FTFS Chap05 P221

# FTFS Chap05 P221 - Chapter 5 The First Law of...

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Chapter 5 The First Law of Thermodynamics 5-221 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and the two tanks come to the same state at the temperature of the surroundings. The final pressure and the amount of heat transfer are to be determined. Assumptions 1 The tanks are stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and thus heat transfer is negligible. 3 There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as ] ) ( ) ( [ ] [ 0) = PE = KE (since ) ( ) ( 1 1 1 1 2 total 2, , 1 , 1 , 2 out out energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in B A B A B A B A u m u m u m U U U Q W U U U Q E E E = = = Δ + Δ = Δ = Δ = + 43 42 1 1 H 2 O 400 kPa H 2 O 200 kPa × B A Q The properties of water in each tank are (Tables A-4 through A-6) Tank A: kJ/kg 1949.3 , 31 . 604 /kg m 0.4625 , 001084 . 0 80 . 0 kPa 400 3 1 1 = = = = = = fg f g f u u v v x P ( ) [ ] () kJ/kg 2163.75 3 . 1949 8 . 0 31 . 604 /kg m 0.3702 001084 . 0 4625 . 0 8 . 0 001084 . 0 1 , 1 3 1 , 1 = × + = + = = × + = + = fg f A fg f A u x u u v x v v Tank B: kJ/kg 2304.9 , 88 . 104 /kg m 43.36 , 001003 . 0 /kg m 0.731 C 25 /kg m 0.731 kg 0.957 m 0.7 kg 0.957 417 . 0 540 . 0 kg 0.417 /kg m 1.1988 m 0.5 kg 0.540 /kg m 0.3702 m 0.2 kJ/kg 2731.2 /kg m 1.1988 C 250 kPa 200 3 3 2 2 3 3 2 , 1 , 1 3 3 , 1 , 1 3 3 , 1 , 1 , 1 3 , 1 1 1 = = = = = = = = = = + = + = = = = = = = = = = = fg f g f t t B A t B B B A A A B B u u v v v T m V v m m m v V m v V m u v T P o o Thus at the final state the system will be a saturated liquid-vapor mixture since v f < v 2 < v g . Then the final pressure must be P 2 = P sat @ 25 ° C = 3.169 kPa Also, kJ/kg 143.60 2304.9 0.0168 104.88 0168 . 0 001 . 0 36 . 43 001 . 0 731 . 0 2 2 2 2 = × + = + = = = = fg f fg f u x u u v v v x Substituting, Q out = -[(0.957)(143.6) - (0.540)(2163.75) - (0.417)(2731.2)] = 2170 kJ 5-208

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Chapter 5 The First Law of Thermodynamics 5-222 Problem 5-221 is reconsidered. The effect of the environment temperature on the final pressure and the heat transfer as the environment temperature varies from 0°C to 50°C is to be investigated. The final results are to be plotted against the environment temperature. "Knowns" Vol_A=0.2 "[m^3]" P_A[1]=400 "[kPa]" x_A[1]=0.8 T_B[1]=250 "[C]" P_B[1]=200 "[kPa]" Vol_B=0.5 "[m^3]" T_final=25 "[C]" "T_final = T_surroundings. To do the parametric study or to solve the problem when Q_out = 0, place this statement in {}." {Q_out=0"[kJ]"} "To determine the surroundings temperature that makes Q_out = 0, remove the {} and resolve the problem." "Solution" "Conservation of Energy for the combined tanks:" E_in-E_out=DELTAE E_in=0 "[kJ]" E_out=Q_out "[kJ]" DELTAE=m_A*(u_A[2]-u_A[1])+m_B*(u_B[2]-u_B[1]) "[kJ]" m_A=Vol_A/v_A[1] "[kg]" m_B=Vol_B/v_B[1] "[kg]" u_A[1]=INTENERGY(Steam,P=P_A[1], x=x_A[1]) "[kJ/kg]" v_A[1]=volume(Steam,P=P_A[1], x=x_A[1]) "[m^3/kg]" T_A[1]=temperature(Steam,P=P_A[1], x=x_A[1]) "[C]" u_B[1]=INTENERGY(Steam,P=P_B[1],T=T_B[1]) "[kJ/kg]" v_B[1]=volume(Steam,P=P_B[1],T=T_B[1]) "[m^3/kg]" "At the final state the steam has uniform properties through out the entire system." u_B[2]=u_final "[kJ/kg]" u_A[2]=u_final "[kJ/kg]" m_final=m_A+m_B "[kg]" Vol_final=Vol_A+Vol_B "[m^3]"
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FTFS Chap05 P221 - Chapter 5 The First Law of...

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