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Transcript of Thermodynamic Properties are Measurements · Thermodynamic Properties are Measurements p,T,v, u...
Thermodynamic Properties are Measurementsp,T,v, u ,h,s - measure directly
-measure by change
Tables
Curve fits Tables
Correlation's, Boyles Law Tables pv=c@T=c limited hand
calculations
Equations of State, pv=RT TablesCalculation Modules
NIST, EES, HYSYMinteractive, callable
PropertyData
vT Tp
vs
∂∂
=
∂∂
sat
sat
sat
sat
Given T and PSuper Heat Region if, p<p @T T>T @p
CompressedLiquid Region if, p>p @ T T<T @ p
Three TablesTemperature Table
at spaced T’sPressure Table
at spaced P’sSuperheat Table
at spaced T and P6 PropertiesTemperaturePressureVolumeInternal EnergyEnthalpyEntropy
Saturation liquid internal energy at 0 C 0. Table BaseSaturated liquid enthalpy at 25 C 104.89 kJ/kgSaturated vapor entropy at 25 C 8.558 kJ/kg KEnthalpy at 20 C, 300 kPa
assume saturated liquid enthalpy at 20 C 83.96 kJ/kg Temperature of saturated vapor with an
internal energy of 2396.1 kJ/kg 15 C Enthalpy of vaporization at 10 C 2477.7 kJ/kg
Table A-4 Metric, TTable A-5 Metric, pTable A-4E English, TTable A-5E English, pTEMPERATURE TABLEPRESSURE TABLE
Table A-6, Metric, T,pTable A-6E, English, T,pSUPERHEAT TABLE
Table A-7 Metric T,pTable A-7E English T,p
kPa 8587.9C 300 @p C 195.04kpa 1400 @T
C 300 and kpa 1400at water
saturation
saturation=
=kpa 362.23C 6 @p
C 15.71kpa 500 @T C 6 and kPa 500at 134aR
saturation
saturation=
=−
C 195.04kpa 1400kPa 8587.9
C 300
superheated
Tp=constant
T
C 6kpa 362.23
kPa 500C 15.71
subcooled
p=constant
v v
Two Phase Real Gas Properties
( )
fg
f
fgf
gf
g
ggff
gf
vvvx
vxvv
xvvx1v
Qualitymm
x
vmvmmv
VVV
−=
×+=
+−=
=
+=
+=
fv gv
fgf
fgf
fgf
vxvv
uxuu
hxhh
×+=
×+=
×+=
/lb.ft 300 vF,90at water of properties theFind 3o =
F90T
o
lb/ft7.467v 3g =
f
f
f
u =58.07h =58.07s =.11165
g
g
g
u =1040.2
h =1107.7
s =2.0083shu
( )vvf −
( )fg vv −
/lbft 300.v 3=lb/ft0161.v 3
f =
f
g f
f f g
f fg
f fg
f fg
o
v vQuality, xv v
300 .0161x .64 (64%)467.7 .0161
u u x (u u )
u u x u
u 58.07 .64 982.2u 686.7 BTU/lb
h h x h
h 58.07 .64 1042.7h 725.4 BTU/lb
s s x s
s .11165 .64 1.8966s 1.3254 BTU/lb R
−=
−
−= =
−
= + × −
= + ×
= + ×=
= + ×
= + ×=
= + ×
= + ×
=
TemperatureTable
Engineering Equation Solver - EESFluid Property Information - 69 fluids availableThermophysical Functions - 25 properties calculatedEquations Window
h=enthalpy(steam, T=200.,P=200) superheated vaporh=enthalpy(steam,T=200.,X=1) saturated vaporu=intenergy(steam,T=200.,X=0.) saturated liquidp=pressure(steam,T=200.,X=0.) saturation pressure
Thermophysical Functionsentropyintenergypressurequalitydensityenthalpyisidealgastemperaturevolume
Function ArgumentsH specific enthalpyP pressureS specific entropyT temperatureU specific internal energyV specific volumeX quality
Table EES Program
2414.29u
1.)X4.,Psteam,intenergy(u 121.3kJ/kgu
0)X4.,Psteam,intenergy(u
g
g
l
l
=
====
===
g2415.2kJ/ku
g121.45kJ/kul
=
=
g
Pressure TableSaturation internalenergy at 4 kPa
Superheat Table
/kg.001004mv
500.)p30.,Tam,volume(stevg125.67kJ/kh
500.)p30.,Tteam,enthalpy(sh
3l
l
l
l
=
====
===
/kg.001004mv
Cliquid@30 saturated vv
kJ/kg 125.75hC30 @ liquid saturatedh h
3l
ol
l
ol
=
=
==Enthalpy and
volume of water at 150 kPa and 30 C
Temperature TableInternal energy of water at 20 MPa uand 300 C
kJ/kg 1333.446u20000.)p300.,Tsteam,intenergy(u
l
l
====
kJ/kg 1332.0C300 @ liquid saturatedu u
l
ol
==
f fg
f fg
10 MPa saturated steam has an enthalpy of 2010 kJ/kg. What is its internal energy?
h h x h
2010 kJ/kg 1407.56 x 1317.1x .4574u u x u
u 1393.04 .4574 1151.4u 1919.69 kJ/kg
EES Programx quality(stea
= +
= +== +
= + ×=
= m,P 10000.,h 2010)x .4576u intenergy(steam,P 10000.,h 2010)u 1919.65 kJ/kg
= === = ==
Steam at 20 C has an enthalpy of 1800 kJ/kg. What is the internal energy?
kJ/kg 1707.25u2319.0.783.95u
ux uu.7x
2454.1x 83.96kJ/kg 1800
hx hh
fgf
fgf
=×+=
+==
+=
+=
SUPERHEAT TABLE
Enthalpy at 700 C and .10 Mpa 3928.2 kJ/kgTemperature at entropy of 8.8642 and .05 Mpa 400 CEnthalpy at .05 MPa and entropy of 10.6662 kJ/kg C 5147.7 kJ/kg
Steam initially at a temperature of 1100 C and a pressure of .10 MPa undergoes a process during which its entropy remains constant to a pressure of .01 MPa. What is the enthalpy and temperature of the steam at the end of the process?
Entropy at 1100 C, .1 MPa 10.1659kJ/kg KEnthalpy at .01 MPa, entropy 10.1659 3705.4 kJ/kgTemperature at .01 MPa, entropy 10.1659 600 C
T.01 MPa
s
Linear Interpolation with 2 Variables
p 25 mPa p 30 MPaT 450 h 2949.7 h 2821.4
27 25For the pressure table entry, 30 25
= =
= = == = = =
−=
−
h@(T 450 C,p 27 MPa) Steam Superheat TableTable Values
p 27 MPa h 2898.4
.4
The desired pressure, 27 kPa, is 40 % of the differencebetween table values. All the other properties at 27 kPa must be at the same difference.
EES h enthalpy(steam,T 450.,p 27000)h 2901.7 kJ/kg= = == .22% difference, table and interpolation
SPREADSHEET WORLD – THERMAL FLUIDS PROPERTIES
123456789
A B C D E F G H I J
T 200 P 15
CALL TTProps("AIR","EE_F", "P", $C$2,"T", $C$1)
P = v = T = u = s = h = X = STATE = ERROR =RETURN 15 16.29669 200 180.5582 1.085135 225.7937 1 erheated va 0
psia ft^3/lbm deg F BTU/lbm BTU/lbm-R BTU/lbm nd
A saturated mixture of 2 kg water and 3 kg vapor in contained ina piston cylinder device at 100 kpa. Heat is added and the piston,initially resting on stops, begins to move at a pressure of 200 kpa.Heating is stopped when the total volume in increased by 20%. Find:
a) the initial and final temperatures.b) the mass of liquid water when the pressure reaches 200 kPa and the
piston starts to move. c) the work done by the expansion.
kJ/kg 1670.622088.2.6417.40u
uxuu /kgm 1.0168v
.001043)(1.694.6.001043v
vxvv99.61T
.6 totalkg 5 vaporkg 3 xkPa, 100at
1
fgf1
31
1
fgf1
=×+=
×+==
−×+=
×+==
==
100 kpa3 kg2 kgQ
3.88
v
1
23
( )
( )
( )
( )
( ) kJ 203.2m 5.08m 6.096kPa 2000W
VVp0pdVpdVW
kJ/kg 65.2988hMPa .2P1.2192,v @h h
MPa .2at 3Point
dsuperheate.8857v
1.2192kg 5
m 6.096v
6.096V1.2VkJ/kg 2816.47h
)MPa .2P1.0168,vh@hkJ/kg 2613.23u
MPa .2P1.0161,vu@u6A Table fromion Interpolat
m 5.081.0161kg 5V
v vMPa, 2 .at 2Point
33
232
3
2
2
1
3
33
3
3
23
2
2
2
2
32
21
g
=−×+=
−+=+=
====
⇒=
==
=×==
====
===−
=×=
=
∫∫
( )
( )233-13-1
3-2
23
uumQW,lyalterative
kJ 9.860Q)47.2816(2988.655Q
hhmQ∆HQ
Vp∆EQW∆EQ
3-2 constant,p
−×−=
=−×=
−==
∆+=+=
=
( )
kJ 05.4713Q) 1670.62-(2613.235Q
uumQ∆U∆EQ
0WW∆EQ
2-1 constant,v
12
=×=
−===
=+=
=
p
3.88
kJ/kg 2988.65h
173.1.1989-1.316231.1989-1.2192 vof ratio
3092.1 1.31623 300 1.2192
2971.2 1.1989 250h v T
1.2192) vMPa, .2p ( @enthalpy 6-A TableSuperheat
hfor ion Interpolat
3
3
=
==
==
p
1
23
water 2vapor 3Properties at 1 given p1 and x1x1 0.6p1 100,000
P = v = T = u = s = h = X = STATE =100000 1.016819 99.62524 1670.481 4.936567 1772.163 0.6 Mixed regioPascals m^3/kg deg C kJ/kg kJ/kg-K kJ/kg nd
Properties at 2 given p2 and v2=v1p2 200,000
P = v = T = u = s = h = X = STATE =200000 1.016819 173.5216 2613.078 7.389259 2816.442 1 erheated vaPascals m^3/kg deg C kJ/kg kJ/kg-K kJ/kg nd
V2=m x v2 5.084094
V3= 1.2 * V2 6.100913v3=V3/m 1.220183p3 200,000
Properties at 3 given p3 and v3P = v = T = u = s = h = X = STATE =
200000 1.220183 259.0808 2744.746 7.742318 2988.782 1 erheated vaPascals m^3/kg deg C kJ/kg kJ/kg-K kJ/kg nd
Q 2-3=m*(h3-h2) 861.7029W 2-3 = Q 2-3-m*(u3-u2) 203.363Q 1-2=m*(u2-u1) 4712.986
v
Spread Sheet World module solution
?
3 kg vapor and 1 kg liquid R-134a is contained in a rigid tank at 20 C.What is the volume of the tank? If the tank is heated until the pressure reaches .6 MPa? What is the quality, and enthalpy of the mixture of liquid and vapor?
kJ/kg 211.27179.71.78779.84h
hxhh
).787(78.7%.0008196.0341
.00008196.027x
12-A Table , 843 page MPa .6 @ vxMPa .6 @ vvvconstantvconstantm constant,V
heating,After
/kgm .027kg 4
m .1082mVv
m .1082V
.0358kg 3.0008157kg 1V
11TableA 842, page C 20 @ vmC 20 @vmV
VVV
2
fgf2
2
fgf12
33
11
31
1
gg ff1
gf1
=×+=
×+=
=−−
=
×+===⇒==
===
=
×+×=
−+×=
+=
T
v
.6 MPa
20 C.5716 MPa
1
2
Q
3 kg of vapor and 2 kg of liquid R-134a is contained in a piston cylinder device.The volume of the vapor is .1074 cubic meters. What is the temperature andpressure? If the cylinder and its contents are heated until volume is .15 cubicmeters what is the quality?
fgf
fg
f
fg
f22
g ff3
2
322
32
3g
33
g1
g1g1
property)(xproperty)(property
(property)property)(propertyx
(83.4%) .834.0008157.0358
.0008157.03v
vvx
C 20 @vxC 20 @ v/kgm .03v
/kgm .035
.15mVv
,.15mVat constant. is pressure theprocess heating theDuring
11TableA 842, page MPa .5716 C, 20 @ /kgm .0358v
/kgm .0358kg 3
m .1074mV
v
×+=
−=
=−−
=−
=
×+==
===
=
−=
===
20 C
T
v
1 2
Q
1. Problem StatementCarbon dioxide is contained in a cylinder
with a piston. The carbon dioxide is compressedwith heat removal from T1,p1 to T2,p2. The gasis then heated from T2, p2 to T3, p3 at constant volume and then expanded without heat transfer to the original state point.
4. Property Diagramstate points - processes - cycle
T1,p1
T3,p3
T2,p2
p
Thermodynamic Problem Solving Technique
QW
QW
2. Schematicv
5. Property Determination
T2,p2
1 2 3 T pvuhs
3. Select Thermodynamic Systemopen - closed - control volume
a closed thermodynamic systemcomposed to the mass of carbondioxide in the cylinder
p
v
2CO
6. Laws of ThermodynamicsQ=? W=? E=? material flows=?
Ideal Gas Law
o o
*
IDEAL (PERFECT) GAS LAW
pv RT pV mRT p - absolute pressure, psia,kPa T - absolute temperature, R, K
RRmolecular weight
==
=
o*
3*
o o
*
*
ft lbf R R 1545.15 lbmole
kJ kPa mR 8.314 orkmole K kmole K
mass moles Molecular Weight
m n Molecular Weight
pv R T pV nR T
=
=
= ×= ×
=
=
2
1
2
1
2
1
2
1
2211
23
TT
pp
TT
vv
LAW CHARLES
vpvp LAW BOLYES
)0 and atm (1 STP at gas of /molemolecules 106.023
liters. 22.4 gasany of mole (1) One
LAW SAVOGADRO'
=
=
×=×
×
=
Co
2
3 o
gage atmosphere
o 3 o
O
What is the mass of 1.2 m of oxygen at 24 Cand a gage pressure of 500 kPa.Atmospheric pressure is 97 kPa
p p p 500 kPa 97 kPa 597 kPa
8.314 kJ/kmole K or kPa m /kmole KR .2598132
= + = + =
= =
( )
3
3
3 o o
o
3 kPa m /kg alsoTableA 1
pV 597 kPa 1.2 mm 9.28 kgRT .259813 kPa m /kg 24 C 273.16 K
What is the volume mass of 1.2 lbm of air at 124 Fand a gage pressure of 500 psia.Atmospheric pressure is 14.
−
×= = =
× +
gage atmosphere
air o
o o
7 psia.
p p p 500 psia 14.7 psia 514 psia
1545.15 ft lbf / lbmole R ft lbfR 53.336 also Table A 1E in BTU and psi units28.97 lbm R
1.2 lbm 53.336 ft lbf/ lbm R 124 F 459.6m R TVp
= + = + =
= = −
× × += =
( )o
2 2
3
9 R514 psia 144 ft /in
V .5047ft×
=
IDEAL GAS EQUATION FORMS - For Air
P v = m R T3
3 OO
3 33 O
O o
32
kPa mkPa m kmole 8.314 K kmole K
kPa m kPa m kPa m kg .287 K R 8.314 / 28.96kg K kmole K
lb ft ft
=
= =
O
O
3 O2 O O
3
ft lbf lbmole 1545.15 R lbmole R
lb ft lbf ft lbf ft lbm 53.35 R R 1545.15 /28.96ft lbm R lbmole R
psi ft lbm
=
= =
= OO O
3 O2 O O
psi lbf ft lbf .3704 R R 1545.15 /144/28.96lbm R lbmole R
lb BTU ft lbf ft lbm .06855 R R 1545.15 /28.96/ft lbm R lbmole R
=
= = 778
( )
( )
/kgm .8417vkPa 101.325
F24K273.15/kgm kPa .287p
RTv
96.28/Kmole kg
m kPa8.314R
kPa 101.325 and C24at air of volumespecific The
/lbft 13.476v/ftin 144lbf/in 14.7
F75R459.69R lbf/lbmft 53.35p
RTv
R lbf/lbmft 35.53 /28.961545.15Rpsia 14.7 and F75at air of volumespecific The
3
2
oo3
O
3
o
3
222
ooair
o
=
+×==
=
=
×+×
==
==
( )
kPa39.117pm 23m 12 kPa 225p
VVp
V T RV p T Rp
VpT R
VpT Rm
m 23kPa 224T273.15.286
VpRTm
constantT constant,mass
pressure? final theisWhat .m 23 of volumea toemperatureconstant t aat expandskPa 225 of pressure a and m 12 of volumeaat initially Air
2
3
3
2
2
11
21
1122
22
2
11
1
31
11
1
3
3
=
=
==
==
×+×
==
==
pm=constT=const
3m12 33m 2v
SPECIFIC HEAT C
( ) ( )
vvp
v
p
vp
vp
vp
vp
vp
constvv
constpp
cR1k Rcc ONLYGASIDEALFOR
cc
k andunitssamewith
Rcc
RdTdTcdTcdu anddh for ngsubsistutiRdTdudh atingdifferenti
RTuh RTpv ngsubstituti
pvuh definitionBy
dTcdu dTcdh
dTc∆udTc∆h
constant assumed are heats specific asgidealan For
dTTc∆udTTc∆h
TuC
Th
=−=−
=
+=
+=+=
+==
+=
==
==
==
∂∂
=
∂∂
=
∫ ∫
∫∫==
IDEAL GAS IMPROVEMENTS
Enthalpy, h, internalenergy, u, and entropy, s. are not absolute but
Differences from a base.
( )
( )
( ) ( )
22E-A 22,-A to17-A 17, -A Tables
Tcc,Tcc
dTTcu
dTTch
vvpp
T
StateBaseTable
v
T
StateBaseTable
p
==
=
=
∫
∫
2-104
( )( ) ( )( )
( )( ) ( )( )
( )
( )
g449.46kJ/k∆h 101.325)p600.,Titrogen,enthalpy(n101.325)p1000.,Titrogen,enthalpy(n∆h
e)EESkJ/kg 448.583kg/kgmolemole/28.0112566kJ/kg∆h
kJ/kgmole 12566kJ/kgmole 17563kJ/kgmole 30,129Kh@600Kh@1000h∆ 18ETableA d)
kJ/kg 415.66001000 kJ/kg 1.039448∆h
kJ/kg 1.039448.5K@800c heat, specific re temperatuc)Room
448.5kJ/kg6001000K kJ/kg 1.121∆Tc∆h
kJ/kgK 1.121K@800c range, re temperatuover theheat specific b)AveragekJ/kg 447.8 8.013kJ/kmole/2 2544∆h
kJ/kmole 125446001000102.873808141600100010.8081
31
6001000.0001571.5600100028.9h∆
dTcTbTac dT,Tch∆ a)
EES. e) 18E,A Table d) 2a,-A Table re temperaturoomat heat specific c) 2b,-A Table re temperatuaverage at theheat specific b)
, 2cA Tableequation heat specific empirical a) :using F)1340 C,(726K 1000 K to600 from heated isit as kJ/kgin nitrogen of ∆h, change,enthalpy theDetermine
OO
Op
p
Op
449335
12
32pp
OOOO
===−===
===−=−=
−=−=
=
=−==
=
==
=−×−−×+
−+−=
+++==
−
−
−−
∫
PRINCIPAL OF CORRERSPONDING STATESCOMPRESSIBILITY FACTOR Z
P
criticalR
criticalR
TTT
)202(p
pP
=
−=
Z is about the same forall gasses at the same reduced temperatureand the same reduced pressure where:
mRTpVZ
RTpvZ
=
=
VAN DER WAALS EQUATION OF STATE - 1873
( )
critical
critical
critical
2critical
2
T2
2
T
2criticalcritical
critical
2
p 8TRb
p 64T R 27a
0dv
p0dvp
va
bvRTp
molecules gas of volumeb
forcesular intermolecva
criticalcritical
==
=
∂=
∂
−−
=
−
−
=−
+ 22)-(2 RTbv
vap 2
critical point
0vp
v
0vp
T
T
=
∂∂
∂∂
=
∂∂
p
v
THERMODYNAMIC PROPERTY MEASURMENT
Thermodynamic properties are independent of path or process and are exact differentials.
Heat and Work are not exact differentials but are dependent on process or path.
gas real afor valueagas, idealan for 0pT
tCoefficienTompson JoulepT
cTu
cTh
equations 60 time,aat 2 properites amic thermodyn6
dvvuds
sudu
v)f(s,u
h
h
vv
pp
sv
=
∂∂
=
∂∂
=
∂∂
=
∂∂
∂∂
+
∂∂
=
=
vT Tp
vs
∂∂
=
∂∂
LawFirst with theused
Course Property Sources
1) Ideal Gas Law with constantspecific heats
2) TablesSteamRefrigerantAir
3) EES CDNIST
for home work convenience
NIST Webbook Propertiesfttp://webbook.nist/gov/chemistry/fluidTemperature Table for Water in .1 degree incrementsfrom 40 to 40 degrees.
Select Units
Select Table Type