Thermodynamic Properties are Measurements p,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 Tables Calculation Modules NIST, EES, HYSYM interactive, callable Property Data v T T p v s ∂ ∂ = ∂ ∂
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Thermodynamic Properties are Measurements · 2010. 5. 16. · Thermodynamic Properties are Measurements p,T,v, u ,h,s - measure directly -measure by change Tables Curve fits Tables
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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
∂∂
=
∂∂
P=1 atm
Q
liquid
vapor
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
STEAM PRESUE AND TEMPERATURE TABLES
f
f
f
suh
SaturatedLiquidLine
g
g
g
s
u
h
SaturatedVaporLineT
fv gv
Three TablesTemperature Table
at spaced T’sPressure Table
at spaced P’sSuperheat Table
at spaced T and P6 PropertiesTemperaturePressureVolumeInternal EnergyEnthalpyEntropy
Solid-Liquid-GasPhase Diagram
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
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
=×+=
+==
+=
+=
STEAM SUPERHEAT TABLE
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
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
SPREAD SHEET WORLDEXCELL ADD-IN40 fluids4 sets of units
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 =
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?
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.
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.