#### Transcript Lecture 07: Enthalpy

```EGR 334 Thermodynamics
Chapter 3: Section 6-8
Lecture 07:
Enthalpy and Internal Energy
Quiz Today?
Today’s main concepts:
• Define enthalpy, h
• Recognize that enthalpy and internal energy are both intensive
state properties that can also be determined on the steam
tables.
• Learn to pull values for enthalpy and internal energy from the
steam tables.
• Learn to use computer tools to find property values.
• Use steam table values to solve 1st Law balance problems.
• Read Chap 3: Sections 9-10
Homework Assignment:
From Chap 3: 35, 41, 42, 45
3
Last time we learned that if you know two of
the three intensive properties:
Temperature, T
Specific Volume, v
and Pressure, p
...then you could find the other properties at
that state.
Other intensive properties include:
Enthalpy, h
Internal energy, u
and
Entropy, s
(which are also found on the steam tables.
h  h(T , v)
or
u  u ( p, v )
or
s  s (T , p )
Sec 3.6.1 : Introducing Enthalpy
4
Internal Energy (U)
Specific Internal Energy (u) : energy / mass
Kinetic Energy
translational
rotational
vibrational
Potential Energy
electric energy of atoms, molecules, or crystals
Chemical Bonds
Enthalpy (H)
Specific Enthalpy (h) : energy / mass
Internal Energy + Work
H  U  PV
h  u  Pv
5
Saturated Steam Table
Enthalpy, h
Internal Energy, u
Notice there are
steam table that we
ignored last time.
For the time being,
we’re going to still ignore
this last quantity.
6
What is enthalpy?
Enthalpy is defined in terms of internal energy and the work of
expansion as
Extensive Form
H  U  pV
Intensive Form
h  u  pv
Your author indicates that this term is defined, simply because it’s
convenient. This combination of terms will show up very frequently,
especially when we begin working with control volumes. It’s a way to
conveniently deal with the internal energy in combination with the
energy associated with expansion or compression of the substance.
Enthalpy is another formulation of energy and is commonly expressed
in units of
kJ
or
ft-lbf.
7
Example 1: Determine property values
Determine specific volume (v), internal energy (u), and enthalpy (h) of
the following state properties of H20.
a) p = 10 bar
T = 179.9 deg C.
b) p = 10 bar
T = 320 deg. C.
c) p = 10 bar
T = 450 deg C.
x = 30%
8
Example 1: Determine property values
Determine specific volume, internal energy, and enthalpy of the
following state properties of H20.
a) p = 10 bar
T = 179.9 deg C.
x = 30%
from table A.3: at 10 bar the saturated temperature is 179.9 which
means the substance is a mixture of liquid and vapor.
9
From table A-3 the following values may be directly read:
quality
x = 30%
p=10 bar
spec. vol.
int. energy
enthalpy
[m3/kg]
[kJ/kg]
[kJ/kg]
vf=0.001127
uf=761.68
hf=762.81
vg=0.1944
ug=2583.6
hg=2778.1
therefore:
v  v f  x (v g  v f )
v  0.0011  0.3(0.1944  0.0011)
v  0.0591 m3 / kg
u  u f  x(u g  u f )
u  761.7  0.3(2583.6  761.7)
u  1308.3 kJ / kg
h  762.8  0.3(2778.1  762.8)
h  1367.4 kJ / kg
h  h f  x(hg  h f )
10
Example 1: Determine property values
Determine specific volume, internal energy, and enthalpy of the
following state properties of H20.
b) p = 10 bar
T = 320 deg. C.
vapor
therefore:
v = 0.2678 m3/kg
u = 2826.1 kJ/kg
h = 3093.9 kJ/kg
11
Example 1: Determine property values
c) p = 10 bar
T = 450 deg C.
This also is a gas or superheated
vapor and will need to be found
by interpolating from table A-4.
v  v1  (T  T1 )
(v2  v1 )
(T2  T1 )
 0.3257  (450  440)
 0.3304 m3 / kg
u  u1  (T  T1 )
(0.3541  0.3257)
(500  440)
(u2  u1 )
(T2  T1 )
 3023.6  (450  440)
(3124.4  3023.6)
(500  440)
 3040.4 kJ / kg
h  h1  (T  T1 )
(h2  h1 )
(T2  T1 )
 3349.3  (450  440)
 3370.8 kJ / kg
(3478.5  3349.3)
(500  440)
12
Linear Interpolation:
yunknown  y1 xknown  x1

y2  y1
x2  x1
yunknown
 y2  y1 
 y1  
 ( xknown  x1 )
 x2  x1 
*
y2
yunknown
y1
*
x1
xknown
x2
Your turn: Find the value for enthalpy at
T = 400 deg. C and p = 50 bar
at 400°C and p1=40 bar
 h1 = 3213.6 kJ/kg
at 400°C and p2=60 bar
 h2 = 3177.2 kJ/kg
hunknown
 h2  h1 
 h1  
 ( pknown  p1 )
 p2  p1 
= 3195.4 kJ/kg
On Tests and Quizzes, you will be expected to pull out data from the printed
steam tables. That’s not your only option when working homework.
Computer Method 1: Web based Steam Table Calculator
at: http://www.dofmaster.com/steam.html
For: T = 400 deg C.
p = 50 bar = 500 kPa
Enthalpy per unit mass:
h = 3195.4 kJ/kg
Computer Method 2: Interactive Thermodynamics (IT)
http://www.wiley.com/college/moran
IT is more than just a look up table. It is also an equation solver
that can automatically look up properties values and attempt to
find a solution using an iterative solver.
h = 3195 kJ/kg
Sec 3.5.2 : Saturation Tables
Example: Use IT to solve this problem.
A cylinder-piston assembly initially contains water at 3 MPa and
300oC. The water is cooled at constant volume to 200 oC, then
compressed isothermally to a final pressure of 2.5 MPa. Find the
specific volume and enthalpy at each of these three states.
17
State 1:
p1 = 3 MPa= 30 bar and
T1 = 300 deg C.
ii) Interpolate between 280 and 320 deg C.
vb  va
v1  va 
(T1  Ta )
Tb  Ta
v1  0.0771 
(0.0850  0.0771)
(300  280)
(320  280)
v1  0.0811 m3 / kg
18
State 2:
v2 = v1 = 0.0811 m3/kg and T2 = 200 deg C.
Recognize that this is in the liquid-vapor mixture range.
Refer to Table A-2
i) Locate vg and vf at T = 200 deg C.
v f  1.1565  103
vg  0.1274
v2  0.0811
ii) Determine quality, x2
v  v f  x (v g  v f )
x

v  vf
vg  v f
0.0811  0.0011565
 0.633  63.3%
0.1274  0.0011565
19
State 3:
p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C.
Recognize that this is in the compressed liquid.
Refer to Table A-5
i) Locate p3=25 bar and T3=200 deg C
v3  1.1555 103 m3 / kg
u3  849 kJ / kg
h3  852.8 kJ / kg
20
Repeat this work, but use IT instead.
Results are the same as
shown yesterday.
Sec 3.6.2: Retrieving u and h Data
Workout Problem
Given Ammonia at 12°C and v = 0.1217 m3/kg
determine P, u and h.
a) Using Tables in Appendix:
b) Using IT:
Sec 3.6.2: Retrieving u and h Data
Workout Problem
Given Ammonia at 12°C and v = 0.1217 m-3/kg
determine P, u and h.
a) Using Tables in Appendix:
Sec 3.6.2: Retrieving u and h Data
Example:
Given Ammonia at 12°C and v=0.1217 m-3/kg
determine P, u and h.
Notice that
vf = 0.001608
Since:
and vg = 0.1923
vf < v < vg
the substance is a liquid-gas mixture
Saturation Properties are:
Tsat = 12 deg C.
Quality if found using:
v  v f  x  vg  v f
Psat= 658.9 kPa


x
v  vf
vg  v f
Sec 3.6.2: Retrieving u and h Data
Example:
Given Ammonia at 12°C and v=0.1217 m-3/kg
determine P, u and h.
x
v  vf
vg  v f
with
vf = 1.608x10-3,
vg = 0.1923 m3/kg
x
x
0.1217  0.001608
0.1923  0.001608

0.63

63%
Sec 3.6.2: Retrieving u and h Data
Example:
Given Ammonia at 12°C and v=0.1217 m-3/kg
determine P, u and h.
Internal Energy:
kJ



u  1  x u f  xug  1  0.63255.16 kJ

0
.
63
1346
.
8
kg
kg
u  942.9 kJ
kg
Enthalpy:
 
 kJ
h  1  x h f  xhg  1  0.63256.2 kJ
kg  0.63 1473.6 kg
h  1023.2 kJ
kg
Check:
h  u  Pv  942.9
kJ
kg
 658.9kPa0.1217
103 N m 2
m3
kJ
kg 103 Nm
kPa
b) Using IT:
Sec 3.8: Appling the Energy Balance using the Property Tables
Workout Problem: (3.60) A rigid, insulated tank fitted with a paddle
wheel is filled with water, initially a two phase liquid-vapor mixture at
20 psi, consisting of 0.07 lb of saturated liquid and 0.07 lb of saturated
vapor. The tank contents are stirred by the paddle wheel until all of
the water is saturated vapor at a pressure greater than 20 psi. Kinetic
and potential energy effects are negligible. For the water, determine
(a) Volume occupied, in ft3,
Pi = 20 psi
(b) Initial temperature, in °F,
mf = 0.07 lb
(c) Final pressure, in psi,
ml = 0.07 lb
(d) Work, in BTU
x
mvapor
mliquid  mvapor

0.07
 0.5
0.07  0.07
27
28
Sec 3.8: Appling the Energy Balance using the Property Tables
29
Example: (3.77) A system consisting of 1 kg or water undergoes a
power cycle composed of the following processes.
Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor.
Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Process 3-4: Isothermal compression with Q34 = -815.8 kJ
Process 4-1: Constant-volume heating
Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and
potential energy effects, determine the thermal efficiency.
30
31
End of Slides for Lecture 07
Sec 3.8: Appling the Energy Balance using the Property Tables
Workout Problem: (3.60) A rigid, insulated tank fitted with a paddle
wheel is filled with water, initially a two phase liquid-vapor mixture at
20 psi, consisting of 0.07 lb of saturated liquid and 0.07 lb of saturated
vapor. The tank contents are stirred by the paddle wheel until all of
the water is saturated vapor at a pressure greater than 20 psi. Kinetic
and potential energy effects are negligible. For the water, determine
(a) Volume occupied, in ft3,
Pi = 20 psi
(b) Initial temperature, in °F,
mf = 0.07 lb
(c) Final pressure, in psi,
ml = 0.07 lb
(d) Work, in BTU
x
mvapor
mliquid  mvapor

0.07
 0.5
0.07  0.07
32
Sec 3.8: Appling the Energy Balance using the Property Tables
(a)
(b)
(c)
(d)
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
x
33
Pi = 20 psi
mf = 0.07 lb
ml = 0.07 lb
mvapor
mliquid  mvapor

0.07
 0.5
0.07  0.07
Sec 3.8: Appling the Energy Balance using the Property Tables
(a)
(b)
(c)
(d)
a)
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
34
Pi = 20 psi
mf = 0.07 lb
ml = 0.07 lb
xi = 0.5
V  1  x v f  xvg
 0.07lb0.01683
 1.407 ft 3
b) T = Tsat = 227.96 °F
ft 3
lb
 0.07lb20.094
ft 3
lb
Sec 3.8: Appling the Energy Balance using the Property Tables
(a)
(b)
(c)
(d)
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
35
Pi = 20 psi
mf = 0.07 lb
ml = 0.07 lb
xi = 0.5
V 1.407 ft 3
ft 3

 10.05
c) v 
m
1.4lb
lb
Interpolate values to find P2 = 42 psi & u2 = 1093 BTU/lb
U  K  PE  Q  W  W  U  mu2  u1 
ui  u f  xug  u f   196.19  0.51082196.19  639.1 BTU
lb
U  0.14lb1093 639.1 BTU
lb  63.6BTU  W
Sec 3.8: Appling the Energy Balance using the Property Tables
36
Example: (3.77) A system consisting of 1 kg or water undergoes a
power cycle composed of the following processes.
Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor.
Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Process 3-4: Isothermal compression with Q34 = -815.8 kJ
Process 4-1: Constant-volume heating
Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and
potential energy effects, determine the thermal efficiency.
Sec 3.8: Appling the Energy Balance using the Property Tables
37
Example: (3.77) A system consisting of 1 kg or water undergoes a
power cycle composed of the following processes.
Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor.
Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Process 3-4: Isothermal compression with Q34 = -815.8 kJ
Process 4-1: Constant-volume heating
Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and
potential energy effects, determine the thermal efficiency.
Sec 3.8: Appling the Energy Balance using the Property Tables
38
Example: (3.77) A system consisting of 1 kg or water undergoes a
power cycle composed of the following processes.
Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor.
Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Process 3-4: Isothermal compression with Q34 = -815.8 kJ
Process 4-1: Constant-volume heating
Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and
potential energy effects, determine the thermal efficiency.
state
1
2
3
P (bar)
10
10
5
T (°C)
Tsat=179.9
v (m3/kg)
v1=0.1944
u (kJ/kg)
2583.6
160
v2= v3=0.3835
3231.8
2575.2
4
160
v4=v1
Sec 3.8: Appling the Energy Balance using the Property Tables
39
Example: (3.77) A system consisting of 1 kg or water undergoes a
power cycle composed of the following processes.
Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor.
W12   PdV  Pm dv  P m v2  v1 
V2
v2
V1
v1
105 N m 2
m3
kJ
1kg 0.3835 0.1944
W12  10bar
bar
kg 103 Nm
W12  189.1kJ
U12  Q12  W12  Q12  U12  W12  mu2  u1   W12
Q12  1kg 3231.8  2583.6 kJ
kg  189.1kJ  837.3kJ
Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Q23  U 23  W23  mu3  u2   0
Q23  1kg 2575.2  3231.8 kJ
kg  656.6kJ
Sec 3.8: Appling the Energy Balance using the Property Tables
40
Example: (3.77) A system consisting of 1 kg or water undergoes a
power cycle composed of the following processes.
Process 3-4: Isothermal compression with Q34 = -815.8 kJ
W34  Q34  U34  Q34  mu3  u2 
v4  v f
0.1944 1.102103
x4 

 0.6317
3
vg  v f
0.3071 1.10210
u4  u f  x4 ug  u f   674.86  0.63172568.4  674.86  1871kJ
kg
W34  Q34  mu3  u4 
 815.8kJ  1kg1871 2575.2 kJ
kg  111.6kJ
Process 4-1: Constant-volume heating
Q41  U 41  W41  mu1  u4   0
Q23  1kg 2583.6  1871 kJ
kg  712.6kJ
Sec 3.8: Appling the Energy Balance using the Property Tables
41
Example: (3.77) A system consisting of 1 kg or water undergoes a
power cycle composed of the following processes.
Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor.
Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Process 3-4: Isothermal compression with Q34 = -815.8 kJ
Process 4-1: Constant-volume heating
Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and
potential energy effects, determine the thermal efficiency.
Q (kJ)
W (kJ)
Process 1-2
837.3
189.1
Process 2-3
-656.6
0
Process 3-4
-815.8
1871
Process 4-1
712.6
0
Total
77.5
77.5
Qin  Q12  Q41
 837.3  712.6  1549.9kJ

Wcycle
Qin
77.5kJ

 0.05  5%
1549.9kJ
```