COLLEGE OF ENGINEERING
CENTRAL PHILIPPINE UNIVERSITY
ILOILO CITY, PHILIPPINES
Tel. Nos. (033) 3291971 (to 79) Local 1085
Fax No. (033) 3203004
Email: chem._engg@cpu.edu.ph
MODULE 2
FIRST LAW OF THERMODYNAMICS
Topic 1 – Introduction
Thermodynamics is both a branch of physics and an engineering science. Engineers are
generally interested in studying systems and how they interact with their surroundings
✓ it is a science which deals with the transformation of all kinds of energy from one form to
another
✓ it is the study of the energy changes accompanying physical and chemical transformations
Examples of these transformations are:
1.
Electrical power generation - converting heat into electrical work
2.
Air conditioning - converting electrical work to cooling
3.
Automotive transportation - converting work into EK
To facilitate this, engineers extend the subject of thermodynamics to the study of systems
through which matter flows. Engineers use principles drawn from thermodynamics and other
engineering sciences (fluid mechanics and heat and mass transfer) in order to analyze and design
things to meet human needs.
Engineers seek to achieve improved design and better performance as measured by factors
such as:
1.
An increase in output of some desired product
2.
A reduced input of a scarce resource
3.
A reduction in total cost
4.
A lesser environmental impact
Fundamental Quantities – discussed in Physics
1.
Time
2.
Length
3.
Mass /Weight
4.
Force
5.
Temperature
•
Temperature defined by James Clerk Maxwell – the body’s thermal state with
reference to its ability to communicate heat to other bodies
•
Temperature is also defined as the measure of coldness and hotness of a body
o An intensive thermodynamic property used to indicate the amount of energy
within the molecules of the substance
2 Arbitrary Scales are Commonly Used
Fahrenheit Scale [Gabriel Fahrenheit (1686 – 1736)]
𝑡℉ =
9
(𝑡 ) + 32
5 ℃
Celsius Scale [Anders Celsius (1701 – 1744)]
𝑡℃ =
5
(𝑡 − 32)
9 ℉
Secondary Quantities
A. Volume - depends on the amount of material considered
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑣𝑜𝑙𝑢𝑚𝑒 =
B. Pressure =
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
; 𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑒
𝑛𝑜𝑟𝑚𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑒𝑥𝑒𝑟𝑡𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
@ STP: P = 14.7 lbf /in2 = 29.92 in Hg = 101.325 N/m2 or Pa
➢ 1 torr = the pressure equivalent of 1 mmHg at 0oC = 133.322 Pa
➢ 1 bar = 105 Pa
Pressure can be expressed as:
1. gauge pressure - pressure of a substance or system measured by a pressure
gauge
𝑃𝑔𝑎𝑢𝑔𝑒 → to express this to absolute units: 𝑃𝑎𝑏𝑠 = ±𝑃𝑔𝑎𝑢𝑔𝑒 + 𝑃𝑎𝑡𝑚
-
(+) is used when the 𝑃𝑔 > 𝑃𝑎𝑡𝑚 or surrounding P of the system
-
(–) is used when the 𝑃𝑔 < 𝑃𝑎𝑡𝑚
The negative of the gauge pressure is also called “vacuum pressure”
2. atmospheric pressure - pressure of the atmosphere at a certain location (i.e. @
the surface of the earth near sea level)
𝑏𝑎𝑟𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ≈ 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
3. fluid pressure or hydrostatic pressure - the force exerted by a column of fluid
per unit area
𝑃=
𝐹𝑔
𝑚𝑔
𝜌𝐴ℎ𝑔
𝜌ℎ
=
=
= ( ) ℎ = 𝛾ℎ
𝐴
𝑘𝐴
𝑘𝐴
𝑘
𝑚
Since: 𝜌 = 𝑉 ; 𝑚 = 𝜌𝑉; 𝑏𝑢𝑡 𝑉 = 𝐴ℎ, 𝑡ℎ𝑢𝑠, 𝑚 = 𝜌𝐴ℎ
where:
m = mass of the column of liquid
k = proportionality constant
V = volume of the column of fluid
Fg = force of gravity or the weight of the column of fluid
Instruments used:
dead weight gauge:
𝐹
𝑃= 𝐴=
𝑚𝑔
𝐴
where:
m = mass of the piston, pan, and weights
g = local acceleration due to gravity
A = cross-sectional area
C. Work is done whenever force acts through a distance:
𝑊 =𝐹x𝑑
For a differential change:
𝑉
𝑑𝑊 = 𝐹𝑑𝐿 = 𝑃𝐴(𝑑 𝐴), since A is constant
∫ 𝑑𝑤 = ∫ 𝑃𝑑𝑉
𝑉2
𝑊 = ∫ 𝑃𝑑𝑉
𝑉1
where: W = work done as a result of a finite process
P
P2
(P2, V2)
P1
(P1, V1)
V2
V1
V
If P increases and V decreases, the work done is a compression process.
If P decreases and V increases, the work done is an expansion process.
D. Energy
The general principle of conservation of energy was established about 1850. The
origin of this principle as it applies to Mechanics was implicit in the work of Galileo (1564
– 1642) and Isaac Newton (1642 – 1726).
There is no concept of work as the product of force × displacement that existed not
until in 1826 when it was introduced by J.V. Poncelet at the suggestion of G.G. Coriolis, a
French Engineer.
Types of energy commonly used in the study of physics are;
1.
energy of motion – work of accelerating the body (Ek) which was introduced
in 1856 by an English Physicist named William Thomson also known as “Lord
Kelvin”
2.
energy of position – work required to raise or elevate the body (Ep) which is
proposed in 1853 by William Rankine, a Scottish Engineer
Applying the principles of conservation of mechanical energy to a free-falling
body, assuming the mechanical processes are frictionless;
∆𝐸𝑘 + ∆𝐸𝑝 = 0
E. Heat
It is a fact that a hot object brought into contact with a cold object becomes warmer,
so we can say that something is transferred from the hot object to the cold one and that
something is called heat.
The theory of which is generally accepted until the middle of the 19th century was
that heat is a weightless and indestructible substance called caloric.
It always flows from a higher temperature to a lower one. The rate of heat transfer
from one body to another is proportional to the temperature difference between 2 bodies.
•
If the temperature difference is big, the rate of heat transfer is also fast. In a
thermodynamic sense, heat is never regarded as being stored within a body. Like
work, it exists only as energy in transit from one body to another or between a
system and a surrounding.
•
When energy in the form of heat is added to a body, it is stored not as heat but as
kinetic, EK and potential, EP energies.
Mr. Joule, in his experiment, was able to show a quantitative relationship that exists
between work and heat. His experiments were in fact, fundamental to understanding of the
first law of thermodynamics.
The question arises as to what happens to the energy between the time it is added
to the water in the form of work and the time it was extracted as heat.
Logic suggests that the energy appeared in another form and is known as internal
energy.
When Mr. Joule stirred the liquid, he observed that the temperature is increased, so
he concluded that the substance has internal energy which caused the temperature to rise.
Stir
H2O
Energy in the form of
heat was extracted
from the system
Energy in the form
of work was added
to the system
(System)
Topic 2 – The Law of Conservation of Energy
The Law of Conservation of Energy states that although energy appears or assumes in
many forms, the total quantity of energy is constant and when energy disappears in one form, it
appears simultaneously in other forms or the total energy of any system and its surroundings
is conserved.
In our study of thermodynamics, we need to know how the following terms are defined:
System → any object, any quantity of matter, any region selected for study and set apart
(mentally) from everything else.
Surroundings → anything not included in the system
Boundary → envelope which encloses the system and separates it from the surroundings
and it may be real, imaginary, rigid or non-rigid.
BOUNDARY
SYSTEM
SURROUNDINGS
Three Types of System
1. Open system – is one which permits matter or energy or both to be exchanged
with its surroundings
2. Closed system – is one which permits only energy and not matter to be
exchanged with its surroundings
3. Isolated system – cannot exchange matter or energy
Example of a closed system
Gas is the system,
cylinder & movable
piston are the
boundaries
Piston
Gas
Cylinder
Example of an open system
Heat
Fuel, air in to
generate heat
ENGINE
Work
Control surface
Exhaust gases out
to perform work
Phase is a homogeneous, physically distinct and mechanically separable portion of the system.
When we say mechanically separable, it means that it can be separated from every phase
by operations such as filtration, sedimentation or decantation.
A system can be homogeneous if it contains only one phase and it is heterogeneous, if it
contains more than a single phase
2 Types of Properties:
1. Extensive properties
Properties that are dependent on the size of the system considered or the amount of
substance present. Examples are volume, internal energy, or mass.
2. Intensive properties
Properties that are independent on the size of the system considered but depends
instead on the concentration of the substance present. Examples are temperature, density,
pressure, molar mass, molar volume).
Thermodynamic functions are of 2 types
1. Point function (or state function)
- quantity which does not depend on the past history of the substance or a quantity
which doesn’t depend on the path it has followed in reaching a given state
Examples are temperature, pressure and internal energy
2. Non - state function
- quantity which depends on the path it has followed
Examples are heat and work
If a change of state had occurred in the system by means of an operation/s, then the system
had undergone a process. A process is reversible when its direction can be reversed to trace the
same path without requiring any additional change in the property produced by the forward force,
otherwise, it is irreversible.
Steps to be followed in solving thermodynamic problems:
1) Indicate the system (Ex. 1lbm of liquid H2O / 1 kg. mole of ideal gas)
2) Specify the unique feature (Ex. isometric process)
3) Draw your diagram
4) Make necessary assumption (Ex. Take the temperature as constant or no heat is exchanged
between system and surrounding or assume no tide nor turbulence has occurred in the
process)
5) Then apply the law of thermodynamics
∆𝐸𝑠𝑦𝑠 + ∆𝐸𝑠𝑢𝑟𝑟 = 0
∆𝑢 + ∆𝐸𝑘 + ∆𝐸𝑝 = ±𝑄 ± 𝑊 (𝑓𝑜𝑟 𝑎 𝑛𝑜𝑛𝑓𝑙𝑜𝑤 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)
∆𝐻 + ∆𝐸𝑘 + ∆𝐸𝑝 = 𝑄 − 𝑊𝑠 (𝑓𝑜𝑟 𝑎 𝑓𝑙𝑜𝑤 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)
∆𝐻 = ∆𝑢 + ∆𝑃𝑉
Q is (+) if heat is added to the system
(–) if heat leaves the system
W is (+) if work is done by the system
(–) if work is done on the system
Substituting:
∆𝐸 + ∆𝐸𝑘 + ∆𝐸𝑝 = ±𝑄 ± 𝑊
In words, the 1st Law states that the total energy of the system is equal to the heat added minus
the work done by the system.
where:
∆𝐸 = change in internal energy
∆𝐸𝑝 = change in potential energy
∆𝐸𝑘 = change in kinetic energy
Q = heat
W = work
if ∆𝐸𝑝 and ∆𝐸𝑘 are negligible
For non-flow process:
∆𝐸 = 𝑄 − 𝑊, for finite changes
𝑑𝐸 = 𝑑𝑄 − 𝑑𝑊, for infinite changes
For any system:
SYSTEM
Wi
Wo
Work done on
the system
Work done by
the system
HE
Work done
against friction
Example 1.
a) Liquid 𝐻2 𝑂 @ 212 oF and 1 atm. What is its enthalpy?
Solution:
Use steam table
𝑈 = 180.02 𝐵𝑇𝑈/𝑙𝑏𝑚
𝑣 = 0.01672 𝑓𝑡 3 /𝑙𝑏𝑚
𝑙𝑏𝑓
𝐵𝑡𝑢
𝐻 = 180.02 𝑙𝑏𝑚 +
𝑖𝑛2
𝑓𝑡3
)
14.7 2 (144 2 )(0.01672
𝑙𝑏𝑚
𝑖𝑛
𝑓𝑡
778 𝑓𝑡−𝑙𝑏𝑓 /𝐵𝑡𝑢
𝐻 = 180.07 𝐵𝑡𝑢/𝑙𝑏𝑚
b) The water of part a) is brought to vapor state @ 400°F and 100 psia
Solution:
Use steam table
V = 4.937
ft 3
lbm
H 2 = 1228.4 Btu
lbm
u2 = H 2 − P2V2 = 1228.4 −
100 144  4.936
= 91.4 Btu
778
Calculate Δu and ΔH:
u = u2 − u1 = 14.7 psia
H = H 2 − H1 = 1048.14 Btu
Example 2.
a) Steam is admitted to the cylinder of a steam engine @ P = k of 300 psia. The bore of the
cylinder is 8 inches and the stroke of the piston is 12 inches. How much work is done
per stroke?
b) If 10 lbm of water is evaporated @ atmospheric pressure with a volume of 288.5 ft3 is
occupied, how much work is done?
Solution:
P is constant = 300 psia
a)
  8   12 
2
V = A L =
    = 0.348 ft
4  12   12 
W = P(V2 − V1 ) = 300
3
lbf
in 2

144
 0.348 ft 3 = _____ ft. lbf
2
in
ft
W = 15,033.6 ft. lbf
P is constant = 14.7 psia
b)
610,696.8 ft. lbf
W = 14.7  288.5 144 = _____
Heat Capacity and Specific Heat
Review:
Heat Energy is basically defined as the energy in motion. It moves from a body of high
temperature to a body of lower temperature.
Heat transfer or heat motion occurs only between bodies or systems mainly because of
temperature difference.
There are 3 modes of heat transfer:
a.
Conduction
b.
Convection
c.
Radiation
Example 1.
How much heat will be required to raise the temperature of 3 moles of 𝐻2 from 0°C to
100°C.
a) @ V = k
b) @ P = k
3
3
𝑐𝑎𝑙
𝑐𝑎𝑙
if 𝐶𝑣 = 2 𝑅 = 2 (1.987 𝑚𝑜𝑙−𝐾) = 3 𝑚𝑜𝑙−𝐾
𝑐𝑎𝑙
𝑐𝑎𝑙
𝐶𝑝 = 𝑅 + 𝐶𝑣 = (1.987 + 3) 𝑚𝑜𝑙−𝐾 = 4.987 𝑚𝑜𝑙−𝐾
Solution:
𝑐𝑎𝑙
a) ∆𝐸 = 𝑄 = 𝑛𝐶𝑣∆𝑇 = (3 𝑚𝑜𝑙𝑒𝑠) (3 𝑚𝑜𝑙−𝐾) (373 − 273)𝐾
= 900 𝑐𝑎𝑙
𝑐𝑎𝑙
b) ∆𝐻 = 𝑄 = 𝑛𝐶𝑝∆𝑇 = (3 𝑚𝑜𝑙𝑒𝑠) (4.987 𝑚𝑜𝑙−𝐾) (373 − 273)𝐾
= 1,496 𝑐𝑎𝑙
But if the molar heat capacity varies with absolute temperature, then the amount of heat is;
𝑇2
𝑄 = ∫ 𝑛𝐶𝑝𝑑𝑇 = ∆𝐻 @ 𝑃 = 𝑘
𝑇1
𝑇2
𝑄 = ∫ 𝑛𝐶𝑣𝑑𝑇 = ∆𝑈 @ 𝑉 = 𝑘
𝑇1
Example 2.
One mole of methane (considered to be a perfect gas) initially at 25°C and 1 atm pressure
is heated at P = k until the volume doubled. The variation of the molar heat capacity with
absolute temperature is 𝐶𝑝 = 5.34 + 11.5𝑥10−3 𝑇, where 𝐶𝑝 is in cal/mol-K. Calculate a) ∆𝐻
and b) ∆𝐸
Given: n = 1 mole
𝐶𝑝 = 5.34 + 11.5𝑥10−3 𝑇
𝑇1 = 25℃
𝑇2 = ?
𝑃1 = 1 𝑎𝑡𝑚
𝑃=𝑘
→
𝑃2 = 1 𝑎𝑡𝑚
𝑉1 = ?
𝑉2 = 2𝑉1
Required:
∆𝐻, ∆𝐸
Solution:
For 𝑉1 and 𝑉2: assuming ideal (perfect) gas
𝑃1 𝑉1 = 𝑛𝑅𝑇1
(1 𝑎𝑡𝑚)(𝑉1 ) = (1 𝑚𝑜𝑙𝑒) (0.08205
𝐿 − 𝑎𝑡𝑚
) (298.15𝐾)
𝑚𝑜𝑙 − 𝐾
𝑉1 = 24.46 𝐿
𝑉2 = 2𝑉1 = 2(24.46𝐿) = 48.93𝐿
For 𝑇2 : using Charles’ Law
𝑉1 𝑉2
=
𝑇1 𝑇2
24.46
48.93
=
298.15
𝑇2
𝑇2 = 596.42 𝐾
a) @ P = k
𝑇2
∆𝐻 = 𝑄 = 𝑛 ∫ 𝐶𝑝𝑑𝑇
𝑇1
596.42
= (1 𝑚𝑜𝑙𝑒) ∫298.15 (5.34 + 11.5𝑥10−3 𝑇)𝑑𝑇
∆𝐻 = 3,123.189 𝑐𝑎𝑙
b) @ P = k
𝑇2
∆𝐸 = ∫ (𝐶𝑝 − 𝑅)𝑑𝑇
𝑇1
𝑇
= ∫𝑇 2( 5.34 + 11.5𝑥10−3 𝑇 − 1.987)𝑑𝑇
1
∆𝐸 = 2,531 𝑐𝑎𝑙
Specific Heat, sp.
It is the amount of heat energy required to raise the temperature of a unit mass substance
by one degree of temperature.
𝑐𝑎𝑙 𝑐𝑎𝑙 𝐵𝑇𝑈 𝐵𝑇𝑈
The units can be 𝑔℃, 𝑔°𝐾, 𝑙𝑏℉ , 𝑙𝑏°𝑅
∆𝑈 + ∆𝐸𝑃 + ∆𝐸𝐾 = 𝑄 − 𝑊; 𝑉 = 𝑘
𝑊 = ∫ 𝑃𝑑𝑉 = 0
∆𝑈 = 𝑄 = ∫ 𝐶𝑣𝑑𝑇
Heat Capacity
𝐶𝑝 = 𝑠𝑝. ℎ𝑒𝑎𝑡 𝑥 𝑀𝑊
Two types:
1. Heat capacity @ constant V
2. Heat capacity @ constant P
@P=k
∆𝐻 = 𝑄 = ∫ 𝐶𝑝𝑑𝑇
∆𝐻 = ∆𝑈 + ∆𝑃𝑉 𝑜𝑟 𝐻 = 𝑈 + 𝑃𝑉
𝛾=
𝐶𝑝
;
𝐶𝑣
𝐶𝑝 − 𝐶𝑣 = 𝑅
Example 3.
A 1.5 kg gaseous nonflow closed system receives heat while the pressure remains
constant at 360 kPaa. The internal energy and temperature are increased 215 kJ and 65 K,
respectively. If the work done is 112 kJ, find a) Cp, b) ∆𝑉, c) Cv.
Given:
m = 1.5 kg
P = 360 kPaa, P = k
∆𝑈 = 215 𝑘𝐽
∆𝑇 = 65 𝐾
W = 112 kJ
Required:
a) Cp
b) ∆𝑉
c) Cv
Solution:
a) ∆𝑈 = 𝑄 − 𝑊
𝑄 = ∆𝑈 + 𝑊 = 215 + 112 = 327 𝑘𝐽
𝐶𝑝 =
∆𝐻
𝑄
327 𝑘𝐽
𝑘𝐽
=
=
= 3.35
𝑚∆𝑇 𝑚∆𝑇 1.5 𝑘𝑔(65)
𝑘𝑔 − 𝐾
𝑊
112
b) 𝑊 = ∫ 𝑃𝑑𝑉 ; ∆𝑉 = 𝑃 = 360
𝑊 = 0.311 𝑘𝐽
c) ∆𝑈 = 𝑚𝐶𝑣∆𝑇
215 𝑘𝐽 = (1.5𝑘𝑔)(𝐶𝑣)(65𝐾)
𝑘𝐽
𝐶𝑣 = 2.205
𝑘𝑔 − 𝐾
Example 4.
The specific heat of air is 𝐶𝑝 = 0.9167 + 2.577𝑥10−4 𝑇 − 3.974𝑥10−8 𝑇 2 where 𝐶𝑝 is
in kJ/kg-K. Find the change in enthalpy when the temperature is increased from 320 K to 1200
K.
Given:
𝐶𝑝 = 0.9167 + 2.577𝑥10−4 𝑇 − 3.974𝑥10−8 𝑇 2
𝑇1 = 320𝐾
𝑇2 = 1200𝐾
Required: ∆𝐻
Solution:
1200
∆𝐻 = ∫
(0.9167 + 2.577𝑥10−4 𝑇 − 3.974𝑥10−8 𝑇 2 )𝑑𝑇
320
∆𝐻 = 956.589 𝑘𝐽/𝑘𝑔
Example 5.
a) 5 moles of N2 @ 80oC is contained in a rigid vessel. How much heat must be added to the
system to raise its temp. to 300oC if the vessel has a negligible heat capacity?
b) If the mass of the vessel is 100 kg and if its heat capacity is 0.5 J/g oC, how much heat is
required?
c) 3 moles of N2 @ 230oC is contained in a piston/cylinder arrangement. How much heat
must be extracted from the system, which is kept at P = k, to cool it to 80oC if the heat
capacity of the piston and cylinder is neglected?
Solution:
Cp = 29.1
J
mol oC
; Cv = 20.8
J
mol oC
N 2 (5 moles ) @ 80o C ⎯
⎯→
300o C

a)
J 

o
u = Q = nCvT = 5 moles  20.8
(300 − 80) C = _____
o
mol C 

b) If vessel has a mass of 100 kg and its heat capacity is 0.5
J
g oC
u = Q = (nCvT )gas + (nCvT )vessel
u = Q = 5(20.8)(300 − 80) + 100 kg 
c)
1000 g
J
o
 0.5 o (300 − 80) C = _____
1 kg
g C
N 2 (3 moles ) @ 230 oC → 80 oC
Q extracted if heat capacity of piston and cylinder is neglected
J 

o
@ P = k: Q = nCpT = 3 moles  29.1
(80 − 230 ) C
o
mol C 

Assignment:
•
Connection of Sign for H. E. - (+) to the system, (-) from the system
•
Difference of Heat and Work
Topic 4 – Ideal Gas
Gas is ideal if
a) 𝑃 → 0
b) Molecules are separated by infinite distance
c) Volume of the molecules becomes infinitesimal
d) Intermolecular forces of attraction → 0
For an ideal gas: 𝑃𝑉 = 𝑅𝑇
V = k; 𝑑𝑈 = 𝐶𝑣𝑑𝑇 → (1)
P = k; 𝑑𝐻 = 𝐶𝑝𝑑𝑇 → (2)
𝐻 = 𝑈 + 𝑃𝑉
𝑑𝐻 = 𝑑𝑈 + 𝑑𝑃𝑉
but 𝑃𝑉 = 𝑅𝑇
𝑑𝐻 = 𝑑𝑈 + 𝑑𝑅𝑇
𝑑𝐻 = 𝑑𝑈 + 𝑅𝑑𝑇 → (3)
substitute (1), (2) in (3)
𝐶𝑝𝑑𝑇 = 𝐶𝑣𝑑𝑇 + 𝑅𝑑𝑇
𝐶𝑝 = 𝐶𝑣 + 𝑅
The Thermodynamic Behavior of Ideal Gases
1.
ΔE of I.G. is independent of volume. It is dependent on temperature.
2.
ΔE of an isolated system is constant.
3.
ΔE of a cyclic process is zero.
4.
ΔE is a state or thermodynamic function because it depends upon the initial and final state
and not on the manner by which the change occurs (or the path followed when undergoing
a change).
5.
ΔH (heat content of the system) is independent of pressure. It is dependent on
temperature; ∆𝐻 = ∆𝐸 + ∆𝑃𝑉
Reversible Non-Flow Processes
System: 1 mol of ideal gas
For finite changes:
u = Q − W
For differential changes:
du = dQ − dW
I.
Isometric (V = k) Process
du = dQ − dW
but V = k, dv = 0
For a finite process:
V2
T2
T2
V1
T1
T1
u = Q −  Pdv ; u = Q =  CvdT ; H =  CpdT
II. Isobaric (P = k) Process
du = dQ − dW
T2
H = Q =  CpdT
T1
V2
W =  Pdv = PV = RT = RT
V1
T2
W = Q − u ; u = T CvdT
1
III. Isothermal (T = k) Process
du = 0 , dH = 0
du = dQ − dW ; dQ = dW = Pdv
For a finite process:
V2
Q = W =  Pdv but PV = RT ; P =
V1
RT
V
dv
V
P
= RT ln 2 = RT ln 1
V1 V
V1
P2
V2
Q = W = RT 
Using Combined Gas Law:
P1V1 P2V2 P1
V2
=
=
T1
T2 ; P2
V1
IV. Adiabatic (Q = 0) Process
- it is one for which there is no heat transfer between the systems and the surroundings
du = dQ − dW ; du = −dW = CvdT
u = −W =  CvdT ; H =  CpdT
du = −W
CvdT = − Pdv but PV = RT , P =
CvdT = − RT
But  =
RT
V
R dv
dv dT
=−
;
T
Cv V
V
Cp
; Cp = Cv
Cv
Cp = Cv + R ;
 = 1+
Cv Cv + R
=
Cv
Cv
R
R
=  −1
;
Cv Cv
V2
dT
dv
(
)
=

−
1
T1 T V1
V
T2
ln
T2
V
= −( − 1) ln 2
T1
V1
Get antilog of both sides:
 V2 
T2
=
V 

T1
 1 
− ( −1)
For ideal gas:
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2
𝑃1 𝑇2 𝑉2
=
𝑃2 𝑇1 𝑉1
𝑇2
𝑃2 +(𝛾−1) 𝑇2 −(𝛾−1)
( )
=( )
𝑇1
𝑃1
𝑇1
𝑇2 1+𝛾+1
𝑃2 𝛾−1
( )
=( )
𝑇1
𝑃1
𝑇2 𝛾
𝑃2 𝛾−1
( ) =( )
𝑇1
𝑃1
Raise both sides by
1

𝑇2
𝑃2
=( )
𝑇1
𝑃1
𝛾−1⁄
𝛾
but
𝑇2
𝑉1 𝛾−1
=( )
𝑇1
𝑉2
𝛾−1⁄
𝛾
𝑉1 𝛾−1
𝑃2
( )
=( )
𝑉2
𝑃1
, 𝑟𝑎𝑖𝑠𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦
1
𝛾−1
1
𝑉1
𝑃2 ⁄𝛾
=( )
𝑉2
𝑃1
or
𝑉1 𝛾 𝑃2
( ) =
𝑉2
𝑃1
𝛾
𝛾
𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝑘
∆𝑈 = −𝑊 = ∫ 𝐶𝑣𝑑𝑇
𝑊=
𝑃1 𝑉1 − 𝑃2 𝑉2
; (𝑓𝑜𝑟 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)
𝛾−1
V. Isobaric process (opposing pressure constant)
P =k
u = Q − P  dv = Q − P(V2 − V1 )
V2
V1
For expansion (opposing pressure zero)
u = Q
Opposing Pressure variable
V2
u = Q −  Pdv
V1
where P = f (v )
Note: P in the equation is not Pgas but the P against which the gas is working
VI. Polytropic Process
- a process when no specific condition other than reversibility is imposed
du = dQ − dW ; du = CvdT
1)
Δu = Q − W
; Δu =  CvdT
dW = Pdv ; dH = CpdT
2)
W =  Pdv ; ΔH =  CpdT
To find the irreversible work:
1) If Wrev is (＋);
𝑊𝑖𝑟𝑟𝑒𝑣 = 𝑊𝑟𝑒𝑣 𝑥 𝑒𝑓𝑓.
2) If Wrev is (－);
𝑊𝑖𝑟𝑟𝑒𝑣 =
𝑊𝑟𝑒𝑣
𝑒𝑓𝑓.
Polytropic Process
A gas in a piston cylinder assembly undergoes an expansion process for which the
relationship between P and V is given by 𝑃𝑉 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. The initial pressure is 3 bars, the
initial volume is 0.1 m3, and the final volume is 0.2 m3.
a) Determine the work for the process in kJ, if 𝑛 = 1.5
b) If the change in specific internal energy of the gas in the process is 𝑈2 − 𝑈1 =
−4.6 𝑘𝐽/𝑘𝑔 and ∆𝐸𝐾 and ∆𝐸𝑃 are negligible, determine the net heat transfer for the
process in kJ.
Basis: 1kg
Given:
𝑃1 = 3 𝑏𝑎𝑟𝑠
polytropic
𝑉1 = 0.1𝑚3
→
𝑃1 =?
𝑉2 = 0.2𝑚3
process
Required: W
Solution:
𝑉2
𝑉2
𝑉2
𝑐𝑜𝑛𝑠𝑡.
𝑑𝑉
𝑑𝑉 = 𝑐𝑜𝑛𝑠𝑡. ∫
𝑛
𝑛
𝑉1
𝑉1 𝑉
𝑊 = ∫ 𝑃𝑑𝑉 = ∫
𝑉1
𝑉21−𝑛 − 𝑉11−𝑛
𝑊 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (
)
1−𝑛
=
𝑃2 𝑉2𝑛 ∙ 𝑉21−𝑛 − 𝑃1 𝑉1𝑛 𝑉11−𝑛
1−𝑛
𝑊=
𝑃2 𝑉2 − 𝑃1 𝑉1
1−𝑛
To solve for 𝑃2 :
𝑉1 𝑛
0.1 1.5
𝑃2 = 𝑃1 ( ) = 3 𝑏𝑎𝑟𝑠 ( ) = 1.06 𝑏𝑎𝑟𝑠
𝑉2
0.5
1.06 𝑏𝑎𝑟𝑠(0.2 𝑚3 ) − 3(0.1) 105 𝑁/𝑚2
1 𝑘𝐽
𝑊=[
][
][ 3
]
1 − 1.5
1 𝑏𝑎𝑟
10 𝑁 − 𝑚
𝑊 = −17.6 𝑘𝐽
∆𝑈 = 𝑈2 − 𝑈1 = −4.6 𝑘𝐽/𝐾
∆𝑈 = 𝑄 − 𝑊; 𝑄 = 𝑊 + ∆𝑈
𝑄 = −17.6 − 4.6 = −22.2 𝑘𝐽
Sample Problem 2-68.
A gas undergoes a thermodynamic cycle consisting of 3 processes;
Process 1-2 compression with 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 from 𝑃1 = 1 𝑏𝑎𝑟, 𝑉1 = 1.6 𝑚3 to
𝑉2 = 0.2 𝑚3 , 𝑈2 − 𝑈1 = 0
Process 2-3 constant pressure to 𝑉3 = 𝑉1
Process 3-1 constant volume 𝑈1 − 𝑈3 = −3,549 𝑘𝐽
There are no significant changes in 𝐸𝐾 and 𝐸𝑃 . Determine the heat transfer and work for
process 2-3 in kJ and calculate the thermal efficiency of the cycle.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑊𝑐𝑦𝑐𝑙𝑒
𝑄𝑖𝑛
Given:
𝑃1 = 1 𝑏𝑎𝑟
𝑉1 = 1.6𝑚3
𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡
𝑈2 − 𝑈1 = 0
𝑃2 = 8 𝑏𝑎𝑟𝑠
𝑉2 = 0.2𝑚3
𝑉=𝑘
𝑈1 − 𝑈3 = −3,549 𝑘𝐽
𝑃=𝑘
𝑃3 = 8 𝑏𝑎𝑟𝑠
𝑉3 = 𝑉1 = 1.6𝑚3
Required:
a) 𝑄23
b) 𝑊23
c) 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
Solution:
Step 2:
𝑉3
𝑊23 = ∫ 𝑃𝑑𝑉 = 𝑃2 (𝑉3 − 𝑉2 )
𝑉2
Solve 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃2 𝑉2 = 𝑃1 𝑉1
𝑃2 = 𝑃1
𝑉1
1.6
= (1 𝑏𝑎𝑟) ( ) = 8 𝑏𝑎𝑟𝑠
𝑉2
0.2
𝑏) 𝑊23 = (8 𝑏𝑎𝑟𝑠)(1.6 − 0.2) = 11.2 𝑏𝑎𝑟 − 𝑚3 𝑥
105 𝑁/𝑚2
1 𝑘𝐽
𝑥 3
1 𝑏𝑎𝑟
10 𝑁. 𝑚
𝑊23 = 1,120 𝑘𝐽
Energy Balance:
𝑄23 = ∆𝑈23 + 𝑊23 = 𝑈3 − 𝑈2 + 𝑊23
For a cycle:
𝑈2 − 𝑈1 + 𝑈3 − 𝑈2 + 𝑈1 − 𝑈3 = 0
0 + 𝑈3 − 𝑈2 − 3,549 = 0
𝑈3 − 𝑈2 = 3,549 𝑘𝐽
∴ 𝑄23 = 3,549 + 1,120 = 4,669 𝑘𝐽
Step 1:
𝑈12 = 𝑄12 − 𝑊12
0 = 𝑄12 − 𝑊12
𝑉2
∴ 𝑄12 = 𝑊12 = ∫ 𝑃𝑑𝑉
𝑉1
but 𝑃 =
𝑐𝑜𝑛𝑠𝑡.
𝑉
𝑉2
𝑑𝑉
𝑉2
= 𝑐𝑜𝑛𝑠𝑡. ln
𝑉1
𝑉1 𝑉
𝑄12 = 𝑊12 = 𝑐𝑜𝑛𝑠𝑡. ∫
But the 𝑐𝑜𝑛𝑠𝑡. = 𝑃1 𝑉1 = 𝑃2 𝑉2
𝑄12 = 𝑊12 = 𝑃1 𝑉1 ln
𝑉2
0.2
= (1)(1.6) ln
= −332.7 𝑘𝐽
𝑉1
1.6
Step 3:
𝑈1 − 𝑈3 = 𝑄31 − 𝑊31 ; 𝑊31 = 0
−3,549 = 𝑄31
𝑄𝑖𝑛 = 4,669 = 𝑄23
𝑊𝑐𝑦𝑐𝑙𝑒 = 𝑊12 + 𝑊23 + 𝑊31
𝑊𝑐𝑦𝑐𝑙𝑒 = −332.7 + 1,120 + 0 = 787.3 𝑘𝐽
𝜂=
787.3
= 16.86%
4,669
Sample Problem 2-69.
A gas undergoes a thermodynamic cycle consisting of the processes:
Process 1-2: P = k
𝑃1 = 1.4 𝑏𝑎𝑟𝑠
𝑉1 = 0.028 𝑚3
𝑊12 = 10.5 𝑘𝐽
Process 2-3: compression with 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡; 𝑈3 = 𝑈2
Process 3-1: V = k
𝑈1 − 𝑈3 = −26.4 𝑘𝐽
There are no significant changes to 𝐸𝐾 and 𝐸𝑃 .
a) Sketch the cycle on a PV diagram
b) Calculate the work net for the cycle in kJ
c) Calculate the heat transfer for the process 1-2, in kJ
Given:
𝑃1 = 1.4 𝑏𝑎𝑟𝑠
𝑉1 = 0.028 𝑚3
𝑃=𝑘
𝑊12 = 10.5 𝑘𝐽
𝑃2 = 1.4 𝑏𝑎𝑟𝑠
𝑉2 = 0.103 𝑚3
𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡.
𝑈3 = 𝑈2
𝑃3 = 5.15 𝑏𝑎𝑟𝑠
𝑉3 = 0.028 𝑚3
𝑉=𝑘
𝑈1 − 𝑈3 = −26.4 𝑘𝐽
Required:
a) 𝑊𝑐𝑦𝑐𝑙𝑒
b) 𝑄12
c) Sketch the PV diagram
Solution:
𝑊12 = 𝑃(𝑉2 − 𝑉1 )
𝑊12
10.5 𝑘𝐽 103 𝑁 − 𝑚
1 𝑏𝑎𝑟
(
)] + 0.028 𝑚3
𝑉2 =
+ 𝑉1 =
[
𝑃
1.4 𝑏𝑎𝑟
1 𝑘𝐽
105 𝑁⁄𝑚2
𝑉2 = 0.103 𝑚3
𝑃2 𝑉2 = 𝑃3 𝑉3
𝑃3 =
1.4(0.103)
= 5.15 𝑏𝑎𝑟𝑠
0.028
𝑄12 − 𝑊12 = 𝑈12
𝑈2 − 𝑈1 + 𝑊12 = 𝑄12
𝑄12 = 10.5 + 26.4 = 36.9 𝑘𝐽
5 𝑁
𝑉3
0.028 10 𝑚2
1 𝑘𝐽
𝑊23 = 𝑃2 𝑉2 ln = 1.4(0.103) ln
∙
∙ 3
𝑉2
0.103 1 𝑏𝑎𝑟 10 𝑁 − 𝑚
𝑊23 = −18.78 𝑘𝐽; 𝑊31 = 0
𝑊𝑐𝑦𝑐𝑙𝑒 = 10.5 − 18.78 + 0 = −8.28 𝑘𝐽
𝑈2 − 𝑈1 + 𝑈3 − 𝑈2 + 𝑈1 − 𝑈3 = 0
𝑈2 − 𝑈1 = 26.4
5.15
(2)
(3)
P
1.4
(1)
0.028
0.103
V
%=
|𝑊𝑐𝑦𝑐𝑙𝑒 | 8.28
=
= 22.43%
𝑄𝑖𝑛
36.9
or 𝑊23 = 𝑄23 − ∆𝑈23
𝑊23 = 𝑄23 − (𝑈3 − 𝑈2 )
𝑄23 = −18.78 𝑘𝐽
𝑊31 = 𝑄31 − ∆𝑈31 ; 𝑊31 = 0
0 = 𝑄31 − (−26.4)
𝑄31 = −26.4 𝑘𝐽
Seatwork:
A closed system undergoes a thermodynamic cycle consisting of the following processes
Process 1-2: Adiabatic compression with 𝑃𝑉 1.4 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, from
𝑃1 = 50𝑙𝑏𝑓/𝑖𝑛2, 𝑉1 = 3𝑓𝑡 3 to 𝑉2 = 1𝑓𝑡 3
Process 2-3: V = k
Process 3-1: P = k, 𝑈1 − 𝑈3 = 46.7 𝐵𝑡𝑢
There are no significant changes in 𝐸𝐾 or 𝐸𝑃 .
a) Calculate the work net for the cycle, in Btu
b) Calculate the heat transfer for process 2-3, in Btu
c) Calculate the thermal efficiency of the cycle
Given:
𝑃1 = 50𝑙𝑏𝑓/𝑖𝑛2
𝑉1 = 3𝑓 3
𝑄=0
𝑉=𝑘
𝑉2 = 1𝑓𝑡 3
𝑃𝑉1.4 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃=𝑘
𝑈1 − 𝑈3 = 46.7 𝐵𝑡𝑢
𝑉3 = 1𝑓 3
Required:
a) 𝑊𝑛𝑒𝑡
b) 𝑄23
c) 𝜂
Solution:
𝑃2 𝑉2𝑛 = 𝑃1 𝑉1𝑛
50
𝑃2 =
𝑙𝑏𝑓
(3)1.4
𝑙𝑏𝑓
𝑖𝑛2
= 232.776 2
1.4
(1)
𝑖𝑛
Process 1-2:
∆𝐸12 = 𝑄12 − 𝑊12 ; 𝑄12 = 0
𝑉2
𝑉2 𝐾
1
1 𝑉 1.4
but 𝑊12 = ∫𝑉 𝑃𝑑𝑉 = ∫𝑉
𝑊12 = 𝑘
𝑊12 = 𝑃1 𝑉11.4 [
𝑊12 = 50
𝑉 𝑑𝑉
𝑑𝑉 = 𝑘 ∫𝑉 2 1.4
1 𝑉
𝑉 −0.4 1
]
−0.4 3
(1)−0.4 − (3)−0.4
]
−0.4
𝑙𝑏𝑓
𝑙𝑏𝑓
144𝑖𝑛2
1 𝐵𝑡𝑢
1.4 (−0.8890)
3
(3)
=
−206.94
∙
𝑓𝑡
∙
∙
2
2
2
𝑖𝑛
𝑖𝑛
1𝑓𝑡
778 𝑓𝑡 − 𝑙𝑏𝑓
𝑊12 = −38.3025 𝐵𝑡𝑢
𝑊12 =
232.776
𝑊12 =
𝑃2 𝑉2 − 𝑃1 𝑉1
1 − 1.4
𝑙𝑏𝑓
𝑖𝑛2
3)
(1𝑓𝑡
(144
) − 50(3)(144)
𝑖𝑛2
𝑓𝑡 2
1 − 1.4
𝑊12 = −
29,799.66
= −38.3025 𝐵𝑡𝑢
778
∆𝐸12 = −𝑊12 = 38.3025 𝐵𝑡𝑢
Process 3-1:
𝑊31 = 𝑃(𝑉1 − 𝑉3 ) = 50(3 − 1)
𝑊31 = 100
𝑙𝑏𝑓
144𝑖𝑛2
1 𝐵𝑡𝑢
3
∙
𝑓𝑡
∙
∙
2
2
𝑖𝑛
𝑓𝑡
778𝑓𝑡 − 𝑙𝑏𝑓
𝑊31 = 18.57 𝐵𝑡𝑢
∆𝐸31 = 𝑄31 − 𝑊31
46.7 = 𝑄31 − 18.51
𝑄31 = 65.21 𝐵𝑡𝑢
Process 2-3:
∆𝐸𝑇 = ∆𝐸12 + ∆𝐸23 + ∆𝐸31
0 = 38.3025 + ∆𝐸23 + 46.7
∆𝐸23 = −85 𝐵𝑡𝑢
but ∆𝐸23 = 𝑄23 − 𝑊23 ; 𝑊23 = 0
∆𝐸23 = 𝑄23 = −85 𝐵𝑡𝑢
𝑊𝑇 = 𝑊12 + 𝑊23 + 𝑊31 = −38.3025 + 0 + 18.51
𝑊𝑇 = −19.793 𝐵𝑡𝑢
𝜂=
|−19.793 𝐵𝑡𝑢|
𝑛𝑒𝑡 𝑊𝑐𝑦𝑐𝑙𝑒
𝑥100 =
𝑥100
𝑄𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑
65.21
𝜂 = 30.35%