this some of notes from my lecturer. hope it can help students outsides there who take same course with me.

1. BMM 2513
THERMODYNAMICS 1
Prof Dr Hj Shahrani Bin Hj Anuar
Dr. Ftwi Yohaness Hagos
Dr. Thamir Khalil Ibrahim
1
Chapter 4. First law in closed system

2. This course focuses on the application of
thermodynamics fundamentals in various
engineering system including properties
of pure substance, perpetual motion
machine, first law, second law and
entropy.
COURSE SYNOPSIS
BMM2513 THERMODYNAMICS 1
2
Chapter-4
Topic 7. First law in closed system

3. BMM2513 THERMODYNAMICS 1
1. Examine the moving boundary work or P dV work commonly
encountered in reciprocating devices such as automotive engines and
compressors.
2. Identify the first law of thermodynamics as simply a statement of the
conservation of energy principle for closed (fixed mass) systems.
3. Develop the general energy balance applied to closed systems.
4. Define the specific heat at constant volume and the specific heat at
constant pressure.
5. Relate the specific heats to the calculation of the changes in internal
energy and enthalpy of ideal gases.
6. Describe incompressible substances and determine the changes in
their internal energy and enthalpy.
7. Solve energy balance problems for closed (fixed mass) systems that
involve heat and work interactions for general pure substances, ideal
gases, and incompressible substances.
OBJECTIVES
Topic 7. First law in closed system
Chapter-4

4. LECTURE SCHEDULE
WEEK TOPIC
1 Introduction and Basic Concept
2 System, boundary, work, heat
3 Energy & general energy analysis
4 Energy and environment impact
5 Pure substances, gas equations
6 Property table and chart
7 First law in closed system
WEEK TOPIC
8 First law steady flow system
9 Second law, heat engine, PMM
10 Second law, heat pump
11 Second law, Carnot cycle principles
12 Entropy, entropy relationships
13 Entropy, Isentropic efficiency
14 Presentation
BMM2513 THERMODYNAMICS 1
4
Chapter-4
Topic 7. First law in closed system

5. i. Definitions of Thermodynamics
ii. Basic Applications of Thermodynamics
iii. System, Boundary and Surrounding
iv. Closed & Open System
v. Properties, Intensive and Extensive Properties
vi. Equilibrium And Quasi-static Equilibrium
vii. State, Path, Process and Cycle
viii. Simple Compressible Substance
ix. Pressure, atmospheric, gauge, absolute
x. Temperature and Zeroth Law Of Thermodynamics
BMM2513 THERMODYNAMICS 1
Thermodynamics concept
Recaptulate
Chapter-2: Energy, energy transfer & general energy analysis

6. v. Properties, Intensive and Extensive Properties
 Properties are any measurable characteristics of a system. eg.
Pressure p, temperature T, volume V, mass m and density ρ.
 Extensive properties are the mass-dependent properties of a
system. i.e. the properties that will vary proportionally with
mass of the system. e.g. volume V
 Intensive properties are the properties that are not dependent
on mass. e.g. temperature T, density ρ.
NB: If any Extensive Property is divided by the mass we would
also obtain an intensive property.
BMM2513 THERMODYNAMICS 1
Thermodynamics concept
Chapter-2: Energy, energy transfer & general energy analysis
Recaptulate

7. FIRST LAW OF THERMODYNAMICS
Energy can be neither created nor destroyed during a process but
can change forms
• Principle of energy conservation
• Sound basis for studying the relationships among the various
forms of energy and energy interactions
• All adiabatic processes between two specified states of a closed
system, the net work done is the same regardless of the nature
of the closed system and the details of the process.
BMM2513 THERMODYNAMICS 1
Essentially
W
Q
ΔU
Single Process
Q – W = ΔU
W1
W2
W3
Q1
Q2
Cyclic Process
ΣQ – ΣW = 0
ΣQ = ΣW
ΔU = 0
Topic 7. First law in closed system
Chapter-4

8. BMM2513 THERMODYNAMICS 1
4.1 Moving Boundary Work
Moving boundary work (P dV work)
The expansion and compression work in a
piston-cylinder device.
Work associated
with a moving
boundary is called
boundary work.
Quasi-equilibrium process
A process during which the system
remains nearly in equilibrium at all times.
Wb is positive  for expansion
Wb is negative  for compression
Gas does a differential
amount of work Wb
as it forces the piston
to move by a
differential amount ds.
Topic 7. First law in closed system
Chapter-4

9. BMM2513 THERMODYNAMICS 1
4.1 Moving Boundary Work
The boundary work done
during a process depends
on the path followed as
well as the end states.
The area under the
process curve on a P-V
diagram is equal, in
magnitude, to the work
done during a quasi-
equilibrium expansion
or compression process
of a closed system.
Cyclic Process
Topic 7. First law in closed system
Chapter-4

10. BMM2513 THERMODYNAMICS 1
4.1 Moving Boundary Work
Polytropic, Isothermal, and Isobaric processes
where : C = constants
n = polytropic exponent)
Polytropic process
Polytropic process schematic and P-V diagram
Question
What is the boundary work for
constant-volume process?
n = ∞
n = 0
n > 1
n = 1
P
V
Topic 7. First law in closed system
Chapter-4

11. Polytropic Process According Relationship PV
n
= C
=
𝑃2 𝑉2
𝑛
𝑉2
1−𝑛
− 𝑃1 𝑉1
𝑛
𝑉1
1−𝑛
1 − 𝑛
𝑊12 =
𝑃2 𝑉2 − 𝑃1 𝑉1
1 − 𝑛
=
𝑚𝑅(𝑇2 − 𝑇1)
1 − 𝑛
; 𝑛 ≠ 1
𝑊12 =
1
2
𝑃𝑑𝑉 = 𝐶
1
2
𝑑𝑉
𝑉 𝑛 = 𝑃1 𝑉1
𝑛
1
2
𝑑𝑉
𝑉 𝑛 =
𝑃1 𝑉1
𝑛
𝑉2
1−𝑛
− 𝑉1
1−𝑛
1 − 𝑛
Boundary Displacement Work from state 1 to state 2 is given,
BMM2513 THERMODYNAMICS 1
4.1 Moving Boundary Work
Topic 7. First law in closed system
Chapter-4

12. 𝑃𝑉
𝑇 1
=
𝑃𝑉
𝑇 2
𝑃1
𝑃2
=
𝑉2
𝑉1
𝑇2 = 𝑇1
isothermal
𝑊12 = 𝑷 𝟏 𝑽 𝟏 𝒍𝒏
𝑽 𝟐
𝑽 𝟏
= 𝒎𝑹𝑻 𝟏 𝒍𝒏
𝑽 𝟐
𝑽 𝟏
KJ 𝑊12 = 𝑷 𝟏 𝑽 𝟏 𝒍𝒏
𝑷 𝟏
𝑷 𝟐
= 𝒎𝑹𝑻 𝟏 𝒍𝒏
𝑷 𝟏
𝑷 𝟐
KJ
𝑊12 =
1
2
𝑃𝑑𝑉 =
1
2
𝐶
𝑉
𝑑𝑉 = 𝐶 𝑙𝑛
𝑉2
𝑉1
= 𝑷 𝟏 𝑽 𝟏 𝒍𝒏
𝑽 𝟐
𝑽 𝟏
KJ
Isothermal Process According Relationship PV = C
Boundary Displacement Work from state 1 to state 2 is given,
For ideal gas
BMM2513 THERMODYNAMICS 1
4.1 Moving Boundary Work
Topic 7. First law in closed system
Chapter-4

13. Isobaric Process According Relationship P = C (n = 0)
𝑾 𝟏𝟐 =
1
2
𝑃𝑑𝑉 = 𝑃
1
2
𝑑𝑉
Boundary Displacement Work from state 1 to state 2 is given,
BMM2513 THERMODYNAMICS 1
4.1 Moving Boundary Work
Topic 7. First law in closed system
Chapter-4
𝑾 𝟏𝟐 = 𝑷 𝑽 𝟐 − 𝑽 𝟏 KJ

14. Isochoric Process According Relationship V = C (n = ∞)
Boundary Displacement Work from state 1 to state 2 is given,
𝑾 𝟏𝟐 =
1
2
𝑃𝑑𝑉 = 𝟎 KJ
Constant volume or isochoric process, 𝑑𝑉 = 0
BMM2513 THERMODYNAMICS 1
4.1 Moving Boundary Work
Topic 7. First law in closed system
Chapter-4

15. BMM2513 THERMODYNAMICS 1
4.2 Energy Balance For Closed Systems
Energy balance for any system
undergoing any process
Energy balance in the rate form
The total quantities are related to the quantities per unit time is
Energy balance per unit mass basis
Energy balance in differential form
Energy balance for a cycle
Topic 7. First law in closed system
Chapter-4

16. BMM2513 THERMODYNAMICS 1
4.2 Energy Balance For Closed Systems
Net Heat Input
Net Work Input
First-law relation for closed systems
when sign convention is used
The first law cannot be proven mathematically, but no process in nature is known to have
violated the first law, and this should be taken as sufficient proof.
Energy balance sign convention
Heat Input & Work Output +ve
Heat Output & Work Input -ve
Topic 7. First law in closed system
Chapter-4

17. BMM2513 THERMODYNAMICS 1
4.2 Energy Balance For Closed Systems
NB: The first law cannot be proven mathematically, but no process in nature is known to
have violated the first law, and this should be taken as sufficient proof.
Topic 7. First law in closed system
Chapter-4
Net Work and Net Heat in Cyclic Processes
When any closed system is taken through a cycle, the net work
delivered to the surroundings is equal to the net heat taken from
the surroundings .i.e 𝑄 = 𝑊
Process 1 − 2: 𝑄12 − 𝑊12 = 𝑈2 − 𝑈1
Process 2 − 3: 𝑄23 − 𝑊23 = 𝑈3 − 𝑈2
Process 3 − 1: 𝑄31 − 𝑊31 = 𝑈1 − 𝑈3
Adding for net value 𝑄 − 𝑊 = ∆𝑈 = 0
Consider an arbitrary cycle an shown
Σ𝑄 = Σ𝑊 ΣΔ𝑈 = 0

18. 4.2 Energy Balance For Closed Systems
Constant-pressure Expansion/Compression Process
General analysis for a closed system undergoing a quasi-equilibrium
constant-pressure process. Q is to the system and W is from the system.
H2 H1
where
𝑄 − 𝑊𝑜𝑡ℎ𝑒𝑟 = 𝐻2 − 𝐻1 𝑸 = 𝑯 𝟐 − 𝑯 𝟏
NB: Microscopic
Wother ≈ 0
BMM2513 THERMODYNAMICS 1
Topic 7. First law in closed system
Chapter-4

19. BMM2513 THERMODYNAMICS 1
4.2 Energy Balance For Closed Systems
Constant-pressure Expansion/Compression Process
HWU b 
For a constant-pressure expansion or compression process
An example of constant-pressure process
Topic 7. First law in closed system
Chapter-4

20. BMM2513 THERMODYNAMICS 1
4.2 Energy Balance For Closed Systems
Topic 7. First law in closed system
Chapter-4

21. BMM2513 THERMODYNAMICS 1
4.3 Specific Heats
Specific heat at constant volume, cv
The energy required to raise the temperature of the unit mass of a substance by
one degree as the volume is maintained constant
Specific heat at constant pressure, cp
The energy required to raise the temperature of the unit mass of a substance by
one degree as the pressure is maintained constant
Constant-volume and constant-pressure
specific heats cv and cp (helium gas).
Topic 7. First law in closed system
Chapter-4

22. BMM2513 THERMODYNAMICS 1
4.3 Specific Heats
 The equations in the figure are valid for any
substance undergoing any process.
 cv and cp are properties.
 cv is related to the changes in internal
energy and cp to the changes in enthalpy.
 A common unit for specific heats is kJ/kg·°C
or kJ/kg·K. Are these units identical?
True or False?
cp is always greater than cv
Formal definitions of cv and cp.
Topic 7. First law in closed system
Chapter-4

23. BMM2513 THERMODYNAMICS 1
4.4 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases
23
Joule showed using this experimental
apparatus that u=u(T)
For ideal gases, u, h, cv, and cp vary with
temperature only.
Internal
energy and
enthalpy
change of
an ideal gas
Topic 7. First law in closed system
Chapter-4

24. BMM2513 THERMODYNAMICS 1
4.4 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases
Ideal-gas constant-pressure specific heats for
some gases (Table A–2c for cp equations).
In ideal-gas tables, 0 K is chosen as
the reference temperature.
 At low pressures, all real gases approach ideal-gas behavior, and therefore their
specific heats depend on temperature only.
 The specific heats of real gases at low pressures are called ideal-gas specific heats, or
zero-pressure specific heats, and are often denoted cp0 and cv0.
 u and h data for a number of gases have been tabulated.
 These tables are obtained by choosing an arbitrary
reference point and performing the integrations by
treating state 1 as the reference state.
Topic 7. First law in closed system
Chapter-4

25. BMM2513 THERMODYNAMICS 1
4.4 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases
Internal energy and enthalpy change when specific heat is taken constant
at an average value
(kJ/kg)
(kJ/kg)
Topic 7. First law in closed system
Chapter-4

26. BMM2513 THERMODYNAMICS 1
4.4 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases
1. Easiest and accurate way is by using
tabulated data of u and h if available
2. From cv or cp relations (Table A-2c) as a
function of temperature and the
integrate. It is inconvenient for hand
calculations but quite desirable for
computerized calculations. The results
obtained are very accurate.
3. Using average specific heats is very simple
and certainly very convenient when
property tables are not available.
Reasonably accurate if the temperature
interval is not very large.
Three ways of calculating u and h
Three ways to calculate u
Topic 7. First law in closed system
Chapter-4

27. Relationship between Specific Heats and Gas
Constant (Cp Cv R)
From definition of enthalpy and substitute gas
equation,
ℎ = 𝑢 + 𝑝𝑣 = 𝑢 + 𝑅𝑇
Its derivative with respective to temperature T
becomes
𝑑ℎ
𝑑𝑇
=
𝑑𝑢
𝑑𝑇
+ 𝑅
Substitute definition for specific heats, becomes
𝑐 𝑝 = 𝑐 𝑣 + 𝑅 𝑘𝐽 𝑘𝑔. 𝐾
Molar basis
𝑐 𝑝 = 𝑐 𝑣 + 𝑅 𝑢 𝑘𝐽 𝑘𝑚𝑜𝑙. 𝐾
Specific heat ratio 𝑘 =
𝑐 𝑝
𝑐 𝑣
> 1.0
Specific ratio varies slightly with
temperature
 1.667 for monatomic gases He, Ar
 1.4 for many diatomic gases e.g. air
at room temperature
cp of ideal gas determined from cv , R.
BMM2513 THERMODYNAMICS 1
4.4 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases
Topic 7. First law in closed system
Chapter-4

28. BMM2513 THERMODYNAMICS 1
4.5 Internal Energy, Enthalpy and Specific Heats of Solids/Liquids
Substance whose specific volume (or density) is constant
Examples: solids and liquids
Incompressible Substance
Topic 7. First law in closed system
Chapter-4

29. BMM2513 THERMODYNAMICS 1
4.5 Internal Energy, Enthalpy and Specific Heats of Solids/Liquids
Incompressible Substance
Internal Energy Changes
Topic 7. First law in closed system
Chapter-4

30. BMM2513 THERMODYNAMICS 1
4.5 Internal Energy, Enthalpy and Specific Heats of Solids/Liquids
Incompressible Substance
Enthalpy Changes
Enthalpy of compressed liquid
Definition
Derivative
Solid vΔP ≈ 0; Tcuh avg
Liquids
Constant-pressure process in heaters (ΔP = 0)
Constant-Temperature in pumps (ΔT = 0)
Tcuh avg
Pvh 
More accurate than
Enthalpy Change
Topic 7. First law in closed system
Chapter-4

31. 1. Moving boundary work
i. Wb for an isothermal process
ii. Wb for a constant-pressure process
iii. Wb for a polytropic process
2. Energy balance for closed systems
i. Energy balance for a constant-pressure expansion or
compression process
3. Specific heats
i. Constant-pressure specific heat, cp
ii. Constant-volume specific heat, cv
4.Internal energy, enthalpy, and specific heats of ideal gases
i. Specific heat relations of ideal gases
5. Internal energy, enthalpy, and specific heats of incompressible
substances (solids and liquids)
BMM2513 THERMODYNAMICS 1
SUMMARY
Topic 7. First law in closed system
Chapter-4

32. BMM2513 THERMODYNAMICS 1
Example 1: Gas in piston-cylinder device
Calculate the work done when a gas expands from 350 kPa and 0.03 m3 to a final
volume of 0.2 m3 according to the relationship PV1.5 = C.
Chapter 4. First law in closed system

33. BMM2513 THERMODYNAMICS 1
Example 2: Gas in piston-cylinder device
Calculate the change in specific volume of air when compress isothermally from 150
kPa and 20oC to 1100 kPa.
Chapter 4. First law in closed system

34. BMM2513 THERMODYNAMICS 1
Example 3: Isothermal expansion of steam in cylinder-piston assembly
A mass of 1 kg of saturated dry steam is expanded isothermally in a cylinder-piston
assembly from 40 bar (4 MPa) to 2 bar (0.2 MPa). Sketch the process in a clearly
labeled P-V diagram and determine:-
a) change in enthalpy in kJ,
b) change in internal energy in kJ, and
c) work done in kJ if the heat supplied to the steam is 857 kJ.
Chapter 4. First law in closed system

35. BMM2513 THERMODYNAMICS 1
Example 3: Isothermal expansion of steam in cylinder-piston assembly (cont’d)
Chapter 4. First law in closed system

36. BMM2513 THERMODYNAMICS 1
Example 4: Isothermal expansion of steam in cylinder-piston assembly
A piston–cylinder device contains 25 g of saturated water vapor that is maintained at a constant
pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of
0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. Determine
the final temperature of the steam.
Chapter 4. First law in closed system
We

37. BMM2513 THERMODYNAMICS 1
Example 4: Isothermal expansion of steam in cylinder-piston assembly (cont’d)
Chapter 4. First law in closed system
P2V2-P1V1

38. Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in
internal energy of air per unit mass, using (a) data from the air Table A–17, (b) the functional
form of the specific heat (Table A–2c), and (c) the average specific heat value Table A–2b.
BMM2513 THERMODYNAMICS 1
Example 4: Isothermal expansion of steam in cylinder-piston assembly (cont’d)
Chapter 4. First law in closed system
(a) Internal energy of air at T1 and T2 from Table A–17
Air cp(T) in Table A–2c expressed in third-degree polynomial