Download p @T

Thermodynamic Properties are Measurements
p,T,v, u ,h,s - measure directly
-measure by change  ∂s  =  ∂p 
 ∂v T
Tables
Curve fits
Property
Data
Tables
Correlation's, Boyles Law
pv=c@T=c
Equations of State, pv=RT
 ∂T  v
Tables
limited hand
calculations
Tables
Calculation Modules
NIST, EES, HYSYM
interactive, callable
P=1 atm
vapor
liquid
Q
Given T and P
Super Heat Region if,
p<p sat @T
T>Tsat @p
Compressed
Liquid Region if,
p>p sat @ T
T<Tsat @ p
STEAM PRESUE AND TEMPERATURE TABLES
T
Saturated
Vapor
Line
Saturated
Liquid
Line
hf
hg
uf
ug
sf
sg
vf
vg
Three Tables
Temperature Table
at spaced T’s
Pressure Table
at spaced P’s
Superheat Table
at spaced T and P
6 Properties
Temperature
Pressure
Volume
Internal Energy
Enthalpy
Entropy
Solid-Liquid-Gas
Phase Diagram
Saturation liquid internal energy at 0 C
0. Table Base
Saturated liquid enthalpy at 25 C
104.89 kJ/kg
Saturated vapor entropy at 25 C
8.558 kJ/kg K
Enthalpy 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-7 Metric T,p
Table A-7E English T,p
Table A-4 Metric, T
Table A-5 Metric, p
Table A-4E English, T
Table A-5E English, p
TEMPERATURE TABLE
PRESSURE TABLE
Table A-6, Metric, T,p
Table A-6E, English, T,p
SUPERHEAT TABLE
R − 134a at 500 kPa and 6 C
Tsaturation@ 500 kpa = 15.71C
psaturation@ 6 C = 362.23 kpa
water at 1400 kpa and 300 C
Tsaturation@ 1400 kpa = 195.04 C
psaturation@ 300 C = 8587.9 kPa
p=constant
T
p=constant
T
300 C
8587.9 kPa
1400 kpa
195.04 C
superheated
15.71C
500 kPa
362.23 kpa
6C
subcooled
v
v
Two Phase Real Gas Properties
V = Vf + V g
mv = m f v f + m g v g
x=
mg
Quality
m
v = (1 − x )v f + xv g
v = v f + x × v fg
v − vf
x=
v fg
h = h f + x × h fg
u = u f + x × u fg
v = v f + x × v fg
vf
vg
Quality, x =
x=
Find the properties of water at 90o F, v = 300 ft 3 /lb.
v − vf
vg − vf
300 − .0161
= .64 (64%)
467.7 − .0161
(v
u = u f + x × (u f − u g )
u = u f + x × u fg
u = 58.07 + .64 × 982.2
u = 686.7 BTU/lb
h = h f + x × h fg
Temperature
Table
g
− vf )
(v f − v )
T
90o F
u g =1040.2
u f =58.07
h f =58.07
s f =.11165
u
h g =1107.7
h
s g =2.0083
s
h = 58.07 + .64 × 1042.7
h = 725.4 BTU/lb
s = s f + x × s fg
s = .11165 + .64 ×1.8966
s = 1.3254 BTU/lb o R
v f = .0161ft 3 / lb
v = 300. ft 3 /lb
v g = 467.7ft 3 / lb
Engineering Equation Solver - EES
Fluid Property Information - 69 fluids available
Thermophysical Functions - 25 properties calculated
Equations Window
h=enthalpy(steam, T=200.,P=200)
h=enthalpy(steam,T=200.,X=1)
u=intenergy(steam,T=200.,X=0.)
p=pressure(steam,T=200.,X=0.)
Thermophysical Functions
entropy
intenergy
pressure
quality
density
enthalpy
isidealgas
temperature
volume
superheated vapor
saturated vapor
saturated liquid
saturation pressure
Function Arguments
H specific enthalpy
P pressure
S specific entropy
T temperature
U specific internal energy
V specific volume
X quality
EES
FLUIDS
FUNCTIONS
Table
Saturation internal
energy at 4 kPa
EES Program
Pressure Table
u l = intenergy(steam, P = 4., X = 0)
u l = 121.45kJ/kg
u l = 121.3kJ/kg
u g = 2415.2kJ/kg
u g = 2414.29
u g = intenergy(steam, P = 4., X = 1.)
Superheat Table
Enthalpy and
volume of water at
150 kPa and 30 C
h l = h saturated liquid @ 30o C
h l = 125.75 kJ/kg
v l = v saturated liquid@30o C
v l = .001004m 3 /kg
h l = enthalpy(steam, T = 30., p = 500.)
h l = 125.67kJ/kg
v l = volume(steam, T = 30., p = 500.)
v l = .001004m 3 /kg
Temperature Table
Internal energy
u l = u saturated liquid @ 300 o C
of water at 20 MPa u = 1332.0 kJ/kg
l
and 300 C
u l = intenergy(steam, T = 300., p = 20000.)
u l = 1333.446 kJ/kg
Solve
Windows
Equations
Windows
Arrays
Windows
Plot Window
10 MPa saturated steam has an enthalpy of
2010 kJ/kg. What is its internal energy?
h = h f + x h fg
2010 kJ/kg = 1407.56 + x 1317.1
x = .4574
u = u f + x u fg
u = 1393.04 + .4574 × 1151.4
u = 1919.69 kJ/kg
EES Program
x = quality(steam,P = 10000.,h = 2010)
x = .4576
u = 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?
h = h f + x h fg
1800 kJ/kg = 83.96 + x 2454.1
x = .7
u = u f + x u fg
u = 83.95 + .7 × 2319.0
u = 1707.25 kJ/kg
STEAM SUPERHEAT TABLE
SUPERHEAT TABLE
Enthalpy at 700 C and .10 Mpa
3928.2 kJ/kg
Temperature at entropy of 8.8642 and .05 Mpa
400 C
Enthalpy 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?
T
Entropy at 1100 C, .1 MPa
10.1659kJ/kg K
Enthalpy at .01 MPa, entropy 10.1659
3705.4 kJ/kg
Temperature at .01 MPa, entropy 10.1659
600 C
.01 MPa
s
Linear Interpolation with 2 Variables
h@(T = 450 C,p = 27 MPa)
Steam Superheat Table
Table Values
p = 25 mPa
T = 450 h = 2949.7
p = 27 MPa
h = 2898.4
p = 30 MPa
h = 2821.4
27 − 25
For the pressure table entry,
= .4
30 − 25
The desired pressure, 27 kPa, is 40 % of the difference
between 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 WORLD
EXCELL ADD-IN
40 fluids
4 sets of units
SPREADSHEET WORLD – THERMAL FLUIDS PROPERTIES
A
B
C
D
E
F
G
1
2
T
200
3
P
15
4
5 CALL
TTProps("AIR","EE_F", "P", $C$2,"T", $C$1)
6
7
P=
v=
T=
u=
s=
h=
8 RETURN
15
16.29669
200
180.5582 1.085135 225.7937
9
psia
ft^3/lbm
deg F
BTU/lbm BTU/lbm-R BTU/lbm
H
X=
1
nd
I
J
STATE = ERROR =
erheated va
0
A saturated mixture of 2 kg water and 3 kg vapor in contained in
a 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.
3 kg vapor
= .6
at100kPa, x =
5 kg total
T = 99.61
v1 = vf + x × vfg
100 kpa
3 kg
v1 = .001043+ .6× (1.694−.001043)
v1 = 1.0168m3/kg
u1 = uf + x × ufg
Q
2 kg
u1 = 417.40+ .6× 2088.2= 1670.62kJ/kg
3.88
v = constant,1 - 2
Q = ∆E + W
p = constant,2 - 3
Point 2 at . 2 MPa, v1 = v 2
Q = ∆E + W
Q = ∆E + p∆V
V2 = 5 kg ×1.0161 = 5.08 m3
Interpolation from Table A − 6
u 2 = u@(v = 1.0161, P = .2 MPa )
W=0
Q = ∆E = ∆U
Q = ∆H
Q = m(u 2 − u1 )
u 2 = 2613.23 kJ/kg
Q = m(h3 − h 2 )
h 2 = 2816.47 kJ/kg
Q = 5 × (2988.65− 2816.47)
Q = 4713.05 kJ
Q2-3 = 860.9 kJ
h 2 = h@(v = 1.0168, P = .2 MPa ))
V3 = 1.2 × V2 = 6.096
alteratively,
3
6.096 m
= 1.2192
5 kg
v g = .8857 ⇒ superheated
v3 =
Q = 5 × (2613.23- 1670.62 )
W1-3 = Q1-3 − m × (u3 − u 2 )
2
Point 3 at .2 MPa
h 3 = h @ (v3 = 1.2192, P = .2 MPa )
3
p
h 3 = 2988.65 kJ/kg
2
3
1
2
W = ∫ pdV + ∫ pdV = 0 + p 2 (V3 − V2 )
(
1
)
W = 0 + 200 kPa × 6.096 m3 − 5.08 m3 = 203.2 kJ
v
3.88
Interpolation for h 3
Superheat Table A - 6
T
250
v
1.1989
1.2192
300 1.31623
enthalpy @ ( p = .2 MPa, v = 1.2192)
h
2971.2
3092.1
1.2192 - 1.1989
= .173
1.31623- 1.1989
h 3 = 2988.65 kJ/kg
ratio of v =
Spread Sheet World module solution
water
2
vapor
3
Properties at 1 given p1 and x1
x1
0.6
p1
100,000
P=
100000
Pascals
2
3
p
1
v=
T=
u=
s=
h=
1.016819 99.62524 1670.481 4.936567 1772.163
m^3/kg
deg C
kJ/kg
kJ/kg-K
kJ/kg
X=
0.6
nd
STATE =
Mixed regio
X=
1
nd
STATE =
erheated va
V2=m x v2
P=
v=
T=
u=
s=
h=
200000 1.016819 173.5216 2613.078 7.389259 2816.442
Pascals m^3/kg
deg C
kJ/kg
kJ/kg-K
kJ/kg
5.084094
V3= 1.2 * V2
v3=V3/m
p3
6.100913
1.220183
200,000
X=
1
nd
STATE =
erheated va
Properties at 2 given p2 and v2=v1
p2
200,000
Properties at 3 given p3 and v3
P=
200000
Pascals
Q 2-3=m*(h3-h2)
W 2-3 = Q 2-3-m*(u3-u2)
Q 1-2=m*(u2-u1)
861.7029
203.363
4712.986
v=
T=
u=
s=
h=
1.220183 259.0808 2744.746 7.742318 2988.782
m^3/kg
deg C
kJ/kg
kJ/kg-K
kJ/kg
v
EES Solution
100 kpa
3 kg
2 kg
Q
2
3
p
1
v
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?
V1 = Vf + Vg
V1 = m f × v f @ 20 C + m g v g @ 20 C page 842, TableA − 11
V1 = 1 kg × .0008157 + 3 kg × .0358
V1 = .1082 m
3
V1 .1082 m 3
v1 =
=
= .027 m 3 /kg
m
4 kg
After heating,
T
20 C
V = constant, m = constant ⇒ v = constant
v 2 = v1 = v f @ .6 MPa + x × v fg @ .6 MPa page 843 , Table A - 12
.027 − .00008196
= .787(78.7%)
.0341 − .0008196
h 2 =h f + x × h fg
x2 =
h 2 = 79.84 + .787 ×179.71 = 211.27 kJ/kg
Q
2 .6 MPa
1
.5716 MPa
v
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 and
pressure? If the cylinder and its contents are heated until volume is .15 cubic
meters what is the quality?
v g1 =
Vg1
m g1
.1074 m 3
=
= .0358 m 3 /kg
3 kg
T
v g = .0358 m /kg @ 20 C, .5716 MPa page 842, TableA − 11
3
During the heating process the pressure is constant.
20 C
at V2 = .15m 3 ,
V2 .15
=
= .03 m 3 /kg
m
5
v 2 = .03 m 3 /kg = v f @ 20 C + x × v f g @ 20 C
1
2
v2 =
v
v − vf
.03 − .0008157
=
= .834 (83.4%)
x2 = 2
v fg
.0358 − .0008157
x=
property − (property)f
(property)fg
property = (property)f + x × (property)fg
Q
Thermodynamic Problem Solving Technique
1. Problem Statement
4. Property Diagram
Carbon dioxide is contained in a cylinder
with a piston. The carbon dioxide is compressed
with heat removal from T1,p1 to T2,p2. The gas
is then heated from T2, p2 to T3, p3 at constant
volume and then expanded without heat transfer to
the original state point.
state points - processes - cycle
T3,p3
p
Q
W
T2,p2
2. Schematic
W Q
T1,p1
v
5. Property Determination
1 2 3
3. Select Thermodynamic System
open - closed - control volume
a closed thermodynamic system
composed to the mass of carbon
dioxide in the cylinder
p
T2,p2
T
p
v
u
h
s
v
6. Laws of Thermodynamics
CO 2
Q=?
W=? E=? material flows=?
Ideal Gas Law
AVOGADRO'S LAW
One (1) mole of any gas = 22.4 liters.
6.023×1023 molecules /mole of gas at
STP (1 atm and 0o C )
BOLYES LAW
p1 × v1 = p 2 × v 2
IDEAL (PERFECT) GAS LAW
pv = RT
pV = mRT
p - absolute pressure, psia,kPa
T - absolute temperature, o R, o K
R*
R=
molecular weight
ft lbf o R
R = 1545.15
lbmole
kJ
kPa m 3
*
R = 8.314
or
kmole o K kmole o K
*
CHARLES LAW
v1 T1
=
v 2 T2
p1 T1
=
p 2 T2
mass = moles × Molecular Weight
m = n × Molecular Weight
pv = R * T
pV = nR * T
water
Ideal Gas
pv = RT
constant specific heat
What is the mass of 1.2 m3 of oxygen at 24o C
and a gage pressure of 500 kPa.
Atmospheric pressure is 97 kPa
p = pgage + patmosphere = 500 kPa + 97 kPa = 597 kPa
8.314 kJ/kmole o K or kPa m3 /kmole o K
= .259813 kPa m3 /kg
R O2 =
32
pV
597 kPa ×1.2 m3
m=
=
= 9.28 kg
RT .259813 kPa m3 /kg × ( 24o C + 273.16o K )
alsoTableA − 1
What is the volume mass of 1.2 lbm of air at 124o F
and a gage pressure of 500 psia.
Atmospheric pressure is 14.7 psia.
p = pgage + patmosphere = 500 psia + 14.7 psia = 514 psia
1545.15 ft lbf / lbmole R
ft lbf
also Table A − 1E in BTU and psi units
= 53.336
28.97
lbm o R
o
o
o
m R T 1.2 lbm × 53.336 ft lbf/ lbm R × (124 F + 459.69 R )
V=
=
514 psia ×144 ft 2 /in 2
p
R air =
V = .5047ft 3
IDEAL GAS EQUATION FORMS - For Air
P
kPa
kPa
lb
ft 2
lb
ft 2
v = m
m
3
m
3
= kmole
= kg
R
kPa m3
8.314
kmole O K
kPa m3
.287
kg O K
T
O
O
K
K
ft lbf
lbmole O R
ft lbf
53.35
lbm O R
ft 3
= lbmole
ft 3
= lbm
psi
ft 3
= lbm
.3704
lb
ft 2
ft 3
= lbm
.06855
O
1545.15
psi lbf
lbm O R
BTU
lbm O R


kPa m3
R
8.314
/
28.96
=


kmoleo K


O
O
R
R
R
O
R
ft lbf


R
=
1545.15
/28.96


lbmole O R


ft lbf


R
=
1545.15
/144/28.96 

O
lbmole R


ft lbf


R
1545.15
/28.96/778 
=

O
lbmole R


The specific volume of air at 75o F and 14.7 psia
R air = 1545.15 /28.96 = 53.35 ft lbf/lbm R
(
RT 53.35 ft lbf/lbm R × 459.69o R + 75o F
v=
=
p
14.7 lbf/in 2 × 144 in 2 /ft 2
)
v = 13.476 ft 3 /lb
The specific volume of air at 24o C and 101.325 kPa
kPa m 3
R = 8.314
/ 28.96
O
kg mole K
(
RT .287 kPa m 3 /kg × 273.15o K + 24o F
v=
=
p
101.325 kPa 2
v = .8417 m 3 /kg
)
Air initially at a volume of 12 m 3 and a pressure of 225 kPa
expands at a constant temperature to a volume of 23 m 3 .
What is the final pressure?
mass = constant,
T = constant
RT1 .286 × (273.15 + T1 )
m=
=
p1V1
224 kPa × 23 m 3
m=
R T1 R T2
=
p1V1 p 2 V2
p2 =
R T2 p1 V1
V
= p1 1
R T1 V2
V2
12 m 3
p 2 = 225 kPa
23 m 3
p 2 = 117.39 kPa
m=const
T=const
p
12 m 3
v
2 3m 3
SPECIFIC HEATS
FOR GASSES
SPECIFIC HEAT
 ∂h 
Cp =  
 ∂T  p =const
 ∂u 
Cv =  
 ∂T  v =const
∆h = ∫ c p (T ) dT
∆u = ∫ c v (T ) dT
For an ideal gas specific heats are assumed constant
∆h = c p ∫ dT
∆u = c v ∫ dT
dh = c p dT
du = c v dT
h = u + pv
By definition
substituting pv = RT
differentiating
h = u + RT
dh = du + RdT
subsistuting for dh and du c p dT = c v dT + RdT
cp = cv + R
with same units and
FOR IDEAL GAS ONLY
k=
cp
cv
cp − cv = R
k −1 =
R
cv
IDEAL GAS IMPROVEMENTS
Enthalpy, h, internal
energy, u, and entropy, s.
are not absolute but
Differences from a base.
T
h=
∫ c (T )dT
p
Table
Base
State
T
u=
∫ c (T )dT
v
Table
Base
State
c p = c p (T ), c v = c v (T )
Tables A - 17, A - 17 to
A - 22, A - 22E
IDEAL GAS WITH VARIBLE SPECIFIC HEAT
2-104
Determine the enthalpy change, ∆h, of nitrogen in kJ/kg as it is heated from
600O K to 1000O K (726O C, 1340O F) using : a) empirical specific heat equation Table A − 2c ,
b) specific heat at the average temperature Table A - 2b, c) specific heat at room temperature
Table A - 2a, d) Table A − 18E, e) EES.
a) ∆h = ∫ cp (T )dT,
cp = a + bT + cT 2 + dT 3
(
∆h = 28.9(1000 − 600 ) + .5(.0001571) 1000 2 − 6001
+
(
)(
) (
)
)(
)
1
1
.8081×10 −5 10003 − 6003 − 2.8738081× 10 −9 1000 4 − 600 4 = 12544 kJ/kmole
3
4
∆h = 2544 kJ/kmole/28.013 = 447.8 kJ/kg
b)Average specific heat over the temperature range, c p @800O K = 1.121 kJ/kgK
∆h = c p ∆T = 1.121 kJ/kg K (1000 − 600 ) = 448.5kJ/kg
c)Room temperature specific heat, c p @800O K = 1.039448.5 kJ/kg
∆h = 1.039448 kJ/kg (1000 − 600 ) = 415.6 kJ/kg
d) TableA − 18E
∆h = h@1000 O K − h@600O K = 30,129 kJ/kgmole − 17563 kJ/kgmole = 12566 kJ/kgmole
∆h = 12566kJ/kgmole/28.013kg/kgmole = 448.58 kJ/kg
e)EES
∆h = enthalpy(nitrogen, T = 1000., p = 101.325) − enthalpy(nitrogen, T = 600., p = 101.325)
∆h = 449.46kJ/kg
PRINCIPAL OF CORRERSPONDING STATES
COMPRESSIBILITY FACTOR Z
Z is about the same for
all gasses at the same
reduced temperature
and the same reduced
pressure where:
P
pv
Z=
RT
pV
Z=
mRT
PR =
TR =
p
p critical
T
Tcritical
(2 − 20)
VAN DER WAALS EQUATION OF STATE - 1873
a 

(2 - 22)
 p + 2 (v − b ) = RT
v 

a
− intermolecular forces
2
v
b − volume of gas molecules
RTcritical
a
p=
− 2
v critical − b v critical
 ∂p 
=0
 
 dv  Tcritical
2
27 R 2 Tcritical
a=
64 p critical
 ∂p 
  =0
 ∂v  T
∂  ∂p 
  =0
∂v  ∂v  T
p
critical point
 ∂ 2p 
 2 
=0
 dv  Tcritical
b=
R Tcritical
8 p critical
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.
u = f(s, v)
 ∂u 
 ∂u 
du =   ds +   dv
 ∂s  v
 ∂v s
6 thermodynamic properites
2 at a time, 60 equations
 ∂h 
  = cp
 ∂T  p
 ∂u 
  = cv
 ∂T  v
 ∂T 
  = Joule Tompson Coefficient
 ∂p  h
 ∂T 
  = 0 for an ideal gas, a value for a real gas
 ∂p  h
used with the First Law
 ∂s   ∂p 
  =

∂
v
∂
T
 T 
v
 ∂T 

 = JT Coefficient
 ∂p  h
h=constant
Course Property Sources
1) Ideal Gas Law with constant
specific heats
2) Tables
Steam
Refrigerant
Air
3) EES CD
NIST
for home work convenience
EQUATION OF STATE ERRORS
nitrogen
NIST Webbook Properties
fttp://webbook.nist/gov/chemistry/fluid
Temperature Table for Water in .1 degree increments
from 40 to 40 degrees.
Select Units
Select Table Type
Select fluid
Set low and high temperature
and temperature increment.