1. EGR 334 Thermodynamics Chapter 3: Section 6-8
Lecture 07:
Enthalpy and Internal Energy
Quiz Today?

2. 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.
Reading Assignment:
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. or or

4. Specific Internal Energy (u) : energy / mass
Sec : Introducing Enthalpy
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

5. Saturated Steam Table Enthalpy, h
For the time being, we’re going to still ignore this last quantity.
Internal Energy, u
Notice there are additional columns on the steam table that we ignored last time.

6. What is enthalpy?
Enthalpy is defined in terms of internal energy and the work of expansion as
Extensive Form
Intensive Form
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 = deg C x = 30%
b) p = 10 bar T = 320 deg. C.
c) p = 10 bar T = 450 deg C.

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 = deg C x = 30%
from table A.3: at 10 bar the saturated temperature is which means the substance is a mixture of liquid and vapor.

9. From table A-3 the following values may be directly read:
quality
spec. vol.
int. energy
enthalpy
[m3/kg]
[kJ/kg]
x = 30%
vf=
uf=761.68
hf=762.81
p=10 bar
vg=0.1944
ug=2583.6
hg=2778.1
therefore:

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 = m3/kg u = kJ/kg h = 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.

12. Linear Interpolation:* y2
yunknown
xknown x1 y1 x2

13. Your turn: Find the value for enthalpy at
T = 400 deg. C and p = 50 bar
at 400°C and p1=40 bar  h1 = kJ/kg
at 400°C and p2=60 bar  h2 = kJ/kg
= kJ/kg

14. 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:
For: T = 400 deg C.
p = 50 bar = 500 kPa
Enthalpy per unit mass:
h = kJ/kg

15. Computer Method 2: Interactive Thermodynamics (IT)
Software you can download to your computer from:
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

16. 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.
i) Start with superheated vapor table A-4
ii) Interpolate between 280 and 320 deg C.

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.
ii) Determine quality, x2

19. p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C.
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
ii) Read the values directly.

20. Repeat this work, but use IT instead.
Results are the same as shown yesterday.

21. Given Ammonia at 12°C and v = 0.1217 m3/kg determine P, u and h.
Sec 3.6.2: Retrieving u and h Data
Workout Problem
Given Ammonia at 12°C and v = m3/kg determine P, u and h.
a) Using Tables in Appendix:
b) Using IT:

22. Given Ammonia at 12°C and v = 0.1217 m-3/kg determine P, u and h.
Sec 3.6.2: Retrieving u and h Data
Workout Problem
Given Ammonia at 12°C and v = m-3/kg determine P, u and h.
a) Using Tables in Appendix:

23. Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h.
Sec 3.6.2: Retrieving u and h Data
Example:
Given Ammonia at 12°C and v= m-3/kg determine P, u and h.
Notice that
vf = and vg =
Since: vf < v < vg
the substance is a liquid-gas mixture
Saturation Properties are:
Tsat = 12 deg C Psat= kPa
Quality if found using:

24. Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h.
Sec 3.6.2: Retrieving u and h Data
Example:
Given Ammonia at 12°C and v= m-3/kg determine P, u and h.
with
vf = 1.608x10-3, vg = m3/kg

25. Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h.
Sec 3.6.2: Retrieving u and h Data
Example:
Given Ammonia at 12°C and v= m-3/kg determine P, u and h.
Internal Energy:
Enthalpy:
Check:

26. b) Using IT:

27. Initial temperature, in °F, Final pressure, in psi, Work, in BTU
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
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
Pi = 20 psi mf = 0.07 lb ml = 0.07 lb

29. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Sec 3.8: Appling the Energy Balance using the Property Tables
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 = 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.

31. End of Slides for Lecture 07

32. Initial temperature, in °F, Final pressure, in psi, Work, in BTU
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
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
Pi = 20 psi mf = 0.07 lb ml = 0.07 lb

33. Initial temperature, in °F, Final pressure, in psi, Work, in BTU
Sec 3.8: Appling the Energy Balance using the Property Tables
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
Pi = 20 psi mf = 0.07 lb ml = 0.07 lb

34. Initial temperature, in °F, Final pressure, in psi, Work, in BTU
Sec 3.8: Appling the Energy Balance using the Property Tables
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
Pi = 20 psi mf = 0.07 lb ml = 0.07 lb xi = 0.5 a)
b) T = Tsat = °F

35. Initial temperature, in °F, Final pressure, in psi, Work, in BTU
Sec 3.8: Appling the Energy Balance using the Property Tables
Volume occupied, in ft3,
Initial temperature, in °F,
Final pressure, in psi,
Work, in BTU
Pi = 20 psi mf = 0.07 lb ml = 0.07 lb xi = 0.5 c)
Interpolate values to find P2 = 42 psi & u2 = 1093 BTU/lb

36. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Sec 3.8: Appling the Energy Balance using the Property Tables
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 = 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.

37. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Sec 3.8: Appling the Energy Balance using the Property Tables
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 = 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.

38. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Sec 3.8: Appling the Energy Balance using the Property Tables
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 = 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 4
P (bar) 10 5
T (°C)
Tsat=179.9
160
v (m3/kg)
v1=0.1944
v2=
v3=0.3835
v4=v1
u (kJ/kg)
2583.6
3231.8
2575.2

39. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Sec 3.8: Appling the Energy Balance using the Property Tables
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

40. Process 3-4: Isothermal compression with Q34 = -815.8 kJ
Sec 3.8: Appling the Energy Balance using the Property Tables
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 = kJ
Process 4-1: Constant-volume heating

41. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Sec 3.8: Appling the Energy Balance using the Property Tables
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 = 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
Process 3-4
-815.8
1871
Process 4-1
712.6
Total
77.5