This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PROBLE M 4] 4.1 An 8ft3 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 constantdiameter 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. (Cantuneed). 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 AIO, 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, oneexit, 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
AlO, 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
fwde. :L..+o.. a re pro w clad  FIVND.‘ De'l'trrw‘nc 4%.. mt: {Ltw “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 11h. tcklm‘h'r. ’03 “ ducked. line.
2 The Cah‘l‘ral Volume I} 4"
.S'ieulf thde.
LADAL‘I’SIS:
w.'+k U; {vanTable Aé, U" ': 0.5796 mus/‘3
. _ A. UT: (oZM‘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:. (AV53
' ‘U'z. J "i
I gm“ 0: = Ly:s (5...... Tvdquﬂm) and 1‘“ Volumdnfc 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 _ 31215 my; +—— ...
View
Full Document
 Spring '06
 staff
 Thermodynamics, ideal gas model, appropriate conversion factor

Click to edit the document details