This preview shows pages 1–7. Sign up to view the full content.
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 DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 5—68 Refrigerant134a is throttled by a capillary tube. The quality of the refrigerant at the exit is to be
determined. Assumptions 1 This is a steadyﬂow 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 steadyﬂow 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 R134a is, from the refrigerant tables (Table A1 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 123493592
x = =——————
h2 = h1 2 h fg 207.38 579E 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 steadyﬂow 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 ASE through A6E),
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 R134a . 12°C 579E 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 steadyﬂow 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 ASE through A6E),
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 steadyﬂow 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 583E [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 steadyow 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 AlE). 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 A4E through A6E) 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 steadyﬂow 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 590 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 A4) 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 A4) 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 521C 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. 5128 R134a from a tank is discharged to an airconditioning line in an isothermal process. The ﬁnal
quality of the R134a 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 uniformﬂow system can be expressed as Mass balance: min "mout = Amsystem
_me = m2 —m1 me =ml _m2 Energy balance:
Em _ Eom = AESystem Liquid Rl34a
%_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 A1 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 0.25kg The ﬁnal state is now ﬁxed. The properties at this state are (Table A1 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 Rl34a in the tank are = 0 000826lm3/k 
T = 24°C “I  g
1 } u1 = 84.44 kJ/kg (Table A1 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 0.25kg _ The ﬁnal state is now ﬁxed. The properties at this state are (Table A1 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) 5124 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 uniformﬂow 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 Al). The properties of air are (Table Al7) 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 uniformﬂow system can be expressed
as Mass balance: min _mout = Amsystem 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 kPam /kgK)(295 K)
3
m2 =K=M=1L946kg RT2 (0.287 kPam3/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. ‘ ...
View
Full
Document
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 .
 Spring '08
 Gomatam

Click to edit the document details