hw1617 - 5—68 Refrigerant-134a is throttled by a...

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Unformatted text preview: 5—68 Refrigerant-134a is throttled by a capillary tube. The quality of the refrigerant at the exit is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. ' ‘ ‘ ' ‘ “ ' ' Analysis There is only one inlet and'one exit, and thus m1 = "'12 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as El _E =>AE 7IO (steady) = 0 in out system ' ' 50°C Em : 1?” Sat. liquid rhhl = mhz . hl = hz since Q§W=Ake§Ape—2'O. The inlet enthalpy of R-134a is, from the refrigerant tables (Table A-1 l), . -12°C Tl = 50°C h =h' =l23.49kJ/k sat. liquid} 1 f g The exit quality is = 0.422 T2 = —12°c hz -hf 12349-3592 x = =———-—-—— h2 = h1 2 h fg 207.38 5-79E Liquid water is heated in a chamber by mixing it with saturated water vapor. If both streams enter at the same rate, the temperature and quality (if saturated) of the exit stream is to be determined. Assumptions 1 This is a steady-flow process-“since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From steam tables (Tables A-SE through A-6E), h] E hf@ 50°F ,=‘ h; = hg@5o psia = 1174.2 Btu/lbm Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady—flow system can be expressed in the rate form as Mass balance: . . - no d T = 50°F min ‘ mout : Amsystem (Stea y) = 0 1 \ min = mout ' I'120 m, m2 = m = 2m (P = 50 psia) ".11 = ".12 = t . T3: x3 Energy balance: S'at' V'apor / m2 = m1 ' ' _ ' $0 (steady) Ein _ Eout " AEsystem = 0 W \ _ - _ ~——J Ra“? 0f "5‘ energy “anSfel’ Rate of change in internal, kinetic, by heat, work, and maSS potential, etc. energies Em = E out n'rlhl + #12)»; = m3h3 (since Q = W =Ake ; Ape ; 0) Combining the two gives 8. mh, + mhz =‘ 2M3 or h3 '=' (h1 + 'h'2 )/ 2 t R-134a .| -12°C 5-79E Liquid water is heated in a chamber by mixing it with saturated water vapor. If both streams enter at the same rate, the temperature and quality (if saturated) of the exit stream is to be determined. Assumptions 1 This is a steady-flow process'since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From steam tables (Tables A-SE through A-6E), hl 2 hf@ 50°F ,= 18.07 Btu/lbm h2 = hg@ 50 pm = 1174.2 Btu/lbm Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: . . . 7Io t d T = 50°F min _ mout : Amsystem (S ea Y) = 0 I \ min = mout ‘ H20 m1 +n'12 =riz3 =2n'1 (P=50 psia) m1=m2 =ri1 73% Energy balance: - Sat. Vapor / m2 = m1 ' ‘ __ ' 90 (steady) _ Ein _ Eout _ AEsystem '_ 0 k—v—J Rate 0f “6‘ energy “3115f” Rate of change in internal, kinetic, by he“, Work, and mass potential, etc. energies Ein = Eout mlhl + mzhz = m3h3 (since = W =Ake E Ape .2. 0) Combining the two gives \. rhh1+ mh2 ; 251713 or h3 ‘=' (k1 + h'z)/ 2 Substituting, h3 = (18.07 + 1174.2)/2 = 596.16 Btu/lbm _ At 50 psia, hf: 250.21 Btu/lbm and kg = 1174.2 Btu/lbm. Thus the exit stream is a saturated mixture since hf< h;, < kg. Therefore, . T3 = Tm 50 p... = 280.99°F and ha -hf = 596.16—250.21=0_374 X3: h fg 924.03 5-83E [Also solved by EES an enclosed CD] Air is heated in a steam heating system. For specified flow rates, the volume flow rate of air at the inlet is to be determined. Assumptions 1 This is a steady-ow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.3704 psia.fi3/lbm.R (Table A-lE). The constant pressure specific heat of air is cp =' 0.240 Btu/lbm-°F (Table -A¥2E). The enthalpies 0f steam at the inlet and the exit states are (Tables A-4E through A-6E) P =30 'a 3 p51 I13 = 1237.9 Btu/lbm T3 = 400°F P4 = 25 psia T4 :21”. }hi 5 f@m.F =180.21Btu/lbm Analysis We take the entire heat exchanger as the system, which is a contr balances for this steady-flow system can be expressed in the rate form as Mass balance (for each fluid stream): 01 volume. The mass and energy - - _ . <90 (steady) _ min _ mout ‘ Amsystem " 0 min = mout litl=lh2=tita and Energy balance (for the entire heat exchanger): E'. _ = 710 (steady) - = 0 in out system Rate of net energy transfer by heat, work, and mass Ein = Eout Rate of change in internal, kinetic, potential, etc. energies mlhl + #13113 = mzhz + m4h4 (since Q = W = Ake 5 Ape ; 0) Combining the two, ma (h2 — h] )= ms (113 414) Solving for ma : m0 = h3—h4 ms 2 [13—174 ms h2 _h1 cp(T2 _Yl) Substituting, ".1 _ (1237.9—180.21)Btu/lbm " _ 15 lbm/ ' =1322 lbm/ ' =22_041b (0.240 Btu/lbm - °F)(130 _ 80)°F ( mm) mm m/s Also, RT1 (0.3704 psia -ft3/1bm -R)(54O R) V1 = —’— — _ 1”1 14.7 pSla = 13.61 ft3/lbm Then the volume flow rate of air at the inlet becomes VI = maul = (22.04 lbm/s)(l3.6l ft3/lbm) = 300 Wis 5-90 An adiabatic open ’feedwater heater mixes steam with feedwater. The outlet mass flow rate and the outlet velocity are to be determined. Assumptions Steady operating conditions exist. Properties From a mass balance m3 =n'i, +2512 =0.2+10=1o.2 kgls 12%;? ' a The specific volume at the exit is (Table A-4) 59°C \A \ ,«\ d’O (steady) = O = Am system Energy balance (for the entire heat exchanger): - - _ ' I 7I0 (steady) - _ Ein _ Eout " AEsystem _ 0 B——fir‘——J Rats Of “Ct encrgy “m5er Rate of change in intemal, kinetic, by heat, work, and mass potential, etc. energies Em = 141,211 + m3h3 = mzhz + 2514/14 (since Q = W = Ake 2 Ape a 0) out Combining the two, ma (h2 — h1 ) = 1515013 — 114) Solving for ma : ma: Substituting, ma = was lbm/min) = 1322 lbm/min = 22.04 lbm/s (0.240 Btu/1bm-°F)(130 — 80)°F Also, ~ . 3 ' RTI _ (0.3704 pSIa ft mnn R)(540 R) =13_61 113 mm P1 14.7 p51a Then the volume flow rate of air at the inlet becomes ()1 = maul =(22.041bm/s)(13.61ft3/1bm)= 300 ft3ls 5—90 An adiabatic open 'feedwater heater mixes steam with feedwater. The outlet mass flow rate and the outlet velocity are to be determined. Assumptions Steady operating conditions exist. Properties From a mass balance m3 =m1 +m2 =0.2+10=1o.2 kgls lggaga The specific volume at the exit is (Table A-4) 50°C \\ 10 kg/s Q) P = 100 kPa 3 0 }U3 sump/60°C =0.001017m3/kg @aloogd’a T3 = 60 C Steam ® 60 C The exit velocity is then 100 kPa / . . 160°C V3 = "W3 = 4m3‘2’3 0.2 kg/s A3 7rD _ 4(102 kg/s)(0.001017 m3/kg) 7r(0.03 nn)2 = 14.68 mls 5-21C Flowing fluids possess flow energy in addition to the forms of energy a fluid at rest possesses. The total energy of a fluid at rest consists of internal, kinetic, and potential energies. The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies. 5-128 R-134a from a tank is discharged to an air-conditioning line in an isothermal process. The final quality of the R-134a in the tank and the total heat transfer are to be determined. Assumptions 1 This is an unsteady, process since the conditions within the device are changing during the process, but it can be analyzed as a uniform—flow process since the state of fluid at the exit remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. Analysis We take thevtank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min "mout = Amsystem _me = m2 —m1 me =ml _m2 Energy balance: Em _ Eom = AESystem Liquid R-l34a %_4 W 5 kg Net energy transfer Change in internal, kinetic, 24°C by heart work, arid mass potential, etc. energies Qin ‘mehe = m2uz _mlul _ Q?“ = "’2"? f’Wl T’I'kbe Combining the two balances: Qm = m2u2 _m1ul +(mi ‘m2 )he. The initial state properties of R—134a in the tank are u, = 0.0008261 m3/kg- u, = 84.44 kJ/kg (Table A-1 1) T1 = 24°C } he = 84.98 kJ/kg x=0 Note that we assumed that the refrigerant leaving the tank is at saturated liquid state, and found the exiting enthalpy accordingly. The volume of the tank is v = mlul = (5 kg)(0.000826lm3/kg) = 0.004131m3 The final specific volume in the container is v 0.004131m3 U2 :— =—-—-——— =0.01652m3/kg' m2 0.25-kg The final state is now fixed. The properties at this state are (Table A-1 1) T2 = 24°C V, = 0.01652 m3/kg x2 = — ————————-— = 0.5061 0 0.031834—0.0008261 fg } V2 _Vf _ 0.01652—0.0008261 u2 =uf +x2ufg =84.44 kJ/kg+(0.5061)(158.65 kJ/kg)=164.73 kJ/kg Substituting into the energy balance equation, Qin = mzuz ‘m1u1+(m1‘m2)hé = (0.25 kg)(164.73 kJ/kg) — (5 kg)(84.44 kJ/kg) + (4.75 kg)(84.98 kJ/kg) = 22.64 kJ : AE system Net energy tIaquel’ Change in internal, kinetic, by heat, work. and mass potential, etcrenergies Qin ‘mehe = mzuz ’miul Qin = mzuz —m1u1+mehe Combining the two balances: Q'm = mzuz ‘m1“1+(m1“m2)he The initial state properties of R-l34a in the tank are = 0 000826lm3/k - T = 24°C “I - g 1 } u1 = 84.44 kJ/kg (Table A-1 1) x = 0 he = 84.98 ch/kg ated liquid state, and found the exiting that the refrigerant leaving the tank is at satur Note that we assumed he volume of the tank is enthalpy accordingly. T v = mlul = (5 kg)(0.000826lm3/kg) = 0.004131m3 The final specific volume in the container is 3 V v 0.004l3lm =0'01652m3fkg _/ U2 : —— ._ m2 0.25-kg _ The final state is now fixed. The properties at this state are (Table A-1 1) T2 =24°C x _ V2 "Vf _ 0.01652—0.000826l _0 5061 } 2 0.031834—0.0008261 I _ 3 ufg “2 ’0'01652‘“ [kg u2 =uf +x2ufg =84.44kJ/kg+(0.506l)(158.65 kJ/kg) = 164.73 kJ/kg Substituting into the energy balance equation, Qin = "12142 ‘m1“1+(m1"m2)hé g = (0.25 kg)(l 64.7 3 kJ/kg) — (5 kg) = 22.64 kJ (84.44 kJ/kg) + (4.75 kg)(84.98 kJ/kg) 5-124 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed to enter the tank until mechanical equilibrium is established. The mass of air that entered and the amount of heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the tank (will be verified). Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-l). The properties of air are (Table A-l7) 1",. = 295 K ——5 h,- = 295.17 kJ/kg T1 = 295 K ——> u1 = 210.49 kJ/kg T2 =350 K ———> u2 =. 250.02 kJ/kg Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and intemal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min _mout = Am-system mi = m2._ m1 Energy balance: Ein _ Eout = AEsystem BTW—J . Net energy “31153? Change in mtemal, kmetic, by hem, work, arid mass potential, etc. energies Qin + mihi = m2u2 ~ mlu1 (since W E ke E pe E 0) The initial and the final masses in the tank are ' 3 m1 = fl! = ._.________.__—(100 “’3” m ) = 2.362 kg RT, (0.287 kPa-m /kg-K)(295 K) 3 m2 =K=M=1L946kg RT2 (0.287 kPa-m3/kg - K)(350 K) Then from the mass balance, m, : m2 — m1 = 11.946 —2.362 = 9.584 kg ([3) The heat transfer during this process is determined from Qin = “mth + "72112 “ mlul = —(9.584 kg)(295.17 kJ/kg)+ (1 1.946 kg)(250.02 kJ/kg)—(2.362 kg)(210.49 kJ/kg) = —339 kJ —> Qout = 339 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction. ‘ ...
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This homework help was uploaded on 04/07/2008 for the course ME 104 taught by Professor Gomatam during the Spring '08 term at Lehigh University .

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hw1617 - 5—68 Refrigerant-134a is throttled by a...

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