E Balance1 Wble

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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 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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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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