L1-5 Energy Balances

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CCB1064 –Principles of Chemical Engineering 101/11/2011

Energy and Energy Balances

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Objectives

At the end of this chapter, you should be able

to understand the following :

• List and define the three components of

total energy of a process system

• Define closed process system, open

process system, isothermal process and

adiabatic process

• Define flow work, shaft work, sp. internal

energy, sp. volume, and sp. enthalpy

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Introduction

• Energy is expensive

• Every chemical process uses energy in some form or

other

• Wasting energy leads to reduced profits in process

plants

• After the sharp increase in energy prices in 1970s, the

need for process intensification to eliminate

unnecessary energy consumption raised

• Account of energy that flows into and out of a process

unit is necessary to determine the overall energy

requirement of the process

• Achieved through performing ENERGY BALANCES

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Introduction

• Typical problems that might be solved include:

– How much power (energy/time) is required to

pump 1250 m3/h of water from a storage tank to a

process vessel?

– How much energy is required to convert 2000 kg

water at 30oC to steam at 180oC?

– How much energy is required to separate the

components by distillation?

– How much energy is required to be removed in an

exothermic process?

– And so on…

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Forms of Energy

• The total energy of a system has three

components:

• Kinetic energy : Energy due to the

translational motion of the system as a whole

relative to some frame of reference (usually the

earth’s surface)

• Potential energy : Energy due to the position

of the system in a potential field (such as

gravitational or electromagnetic field).

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Forms of Energy

• Internal energy :

All energy possessed by a system due to

– the motion of molecules relative to the

center of mass of the system,

– to the rotational and vibrational motion

and the electromagnetic interactions of

the molecules,

– to the motion and interactions of the

atomic and subatomic constituents of the

molecules.

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Classification of Systems

Closed system

• No mass is transferred across its boundaries while

the process is taking place

• Energy may be transferred between such a system

and its surroundings

• Example: Batch processes

Open system

• Both mass and energy are transferred across its

boundaries while the process is taking place

• Example: Continuous processes

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Transfer of energy

• Suppose a process system is closed

• Energy may be transferred between such a system

and its surroundings in two ways:

• As heat and work

• As Heat, or 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 lower temperature one.

• Heat is defined as positive when it is transferred to

the system from surroundings

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Transfer of energy

• As work, or energy that flows in response to any

driving force other than a temperature difference,

such as a force, a torque, or a voltage.

• For example, if a gas in a cylinder expands and

moves a piston against a restraining force, the

gas does work on the piston (energy is transferred

as work from the gas to its surroundings, which

include the piston).

• In this text, work is defined as positive when it is

done by the system on the surroundings.

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Units of Energy

• The terms “work” and “heat” refer only to energy

that is being transferred

• Energy, like work, has units of force times distance:

for example, joules (N.m), ergs (dyne.cm), and ft.lbf

• It is also common to use energy units defined in

terms of the amount of heat that must be transferred

to a specified mass of water to raise the

temperature of the water by a specified temperature

interval at a constant pressure of 1 atm

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Units of Energy

Unit Symbol Mass of

Water

Temperature

Interval

Kilogram – calorie or kilocalorie kcal 1 kg 15°C to 16°C

Gram – calorie or calorie cal 1 g 15°C to 16°C

British thermal unit Btu 1 lbm 60°F to 61°F

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First law of thermodynamics

• The principle that underlies all energy balances is

the law of conversion of energy, which states that

energy can either be created nor destroyed

• The rate at which energy (kinetic+ potential +

internal) is carried into a system by the input

streams, plus the rate at which it enters as heat,

minus the rate at which it is transported out of the

system by the output streams, minus the rate at

which it leaves as work, equals the rate of

accumulation of energy in the system

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First law of thermodynamics

streamsoutput

through system the

ofout energy of Rate

work

as system theleaves

t energy tha of Rate

heat as

system theinto

energy of Rate

streamsinput

through system theinto

internal)potential(kinetic

energy of Rate

system ahin energy wit of

on accumulati of Rate

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Kinetic and Potential Energy

The 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

2

2

1muEk

If a fluid enters a system with a mass flow rate m (kg/s) and

uniform velocity u (m/s), then

2

2

1umEk

kE (J/s) may be thought of as the rate at which kinetic energy is

transported into the system by the fluid.

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Kinetic and Potential Energy

The gravitational potential energy of an object of mass m

is

g is the acceleration due to gravity

z is the height of the object above a reference plane

If a fluid enters a system with mass flow rate

Change in potential energy:

mgzEp

m

gzmEp

1212 zzgmEE pp

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Energy Balances on a Closed System

• An integral energy balance may be derived for a

closed system between two instants of time

accumulation = input – output …(1)

• For a closed system, input and output terms can be

eliminated, since no mass crosses the boundaries

of a closed system

• Eqn (1) may be written as

Final system

energy

Initial system

energy

Net energy

transferred to the

system (in – out)

- =

…(2)

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initial system energy = Ui + Eki + Epi

final system energy = Uf + Ekf + Epf

energy transferred = Q – W

Eqn. (2) becomes

(Uf – Ui ) + (Ekf – Eki ) + (Epf – Epi ) = Q – W

or, if the symbol Δ is used to signify (final – initial),

WQEEU pk

• The basic form of the first law of thermodynamics

for a closed system

…(3)

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• When applying this equation to a given process, you should be aware of the following points:

– The internal energy of a system depends almostentirely on the chemical composition, state ofaggregation (solid, liquid, or gas) andtemperature of the system materials.

– It is independent of pressure for ideal gases andnearly independent of pressure for liquids andsolids.

– If no temperature changes, phase changes, orchemical reactions occur in a closed systemand if pressure changes are less than a fewatmospheres, then ΔU ≈ 0.

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• When applying this equation to a given process,

you should be aware of the following points:

– If a system is not accelerating, then ΔEk = 0.

– If a system is not rising or falling, then ΔEp= 0.

– 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.

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• When applying this equation to a given process,

you should be aware of the following points:

– 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.

– Examples of the first type of work are motion of

a piston or rotation of a shaft that projects

through the system boundary.

– If there are no moving parts or electrical currents

or radiation at the system boundary, then W = 0.

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Energy Balances on a Open System at

Steady state

• An open process system by definition has

mass crossing its boundaries as the

process occurs.

• Work must be done on such a system to

push mass in, and work is done on the

surroundings by mass that emerges.

• Both work terms must be included in the

energy balance

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Flow Work and Shaft Work

The net rate of work done by an open system on its

surroundings may be written as

fls WWW

where sW shaft work, or rate of work done by the process

fluid on a moving part within the system (e.g., a pump rotor)

flW 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.

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Flow Work and Shaft WorkTo derive an expression for flW , consider the single – inlet – single –

outlet system shown here.

)/( 3 smVin )/( 3 smVout

)/( 2mNPin )/( 2mNPout

PROCESS

UNIT

The fluid that enters the system has work done on it by

the fluid just behind it at a rate

)/()/()/( 32 smVmNPsmNW ininin

while the fluid leaving the system performs work on the

surroundings at a rate outoutout VPW

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Flow Work and Shaft Work

The net rate at which work is done by the system at the inlet

and outlet is therefore

ininoutoutfl VPVPW

If several input and output streams enter and leave the

system, the VP products for each stream must be summed to

determine flW .

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Specific Properties and Enthalpy

• Properties of a process material are either

extensive (proportional to the quantity of the

material) or intensive (independent of the quantity)

• Kinetic energy, potential energy, and internal

energy are extensive properties

• A specific property is an intensive quantity

obtained by dividing an extensive property by the

total amount of the process material

– Specific volume

– Specific kinetic energy

– Symbol ˆ denote a specific property

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Specific Enthalpy

• A property that occurs in the energy balance

equation for open systems is the specific enthalpy,

defined as

VPUH ˆˆˆ

where P is total pressure and U and V are specific internal energy and specific volume.

…(4)

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Steady – State Open – System Energy

Balance

If jE denotes the total rate of energy transport by the jth input

or output stream of a process, and Q and W are defined as the

rates of flow of heat into and work out of the process, then

stream soutput

stream sinput

jj WQEE

If jm , kjE , pjE , and jU are the flow rates of mass, kinetic

energy, potential energy, and internal energy for the jth process

stream, then the total rate at which energy is transported into or

out of the system by this stream is

pjkjjj EEUE

…(5)

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Balance

The total work W done by the system on its surroundings

jjpj

jjkj

jjj

gzmE

umE

UmU

2/

ˆ

2

j

j

jjj gzu

UmE2

ˆ2

streamsoutput

streamsinput

jjjjjjs VPmVPmWW ˆˆ

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Balance

Substituting in Eqn (5),

streamsoutput

s

streamsinput

j

j

jjjjj

j

jjjj WQgzu

VPUmgzu

VPUm 2

ˆˆ2

ˆˆ22

streamsinput

jj

streamsoutput

jj HmHmH ˆˆ

stream sinput

jj

stream soutput

jjk umumE 2/2/ 22

stream sinput

jj

stream soutput

jjp gzmgzmE

spk WQEEH

where

…(6)

jjjj VPUH ˆˆˆ

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BalanceIf a process has a single input stream and a single output stream

and there is no accumulation of mass in the system (so that

mmm outin ), the expression for H simplifies to

HmHHmH inoutˆˆˆ

If jH is the same for all streams, then

streamsoutput

streamsinput

jj mmHH ˆ

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Tables of Thermodynamic Data

Reference States

• It is not possible to know the absolute

values of for a process material

• Only the change in

corresponding to a specific change of state

can be determined

• A convenient way to tabulate measured

changes is to choose a temperature,

pressure and state of aggregation as a

reference state

HU ôr ˆ

)ˆ(în or ˆ ˆ HH)U(ΔU

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Tables of Thermodynamic Data

• 0 oC and 1 atm is one of the reference

states

• is a state property that depends

only on the state of the system and not on

how the system reached that state

Steam Tables: Properties of saturated liquid

water, saturated steam, and superheated

steam are tabulated in steam tables.

HU ôr ˆ

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Steam Tables


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Steam Tables


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Superheated Steam Tables


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Energy Balance Procedures

• A properly drawn and labeled flowchart is essential

for the efficient solution of energy balance

problems.

• When labeling the flowchart, be sure to include all

of the information you will need to determine the

specific enthalpy of each stream component,

including known temperatures and pressures.

• In addition, show states of aggregation of process

materials when they are not obvious: do not simply

write H2O, for example, but rather H2O(s), H2O(l),

or H2O(v), according to whether water is present as

a solid, a liquid, or a vapor.

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Example 1 -Energy Balance on a One –

Component Process

Two streams of water are mixed to form the feed to a boiler. Process data are as follows:

Feed stream 1 120 kg/min @ 30°C

Feed stream 2 175 kg/min @ 65°C

Boiler pressure 17 bar (absolute)

The exiting steam emerges from the boiler through a 6-cm ID pipe.

Calculate the required heat input to the boiler in kJ/min. if the emerging steam is saturated at the boiler pressure.

Neglect the kinetic energies of the liquid inlet streams

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Example 2:Energy Balance on a Two –

Component Process

A liquid stream containing 60.0 wt% ethane and 40.0% n-butane

is to be heated from 150K to 200K at a pressure of 5 bar.

Calculate the required heat input per kilogram of the mixture,

neglecting potential and kinetic energy changes, using tabulated

enthalpy data for C2H6 and C4H10 and assuming that mixture

component enthalpies are those of the pure species at the same

temperature.

Data:

kJ/kg 0.30ˆ

kJ/kg 2.130ˆ

kJ/kg 3.314ˆ

kJ/kg 5.434ˆ

K 150 ,

K 200 ,

K 150 ,

K 200 ,

104

104

62

62

HC

HC

HC

HC

H

H

H

H

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Example 3: Energy Balance on Steam

System

A 10.0-m3 tank contains steam at 275 oC and 15.0

bar. The tank and its contents are cooled until the

pressure drops to 1.2 bar. Some of the steam

condenses in the process.

(a). How much heat was transferred from the tank?

(b). What is the final temperature of the tanks

contents?

(c). How much steam condensed (kg)?

Home work!!!


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• You have learnt

– Forms of energy

– Specific properties

– Energy balance on a closed system

– Energy balance on an open system

Conclusions

Documents

L1-5 Energy Balances