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Hie hitlanahan_?5. #2 uesday, February 21, 2023 6:39 PM Problem 3.25 Complete this table for H20: Eo P, kPa h, kI/kg x Phase description F 200 0.7 2. 140 1800 2. 950 0.0 4. 80 500 5. 800 3162.2 P= 200 Ka, X, =F Gat = o.21°C he = $04. 71 hr? het ¥ Chey) hey > a2al-6 $04.71 + (7) C22o1-6) hy? 2046. €3 KY/Xy Sou.7) & 1S4).12 A045 -33 Pride. : Sat liquid Vapool Nike 2. kat = 140%, = oo K\/M Psat = 361-53. Kfa he > 584.1¢ h) = he th Che) hey = 144.3 Koo = 584.14 +X C21Ha3 Ge = 5S 1alo-84 = x, (2144.3) xX, = -56S Phase * Sat guid vapoot mixte 3. Pout = 160 Ken, Xo Tent = 177-66 °C he = 752.74 hg> he + x3 hf) hy = 2022-4 = 52-24 +0 © 2032.4) hy = 752-74 Ky hg > 752-74 Phases Sat ligvid Uy. T : 40°C fect = S00 Kon Taf = 151-93 > Y Phase: Compressed \qvid hy= 335-24 KS/ty Xy> Canned define f. Coat = 400 Kea, hs? 3-2 Kj/ Hy Teep 2 360°C Phase * Super) htated Stan Xo: Cannot be defined- Problem 3.26E Complete this table for refrigerant-134a: Pow P, psia h, Btu/lbm sx Phase description I. 80 78 2 15 0.6 3. 10 70 Y. 180 129.46 5: 110 1.0 f= Bo pia, hy? 76 biv/ bm Tout = 65.9096 he = 33.344 hy > 33-3444 x,C 74.904) > Fe hey * 78-604 XC 78-664) = 14bob x)= Gb y? bbb Chase: 544 Iigyid Vapor nishle 2. qh? °F, X78 Osuin = 25-754 P5ia hg? 6-494 t (4) C 69.379) ° he = 16-464 = 16-444 + 63.0262 hey? 94-377 has 64-415 hy = 64.419 Gte/ Ibn frase: Set liquid -Vapol Matte 3 I? lo°P erty jo in fuse Compressed INyold, Tap 256.3°> °F hg * 18:38 Gh/ bea x5 = camot deflor compresses llgatd L. fz lo Foi hy 7 1A%46 Btu/ion Ty= bo? F Print : Sutel heated Xy = Cannot aefing fol Suphhented 5&7 lloF, Xz = | 6 = 161.18 P5ia he © 117-23 Gtu/ ibm Prost: Sat. Vofoul Problem 3.27 Complete this table for refrigerant-134a: Ta P, kPa u, kJ/kg Phase description 20 95 —12 Saturated liquid 400 300 8 600 J= 40°C | U2 45 K/Ky Coat = 572.07 Kf4 Uf= 7¢-%6 Ug = P4192 State = Satlated ninnle 2 %s-p %, Godt 2 Satvloted Mqutd Baty : 195-37 Kfar Up = 3578 Kit 3, 6: Yoo KPa, U= 300 Ky/Ky Phase: Sufpthented 2 Boo= 244.53 _ T= Got (Sem ie (46-90) T= 96.a2°C Ly. ly? 3%, f, = boo Ker ruse + Sup Cooled Uy 2 62:34 KI/ky Problem 3.30 A piston—cylinder device contains 0.85 kg of refrigerant- 134a at —10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a. FIGURE P3-30 |. Car X Ae T Mpisen <9 = Px Ae Ap = Fx bp" Fear + Meson X_4 = Pring) Ae 4X0? pa + IAX4$1 = inal Fx -as? Spor tN = Peinal 7954 + 0695 a ee Oe 0YG | $4,000 + 9347-556 > Reina [Ginn 2 10.348 pa | a 7+ Wwe, T= 18°C et 90.348 Kea Viz 2302 n/Ka y “a? 2544 7 /Kg Viz MV = -45X -230% = IG5¢? w? 2 nvg = -65 x 2544 = 1624 nd Va Vp = -AMbaY - 19567 = 3 T2-l0°C hy = 247-26 KS/Kg F715 , hy 2 Qoe.le KI/Kg AH = Hy-Hy = h-hh) ~ 45C 264 -Qu7. 76) [4H = 17:34 K5/xg) Problem 3.44 Water initially at 200 kPa and 300°C is contained in a piston—cylinder device fitted with stops. The water is allowed to cool at constant pressure until it exists as a saturated vapor and the piston rests on the stops. Then the water continues to cool until the pressure is 100 kPa. On the T-v diagram, sketch, with respect to the saturation lines, the process curves passing through the initial, intermediate, and final states of the water. Label the 7, P, and v values for end states on the process curves. Find the overall change in internal energy between the initial and final states per unit mass of water. Q FIGURE P3-44 (= oo Kf B\2 doe Kfar 63 = loo Kf 7) = 300°C Satulatt) \afoo f VE = 001 04 hy Uy? bay A ¥/uy Supc(htattd Stem whl U37 .9457 M>/Ky f =Ao0 Kyo, Togp = 20.2% / V3 = Uf t+ x Cuy-uf) > Tot U= 2408-4 KI/Ky 572 00104 + 1-644- .00/04) V, > 23162 M%/mng X= 52.26g Sablatid Stan inble U3 = up ¢ x Uy-vF) 17-4 + -$296 02505-6-417-6) Uy 2 4/7-4 + _Jogl. al [uy * 1504594 Kit) Va= 9529.) KT/Mg Va= - 9957 Nk, Au = lu,- uv, 1409.4 - 508.544 Au = [390.206 K/h Problem 3.63 The spring-loaded piston—cylinder device shown in Fig. P3—63 is filled with 0.5 kg of water vapor that is initially at 4 MPa and 400°C. Initially, the spring exerts no force against the piston. The spring constant in the spring force relation F = kx is k = 0.9 kN/cm and the piston diameter is D = 20 cm. The water now undergoes a process until its volume is one-half of the original volume. Calculate the final temperature and the specific enthalpy of the water. Answers: 220°C, 1721 kJ/kg Spring Dt= h-F, = Ka-x) Ke a Ku/m > 4mpA = KC Ay - Axi) d= 2om 1)? Hore = K ft Supt heated table at p24 Mpa and 7% 2 Yac% ALB-Uy “Kn Cy-v) Vz -07343m9/My yz SEE: opcais A ere gepceny 40 hth, Ca ore) 412 PEP H= Ka (yu) E Ty = 2lé.4/ 232 - 2252 ) (293.45- at6-34) 2500 - 2250 Vy= -09604Y4 n'Mhy he = 943.55 Ky/Kg hy = 2401 Ky /hy My = Un -Up = -036715- -o0)lG0 —————— Vy-VF OBLoqy- . 00 [140 X7> .4l¢ y hae he + aC bq-he) 43.55 + ule X4ol- 943-55) 
