7-88 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinder is turned on, and air is heated for 15 min at constant pressure. The entropy change of air during this process is to be determined for the cases of constant and variable specific heats. Assumptions At specified conditions, air can be treated as an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis The mass of the air and the electrical work done during this process are ( ) ( ) ( ) ( ) ( )( ) kJ 180 s 60 15 kJ/s 0.2 kg 0.4325 K 290 K /kg m kPa 0.287 m 0.3 kPa 120 in e, in e, 3 3 1 1 1 = ¥ = D = = ⋅ ⋅ = = t W W RT P m & V The energy balance for this stationary closed system can be expressed as ) ( ) ( 1 2 1 2 in e, out b, in e, energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in T T c h h m W U W W E E E p - @ - = æÆ æ D = - D = - 43 42 1 43 42 1 since DU + W b = DH during a constant pressure quasi-equilibrium process. (a) Using a constant c p value at the anticipated average temperature of 450 K, the final temperature becomes Thus, ( )( ) K 698 K kJ/kg 1.02 kg 0.4325 kJ 180 K 290 in e, 1 2 = ⋅ + = + = p mc W T T Then the entropy change becomes ( ) ( )( ) kJ/K 0.387 K 290 K 698 ln K kJ/kg 1.020 kg 0.4325 ln ln ln 1 2 avg , 0 1 2 1 2 avg , 1 2 sys = ˜ ˜ ¯ ˆ Á Á Ë Ê ⋅ = = ˜ ˜ ¯ ˆ Á Á Ë Ê - = - = D T T mc P P R T T c m s s m S p p à (b) Assuming variable specific heats, ( ) kJ/kg 706.34 kg 0.4325 kJ 180 kJ/kg 290.16 in e, 1 2 1 2 in e, = + = + = æÆ æ - = m W h h h h m W From the air table (Table A-17, we read s 2 o = 2.5628 kJ/kg·K corresponding to this h 2 value. Then, ( ) ( )( ) kJ/K 0.387 K kJ/kg 1.66802 2.5628 kg 0.4325 ln 1 2 0 1 2 1 2 sys = ⋅ - = - = ˜ ˜ ¯ ˆ Á Á Ë Ê + - = D o o o o s s m P P R s s m S à 7-92 One side of a partitioned insulated rigid tank contains an ideal gas at a specified temperature and pressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. The total entropy change during this process is to be determined. Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relations apply. Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as AIR 0.3 m 3 120 kPa 17°C W e
Transcript
7-88 An insulated cylinder initially contains air at a specified state. A resistance heaterinside the cylinder is turned on, and air is heated for 15 min at constant pressure. Theentropy change of air during this process is to be determined for the cases of constant andvariable specific heats.Assumptions At specified conditions, air can be treated as an ideal gas.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Analysis The mass of the air and the electrical work done during this process are
( )( )( )( )
( )( ) kJ 180s 6015kJ/s 0.2
kg 0.4325K 290K/kgmkPa 0.287
m 0.3kPa 120
ine,ine,
3
3
1
11
=¥=D=
=⋅⋅
==
tWW
RT
Pm
&
V
The energy balance for this stationary closed system can be expressed as
)()( 1212ine,outb,ine,
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
TTchhmWUWW
EEE
p -@-=æÆæD=-
D=-4342143421
since DU + Wb = DH during a constant pressure quasi-equilibrium process.(a) Using a constant cp value at the anticipated average temperature of 450 K, the finaltemperature becomes
Thus,( )( )
K 698KkJ/kg 1.02kg 0.4325
kJ 180K 290ine,
12 =⋅
+=+=pmc
WTT
Then the entropy change becomes
( )
( )( ) kJ/K0.387 K 290
K 698ln KkJ/kg 1.020kg 0.4325
lnlnln1
2avg,
0
1
2
1
2avg,12sys
=˜¯
ˆÁÁË
Ê⋅=
=˜˜¯
ˆÁÁË
Ê-=-=D
T
Tmc
P
PR
T
TcmssmS pp
Ã
(b) Assuming variable specific heats,
( ) kJ/kg 706.34kg 0.4325
kJ 180kJ/kg 290.16ine,
1212ine, =+=+=æÆæ-=m
WhhhhmW
From the air table (Table A-17, we read s2o = 2.5628 kJ/kg·K corresponding to this h2
7-92 One side of a partitioned insulated rigid tank contains an ideal gas at a specifiedtemperature and pressure while the other side is evacuated. The partition is removed, andthe gas fills the entire tank. The total entropy change during this process is to bedetermined.Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relationsapply.Analysis Taking the entire rigid tank as the system, the energy balance can be expressedas
AIR0.3 m3
120 kPa17°CWe
)(0
12
12
12
energies etc. potential, kinetic, internal,in Change
system
mass and work,heat,by nsferenergy traNet
outin
TT
uu
uumU
EEE
=
=
-=D=
D=-4342143421
since u = u(T) for an ideal gas. Then the entropy change of the gas becomes
( )( ) ( )
kJ/K28.81
2ln KkJ/kmol 8.314kmol 5
lnlnln1
2
1
20
1
2avg,
=
⋅=
=˜˜¯
ˆÁÁË
Ê+=D
VV
VV
v uu NRRT
TcNS
Ã
This also represents the total entropy change since the tank does not contain anythingelse, and there are no interactions with the surroundings.
7-95 An insulated rigid tank contains argon gas at a specified pressure andtemperature. A valve is opened, and argon escapes until the pressure drops to aspecified value. The final mass in the tank is to be determined.Assumptions 1 At specified conditions, argon can be treated as an ideal gas. 2 Theprocess is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropicrelations of ideal gases apply.Properties The specific heat ratio of argon is k = 1.667 (Table A-2).Analysis From the ideal gas isentropic relations,
( )
( ) K 0.219kPa 450
kPa 200K 303
1.6670.6671
1
212 =˜
¯
ˆÁÁË
Ê=˜
¯
ˆÁÁË
Ê=
- kk
P
PTT
The final mass in the tank is determined from the ideal gas relation,
( )( )( )( )
( ) kg 2.46===æÆæ= kg 4K 219kPa 450
K 303kPa 2001
21
122
22
11
2
1 mTP
TPm
RTm
RTm
P
P
VV
7-99 Nitrogen is compressed in an adiabatic compressor. The minimum work input is tobe determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Theprocess is adiabatic, and thus there is no heat transfer. 3 Nitrogen is an ideal gas withconstant specific heats.Properties The properties of nitrogen at an anticipated average temperature of 400 K arecp = 1.044 kJ/kg·K and k = 1.397 (Table A-2b).Analysis There is only one inlet and one exit, and thus mmm &&& == 21 . We take thecompressor 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
outin
energies etc. potential, kinetic, internal,in change of Rate
(steady) 0system
mass and work,heat,by nsferenergy tranet of Rate
outin 0
EE
EEE
&&
444 344 21&
43421&&
=
=D=- ä
IDEAL GAS5 kmol40°C
ARGON4 kg
450 kPa30°C
Nitrogencompress
or
600kPa
120kPa
30°C
)( 12in
2in1
hhmW
hmWhm
-=
=+
&&
&&&
For the minimum work input to the compressor, theprocess must be reversible as well as adiabatic (i.e.,isentropic). This being the case, the exittemperature will be
K 479kPa 120
kPa 600K) (303
397.1/397.0/)1(
1
212 =˜
¯
ˆÁË
Ê=˜
¯
ˆÁÁË
Ê=
- kk
P
PTT
Substituting into the energy balance equation gives kJ/kg 184=-⋅=-=-= )K303K)(479kJ/kg 044.1()( 1212in TTchhw p
7-108C The work associated with steady-flow devices is proportional to the specificvolume of the gas. Cooling a gas during compression will reduce its specific volume, andthus the power consumed by the compressor.
7-109C Cooling the steam as it expands in a turbine will reduce its specific volume, andthus the work output of the turbine. Therefore, this is not a good proposal.
7-118 Liquid water is pumped by a 70-kW pump to a specified pressure at a specifiedlevel. The highest possible mass flow rate of water is to be determined.Assumptions 1 Liquid water is an incompressible substance.2 Kinetic energy changes are negligible, but potential energychanges may be significant. 3 The process is assumed to bereversible since we will determine the limiting case.Properties The specific volume of liquid water is given to bev1 = 0.001 m3/kg.Analysis The highest mass flow rate will be realized whenthe entire process is reversible. Thus it is determined fromthe reversible steady-flow work relation for a liquid,
Thus,( ) ( ){ }
Ô
Ô˝¸
ÔÓ
ÔÌÏ
˜˜¯
ˆÁÁË
Ê+˜
˜¯
ˆÁÁË
Ê
⋅-=
-+-=˜¯
ˆÁË
ÊD+D+= Ú
222
33
1212
2
1
0in
/sm 1000
kJ/kg 1)m 10)(m/s 9.8(
mkPa 1
kJ 1kPa)1205000)(/kgm 0.001(kJ/s 7 m
zzgPPmpekedPmW
&
&&& v v Ã
It yieldskg/s 1.41=m&
7-128 Steam is expanded in an adiabatic turbine with an isentropic efficiency of 0.92.The power output of the turbine is to be determined.
T
s
2
1
600kPa
120kPa
PUMP
P2 = 5MPa
WaterP1 = 120
kPa
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kineticand potential energy changes are negligible. 3 The device is adiabatic and thus heattransfer is negligible.Analysis There is only one inlet and one exit, and thus mmm &&& == 21 . We take the actualturbine as the system, which is a control volume since mass crosses the boundary. Theenergy balance for this steady-flow system can be expressed in the rate form as
outin
energies etc. potential, kinetic, internal,in change of Rate
(steady) 0system
mass and work,heat,by nsferenergy tranet of Rate
outin 0
EE
EEE
&&
444 3444 21&
43421&&
=
=D=- Ã
)(
0)ÄpeÄke (since
21out,
2out,1
hhmW
QhmWhm
a
a
-=
@@@+=
&&
&&&&
From the steam tables (Tables A-4 through A-6),
kJ/kg 2335.7)3.2335)(8763.0(289.27
8763.08234.6
9441.09235.6
kPa 30
KkJ/kg 6.9235
kJ/kg 3231.7
C004
MPa 3
22
22
12
2
1
1
1
1
=+=+=
=-
=-
=
˛˝¸
=
=
⋅=
=
˛˝¸
°=
=
fgsfs
fg
fss
s
s
hxhhs
ssx
ss
P
s
h
T
P
The actual power output may be determined by multiplying the isentropic power outputwith the isentropic efficiency. Then,
