e balance1 wble
TRANSCRIPT
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MASS AND ENERGYBALANCE
UGPA1193
TAN KEE LIEW, Mr.
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Energy Balances on Closed Systems
Energy Balances on Open Systems at Steady State
Energy Balances on Non-reactive Processes
Energy Balances on Reactive Processes
Topics
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Forms of Energy
The total energy of a system has three components :
sometimes known as heat balance.
application of the law of conservation of energy to physical systems.
(1) Kinetic energy
Energy due to the translational motion of the system as a whole relative tosome frame of reference (usually the earths surface) or due to rotation of the
system about some axis.
(2) Potential energy
Energy due to the position of the system in a potential field (such as a
gravitational or electromagnetic field).
(3) Internal energy
All energy possessed by a system other than kinetic and potential energy,
such as energy due to the motion of molecules relative to the center of mass
of the system, and due to the rotational and vibrational motion of the
molecules.
Energy balance
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Energy Balances on Closed Systems
Energy may be transferred between a closed system and its surrounding
in two ways :
1. As Heat
2. As Work
Energy that flows as a result of temperature difference between a system and its
surroundings.
Direction of flow from a higher temperature to a lower one.
Heat is defined as positive when transferred to the system from the surroundings.
Energy that flows in response to any driving force other than a temperature
difference, such as a force, torque or voltage.
Example : If a gas in a cylinder expands and moves a piston against a resisting
force, the gas does work on the piston (Energy is transferred as work from the
gas to its surroundings, which include the piston).
Work is defined as positive when it is done by the system on the
surroundings. (Note : Other references might use the opposite sign convention)
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U+Ek+Ep= Q -W
General balance equation for a closed system
where signifies change (=final-initial) , U =internal energy, Ek=kinetic energy,
Ep=potential energy, Q=heat transferred to the system from the surroundings, W=work done by the system on its surroundings.
(Equation 7.3-4)
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Example 1:
Energy Balance on a Closed System
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U+Ek+Ep= Q-W
Ek=0 (the system is stationary)
Ep=0 (no vertical displacement)
W=0 (no moving boundaries)
U = Q
Q= 2.00 kcal
U = 2.00 kcal 103 cal
kcal
4.1858 J
1 cal8372 J=
The gas thus gains 8372 J of internal energy in going from 25 to 100oC.
1)
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U+Ek+Ep= Q-W
Ek=0 (the system is stationary at the initial and final states)
Ep=0 (assumed negligible by hypothesis)
U=0 (U depends only on T for ideal gas, and T here does not change)
0= Q -W
W=+100J (Why positive ?)
Q = 100 J
Thus an additional 100J of heat is transferred to the gas as it expands and
reequilibrates at 100oC.
2)
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Energy Balances on Open Systems at Steady State
flS WWW
where
SW
flW
= shaft work, or rate of work done by the process fluid on a moving
part within the system (eg., a pump rotor)
= flow work, or rate of work done by the fluid at the system outlet
minus the rate of work done on the fluid at the system inlet
The net rate of work done by an open system on its surroundings is :
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Derivation of flW
We initially consider the single inlet single outlet system as shown below.
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Specific Properties and Enthalpy
We use the symbol ^ to denote a specific property
Specific volume,
)(
)()/(
kgm
LVkgLV
Specific internal energy
Specific enthalpy
)(
)()/(
kgm
JUkgJU
)/(
)/(
)/(skgm
sJU
kgJU
VPUH
Specific property is an intensive quantity obtained by dividing an extensive
property(or its flowrate) by the total amount(or flowrate) of the process material.
Examples :
)/(
)/(
)/( skgm
sLV
kgLV
or
or
where P is total pressure
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Example 2 : Calculation of enthalpy
Solution :
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The Steady State Open System Energy Balance
General balance equation :
Spk WQEE
where denotes total output minus total input
(7.4-14a)
(7.4-14b)
(7.4-14c)
If the process has a single input and single output stream and there is noaccumulation of mass in the system (so that ) , the
expression for in equation (7.4-14a) simplifies to
mmm outin
(7.4-15)
(7.4-16)
Subscript j denotes
species,
u = velocity
z= vertical
displacement
g= gravitational
constant (9.81m/s2)
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Example 3 : Energy Balance on a Turbine
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kWWs 70
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Reference State and State Property
It is impossible to know the absolute value of or for a process
material.
We can only determine the change in or (ie or ).
All the terms (ie, heat, work, changes in kinetic and potential energy) in the
general balance equation can be known, enabling the determination of
or .
Note that
A convenient way to tabulate measured changes in or is to
choose a temperature, pressure and state of aggregation as a reference
state, and to list or for changes from this state to a series of
other states.
For example, the enthalpy changes for CO going from a reference state of
0oC and 1 atm to two other states are measured, with the following results:
U
H
U
H
U
H
U
H
)(
VPUH U
H
U
H
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Since cannot be known absolutely, for convenience, we may assign a
value = 0 to the reference state ; then
and so on.
A table may then be constructed for CO at 1 atm :
T(oC) (J/mol)
0 0
100 2919
500 15060
From the table, we should say that the specific enthalpy of CO at 100oC and 1
atm relative to CO at 0oC and 1 atm is 2919 J/mol.
Note that the value 2919 J/mol for does not mean that the absolute value of
the specific enthalpy of CO at 100oC and 1 atm is 2919 J/mol.
Enthalpy is very much of a concern for process engineers !!!
1
2
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If another reference state had been used to generate the specific
enthalpies of CO at 100oC and 500oC , and would have different
values, but would still be 12141 J/mol.1
2
12
This convenient result is due to the fact that : , like is a stateproperty.
A state property is the property of a system component whose value
depends only on the state of the system (temperature, pressure, phase,
and composition) and not on how the system reached that state.
H
U
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Steam Tables
Water is often used in the process industries :
(i) as process material
(ii) for heat exchange purposes.
Steam is used to generate electrical power.
Chemical and mechanical engineers often find the need to know the physical
properties of water in various phases.
Physical properties of liquid water , saturated steam and superheated steam
are compiled in the Steam Tables, a standard reference for engineers.
Phase diagram for water
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Example 5 : Energy Balance on One-Component Process
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Observe that the kinetic energy change is only a small fraction roughly 0.8% -
of the total energy requirement for the process. This is a typical result, and it is
common to neglect kinetic and potential energy changes relative to enthalpy
changes for processes that involve phase changes, chemical reactions, or
large temperature changes.
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Example 6 : Energy Balance on a Two-Component Process
A liquid stream containing 60.0wt% ethane and 40.0% n-butane is to be heated
from 150 K to 200 K at a pressure of 5 bar. Calculate the required heat input per
kilogram of the mixture, neglecting potential and kinetic energy changes. Assumethat mixture component enthalpies are those of the pure species at the same
temperature. The specific enthalpy values at 5 bar are tabulated as follows :
Temperature (K) (kJ/kg)
n-butane 150 30.0
200 130.2
Ethane 150 314.3
200 434.5
i
Basis : 1 kg/s mixture
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(kJ/s)
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Example 7 : Simultaneous Material and Energy Balances
Saturated steam at 1 atm is discharged from a turbine at a rate of 1150 kg/h.
Superheated steam at 300oC and 1 atm is needed as a feed to a heat
exchanger ; to produce it, the turbine discharge stream is mixed with
superheated steam available from a second source at 400oC and 1 atm. The
mixing unit operates adiabatically. Calculate :
(i)The amount of superheated steam at 300oC produced.
(ii)The required volumetric flowrate of the 400oC steam
Solution :
Specific enthalpies of the two feed streams and the product stream are
obtained from the steam tables and labeled on the flowchart.
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There are two unknown quantities in this process- and , and onlyone material balance equation. Therefore mass balance and energy
balances must be solved simultaneously to determine the two flowrates .
1
m 2
m
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2243.6 kg/h
3393.6 kg/h
2243.6 kg 6977.6 m3/h
In absence of the specific volume data, the ideal gas equation of state
(PV=nRT) could be used to approximate the volumetric flowrate.
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Energy Balances on Closed Systems
Energy Balances on Open Systems at Steady State
End of