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What Thermo’s all about?
Imagine how the world changed:
- Four horses pulling a city Omnibus
- Wind pushing ships a cross the ocean
- Bucket and rope to lift water out of mine.
Instead all of this “engines” existed that could do all these jobs at much lower cost.

In the 19th century, it come “Industrial Revolution”.
Steam Engines, Internal Combustion Engines, Electric Motors, and many other types
of engines.
Goal: “Ability to do Work”
Thermo was born here to describe the heat engines “Power developed from heat”.

Thermo was then developed to study energy.
How much of it is in coal, wood, running water, in steam at low and at high pressure,
and at low and high temperature? This is the concern of the
“First Law of Thermodynamics”

It was discovered early that you couldn’t always transfer all of one kind to another.
(Energy to work is not possible).
How efficiently can energy be converted from one form to another?
“Sadi Carnot” answered leading to the
“Second Law of Thermodynamics”

1st and 2nd laws are the main business of thermodynamics.
Thermo: Is the science that deals with the transformation of all kinds of energies from
one form to another.

It is important to note that these laws have No proof in the mathematical sense. Their
validity lies in the absence of contrary results.

1st law: Energy interchange. (How much of this kind of energy is equivalent to that
kind of energy?).

2nd law: Energy amount and direction possible. (What changes are possible, what are
not, and which direction?).

READ Ch.1, Introduction to Chemical Eng. Thermodynamics / Smith.
Chapter 2
First Law of Thermodynamics
Energy can’t be created or destroyed. You can only change it from one form to
another, or you can only add it to the system from the surroundings.
Other forms: Introduction to Chemical Eng. Thermodynamics / Smith / p22.

Energy Balance
Closed system (No composition change)
U + K + P = Q + W
Units: Energy (J)
SI
(ft.lbf) EE System.
For units check: App. A
W
Q
+VE
System
System
W
Q
-VE
Surroundings

Thermodynamics State And State Functions
State Functions: “Exact Differential”
F= state1
= F(state2) – F(state1)
Independent of path depends only on the state of the material, like “height”.
state2
dF
(State Function = Point Function) Any state function can be expressed
mathematically as a function of thermodynamic coordinates (P, T). Their value
can be identified with points on a graph.
Path Function: Depend on the path, like distance.
Work and Heat, are not properties, they represent the energy changes between
surr. and system. They are associated with areas rather than points on the graph.
They considered as path functions.
Intensive and Extensive Properties:
Extensive Property: Value of property depends on size of the system (Mass, Vol.,
Internal Energy). (Additive)
Intensive Property: Value of property does not depend on size of the system (P, T,
Specific Vol.). (Not additive)
Properties are usually tabulated on a molar basis.

Equilibrium
“Static Conditions” or No change.
Thermodynamically, indicates the absence of any tendency toward change on a
macroscopic scale.
If all forces in “exact” balance, then the system at equilibrium.
If a driving force(s) exist but due to a resistance, no change in state occurs, the
system is still considered in a non-equilibrium state.
Different driving forces to bring different kinds of changes:
1. Mechanical forces: (Equal pressure, mech. equilibrium)
2. Thermal forces:
( Equal temp., Thermal equilibrium)
3. Chemical potentials: (Chemical rxn, Chemical equilibrium)
(Phase change, Phase equilibrium)
At equilibrium, all such forces are in balance.

The Phase Rule
Degrees of freedom (F): The number of independent variables that must be fixed
to establish the intensive state of any system.
F = 2-  + N
Where:
 = No. of phases.
N= No. of Components.
If F = 0, the system is Invariant
max= 2 + N
If N =1, then max = 3 (Triple point)
Ex. 2.5

The Reversible Process
A process is reversible when its direction can be reversed at any point by an
infinitesimal change in external conditions.
Ex.: Removing powder particles from piston-cylinder arrangement. (No friction).
dW = -P dV
 PdV
v2
Wrev = The reversible Process = Ideal Process
Ex. 2.6
v1

Constant–V and Constant-P Processes
For a closed system
d(nU) = dQ + dW
For a mechanically reversible, non flow process
dW= -P d(nV)
Combine
d(nU) = dQ - P d(nV)
(2.6)
(1.2)
(2.8)
Constant Volume process
Work is zero
dQ = d(nU)
Integrate
Q= n U
(const V)
(const V)
(2.9)
(2.10)
For mechanically reversible, constant volume, non flow process
Heat transfer = Internal energy change of the system
Constant Pressure process
From definition of enthalpy (H)
H = U + PV
(2.11)
Enthalpy is an extensive property (like V and U) depends on mass. Its unit is (J).
nH = nU + P(nV)
For infinitesimal constant pressure change
d(nH) = d(nU) + P d(nV)
Substitute d(nU) from eq’n 2.8, you get;
dQ = d(nH)
(const P)
(2.12)
Integrate
Q = n H
(const P)
(2.13)
For mechanically reversible, constant pressure, non flow process
Heat transfer = Enthalpy change of the system

Heat Capacity
The amount of heat to be added or removed from the body to change its temperature
by 1oC (1k).
C = dQ/dT
Heat Capacity at Constant Volume
CV = (dU/dT)V
dU = CV dT
Integrate
U =
(const V)
 C dT
T2
T1
V
For rev., const.- volume process
Q = n U = n
 C dT
T2
T1
V
(const V)
Heat Capacity at Constant Pressure
CP = (dH/dT)P
dH = CP dT
Integrate
H =
For rev., const.- pressure process
(const P)
 C dT
T2
T1
Q = n H = n
P
 C dT
T2
T1
P
(const P)

Mass And Energy Balances for Open Systems
Mass Balance for open system
dmcv/dt + (m.)fs = 0
(m.)fs = m.3 – m.1 - m.2
m. = u A 
dmcv/dt + ( u A )fs = 0
Continuity Equation
Accumulation term
Where
Special case: Steady State (Conditions within the control volume do not change
with time).
( u A )fs = 0
For a single entrance and single exit streams, the mass flow rate is the same:
m. = u2 A2 u1 A1 
Because specific volume is reciprocal of density
m. = u1 A1 /V1 u2 A2 /V2u A /V
Continuity Equation
Energy Balance for Steady-State Flow system
Open system (steady state d(mU)cv/dt) = 0)

H + u2/2gc + g/gc z = Q + W

Enthalpy
H = U + PV
Work =
Ws = Shaft work
+ PV work
W = Ws + P1V1 - P2V2
U + K + P = Q + Ws + P1V1 - P2V2
Arrange
(U2+ P2V2) - (U1+ P1V1) + K + P = Q + Ws
(H) + K + P = Q + Ws
K= ½ u2
P = g Z
EES
Special case:
H + u2/2 + g z = Q + Ws
H + u2/2gc + g/gc z = Q + Ws
If K = 0
P = 0
H = Q + Ws