The document outlines a thermodynamics learning plan over six weeks, covering basic principles, the first law of thermodynamics, and ideal gas behavior. It includes various lesson topics, activities, and mathematical formulations related to force, mass, acceleration, pressure, and energy transformations. Assessment methods and specific calculations related to course content are also detailed.

1. ME-THER
THERMODYNAMICS

2. Learning Plan – Module 1
Week 1 – 6 ( 18 hrs Sept 1 – October 9)
Week 1 – 2 (6 hrs Sept 2 and Sept 9) - Grading policies, course
syllabus, expectations.
• 1 - Basic Principles, Concepts and Definitions
Week 3 – 4 (6 hrs Sept 16 and 23)
• 2 - First Law of Thermodynamics – Conservation of Energy
Week 5 – 6 (6 hrs Sept 30 and October 7)
• 3 - Ideal Gas
Week 5 Oct 7 W– Preliminary Examination

3. Learning Plan – Module 1
Week 1 – 2 (6 hrs Sept 2 – Sept 9)
Lesson 1 - Basic Principles, Concepts and Definitions
Thermodynamics is a branch of physical sciences that treats various
phenomena of energy and related properties of matter, especially
the laws of transformation of heat into other forms of energy and
vice versa.
Systems of units
Newton’s law states that acceleration of a body is directly
proportional to force acting on it and inversely proportional to mass.
a = k F/m F = ma / k k = ma / F
where k is a proportional constant

4. Learning Plan – Module 1
cgs system: 1 dyne accelerates 1 gram mass at 1cm/sec2
mks system: 1 newton accelerates 1 kilogram mass at 1 m/sec2
fps system: 1 pound-force accelerates 1 slug mass at 1 ft/sec2
1g 1 dyne 1kg 1 newton 1 slug 1 lbf
a a a
1 cm/sec2 1 m/sec2 1 ft/sec2
k = 1 g-cm / dyne-sec2 k = 1 kg-m/N-sec2 k = 1 slug-ft/lbf-sec2
Question: How many dynes in 1 newton, where 1N = 1 kg-m / sec2
Answer : 1 x 105 dynes

5. Learning Plan – Module 1
Acceleration values
1 gram force accelerates a 1 gram mass at 980.66 cm/sec2
1 kg force accelerates a 1 kg mass at 9.8066 m/sec2
1 lb force accelerates a 1 lb mass at 32.174 ft/sec2
1gm 1 gf 1kgm 1 kgf 1 lbm 1 lbf
a a a
980.66 cm/sec2 9.8066 m/sec2 32.174 ft/sec2
k= 980.66 gm-cm/gf -s2 k= 9.8066 kgm-m/ kgf -s2 k= 32.174 lbm-ft/lbf-s2
Question: If 1 N = 1 kg-m/sec2 and 1 kgf = 9.8066 kgm-m/sec2, how
many Newtons in 1 kgf ? Answer : 1 kgf = 9.8066 N

6. Learning Plan – Module 1
Solution: Relation between Kilogram force (Kgf) and Newton (N)
k = 1 kg-m/N-sec2
k = 9.8066 kg-m/ kgf - s2
Equating k’s: 1 kg-m/N-sec2 = 9.8066 kg-m/ Kgf - s2
Therefore, 1 Kgf = 9.8066 N
Or, 1 Kgf = 9.8066 kg-m/ s2 x 1 N / 1 kg-m/ s2 = 9.8066 N
Question: How many Pound mass (lbm) is 1 slug? Ans: 32.174 lbm
k = 32.174 lbm-ft/lbf - s2
k = 1 slug-ft/lbf - s2
Equating k’s: 1 slug -ft/lbf - s2 = 32.174 lbm- ft/lbf - s2

7. Learning Plan – Module 1
Acceleration
A unit of force is one that produces 1 unit acceleration in a body of
1 unit of mass. F = mass x acceleration
1g 1 dyne 1kg 1 newton 1 slug 1 lbf
a = 1 cm/sec2 a = 1 m/sec2 a = 1 ft/sec2
1 dyne = 1 g x 1 cm/sec2 1 lbm 1 poundal
1 Newton = 1 kg x 1 m/sec2
1 lbf = 1 slug x 1 ft/sec2 a = 1 ft/sec2
1 poundal = 1 lbm x 1 ft/sec2

8. Learning Plan – Module 1
Mass and Weight
The mass of a body is the absolute quantity of matter in it.
The weight of a body means the force of gravity on the body.
F = ma / k or
m / k = F / a = Fg / g
where g = acceleration due to gravity produced by force Fg
a = acceleration produced by another force F
At or near the surface of the surface of the earth, k and g are
numerically equal, so are mass and Fg
End of lesson 1 / week 1 – next Activity #1 / week 1 in
Asynchronous period

9. Learning Plan – Module 1
Activity #1 – Lesson 1 / Week 1 Due: Sept 9, 2020 (week 2)
Problems:
1. What is the weight ( Kgf ) of a 66 kg-mass man at standard
condition? m = 66 kgm g = 9.8066 m/s2
2. What is the mass ( lbm ) of an object at standard condition which
weighs 50 lb ? Fg = 50 lbf g = 32.174 ft/s2
3. Five masses where the acceleration due to gravity is 30.5 ft/s2
are as follows: m1 = 500 gram mass, m2 = weighs 800 gram
force, m3 = 15 poundals, m4 = 3 lbf and m5 = 0.01 slug mass.
What is the total mass expressed in a) grams b) pounds c) slugs
30% presentation = 30
70% content / solution – accuracy = 70

10. Learning Plan – Module 1
Week 2 (3 hrs Sept 9 )
Lesson 1 - Basic Principles, Concepts and Definitions
Specific Volume, Density and Specific Weight
The density, ρ of a substance is its mass (not weight) per unit volume
ρ = m / V (rho)
The specific volume, ν is the volume per unit mass.
ν = V / m = 1/ ρ ( nu)
The specific weight , ɣ of a substance is the force due to gravity per
unit volume. ɣ = Fg / V (gamma)
Since the specific weight is to local acceleration of gravity , the
density is to standard acceleration. ɣ / g = ρ / k

11. Learning Plan – Module 1
ɣ / g = ρ / k
ρ = ɣ k /g
ɣ = ρ g / k
At or near the surface of the earth, k and g are numerically equal, so
are ρ and ɣ
Example: What is the specific weight of water at standard
condition.
g = 9.8066 m/s2 ρ = 1000 kg / m3
ɣ = ρ g / k
= 1000 kg / m3 x 9.8066 m /s2
9.8066 kg- m/ kgf - s2
= 1000 kgf /m3

12. Learning Plan – Module 1
Pressure
The standard reference atmospheric pressure is: Po
760 mm Hg or
29.92 inches Hg at 32 oF or
14.696 psi absolute (psia) or 14.7 psia
1 atmosphere (atm)
Measuring Pressure
1. By using manometers
2. By using pressure gages

13. By using manometers

14. Learning Plan – Module 1
By using pressure gages

15. Learning Plan – Module 1

16. Learning Plan – Module 1
Gage pressure

17. Learning Plan – Module 1
Solution
pg = ρ g hg / k P = F / A = N/m2
2
= 9.65 m/s2 ( 1878 kg / m3) (30 m)
1 kg-m/ N- s2
= 543,680 N/m2 = 543.68 KPa (gage)
NOTE : 1 PASCAL = 1 N / m2
1 KILO PASCAL (kPa) = 1000 N / m2
Pabs = Po + Pg = Po + 543.68 kPa = Po + 543.68 kPa (abs)

18. Learning Plan – Module 1
Atmospheric Pressure

19. Learning Plan – Module 1
Problem 1: A vertical column of water will be supported to what
height by standard atmospheric pressure?
P = ɣ h

20. Learning Plan – Module 1
Atmospheric Pressure
Problem
Sp gr of Hg = 13.6
P = ɣ h
ɣHg = sg ɣw

21. Learning Plan – Module 1
Absolute Pressure

22. Learning Plan – Module 1
Absolute Pressure
Ans : 54.5 psia = 375.78 kPa

23. Learning Plan – Module 1
Absolute Pressure
1 lbf = 1N / 0.2248
k = 1
1

24. Learning Plan – Module 1
Absolute Pressure
1 lbf = 1N / 0.2248
2
1

25. Learning Plan – Module 1
Temperature

26. Learning Plan – Module 1
Temperature

27. Learning Plan – Module 1
Temperature

28. Learning Plan – Module 1
Conservation of Mass

29. Learning Plan – Module 1
Conservation of Mass
where ρ = 1/ ν
m =

30. Learning Plan – Module 1
Conservation of Mass

31. Learning Plan – Module 1
Conservation of Mass
m = V ρ
m = A Ʋ ρ

32. Learning Plan – Module 1
Conservation of Mass
ρ = m / V
h = V / A

33. Learning Plan – Module 1
ACTIVITY 2
SOLVE THE FOLLOWING PROBLEMS 1 TO 11 AS YOUR
ASYNCHRONOUS ACTIVITY FOR WEEK 3 (SEPT 16) AND WEEK
4 (SEPT 23)
DUE ON SEPT 25, 2020 FRIDAY 5PM

34. Learning Plan – Module 1

35. Learning Plan – Module 1

36. Learning Plan – Module 1

37. Learning Plan – Module 1
DUE ON SEPT 25, 2020 FRI 5PM

38. Learning Plan – Module 1
Week 3 – 4 - Sept 16 and 23
• Lesson 2 - First Law of Thermodynamics

39. Thermodynamics
General Engineering
OCM

40. OCM
Thermodynamics
General Engineering
 States that the energy of an isolated system
remains constant.
 Energy cannot be created or destroyed in an
isolated system.
 It can only be transformed from one form to
another.

41. OCM
Thermodynamics
General Engineering
 Energy or stored capacity for performing work
possessed by a moving body, by virtue of its
momentum = F x d = N - m = Joule
KE = ½ mV2 kg – m2/s2 = kg-m/s2 - m
ΔKE = K2 – K1 = ½ m (V2
2 – V1
2)
Where ΔKE = change in Kinetic energy
V1
m m V2

42. OCM
Thermodynamics
General Engineering
 Energy due to its position or elevation
PE = mgh = kg-m/s2 – m
= N – m = Joule
ΔPE = P2 – P1 = mg (h2 – h1)
Where ΔPE = change in potential energy
h
Fg = mg
2
1

43. OCM
Thermodynamics
General Engineering
 Energy stored within a body or substance by
virtue of the activity and configuration of its
molecules and of the vibration of the atoms
within the molecules.
ΔU = U2 – U1 (m mass)

44. OCM
Thermodynamics
General Engineering
 Energy that is transferred between two
systems by a virtue of a temperature
difference
 Q is positive when heat is added to the body.
 Q is negative when heat is rejected by the
body.

45. OCM
Thermodynamics
General Engineering
 Three mechanisms:
1. Conduction – transfer of energy from the
more energetic particles of a substance to the
adjacent less energetic ones as a result of
interaction between particles

46. OCM
Thermodynamics
General Engineering
 Three mechanisms:
2. Convection – transfer of energy between a
solid surface and the adjacent fluid that is in
motion. It includes combined effects of
conduction and fluid motion

47. OCM
Thermodynamics
General Engineering
 Three mechanisms:
3. Radiation – transfer of energy due to the
emission of electromagnetic waves or
photons

48. OCM
Thermodynamics
General Engineering
 Three mechanisms:

49. OCM
Thermodynamics
General Engineering
 Energy transfer associated with a force acting
through a distance. W = F x d = Joules
 W is positive when work is done by the system.
 W is negative when work is done on the system.
W = ʃ1
2 p dV = F x d ( N-m)
Where p s pressure (N/m2) or Pascal and dV is
differential Volume (m3) = W (N-m) or Joule

50. OCM
Thermodynamics
General Engineering
 Flow work or flow energy is work done in
pushing a fluid across a boundary, usually into
or out of a system.
 Wf = FL = pAL
 Wf = pV
 ΔWf = Wf2 – Wf1 = p2V2 – p1V1
 Δ Wf = change in flow work
L
Area of surface
F V
system
Flow Work
Boundary
p

51. OCM
Thermodynamics
General Engineering
 Characteristics of steady flow system
◦ There is neither accumulation nor diminution of
mass within the system
◦ There is neither accumulation nor diminution of
energy within the system
◦ The state of the working substance at any point in
the system remains constant

52. OCM
Thermodynamics
General Engineering
 Energy entering System = Energy Leaving System
 P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
 Q = (P2 - P1)+(K2 - K1)+ (Wf2 - Wf1)+(U2 - U1)+ W
 Q = ΔP+ ΔK + ΔWf + ΔU+ W
Where Q = Heat
P = Potential energy = mgh
K = Kinetic energy = ½ mV2
Wf = Flow work or Flow energy = pV = FL = pAL
U = Internal energy
W = Work = F d = pAL

53. OCM
Thermodynamics
General Engineering
P1
K1
Wf1
U1
Q
2
2
System
1
h1
h2
W
U2
Wf2
K2
P2
Datum level
1

54. OCM
Thermodynamics
General Engineering
 It is a thermodynamic quantity equivalent to the
total heat content of a system.
H = U + pV
Q = ΔU + W
Q = ΔH = H2 – H1
 The steady flow energy equation becomes
P1 + K1 + (Wf1+U1)+ Q = P2 + K2 + (Wf2+U2)+ W
P1 + K1 + H1 + Q = P2 + K2 + H2 + W
Q = ΔP+ ΔK + ΔH + W

55. OCM
Thermodynamics
General Engineering
1. During a steady flow process, the pressure of the working substance
drops from 200 to 20 psia. The speed increases from 200 to 1000 ft/sec.
The internal energy U of the open system decreases by 25 Btu/lb. the
specific volume increases from 1 to 8 ft3/lb. No heat is transferred.
a. Sketch an energy diagram.
b. Determine the Work per lb. Is it done on ( - ) or by (+) the substance?
c. Determine the Work in Hp for 10 lb per min. (1Hp = 42.4 Btu/min)
W
p1 = 200 psia p2 = 20 psia
v1 = 200 fps K1 K2 v2 = 1000 fps
ν1 = 1 ft 3 /lb Wf1 Wf2 ν2 = 8 ft 3 /lb
ΔU = -25 Btu/lb U1 U2 Q = 0
Energy System
P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
+
_

56. OCM
Thermodynamics
General Engineering
Assume 1 lbm of substance
K1= ½ mV2/k = (200 ft/sec)2/ 2 (32.174 lbm-ft/lbf-sec2) (778 lbf-ft/Btu)
= 0.80 Btu/lbm
K2 = ½ mV2/k = ( 1000) 2 / 2 (32.174) (778) = 19.97 Btu/lbm
Wf1 = p V = (200 lbf/in2) ( 144 in2/ft2) (1 ft3/lbm) / 778 ft-lbf/Btu
= 37.02 Btu/lbm
Wf2 = p V = 20 (144) (8) / 778 = 29.61 Btu / lbm
No heat is transferred, thus, Q = 0, assume same elevation, no ΔP
P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
K1 + Wf1 + U1 = K2 + Wf2 + U2 + W
0.8 + 37.02 = 19.97 + 29.61 + (U2 - U1 ) + W
0.8 + 37.02 = 19.97 + 29.61 + (- 25 ) + W
W = + 13.24 Btu / lbm (positive so W is done by the substance)
W = 13.24 Btu/lbm ( 10 lbm /min) / 42.4 Btu / min / Hp = 3.12 Hp

57. OCM
Thermodynamics
General Engineering
2. Steam is supplied to a fully loaded 100 hp turbine at 200 psia with U1 =
1163.3 Btu/lb, V1 = 2.65 ft3/lb and v1 = 400 ft/sec. Exhaust is at 1 psia with
U2 = 925 Btu/lb, V2 = 294 ft3/lb and v2 = 1100 ft/sec. The heat loss from
the steam in the turbine is 10 Btu/lb. Neglect Potential energy change,
determine ,
a. Work per lb of steam
b. Steam flow rate in lb/hour (1Hp = 2544 Btu/hr) = 42.4 x 60
P1 + K1 + Wf1 + U1 + Q = P2 + K2 + Wf2 + U2 + W
K1 + Wf1 + U1 + Q = K2 + Wf2 + U2 + W
K1= ½ mV2/k = 400 2 /2 (32.174)(778) = 3.2 Btu/lbm
K2 = ½ mV2/k = 1100 2 / 2 (32.174) (778) = 24.17 Btu/lbm
Wf1 = p V = (200) ( 144) (2.65) / 778 = 98.10 Btu/lbm
Wf2 = p V = (1) (144) (294) / 778 = 54.42 Btu/lbm
3.2 + 98.10 + 1163.3 + (- 10) = 24.17 + 54.42 + 925 + W
W = 251.01 Btu / lbm
Steam Flow = 100 Hp (2544 Btu/hr /Hp) / 251.01 Btu/lbm = 1014 lbm/hr

58. OCM
Thermodynamics
General Engineering
1. Steam enters a turbine with an enthalpy of 1292 Btu/lb
and leaves with an enthalpy of 1098 Btu/lb. The
transferred heat is 13 Btu/lb. What is the Work in
Btu/min and in Hp for a flow of 2 lb/sec. Ans. 512.3
Hp
2. It could be (1292-1098)-13 *2 = 362 BTU/sec ->21720 BTU/min
2. A thermodynamic steady flow system receives 4.56
kg/min of fluid (mass flow rate) where p1 = 137.9 kPa, ν1
= 0.0388 m 3/kg, v1 = 122 m/sec and U1 = 17.16
kJoules/kg. The fluid leaves the system at a boundary
where p2 = 551.6 kPa, ν2 = 0.193 m 3/kg, v2 = 183 m/sec
and U2 = 52.8 kJoules/kg. During passage through the
system, the fluid leaves 3,000 Joules/sec of Heat.
Determine the Work.
Ans. W = - 486 kJoules / min

60. OCM
Thermodynamics
General Engineering
 It is the quantity of heat required to change
the temperature of unit mass through one
degree.
 Ration of specific heats
k = cp/cv > 1

61. OCM
Thermodynamics
General Engineering
 Based on Joule’s Law,
ΔU = mcv (T2 – T1)
whether the volume remains constant or not.

62. OCM
Thermodynamics
General Engineering
 Based on Joule’s Law,
ΔH = mcp (T2 – T1)
whether the pressure remains constant or
not.

63. OCM
Thermodynamics
General Engineering
 H = U + pV
 pV = RT
 cp dT = cv dT + R dT
 cp = cv + R
 cv = R/ k – 1
 cp =kR/ k - 1

64. Learning Plan – Module 1
Week 5 – 6 Sept 30 and Oct 7
• Preliminary Examination – Oct 7 W
• Lesson 3 - Ideal Gas

65. Thermodynamics
General Engineering
OCM

66. OCM
Thermodynamics
General Engineering
 Vapor phase of a
substance when it
is above the critical
temperature
 Has a single
defined
thermodynamic
state at room
temperature
 A gas that is not far
from the state of
condensation
 Mixture of two
phases at room
temperature

67. OCM
Thermodynamics
General Engineering

68. OCM
Thermodynamics
General Engineering
Simply because it conforms to
the simple perfect gas laws!

69. OCM
Thermodynamics
General Engineering
 Air
 Nitrogen
 Oxygen
 Hydrogen
 Helium
 Argon
 Neon
 Krypton
 Heavier gases (carbon dioxide)

70. OCM
Thermodynamics
General Engineering
 Any equation that relates the pressure,
temperature and specific volume of a
substance.

71. OCM
Thermodynamics
General Engineering

72. OCM
Thermodynamics
General Engineering
 It is different for each gas.
 R = Ru/M
where: Ru = Universal gas constant
M = Molar mass/mol. Weight

73. OCM
Thermodynamics
General Engineering
 Same for all substances
 Ru Values: 8.314 kJ/kMol.K
8.314 kPa.m3/kmol.K
0.08314 bar.m3/kmol.K
1.986 Btu/lbmol.R
10.73 psia.ft3/lbmol.R
1545 ft.lbf/lbmol.R

74. OCM
Thermodynamics
General Engineering
 Mass of one mole, also called as gram-mole
(gmol), of a substance in grams
 m = MN
where: m = mass of system
M = molar mass
N = mole number

75. OCM
Thermodynamics
General Engineering
1. Determine the mass of the air in a room
whose dimensions are 4m x 5m x 6m at 100
kPa and 25 degC. Gas constant of air is 0.287
kPa.m3/kg.K.

76. OCM
Thermodynamics
General Engineering
2. The volume of a 7 x 13 ft tank is 543.3 cu.ft.
It contains air at 180 psig and 75 degF. How
many 1 cu.ft. drums can be filled to 60 psig
and 70 degF if its assumed that the air
temperature in the tank remains at 75 degF.
The drums have been sitting around in the
atmosphere which is at 14.7 psia and 70
degF.

77. OCM
Thermodynamics
General Engineering
3. Two vessels X and Y of different sizes are
connected by a pipe with a valve. Vessel X
contains 164 L of air at 3246.87 kPa, 90
degC. Vessel Y, of unknown volume, contains
air at 70.5 kPa, 5.6 degC. The valve is opened
and when the properties have been
determined, it is found that pm = 1500 kPa,
tm = 40.5 degC. What is the volume of vessel
B?

78. OCM
Thermodynamics
General Engineering
 Degree of disorder or randomness in the
system.
 It is the property of a substance which
remains constant if no heat enters or leaves
the substance, while it does work or alters its
volume, but which increases or diminishes a
small amount of heat enter or leave.
ΔS = mc ln(T2/T1)

79. OCM
Thermodynamics
General Engineering
4. For a certain ideal gas R = 25.8 ft.lb/lb.R
and k = 1.09,
a) What are the values of cp and cv ?
b) What mass of this gas would occupy a
volume of 15 cu.ft at 75 psia and 80 F?
c) If 30 BTU are transferred to this gas at
constant volume in (b), what are the
resulting temperature and pressure?

80. OCM
Thermodynamics
General Engineering
5. The specific heat of superheated steam at
approximately 150 kPa can be determined by
the equation
What is the enthalpy change between 300 ºC
and 700 ºC for 3 kg of steam? Compare with
the steam table (2565 kJ).

81. OCM
Thermodynamics
General Engineering
6. A 10 ft3 tank contains gas at a pressure of
500 psia, temperature of 85 degF and a
weight of 25 pounds. A part of the gas was
discharged and the temperature and pressure
changed to 70 degF and 300 psia,
respectively. Heat was applied and the
temperature was back to 85 degF. Find the
final weight, volume and pressure of the gas.

82. OCM
Thermodynamics
General Engineering
7. A motorist equips his automobile tires with
a relief-type valve so that the pressure inside
the tire will never willexceed 240 kPa (gage). He
starts a trip with a pressure of 200 kPa (gage) and
a temperature of 23 degC in the tires.During the
long drive, the temperature of the air in the tires
reaches 83 degC. Each tire contains 0.11 kg of air.
Determine (a) the mass of air escaping each tire, (b)
the pressure of the tire when the temperature
returns to 23degree Celsius.
Use R = 287.08 N.m/kg.K

83. OCM
Thermodynamics
General Engineering
8. An automobile tire contains 3730 cu in. of air at
32 psig and 80 degF. R = 53.342 lb.ft/lb.R
(a) What mass of air is in the tire?
(b) In operation, the air temperature increases to 145
degF. If the tire is inflexible, what is the resulting
percentage increase in gage pressure?
(c) What mass of the 145 degF air must be bled off to
reduce the pressure back to its original value?

84. OCM
Thermodynamics
General Engineering
Guidelines:
 With Front Page
 Margin: 1x1x1x1 (Visible)
 Engineering Lettering
 Problem Statement and Solution
 Box your final answers and round off up to two (2)
decimal places.
 CHAPTER 3: REVIEW PROBLEM
No. 6 (A 6 m3 tank…) and
No. 8 (A spherical balloon..)