Chapter 7
Energy and Energy
Balance
By : Miss Noorulanajwa Diyana Yaacob
noorulnajwa@unimap.edu.my
Content




Components of total energy, energy
balance on a closed and open
system, energy or power conversion
to other units
First law of thermodynamics
Enthalpies and kinetic energy of a
body of mass moving
Enthalpy data and steam tables
2
Forms of Energy
Three component of total energy of a system
 Kinetic energy (Ek)
 energy due to the translational motion of the system as
a whole relative to some frame of reference (usually the
earth’s surface) or to rotation of the system about some
axis.
 Potential energy (Ep)
 energy due to the position of the system in a potential
field (such as a gravitational or electromagnetic field).
 Internal energy (U)
 all energy possessed by a system other than kinetic and
potential energy; or
 Energy due to translation, rotation, vibration &
electromagnetic interactions of the molecules, atom and
subatomic particle within the system.
3
Transfer of Energy
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

In closed system (i.e. no mass is transferred across the
system boundaries while the process is taking place), energy
may be transferred between such a system and its
surroundings in two ways as heat or work.
Heat
 Energy that flows as a result of temperature difference
between a system and its surroundings.
 The direction of flow is always from a higher temperature
to a low one.
 Heat is defined as positive when its transferred to the
system from the surroundings.
Work
 energy that flows in response to any driving force other a
temperature difference, such as a force, a torque or a
voltage
 Work is defined as positive when it is done by the system
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on the surroundings.
First Law of Thermodynamics
Law
of conservation of energy, which state
that energy can neither be created nor
destroyed.
General
form of first law of thermodynamics
Inlet Energy + Heat - Outlet Energy –
Work = Accumulation
Inlet energy and outlet energy is
summation/total of all energy such as potential,
kinetic and internal energy
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Kinetic Energy Equation (Ek)

Kinetic energy, Ek (J) of an object of mass m (kg)
moving with velocity u (m/s) relative to the surface of
the earth is
1
E k  mu 2
2


If the fluid enters a system with a mass flow rate m
(kg/s) and uniform velocity u (m/s), the rate at which
kinetic energy E k (J/s) is transported into the system is
1

E k  m u 2
2
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Potential Energy Equation (Ep)

Gravitational potential energy, Ep
E p  mgz
 gz
E p  m

if the fluid enters a system with a
 (kg/s) and an
mass flow rate m
elevation z relative to the potential
energy reference plane.
E  E  E  m g ( z  z )
p
p2
p1
2
1
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ERT104/4
ppkas@UniMAP
Let’s do it
Water flows into a process unit through a 2
cm ID pipe at a rate of 2 m3/h. Calculate the
kinetic energy transport in this stream in unit
J/s.
Solution:
Ek= 0.870 N.m/s= 0.870 J/s
8
Class
Discussion
9
Let’s try

Crude oil is pumped at a rate of
15.0 kg/s from a point 220 meters
below the earth’s surface to a point
20 meters above ground level.
Calculate the attendant rate of
increase of potential energy.
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Energy Balances on Closed
System


Closed system
 no mass is transferred across the system boundaries
while the process is taking place
Energy balance
Final System Energy – Initial System Energy
= Net Energy Transferred to the System
Initial energy system
Final energy system
Net energy transfer
= Ui + Eki + Epi
= Uf + Ekf + Epf
= Q-W
(Uf-Ui) + (Ekf-Eki) + (Epf-Epi) = Q-W
U  E k  E p  Q  W
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Energy Balances on Closed
System

When applying energy balance equation to a given process, the
following point must be aware;
1. The internal energy of a system depends almost entirely on
the chemical composition, state of aggregation (solid, liquid,
or gas), and temperature of the system materials. If no
temperature changes, phase changes, or chemical reactions
occur in a closed system and if pressure changes are less
than a few atmospheres, then ∆U ≈ 0.
2. If a system is not accelerating, then ∆Ek = 0. If a system is
not rising or falling, then ∆Ep = 0.
3. If a system and its surroundings are at the same
temperature or the system is perfectly insulated, then Q = 0.
The process is then termed adiabatic.
4. Work done on or by a closed system is accomplished by
movement of the system boundary against a resisting force
or the passage of an electrical current or radiation across the
system boundary. If there no moving parts or electrical
current at the system boundary, then W = 0.
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Class
Discussion
13
Let’s try


A gas is contained in a cylinder fitted with a movable
piston. The initial gas temperature is 25°C. The cylinder is
placed in boiling water with the piston held in a fixed
position. Heat in the amount of 2.00 kcal is transferred to
the gas, which equilibrates at 100°C. The piston is then
released and the gas does 100J of work in moving the
piston to its new equilibrium position. The final gas
temperature is 100°C.
Write the energy balance equation for each of the two
stages of this process, and each case solve for the
unknown energy term in equation. Consider the gas in the
cylinder to be the system, neglect the change in potential
energy, and assume the gas behaves ideally. Express all
energies in joules.
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Energy Balances on Open
System
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

In open system, mass is transferred across the system
boundaries while the process is taking place.
Therefore work must be done on open system to push mass in
and work is done on the surrounding by mass that emerges
from the systems.
Both work terms must be include in the energy balance for open
system
The net work done by an open system
W  W s  W fl
W s
W fl
and
- shaft work
rate of work done
within the system
- flow work
rate of work done
rate of work done
W fl  W out  Win  PoutVout  PinVin
by the process fluid on a moving part
such as a pump rotor.
by the fluid at the system outlet minus
by the fluid at the system inlet.
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Energy Balances on Open
System


^ symbol is used to denote the specific property
(property divided by mass or by mole) such as
specific internal energy (Û kJ/kg), specific volume ( Vˆ
m3/kg) and so on.
One important property for energy balance on open
system is specific enthalpy (Ĥ kJ/kg).
Hˆ  Uˆ  PVˆ

Sometimes, universal gas law constant can be used
as a conversion factor to evaluate specific enthalpy.
16
Class
Discussion
17
Let’s try

The specific internal energy of helium at
300K and 1 atm is 3800 J/mol, and the
specific molar volume at the same
temperature and pressure is 24.63 L/mol.
Calculate the specific enthalpy of helium at
this temperature and pressure, and the
rate at which enthalpy is transported by a
stream of helium at 300K and 1 atm with a
molar flow rate of 250 kmol/h
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Energy Balances Equation for
Open System
H  E k  E p  Q  W s
 m Hˆ
H 
j
j
output
stream
E k 

m ju j
2
2
 m gz
j
output
stream
j
j
input
stream
output
stream
E p 
 m Hˆ



m ju j
2
input
stream
j

 m gz
j
input
stream
2
j
19
Class
Discussion
20
Let’s try

Five hundreds kilograms per hour of steam
drives a turbine. The steam enters that
turbine at 44 atm and 450°C at a linear
velocity of 60 m/s and leaves at a point 5m
below the turbine inlet at atmospheric
pressure and a velocity of 360 m/s. The
turbine delivers shaft work at a rate of 70
Kw, and the heat loss from the turbine is
estimated to be 104 kcal/h. Calculate the
specific enthalpy change associated with
the process
21
Reference States and State
Properties
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
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It is not possible to know the absolute value of Û and Ĥ for a process
material, but we can determine the change in ΔÛ and change in ΔĤ
corresponding to a specific change of state (temperature, pressure,
phase).
A convenient way to tabulate ΔÛ and ΔĤ is to choose a temperature,
pressure and state of aggregation (i.e. phase) as a reference state.
Since Ĥ cannot be known absolute, for convenience we may assign a
value Ĥo=0 to be reference state. Then ΔĤ1= Ĥ1-0; ΔĤ2= Ĥ2-0 and so
on.
Some enthalpy tables give the reference states on which the listed
values of Ĥ are based and others do not.
However, we do not have to know the reference state to calculate ΔĤ
for the transition from one state to another state since the value are
based on the same reference in the table.
Ĥ and Û can be said as state properties
 Property whose change of value in any process depend only on it
initial and final states and do not depend on the path take to
reach the state.
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Class
Discussion
Example
7.5-1
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Steam Table
Class
Discussion
Example
7.5-2
24
Class
Discussion
Example
7.5-3
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Energy Balance Tips
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
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When labeling flowchart, write down together the
temperature, pressure and state of aggregation of the
process material.
Normally (depend on the process description) for chemical
process unit; shaft work, kinetic and potential energy
change tend to be negligible compared to heat flows,
internal energy and enthalpy changes.
Then simplified energy balance become
Closed System:
Q  U
Open System:
Q  H
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THANK
YOU
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