201-5 h3= 99g.788 kJ kg 9₂-3= m (hz-12) 2 x 1999.788 - Pos 8 49.33) ₂-3 = 300.9 kJ closer to 280 KS AS P8
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