# Ch4solutions - 4-4C Energycan be transferred o or frOO...

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4-4C Energy can be transferred to or frOO) a control volume as heat.various fOnDS of work. and by mass transport. 4-11 The radiator of a Sleamheating system is initially filled with superheated steam. The valves are closed. and steam is allowed to cool until the pressure drops to a specified value by b"ansferring heat to the room. The amount of heat transfer is to be determined. and the process is to be shown on a P-v diagram. Assumptions 1 The tank is stationary am thus the kinetic and potential energy changes are zero. 2 There are no work interactions. Analysis We take the radiator as the sys~m. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constarc and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E ill - E 0lIl = fj.E IySICID . ~~~ ~~~ ci llll~ Net ~ --- a-.e a. w.-I. kiIICtic. by hal, work,8nd mass pcxenIIaJ, clc.cnerJies '. -Q. =~U = m(u2 -uJ) (sinceW - KE - PE- 0) . -fz!ouc = DoU = m(U2 -UJ) (SmCe W = KJ:; = Pc - 0) ! QQUI=m(ul-U2) I Using data from the steam tables (Tables A-4 ! through A-6), ". some properties are determined to be p. = 300kPa } VI = O. 7964m 3~g ~ = 250.C UI =2728.7kJ~~ kP V I = 0.001043, v A' = 1.6940m3 /kg P2=100 a~ J U I = 417.3", U Ig = 2088.7kJ/kg I I Noting that VI = V2 and Vr < V2 < VI ' thejmass and the final internal en? becomes VI 0.020m3 1 p m=-= 3=Of251 kg VI 0.7964 m /kg I X2 =~~= 0.7964-0.~1043=0.470 v II 1.6940 - O. 1043 U2 =ul +X2ufg =417.36+ 0.470x2088.7)=1399.0kJ/kg I . Substituting, QOUI =m(ul -U2) = (0.0251 kg)(2728.7 - 399.0) kJ/kg Substituting, = 33.4 kJ

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, 4-14 An insulatedrigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined. and the pr<x:ess is to be shown on a P-v diagram. Asswnptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 1. The device is well-insulated and thus heattrans&r is negligible. 3 The energy stored in the resistance wires. and the heat transferred to the tank itself is negligible. Analysis We take the contents of the tank as the system. This is a closedsystem since no mass enters or leaves. Noting that the volume of the systemis constant and thus there is no boundarywork. the energy balance for this stationary closed system Ca4 be expressed as .. E in - E out = I1E System ~-~ ~-~ Net enCI'IY Ir3RSfer ChInle ia internal. kinetic. by Ileal. .-k. aIM!mass poccnlial. etc. energIeS W"ln =~U =m(u2 -u,) (sinceQ = KE = PE =0) VJJ}J =m(U2 -U,) W~ . The properties of water are (Tables A4 thr~ugh A-6) ~ PI: 100kPa } Vf =0.001043, vJ = 1.694Om3/kg xI = 0.25 U f C 417.36, u.. = 2088.7kJ/kg T vI =vf +x,vfg = 0.001043+ [0.21 x(I.6940-0.001043)]= 0.42428m3/kg UI =Uf +X'"jg = 417.36+(0.25)(2088.7)= 939.5kJ/kg /'" v2 = vI = 0.42428m3!kg } = = 2556 7kJ/k "2 u ~4 2 428 J/k . g sat. vapor ~¥. m I Substituting, (IIOY)(8A)6I = (Skg)(2556. 7 -~9_5) kJ/I,.J~~ ~ ) '~ IkJ/s
4.18 A cylinder is initially filled with R-I ~a at a specified state. The refrigerant is cooled at constant pressure. The amount of heatloss is to be determined. and the process is to be shown on a T -y diagram.

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