3 Mass Balance Agro1

8/19/2019 3 Mass Balance Agro1

1/28

Mass and Energy Balances 1

Supratomo


2/28


• The study of process engineering is an attempt to combine all forms of physical

processing into a small number of basic operations, which are called unit

operations

• The essential concept is therefore to divide physical food processes into basicunit operations, each of which stands alone and depends on coherent physical

principles

• Because of the dependence of the unit operation on a physical principle, or a

small group of associated principles, quantitative relationships in the form of

mathematical equations can be built to describe them. The equations can beused to follow what is happening in the process, and to control and modify the

process if required.

• Two very important laws which all unit operations obey are the laws of

conservation of mass and energy.


3/28


BASIC PRINSIPLES


4/28


CONSERVATION of MASS

Mass can be neither created nor destryed. However , its

composition can be altered from one form to another.

Rate of mass

entering

through the

boundary of asystem

Rate of mass

exiting through

the boundary

of a system

Rate of mass

accumulation

within the

system

- =

∑=

=n

1i

.

inlet

.

imm ∑=

=n

1i

.

outlet

.

ommdt

mm

system

onaccumulati

d =

dt

mmm

system

outletinlet

d =−


5/28


CONSERVATION of MASS for an Open System

Consider a section of pipe used in transporting a fluid.

dAvdm o ρ = ∫∫ ==v

o dVdAvm ρ ρ A

∫∫∫ =V

oo dVdt

d dAv-dAv ρ ρ ρ

outlet inlet A A


6/28


∫∑∑ = Vo

outleto

inletdVdt

d

dAv-dAv ρ ρ ρ

For a uniform flow :

For a steady state flow flow does not change with time

dAvdAv ooutlet

o

inlet∑∑ = ρ ρ

If we are dealing with and incompressible fluid there is no change in density.

dAvdAvoutlet

o

inlet

o ∑∑ =


7/28


CONSERVATION of MASS for a Closed System

A closed system mass cannot cross system bounderies. Therefore,

there is no time rate of change of mass in the system.

0

dt

msystem =d

msystem = constant


8/28

Discusion


Which of the following statements are true and which are false?1. The mass balance is based on the law of conservation of mass.

2. Mass balance may refer to total mass balance or component

mass balance.

3. Control volume is a region in space surrounded by a control

surface through which the fluid flows.

4. Only streams that cross the control surface take part in the mass

balance.

5. At steady state, mass is accumulated in the control volume.

6. In a component mass balance, the component generation termhas the same sign as the output streams.

7. It is helpful to write a mass balance on a component that goes

through the process without any change.


9/28


Laws of Thermodynamics

1st Law

Energy can be neither created nor detroyed but can

be transformed from one form to another

2nd Law by Rudolf Clausius and Lord Kelvin

No process is possible where sole result is the

removal of heat from a reservoir (system) at one

temperature and the absorption of an equal quantity

of heat by reservoir at a higher temperature.

The 2nd law of thermodynamics help explain why heat

always flow from a hot object to a cold object.

CONSERVATION of ENERGY


10/28


Forms of Energy

Potensial Energy of a system is by vitue of its location with respect to the

gravitational field.

mgh E PE =Kinetic Energy of an object is due to its velocity.

2

2

1

mv E KE =

Internal Energy is due to the microscopic nature of the system . They move in

random direction, collide each other, vibrate and rotate.

T mc E pi ∆=

Total Energy of a system can be written in the form of an equation as :

Etotal = EKE + EPE + Eelectrical + Echemical + …….. + Ei


11/28


ENERGY BALANCE

The first law of thermodynamics states that energy can be

neither created nor destryed.

Total energy

entering the

system

Total energy

leaving the

system

Change in the

total energy at

the system- =

systemoutin Ê ÊÊ ∆=−For a system is in a steady state, there is no change in the energy of the

system with time.

0ˆˆˆ =∆=− systemout in E E E

out in E E ˆˆ =


12/28


Energy Balance for an Open System

Movement of a liquid volume

Force to push a liquid through the system boundary is : F = P.A

A : cross-sectional area

P : pressure of the fluid

The work done on the fluid element is : Wmass flow = F.L = P.A.L = PV

The total energy of the fluid element :

ET = EPE + EKE + Ei + PV PV E mv

mgz E iT +++=2

2


13/28


The overall energy balance equation for a system with one inlet (point 1)

and one outlet (point 2) is:

The overall energy balance equation for a system at steady state with more

than two streams can be written as:


14/28


Where :

In most of the cases, the overall energy balance ends up as an enthalpy balancebecause the terms of kinetic and potential energy are negligible compared to the

enthalpy term, the system is assumed adiabatic (Q = 0), and there is no shaft work

(Ws = 0). Then:


15/28

Discusion


Which of the following statements are true and which are false?1. The energy in a system can be categorize as internal energy,

potentialenergy, and kinetic energy.

2. A fluid stream carries internal energy, potential energy, and

kinetic energy.

3. A fluid stream entering or exiting a control volume is doing PV

work.

4. The internal energy and the PV work of a stream of fluid make

up the enthalpy of the stream.

5. Heat and shaft work may be transferred through the control

surface to or from the control volume.

6. Heat transferred from the control volume to the surroundings isconsidered positive by convention.

7. For an adiabatic process, the heat transferred to the system is

zero.


16/28


Steps in problem solving a Material Balance

1. Collect all known data on mass and composition of all inlet and exit streamsfrom the statement of the problem.

2. Draw a block diagram, indicating the process, with inlet and exit streams

properly identified. Draw the system boundary.

3. Write all available data on the block diagram

4. Select a suitable basis (such as mass or time) for calculations. The selection

of basis depends on the convenience of the computations.

5. Using equation of mass balance, write material balance in term of the

selected basis for calculating unknowns. For each unknown, an independent

material balance is required.

6. Solve material balances to determine the unkowns.


17/28

17

Example 1.

A wet food product contains 70 % water. After drying, it is found that 80 % of

original water has been removed. Determine (a) mass of water removed per

kilogram of wet food, and (b) composition of dried food.

Given :

Initial water content : F = 70 %

Water removed : R = 80 % of original water content

Feed (F)intial water

content 70 %

R : water removed 80 % oforiginal water content

Product (P)composition ?

Solution :

1. Select basis 1 kg wet food

2. Mass of water in inlet stream : Fwater = 70 % x 1 kg = 0.70 kg

3. Water removed in drying : R = 80 % x Fwater = 0.8 x 0.7 = 0.56 kg H2O / kg ofwet food material

4. Write material balance on water : Fwater – Pwater = R (a) Pwater = 0.14 kg

5. Write material balance on solid : Fsolid = Psolid Psolid = 0.3 kg

6. (b) the dried food contains 0.14 kg water and 0.3 kg solids

dryer


18/28

18


19/28


Example 2.

Potato flakes (moisture content (mc) 75 % wet basis) are being dried in a

concurrent flow drier. The mc of the air entering the drier is 0.08 kg H2O/kg dry air.

The mc leaving the drier is 0.18 kg H2O/kg dry air. The air flow rate in the drier is

100 kg dry air per hour. At steady state, calculate the following :

a. What is the mass flow rate of dried potatoes ?

b. What is the mc, dry basis, of dried potatoes exiting the drier ?

Inlet air 0.08 kg H2O/kg dry air 100 kg/h dry air

Feed50 kg wet potato flakes/h

MC 75 % wet basis

Exit air 0.18 kg H2O/kg dry air

Productmoisture content ?

dryer

Given :

Weight of potato flakes entetring the dryer F = 50 kg with 75 % of MC wb

Moisture content of air entering the dryer is 0.08 kg H2O/kg dry air and the flow

rate is 100 kg dry air/h


20/28


Solution :

1. Basis = 1 hour

2. Mass of air entering the dryer = mass of dry air + mass of water

I = 100 + 0.08 x 100 = 108 kg

3. Mass of air leaving the dryer = mass of dry air + mass of water

E = 100 + 0.18 x 100 = 118 kg

4. Total balance on the dryer

I + F = E + P 108 + 50 = 118 + P (a) P = 40 kg/h

5. Solid balance on the dryer Mass of dry solids entering the dryer = mass of dry solid leaving the dryer

(1 – 0.75) F = y P

6. Therefore, moisture content (wet basis) of the product : 1 – 0.3125 =0.6875 kg

7. The dry basis MC of the Product

kg3125.040

5025.0==

x y

solidsdryO/kgHkg2.23125.0

6875.0

solidsdryof mass

water of mass2 _ ===basisdry MC


21/28

21


22/28


Example 3.

A tubular water blancher is being used to process lima beans. The product

mass flow is 860 kg/h. it is found that the theoretical energy consumed

for blanching process amounts to 1.19 GJ/h. the energy lost due to lack

of insulation around the blancher is estimate to be 0.24 GJ/h. if the totalenergy input to blancher is 2.71 GJ/h,

a. Calculate the energy required to reheat water

b. Determine the percent energy associated with each stream

Water

blancher

Energy losses from

surface = 0.24 GJ/h

Energy input =

2.71 GJ/h

Energy leaving the

product = 1.19 GJ/h

Energy leaving with water ?

Given

Product mass flow rate = 860 kg/h

Theoritical energy consumed for blanching process = 1.19 GJ/h

Energy loss due to lack of insulation = 0.24 GJ/h

Energy input to blancher = 2.71 GJ/h


23/28


Solution

Select 1 hour as a basis

Write energy balance

E input to blancher = E out with product + E out with water + E losses fromsurfaces

2.71 = 1.19 + Eo_water + 0.24 Eo_water = 1.28 GJ/h

a. E required to reheat water = E input – E o_water = 2.71 – 1.19 = 1.43 GJ/h

b. Percent energy associate with :

- Product (out) = 1.19/2.71 x 100 % = 43.91 %

- Water (out) = 1.28/2.71 x 100 % = 47.23 %

- Losses = 0.24/2.71 x 100 % = 8.86 %


24/28


Example 4.

Steam is used for peeling of potatoes in a semicontinuous operation. Steam is

supplied at the rate of 4 kg per 100 kg of unpeeled potatoes. The unpeeled

potatoes enter the system with a temperature of 17 oC, and peeled pototoes

leave at 35o

C. A specific heat of unpeeled potatoes, waste stream, and peeledpotatoes are 3.7, 4.2 and 3.5 kJ/(kg.K), respectively. If the heat content

(assuming 0 oC reference temperature) of the steam is 2750 kJ/kg, determine

the quantities of the waste stream and peeled potatoes from the process.


25/28


Given : up unpeeled potatoes pp peeled potatoes waste steam wt

Mass flow of steam = 4 kg per 100 kgup

Tup = 17o

C Tpp = 35o

C Twt = 60o

CCpup = 3.7 kJ/kg-K Cppp = 3.5 kJ/kg-K Cpwt = 4.2 kJ/kg-K

Heat content of steam (H) = 2750 kJ/kg

Assuming 0 oC as temperature’s reference

Potato peeler

F = 100 kg

Tf = 17oC

W = ?

Tw = 60oC

Hs = 2750 kJ/kgS= 4 kg

P = ?

Tp = 35oC


26/28


Solution

Select 100 kg of up as a basis

From mass balance

F + S = W + P

100 + 4 = W + PW = 104 – P (I)

From energy balance

Einput_up + Einput_s = Eout_ws + Eout_pp

F Cpup ΔT + SH = W Cpwt ΔT + P Cppp ΔT

100 x 3700 x (17-0) + 4 x 2750 = W x 4200 x (60-0) + P x 3500 x (35-0)

6290 + 11000 = 252 W + 122.5 P

(I) 17290 = 252 (104 – P) + 122.5 P P = 68.87 kg

W = 35.14 kg


27/28



28/28


1. 10 kg of food at a moisture content of 320% dry basis is dried to 50% wet

basis. Calculate the amount of water removed.

2. A batch of 5 kg of food product has a moisture content of 150% dry basis.Calculate how much water must be removed from this product to reduce its

moisture content to 20% wet basis.

3. A liquid product with 10% product solids is blended with sugar before being

concentrated (removal of water) to obtain a final product with 15% product

solids and 15% sugar solids. Determine the quantity of final product obtained

from 200 kg of liquid product. How much sugar is required? Compute mass of

water removed during concentration.

4. 1000 kg/h of milk is heated in a heat exchanger from 45 oC to 72 oC. Water is

used as the heating medium. It enters the heat exchanger at 90 oC and leaves

at 75 oC. Calculate the mass flow rate of the heating medium, if the heat losses

to the environment are equal to 1 kW. The heat capacity of water is given equalto 4.2 kJ/kgoC and that of milk 3.9 kJ/kgoC.

5. How much saturated steam with 120.8 kPa pressure is required to heat 1000

g/h of juice from 5 oC to 95 oC? Assume that the heat capacity of the juice is 4

kJ/ kgoC.

Exercises

Documents

3 Mass Balance Agro1