HW11soln

HW11soln - PROBLE M 4-] 4.1 An 8-ft3 tank contains air at...

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Unformatted text preview: PROBLE M 4-] 4.1 An 8-ft3 tank contains air at an initial temperature of 80°F and initial pressure of 100 lbf/in.2 The tank develops a small hole, and air leaks from the tank at a constant rate of 0.03 lb/s for 90 3 until the pressure of the air remaining in the tank is 30 lbinn.2 Employing the ideal gas model, determine the final temperature, in °F, of the air remaining in the tank. KNOWN: Air at specified initial temperature and pressure leaks from rigid tank until a final specified pressure is attained by the air remaining in the tank. FIND: Final temperature of air remaining in tank. SCHEMATIC AND GIVEN DATA: Initial State — State 1 Final State — State 2 Process 1 —» 2 12::— Ar = 90 s 4—-> m, = 0.03 lbfs V,=8fi3 V,=V,=8fi3 T = 80°F 2 30 lbf/in.3 p: = 100 lbfi’in? p2 ENGINEERING MODEL: 1. The centrol volume is defined by the dashed line on the accompanying diagram. 2. Air can be modeled as an ideal gas. ANALYSIS: The ideal gas model can be applied to the final state, state 2, to determine the temperature of the air remaining in the tank. 1921/2 3 mzRT 2 Solving for temperature yields T2 : P2 V2 m2 R Pressure and volume are known at state 2. The mass in the tank at state 2, m2, equals the initial mass in the tank, m1, less the mass that leaks from the tank. Since the mass flow rate, me , is constant, the amount of mass that leaks from the tank is m, Ar = (0.03 lb/s)(90 s) z 2.7 lb The initial mass, m I, is obtained using the ideal gas equation of state PROBLEM 4.1 (Can‘b'nuecl) The gas constant, R, is the universal gas constant divided by the molecular weight of air. Temperature must be expressed on an absolute scale, T1 = 80 0F = 540°R. Substituting values and applying the appropriate conversion factor yield [100%](3 ft3) - 2 m]: f "Rf 144 2 4.0 lb 15451—5 {'1 0R lfi m0 ' _ o ——w——lb (540 R) 28.97 lbmol Collecting results m2 = 4.01b—~2.7 1b = 1.3 lb Substituting mg to solve for T2 yields _ [309:](3113) - 2 T2_ “1' ‘ 144}? =49s.5°R=33.s°F 1545 ——fi'lbt 1“ (1.3113 —1bm‘iL‘°R 28.97 lbmol Note the need to convert the final temperature from 0R to OF to provide the answer in the requested units. PROBLEM 4.11 4.12 Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at —4°C and quality of 20% at a velocity of 7 mfs. At the exit, the refrigerant is a saturated vapor at a temperature of ~—4°C. The evaporator flow channel has constant diameter. If the mass flow rate of the entering refrigerant is 0.1 kg/s, determine (a) the diameter of the evaporator flow channel, in cm. (In) the velocity at the exit, in mfs. KNOWN: Refrigerant 134a flows through a constant-diameter evaporator entering as a saturated mixture at given temperature, quality, and velocity and exiting as a saturated vapor at a given temperature. FIND: Determine the diameter of the flow channel and the velocity at the exit. SCHEMATIC AND GIVEN DATA: T1 = 4°C x1 = 1.0 (saturated vapor) ENGINEERING MODEL: 1. The control volume shown on the accompanying figure is at steady state. ANALYSIS: (a) The diameter of the flow channel can be determined from the mass flow rate at the inlet, state 1 l m : A1V1=(7r/4)d12Vl:> d1: My, 5 1 v1 v1 7er Apply the quality relation to determine the specific volume at state 1. From Table A- l 0, vfl = 0.0007644 [if/kg, vgl = 0.0794 m3/kg. Substituting to determine specific volume V1 : Vn + x1(Vg1— Vn) v1 = 0.0007644 m3/kg + (0.20)(0.0794 m3/kg 4 0.0007644 m3/kg) = 0.01649 m3/kg Peon L. EM 4H7. (Cantu-need). Substituting, applying the appropriate conversion factor, and solving for the diameter 3. % z{0.15%} 0.016493] S kg 104 cm2 is) 2 m (b) The exit flow velocity can be determined from the mass flow rate being equal at inlet and exit: d1: = 1.732 cm A "-1 = 1V1 = szz V1 V2 Since the diameter is constant throughout the channel, the inlet and exit areas are the same. Since the refrigerant is a saturated vapor at the exit, from Table A-IO, v; = vgz = 0.0794 m3/kg. Solving for the exit velocity 3 0.0794? in = “[v4] = [73] —g, = 33.7 m/s v1 S 0.016491“— kg As an alternative solution, the exit flow velocity can be determined from the mass flow rate at the exit, state 2 m I sz2 2 (yr/4)d§V2 2) V2: 4m,v, 2 V2 V2 739122 The mass flow rate is the same at the inlet and the exit based on the mass rate balance for one- inlct, one-exit, steady flow. The diameter is the same at the inlet and exit since the diameter is constant through the evaporator. Since the refrigerant is a saturated vapor at the exit, from Table A-lO, v2 = vgz = 0.0794 m3/kg. Substituting values and applying the appropriate conversion factor 3 {0135100794 3] 3 kg 7r(1.732 cm)2 104 cm2 m2 = 33.7 mfs V2: PROBLEM ‘H4 {(#30le W‘kr ‘mpur- {llama 'Hu‘ousln o. Pupins confl'gurafi‘on MPH“; om mid: and: +uru WH’S- Shady $511 “P‘r‘fi'fi “*4 fwd-e. :L..+o.. a re pro w clad - FIVND.‘ De'l'trrw‘nc 4%.. mt: {Lt-w “it 413*“ “1‘31" “"d 3““ "r ‘ +k._m.‘f;, an a; Isa/.3. _ _ 1' - ' fififl'BMfiT'C - 4W?“ Dfi'ml EMQ‘UEEKIMG MODEL: |. The canhul Volume is SMWH I'n 11-h.- tc-klm‘h'r. ’03 “ ducked. line. 2- The Cah‘l‘ral Volume I} 4"- .S'i-eulf thde. LADAL‘I’SIS: w.'+k U; {van-Table A-é, U" ': 0.5796 mus/‘3 . _ A. UT: (o-ZM‘XBO His) =10.5SL£§_ 4.__. w“- ___. ——————‘ 5 ’0; (9.51% m‘lna) M SW7 5+:ch Ml=-v~'«a.+w'~3 J where. - __(mr-)1. “3.3:.- (AV-53 ' ‘U'z. J "i I gm“ 0-: = Ly:s (5...... T-vdquflm) and 1‘“ Volumd-nfc WWW “"f 910.1(Ieg CMMI V'Dlum which), we I'LAH $13413 - "Than: I l l M = m + W3 , 7. 1"“ I 2‘ t:lM2, 1 513mg: rm _ 3121-5 my; +-—-—- ...
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This note was uploaded on 12/19/2011 for the course ME 302 taught by Professor Staff during the Spring '06 term at Cal Poly.

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HW11soln - PROBLE M 4-] 4.1 An 8-ft3 tank contains air at...

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