lect-1.3 material and energy balances

9/9/2017 DAN Dharmaena

MATERIAL AND ENERGY

BALANCES

Material quantities, as they pass through food

processing operations, can be described by

material balances.

Such balances are statements on the

conservation of mass. Similarly, energy quantities

can be described by energy balances, which are

statements on the conservation of energy.

Lect-1.3


If there is no accumulation, what goes into a

process must come out. This is true for both batch

operation and for continuous operation over any

chosen time interval.

Material and energy balances are very important in

the food industry. Material balances are

fundamental to the control of processing,

particularly in the control of yields of the products.


The increasing cost of energy has caused the food

industry to examine means of reducing energy

consumption in processing.

Energy balances are used in the examination of

the various stages of a process, over the whole

process and even extending over the total food

production system from the farm to the consumer's

plate.


Material and energy balances can be

simple, at times they can be very

complicated, but the basic approach is

general.

Experience in working with the simpler

systems such as individual unit operations will

develop the facility to extend the methods to

the more complicated situations, which do

arise.


BASIC PRINCIPLES

If the unit operation, whatever its nature is

seen as a whole, it may be represented

diagrammatically as a box, as shown in Fig.

2.1. The mass and energy going into the box

must balance with the mass and energy

coming out.


Figure 2.1. Mass and

energy balance


The law of conservation of mass leads to what

is called a mass or a material balance.

Mass In = Mass Out + Mass Stored


ΣmR = ΣmP + ΣmW + ΣmS

ΣmR = mR1 + mR2 + mR3 +..... = Total Raw Materials.

ΣmP = mP1 + mP2 + mP3 + .... = Total Products.

ΣmW = mW1 + mW2 + mW3 + ....= Total Waste Products.

ΣmS = mS1 + mS2 + mS3 + ... = Total Stored Materials.

Raw Materials = Products + Wastes + Stored Materials.

If there are no chemical changes occurring in the plant, the

law of conservation of mass will apply also to each

component, so that for component A:


mA in entering materials = mA in the exit materials + mA stored in plant.

For example, in a plant that is producing sugar,

if the total quantity of sugar going into the plant

in sugar cane or sugar beet is not equalled by

the total of the purified sugar and the sugar in

the waste liquors, then there is something

wrong. Sugar is either being burned (chemically

changed) or accumulating in the plant or else it

is going unnoticed down the drain somewhere.

In this case:


(mA ) = (mAP + mAW + MAS+ mAU)

where mAU is the unknown loss and needs to be

identified.

So that the material balance is now:

Raw Materials = Products + Waste Products +

Stored Products + Losses

Where, Losses are the unidentified materials.


Just as mass is conserved, so is energy conserved in food

processing operations. The energy coming into a unit

operation can be balanced with the energy coming out and

the energy stored.

Energy In = Energy Out + Energy Stored

ΣER = ΣEP + ΣEW + ΣEL + ΣES

where:

ΣER = ER1 + ER2 + ER3 + ….…. = Total Energy Entering

ΣEP = EP1 + EP2 + EP3 + ……= Total Energy Leaving with Products

ΣEW = EW1 +EW2 + EW3 + .= Total Energy Leaving with Waste Materials

ΣEL = EL1 + EL2 + EL3 + …….... = Total Energy Lost to Surroundings

ΣES = ES1 + ES2 + ES3 + …..….. = Total Energy Stored


Energy balances are often complicated because

forms of energy can be interconverted, for example

mechanical energy to heat energy, but overall the

quantities must balance.

Mass Balance

Consider part of a flow system, such for example as that

shown in the following figure .


This consists of a continuous pipe that changes its diameter,

passing into and out of a unit of processing plant, which is

represented by a tank. The processing equipment might be,

for example, a pasteurizing heat exchanger. Also in the

system is a pump to provide the energy to move the fluid.


we can apply the law of conservation of mass to

obtain a mass balance.

Once the system is working steadily, and if there is

no accumulation of fluid in any part the system, the

quantity of fluid that goes in at section 1 must come

out at section 2.

If the area of the pipe at section 1 is A1 , the velocity

at this section, v1 and the fluid density p1, and if the

corresponding values at section 2 are A2, v2, p2, the

mass balance can be expressed as


ρ1A1v1 = ρ2A2v2 - (3.4)

If the fluid is incompressible ρ1 = ρ2 so in this case

A1v1 = A2v2 – (3.5)

3.5 Equation is known as the continuity equation for

liquids and is frequently used in solving flow problems.

It can also be used in many cases of gas flow in which

the change in pressure is very small compared with the

system pressure, such as in many air-ducting systems,

without any serious error.

9/9/2017 DAN Dharmasena

EXAMPLE1. Velocities of flow

Whole milk is flowing into a centrifuge through a full

5 cm diameter pipe at a velocity of 0.22 m s-1, and

in the centrifuge it is separated into cream of specific

gravity 1.01 and skim milk of specific gravity 1.04.

Calculate the velocities of flow of milk and of the

cream if they are discharged through 2 cm diameter

pipes. The specific gravity of whole milk of 1.035.


From eqn. (3.4):

p1A1v1 = p2A2v2 + p3A3v3

where suffixes 1, 2, 3 denote respectively raw milk,

skim milk and cream.

A1v1 = A2v2 + A3v3 and from this equation

v2 = (A1v1 - A3v3)/A2 ------------a


This expression can be substituted for v2 in the mass

balance equation to give:

p1A1v1 = p2A2(A1v1 – A3v3)/A2 + p3A3v3

p1A1v1 = p2A1v1 - p2A3v3 + p3A3v3.

So A1v1(p1 - p2) = A3v3(p3 - p2) (b)


From the known facts of the problem we have:

A1 = (¶/4) x (0.05)2 = 1.96 x 10-3 m2

A2 = A3 = (¶/4) x (0.02)2 = 3.14 x 10-4 m2

v1 = 0.22 m s-1

ρ1 = 1.035 x ρw, ρ2 = 1.04 x ρw , ρ3 = 1.01 x ρw

Where, ρw is the density of water.

Substituting these values in eqn. (b) above we obtain:

-1.96 x 10-3 x 0.22 (0.005) = -3.14 x 10-4 x v3 x (0.03)

So v3 = 0.23 m s-1


Also from eqn. (a) substitute 0.23 m s-1 for v3,

v2 = [(1.96 x 10-3 x 0.22) - (3.14 x 10-4 x 0.23)] /

3.14 x 10-4

= 1.1 m s-1

Move to material balance worked example


End


Typical content

Units and dimensions

Engineering Math

Mass and energy balance

Fluid flow theory and applications

Heat transfer theory and applications

Drying, Evaporation

Separation (cleaning, Mechanical separation, Filtration including reverse

osmosis)

Size reduction

Materials handling

Mixing

Process control and automation

lect-1.3 material and energy balances

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