Dasar Keteknikan Pengolahan
Pangan
Sudarminto Setyo Yuwono
Apple Cooling
qfrig
INTRODUCTION
• 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
• Food processes may seem bewildering in their diversity, but careful
analysis will show that these complicated and differing processes
can be broken down into a small number of unit operations
• Important unit operations in the food industry are fluid flow, heat
transfer, drying, evaporation, contact equilibrium processes (which
include distillation, extraction, gas absorption, crystallization, and
membrane processes), mechanical separations (which include
filtration, centrifugation, sedimentation and sieving), size reduction
and mixing.
DIMENSIONS AND UNITS
• All engineering deals with definite and measured
quantities, and so depends on the making of
measurements
• To make a measurement is to compare the unknown with
the known
• record of a measurement consists of three parts: the
dimension of the quantity, the unit which represents a
known or standard quantity and a number which is the ratio
of the measured quantity to the standard quantity
Dimensions
• These dimensions are length, mass, time and
temperature.
• For convenience in engineering calculations, force
is added as another dimension
• Dimensions are represented as symbols by: length
[L], mass [M], time [t], temperature [T] and force [F].
•
example:
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Length = [L], area = [L] 2, volume = [L] 3.
Velocity = length travelled per unit time=[L]/[t]
Pressure
= force per unit area=[F]/[L] 2
Energy
= force times length=[F] x [L].
Power
= energy per unit time=[F] x [L]/[t]
h=[F] x [L]/[L]2 [t] [T] =[F] [L] -1 [t] -1 [T] -1
Units
• the metre (m) is defined in terms of the wavelength of light;
• the standard kilogram (kg) is the mass of a standard lump of
platinum-iridium;
• the second (s) is the time taken for light of a given wavelength to
vibrate a given number of times;
• the degree Celsius (°C) is a one-hundredth part of the temperature
interval between the freezing point and the boiling point of water at
standard pressure;
• the unit of force, the newton (N), is that force which will give an
acceleration of 1 m sec-2 to a mass of 1kg;
• the energy unit, the newton metre is called the joule (J),
• the power unit, 1 J s-1, is called the watt (W).
Dimensionless Ratios
•
It is often easier to visualize quantities if they are expressed in ratio form and ratios have the
great advantage of being dimensionless
•
For example, specific gravity is a simple way to express the relative masses or weights of
equal volumes of various materials. The specific gravity is defined as the ratio of the weight
of a volume of the substance to the weight of an equal volume of water
SG = weight of a volume of the substance/ weight of an equal volume of water .
Dimensionally, SG=[F]/ [L]-3 divided by[F]/ [L]-3 = 1
it gives an immediate sense of proportion
This sense of proportion is very important to food technologists as they are constantly
making approximate mental calculations for which they must be able to maintain correct
proportions
Another advantage of a dimensionless ratio is that it does not depend upon the units of
measurement used, provided the units are consistent for each dimension
Dimensionless ratios are employed frequently in the study of fluid flow and heat flow. These
dimensionless ratios are then called dimensionless numbers and are often called after a
prominent person who was associated with them, for example Reynolds number, Prandtl
number, and Nusselt number
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Neraca Massa
• Tahapan perhitungan:
– Gambar diagram
– Tulis reaksi kimia jika ada
– Tulis dasar-dasar perhitungan
– Hitung neraca massanya
Mass Balance
• soda api (NaOH), sebanyak 1000 kg/jam larutan
mengandung 7,5% NaOH di pekatkan pada
evaporator yang pertama sehingga kadarnya
menjadi 20%. Larutan ini lalu dipekatkan pada
evaporator yang kedua yang menghasilkan pekatan
berkadar 60% NaOH. Hitung NaOH yang
dihasilkan.
Dikerjakan dan dikumpulkan
• Proses produksi jam buah dilakukan dengan cara
memekatkan bubur buah dari kadar padatan 10%
menjadi 35%. Pemekatan dilakukan dalam 2 tahap
evaporator. Pada evaporator yang pertama kadar
padatan meningkat menjadi 22%. Hitung jam buah
yang dihasilkan untuk tiap 100 kg bubur buah yang
dipakai.
Neraca massa jika terjadi reaksi kimia
• Beberapa proses pengolahan kemungkinan terjadi
reaksi kimia
– Fermentasi
– Pembakaran
• Dasar perhitungan bukan dari massa tetapi dari
perubahan mol
• Setelah itu baru dikonversikan ke massa
contoh
• Bahan bakar mengandung 5 %mol H2, 30 %mol
CO, 5 %mol CO2, 1 %mol O2, dan 59 %mol N2.
Dibakar dengan media udara. Untuk 100 kg mol
bahan bakar hitung mol gas buang dan
kamponennya, jika :
• A. Pembakaran sempurna, udara pas
• B. Pembakaran 90% sempurna
• C. Udara berlebih 20%, pembakaran sempurna
80%
contoh
• Larutan NaOH diproduksi dengan cara
menambahkan larutan Na2CO3 10% ke dalam
aliran bubur Ca(OH) 2 25%. Bagaimana komposisi
bubur akhir jika reaksi 90% sempurna dan 100%
sempurna. Gunakan dasar 100 kg aliran bubur
Ca(OH)2
• Na2CO3 + Ca(OH) 2 => 2NaOH + CaCO3
• Ca(OH) 2= 74,1; Na2CO3 = 106
Recycle
• Pada suatu proses produksi sodium sitrat,
1000kg/jam larutan sodium sitrat berkadar 20%
dipekatkan di suatu evaporator bersuhu 353K
sehingga diperoleh kadar 50%. Larutan lalu
dimasukkan ke kristalizer yang bersuhu 303K
sehingga diperoleh kristal Na sitrat berkadar 95%.
Larutan jenuh yang mengandung 30% Na sitrat lalu
direcycle ke evaporator. Hitung berapa laju aliran
recycle dan produk yang dihasilkan.
Harap dikerjakan
• Pada industri gula, larutan gula 1000 kg/jam
berkadar 30% dipekatkan hingga berkadar 60%.
Larutan tersebut lalu dimasukkan ke kristalizer
sehingga diperoleh kristal gula berkadar air 5%.
Larutan jenuh berkadar gula 40% selanjutnya
direcycle ke evaporator lagi. Hitung jumlah larutan
yang direcycle dan gula yang dihasilkan.
FLUID FLOW THEORY
• Many raw materials for foods and many finished
foods are in the form of fluids.
• Thin liquids - milk, water, fruit juices,
Thick liquids - syrups, honey, oil, jam,
Gases - air, nitrogen, carbon dioxide,
Fluidized solids - grains, flour, peas.
• The study of fluids can be divided into:
– the study of fluids at rest - fluid statics, and
– the study of fluids in motion - fluid dynamics.
FLUID STATICS
• very important property : the fluid pressure
• Pressure is force exerted on an area
• force is equal to the mass of the material multiplied by
the acceleration due to gravity.
• mass of a fluid can be calculated by multiplying its volume
by its density
• F = mg = Vρg
• F is force (Newton) or kg m s-2, m is the mass, g the
acceleration due to gravity, V the volume and ρ the
density.
The force per unit area in a fluid is called the fluid pressure. It is exerted
equally in all directions.
• F = APs + ZρAg
• Ps is the pressure above the surface of the
fluid (e.g. it might be atmospheric pressure
• total pressure P = F/A = Ps + Zρg
• the atmospheric pressure represents a
datum P = Zρg
EXAMPLE . Total pressure in a tank of
peanut oil
• Calculate the greatest pressure in a spherical tank,
of 2 m diameter, filled with peanut oil of specific
gravity 0.92, if the pressure measured at the
highest point in the tank is 70 kPa.
• Density of water
= 1000 kg m-3
Density of oil
= 0.92 x 1000 kg m-3 = 920 kg m-3
Z =greatest depth
=2m
and
g = 9.81 m s-2
Now P = Zρg
= 2 x 920 x 9.81 kg m-1 s-2
= 18,050 Pa = 18.1 kPa.
• To this must be added the pressure at the surface of 70 kPa.
• Total pressure
= 70 + 18.1 = 88.1 kPa.
• the pressure depends upon the pressure at the top of the tank, the
depth of the liquid
Expressing the pressure
• absolute
pressures
• gauge
pressures
• head
EXAMPLE. Head of Water
• Calculate the head of water equivalent and mercury to standard
atmospheric pressure of 100 kPa.
• Density of water = 1000 kg m-3, Density of mercury = 13,600 kg m-3
g = 9.81 m s-2
and pressure = 100 kPa
= 100 x 103 Pa = 100 x 103 kg m-1s-2.
Water
Z
Mercury Z
= P/ ρ g
= (100 x 103)/ (1000 x 9.81)
= 10.2 m
= (100 x 103)/ (13,600 x 9.81)
= 0.75m
FLUID DYNAMICS
• In most processes fluids have to be moved
• Problems on the flow of fluids are solved by
applying the principles of conservation of mass and
energy
• The motion of fluids can be described by writing
appropriate mass and energy balances and these
are the bases for the design of fluid handling
equipment.
Mass Balance
• ρ1A1v1 = ρ2A2v2
• incompressible
ρ1 = ρ2
so in this case
• A1v1 = A2v2
(continuity equation)
• area of the pipe at
section 1 is A1 , the
velocity at this section,
v1 and the fluid density
ρ1 , and if the
corresponding values
at section 2 are A2, v2,
ρ2
EXAMPLE. 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.
Solving
• ρ1A1v1 = ρ2A2v2 + ρ3A3v3
• where suffixes 1, 2, 3 denote respectively raw milk, skim milk and
cream.
• since the total leaving volumes equal the total entering volume
• A1v1 = A2v2 + A3v3
• v2 = (A1v1 - A3v3 )/A2
• ρ1A1v1 = ρ2A2(A1v1 – A3v3)/A2 + ρ3A3v3
• ρ1 A1v1 = ρ2 A1v1 - ρ2 A3v3 + ρ3 A3v3
• A1v1(ρ1 - ρ2 ) = A3v3(ρ3 - ρ2 )
• A1 = (π/4) x (0.05)2 = 1.96 x 10-3 m2
• A2 = A3 = (π/4) x (0.02)2 = 3.14 x 10-4 m2
v1 = 0.22 m s-1
ρ1 = 1.035, ρ2 = 1.04, ρ3 = 1.01
• A1v1(ρ1 - ρ2 ) = A3v3(ρ3 - ρ2 )
• -1.96 x 10-3 x 0.22 (0.005) = -3.14 x 10-4 x v3 x (0.03)
v3 = 0.23 m s-1
• v2 = (A1v1 - A3v3 )/A2
• v2 = [(1.96 x 10-3 x 0.22) - (3.14 x 10-4 x 0.23)] / 3.14 x 10-4
= 1.1m s-1
Energy Balance
• Referring Fig. before we shall consider the changes in the total energy of unit
mass of fluid, one kilogram, between Section 1 and Section 2.
• Firstly, there are the changes in the intrinsic energy of the fluid itself which include
changes in:
(1) Potential energy = Ep = Zg (J)
(2) Kinetic energy = Ek = v2/2 (J)
(3) Pressure energy = Er = P/ρ (J)
• Secondly, there may be energy interchange with the surroundings including:
(4) Energy lost to the surroundings due to friction = Eƒ (J).
(5) Mechanical energy added by pumps = Ec (J).
(6) Heat energy in heating or cooling the fluid
• In the analysis of the energy balance, it must be remembered that energies are
normally measured from a datum or reference level.
• Ep1 + Ek1 + Er1 = Ep2 + Ek2 + Er2 + Ef - Ec.
• Z1g + v12/2 + P1/ρ1 = Z2g + v22/2 + P2/ρ2 + Ef - Ec.
• Zg + v2/2 + P/ρ = k
Persamaan Bernouilli
Water flows at the rate of 0.4 m3 min-1 in a 7.5 cm diameter pipe at a pressure of 70
kPa. If the pipe reduces to 5 cm diameter calculate the new pressure in the pipe.
Density of water is 1000 kg m-3.
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Flow rate of water = 0.4 m3 min-1 = 0.4/60 m3 s-1.
Area of 7.5 cm diameter pipe = (π/4)D2
= (π /4)(0.075)2
= 4.42 x 10-3 m2.
So velocity of flow in 7.5 cm diameter pipe,
v1 = (0.4/60)/(4.42 x 10-3) = 1.51 m s-1
Area of 5 cm diameter pipe = (π/4)(0.05)2
= 1.96 x 10-3 m2
and so velocity of flow in 5 cm diameter pipe,
v2 = (0.4/60)/(1.96 x 10-3) = 3.4 m s-1
Now
Z1g + v12/2 + P1 /ρ1 = Z2g + v22/2 + P2 / ρ2
and so
0 + (1.51)2/2 + 70 x 103/1000 = 0 + (3.4)2/2 + P2/1000
0 + 1.1 + 70 = 0 + 5.8 + P2/1000
P2/1000 = (71.1 - 5.8) = 65.3
P2 = 65.3k Pa.
Water is raised from a reservoir up 35 m to a storage tank through a 7.5 cm diameter pipe. If it
is required to raise 1.6 cubic metres of water per minute, calculate the horsepower input to a
pump assuming that the pump is 100% efficient and that there is no friction loss in the pipe.
1 Horsepower = 0.746 kW.
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Volume of flow, V = 1.6 m3 min-1 = 1.6/60 m3 s-1 = 2.7 x 10-2 m3 s-1
Area of pipe, A = (π/4) x (0.075)2 = 4.42 x 10-3 m2,
Velocity in pipe, v = 2.7 x 10-2/(4.42 x 10-3) = 6 m s-1,
And so applying eqn
Z1g + v12/2 + P1/ρ1 = Z2g + v22/2 + P2/ρ2 + Ef - Ec.
Ec = Zg + v2/2
Ec = 35 x 9.81 + 62/2
= 343.4 + 18
= 361.4 J
Therefore total power required
= Ec x mass rate of flow
= EcVρ
= 361.4 x 2.7 x 10-2 x 1000 J s-1
= 9758 J s-1
and, since 1 h.p. = 7.46 x 102 J s-1,
required power = 13 h.p.
VISCOSITY
• Viscosity is that property of a fluid that gives rise to forces
that resist the relative movement of adjacent layers in the
fluid.
• Viscous forces are of the same character as shear
forces in solids and they arise from forces that exist
between the molecules.
• If two parallel plane elements in a fluid are moving relative
to one another, it is found that a steady force must be
applied to maintain a constant relative speed. This force is
called the viscous drag because it arises from the action of
viscous forces.
If the plane elements are at a distance Z apart, and if their relative velocity is v, then
the force F required to maintain the motion has been found, experimentally, to be
proportional to v and inversely proportional to Z for many fluids. The coefficient of
proportionality is called the viscosity of the fluid, and it is denoted by the symbol µ
(mu).
From the definition of viscosity we can write
F/A = µ v/Z
Unit of Viscosity
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N s m-2 = Pascal second, Pa s,
The older units, the poise and its sub-unit the centipoise,
1000 centipoises = 1 N s m-2, or 1 Pa s.
the viscosity of water at room temperature 1 x 10-3 N s m-2
acetone, 0.3 x 10-3 N s m-2;
tomato pulp, 3 x 10-3;
olive oil, 100 x 10-3;
molasses 7000 N s m-3.
Viscosity is very dependent on temperature decreasing sharply as
the temperature rises. For example, the viscosity of golden syrup is
about 100 N s m-3 at 16°C, 40 at 22°C and 20 at 25°C.
Newtonian and Non-Newtonian Fluids
• F/A = µ v /Z = µ(dv/dz) = t
 t = k(dv/dz)n power-law equation
• Newtonian fluids (n = 1, k = µ )
• Non-Newtonian fluids (n ≠ 1)
• (1) Those in which n < 1. The viscosity is apparently
high under low shear forces decreasing as the shear
force increases. Pseudoplastic (tomato puree)
• (2) Those in which n > 1. With a low apparent viscosity
under low shear stresses, they become more viscous
as the shear rate rises. Dilatancy (gritty slurries such as
crystallized sugar solutions).
• Bingham fluids have to exceed a particular shear stress
level (a yield stress) before they start to move.
• Food : Non-Newtonian
STREAMLINE AND TURBULENT FLOW
• STREAMLINE, flow is calm, in slow the pattern and
smooth
• TURBULENT, the flow is more rapid, eddies develop and
swirl in all directions and at all angles to the general line of
flow.
rv2D/mv = Dvr/m =Reynolds number (Re), dimensionless
• D is the diameter of the pipe
• For (Re) < 2100 streamline flow,
For 2100 < (Re) < 4000 transition,
For (Re) > 4000 turbulent flow.
EXAMPLE . Flow of milk in a pipe
Milk is flowing at 0.12 m3 min-1 in a 2.5-cm diameter pipe. If the
temperature of the milk is 21°C, is the flow turbulent or streamline?
• Viscosity of milk at 21°C = 2.1 cP = 2.10 x 10-3 N s m-2
Density of milk at 21°C = 1029 kg m-3.
Diameter of pipe = 0.025 m.
Cross-sectional area of pipe = (p/4)D2
= p/4 x (0.025)2
= 4.9 x 10-4 m2
Rate of flow = 0.12 m3 min-1 = (0.12/60) m3 s-1 = 2 x 10 m3 s-1
• So velocity of flow = (2 x 10-3)/(4.9 x 10-4)
= 4.1 m s-1,
and so (Re) = (Dvr/m)
= 0.025 x 4.1 x 1029/(2.1 x 10-3)
= 50,230
and this is greater than 4000 so that the flow is turbulent.
ENERGY LOSSES IN FLOW
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Friction in Pipes
Energy Losses in Bends and Fittings
Pressure Drop through Equipment
Equivalent Lengths of Pipe
Friction in Pipes
• Eƒ : the energy loss due to friction in the pipe.
• Eƒ : proportional to the velocity pressure of the fluid and to a factor related
to the smoothness of the surface over which the fluid is flowing.
• F/A = f rv2/2
• F is the friction force, A is the area over which the friction force acts, r is
the density of the fluid, v is the velocity of the fluid, and f is a coefficient
called the friction factor (depends upon the Reynolds number for the flow, and
upon the roughness of the pipe).
•

•
•
P1 - P2 = (4frv2/2)(L1 - L2)/D
DPf = (4frv2/2) x (L/D)
(Fanning-D'Arcy equation)
Eƒ = DPf/r = (2fv2)(L/D)
L = L1 - L2 = length of pipe in which the pressure drop, DPf = P1 - P2 is the
frictional pressure drop, and Eƒ is the frictional loss of energy.
Friction factors in pipe (Moody graph)
predicted f
• f = 16/(Re) streamline flow, Hagen-Poiseuille
equation 0 < (Re) < 2100
• ƒ = 0.316 ( Re)-0.25/4 ( Blasius equation for smooth
pipes in the range 3000 < (Re) < 100,000)
• roughness ratio = Roughness factor (e)/pipe
diameter
(turbulent region)
•
ROUGHNESS FACTORS FOR PIPES
Material
Roughness factor
(e)
Riveted steel
0.001- 0.01
Concrete
0.0003 - 0.003
Wood staves
0.0002 - 0.003
Cast iron
0.0003
Material
Galvanized
iron
Asphalted
cast iron
Commercial
steel
Roughness factor
(e)
0.0002
0.001
0.00005
Drawn tubing Smooth
EXAMPLE Pressure drop in a pipe
Calculate the pressure drop along 170 m of 5 cm diameter horizontal steel pipe
through which olive oil at 20°C is flowing at the rate of 0.1 m3 min-1
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Diameter of pipe = 0.05 m,
Area of cross-section A
= (π/4)D2
= π /4 x (0.05)2
= 1.96 x 10-3 m2
From Appendix 4,
Viscosity of olive oil at 20°C = 84 x 10-3 Ns m-2 and density = 910 kg m-3,
and velocity = (0.1 x 1/60)/(1.96 x 10-3) = 0.85 m s-1,
Now
(Re) = (Dvρ/µ)
= [(0.05 x 0.85 x 910)/(84 x 10-3)]
= 460
so that the flow is streamline, and from Fig. moody, for (Re) = 460
f = 0.03.
Alternatively for streamline flow from f = 16/(Re) = 16/460 = 0.03 as before.
And so the pressure drop in 170 m,
DPf = (4frv2/2) x (L/D)
= [4 x 0.03 x 910 x (0.85)2 x 1/2] x [170 x 1/0.05]
= 1.34 x 105 Pa
= 134 kPa.
Thermal
conductivity
Water
Sucrose 20% soln.
Specific heat
Density
Viscosity
Temperature
(J m-1 s-1 °C-1)
(kJ kg-1 °C-1)
(kg m-3)
(N s m-2)
(°C)
0.57
4.21
1000
1.87 x 10-3
0
4.21
987
0.56 x 10-3
50
0.68
4.18
958
0.28 x 10-3
100
0.54
3.8
1070
1.92 x 10-3
20
0.59 x 10-3
80
6.2 x 10-3
20
5.4 x 10-3
80
60% soln.
Sodium chloride 22%
soln.
0.54
3.4
1240
2.7 x 10-3
2
Olive oil
0.17
2.0
910
84 x 10-3
20
Rape-seed oil
900
118 x 10-3
20
Soya-bean oil
910
40 x 10-3
30
Tallow
900
18 x 10-3
65
1030
2.12 x 10-3
20
Milk (skim)
1040
1.4 x 10-3
25
Cream 20% fat
1010
6.2 x 10-3
3
30% fat
1000
13,8 x 10-3
3
Milk (whole)
0.56
3.9
Energy Losses in Bends and Fittings
• energy losses due to altering the direction of flow,
fittings of varying cross-section
• This energy is dissipated in eddies and additional
turbulence and finally lost in the form of heat.
• Eƒ = kv2/2
Losses in fittings
• Ef = (v1 - v2)2/2
• Ef = kv22/2
Losses in sudden enlargements
Losses in sudden contraction
FRICTION LOSS
FACTORS IN FITTINGS
k
Valves, fully open:
gate
0.13
globe
6.0
angle
3.0
90° standard
0.74
medium sweep
0.5
long radius
0.25
square
1.5
LOSS FACTORS IN
CONTRACTIONS
Elbows:
Tee, used as elbow
1.5
Tee, straight through
0.5
Entrance, large tank to pipe:
sharp
0.5
rounded
0.05
D2/D1
0.1
0.3
0.5
0.7
0.9
k
0.36
0.31
0.22
0.11
0.02
FLUID-FLOW APPLICATIONS
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Two practical aspects of fluid flow in food technology :
measurement in fluids: pressures and flow rates, and
production of fluid flow by means of pumps and fans.
Pumps and fans are very similar in principle and usually
have a centrifugal or rotating action
• a gas : moved by a fan,
• a liquid: moved by a pump.
MEASUREMENT OF PRESSURE IN A
FLUID
• Method :
– Piezometer ("pressure measuring") tube
– U-tube
– Pitot tube
– Pitot-static tube
– Bourdon-tube
• P = Z1r1g
EXAMPLE. Pressure in a vacuum evaporator
The pressure in a vacuum evaporator was measured by using a U-tube containing
mercury. It was found to be less than atmospheric pressure by 25 cm of mercury.
Calculate the extent by which the pressure in the evaporator is below atmospheric
pressure (i.e. the vacuum in the evaporator) in kPa, and also the absolute pressure in
the evaporator. The atmospheric pressure is 75.4 cm of mercury and the specific
gravity of mercury is 13.6, and the density of water is 1000 kg m-3.
• We have P = Zrg
= 25 x 10-2 x 13.6 x 1000 x 9.81
= 33.4 kPa
• Therefore the pressure in the evaporator is 33.4 kPa below atmospheric pressure
and this is the vacuum in the evaporator.
• For atmospheric pressure:
• P = Zrg
• P = 75.4 x 10-2 x 13.6 x 1000 x 9.81
= 100.6 kPa
Therefore the absolute pressure in the evaporator
= 100.6 - 33.4
= 67.2 kPa
MEASUREMENT OF VELOCITY IN A FLUID
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•
•
Pitot tube and manometer :
Z1g + v12/2 + P1/r1 = Z2g + v22/2 + P2/r1
Z2 = Z + Z'
Z' be the height of the upper liquid surface in the pipe above the datum,
Z be the additional height of the fluid level in the tube above the upper liquid
surface in the pipe;
Z' may be neglected if P1 is measured at the upper surface of the liquid in the pipe,
or if Z' is small compared with Z
v2 = 0 as there is no flow in the tube
P2 = 0 if atmospheric pressure is taken as datum and if the top of the tube is open
to the atmosphere
Z1 = 0 because the datum level is at the mouth of the tube.
v12/2g + P1/r1 = (Z + Z')g  Z.
Pitot-static tube
Z = v2/2g
EXAMPLE . Velocity of air in a duct
Air at 0°C is flowing through a duct in a chilling system. A Pitot-static
tube is inserted into the flow line and the differential pressure head,
measured in a micromanometer, is 0.8 mm of water. Calculate the
velocity of the air in the duct. The density of air at 0°C is 1.3 kg m-3.
• Z = v12/2g
 r1Z1 = r2Z2.
• Now 0.8 mm water
= 0.8 x 10-3 x
= 0.62 m of air
• v 12
= 2Zg
= 2 x 0.62 x 9.81
= 12.16 m2s-2
• Therefore v1 = 3.5 m s-1
1000
1.3
Venturi and orifice meters
•
v12/2 + P1/r1 = v22/2+ P2/r2 (Bernouilli's equation)
•
A1v1 = A2v2

r1 = r2 = r

v12/2 + P1/r = (v1A1/A2)2/2 + P2/r
(mass balance, eqn)
v12 = [2(P2 -P1)/r] x A22/(A22 -A12)

(P2 -P1)/r = gZrm /r

Z = (P2 -P1)/rm g

v1 = C √[2(P2 -P1 )/r]x A22/(A22 -A12 )

In a properly designed Venturi meter,
C lies between 0.95 and 1.0.
Pompa dan Fan
• mechanical energy from some other source is converted
into pressure or velocity energy in a fluid.
• The food technologist is not generally much concerned
with design details of pumps, but should know what
classes of pump are used and something about their
characteristics.
• The efficiency of a pump is the ratio of the energy supplied
by the motor to the increase in velocity and pressure
energy given to the fluid.
Jenis Pompa
• Positive Displacement Pumps
• the fluid is drawn into the pump and is then forced
through the outlet
• Positive displacement pumps can develop highpressure heads but they cannot tolerate throttling or
blockages in the discharge.
Jet Pumps
• a high-velocity jet is produced in a Venturi nozzle, converting the
energy of the fluid into velocity energy.
• This produces a low-pressure area causing the surrounding fluid to
be drawn into the throat
• Jet pumps are used for difficult materials that cannot be satisfactorily
handled in a mechanical pump.
• They are also used as vacuum pumps.
• Jet pumps have relatively low efficiencies but they have no moving
parts and therefore have a low initial cost.
Air-lift Pumps
•
•
•
•
•
air or gas can be used to impart energy to the liquid
The air or gas can be either provided from external sources or produced by boiling within the
liquid. Examples of the air-lift principle are:
Air introduced into the fluid as shown in Fig. 4.3(e) to pump water from an artesian well.
Air introduced above a liquid in a pressure vessel and the pressure used to discharge the
liquid.
Vapours produced in the column of a climbing film evaporator.
In the case of powdered solids, air blown up through a bed of powder to convey it in a
"fluidized" form.
A special case of this is in the evaporator, where boiling of the liquid generates the gas
(usually steam) and it is used to promote circulation. Air or gas can be used directly to
provide pressure to blow a liquid from a container out to a region of lower pressure.
Air-lift pumps and air blowing are inefficient, but they are convenient for materials which will
not pass easily through the ports, valves and passages of other types of pumps.
Propeller Pumps and Fan
• Propellers can be used to impart energy to fluids
• They are used extensively to mix the contents of tanks and
in pipelines to mix and convey the fluid.
• Propeller fans are common and have high efficiencies.
• They can only be used for low heads, in the case of fans
only a few centimetres or so of water
Centrifugal Pumps and Fans
• The centrifugal pump converts rotational energy into velocity and
pressure energy
• The fluid to be pumped is taken in at the centre of a bladed rotor and
it then passes out along the spinning rotor, acquiring energy of
rotation. This rotational energy is then converted into velocity and
pressure energy at the periphery of the rotor.
• Centrifugal fans work on the same principles. These machines are
very extensively used and centrifugal pumps can develop moderate
heads of up to 20 m of water. They can deliver very large quantities
of fluids with high efficiency.
Gambar jenis-jenis pompa
• EXAMPLE Centrifugal pump for raising water
Water for a processing plant is required to be stored in a reservoir to
supply sufficient working head for washers. It is believed that a
constant supply of 1.2 m3 min-1 pumped to the reservoir, which is 22
m above the water intake, would be sufficient. The length of the pipe
is about 120 m and there is available galvanized iron piping 15 cm
diameter. The line would need to include eight right-angle bends.
There is available a range of centrifugal pumps whose characteristics
are shown in Fig. 4.4. Would one of these pumps be sufficient for the
duty and what size of electric drive motor would be required?
Reynold number
• Assume properties of water at 20°C are density 998 kg m-3, and viscosity 0.001 N
s m-2
• Cross-sectional area of pipe A = (π/4)D2
= π /4 x (0.15)2
= 0.0177 m-2
Volume of flow V = 1.2 m3 min-1
= 1.2/60 m3 s-1
= 0.02 m3 s-1.
•
Velocity in the pipe = V/A
= (0.02)/(0.0177)
= 1.13 ms-1
•
Now (Re) = Dvρ/µ
•
= (0.15 x 1.13 x 998)/0.001
= 1.7 x 105
so the flow is clearly turbulent.
friction loss of energy
From Table 3.1, the roughness factor ε is 0.0002 for galvanized
iron
and so
roughness ratio ε /D = 0.0002/0.15 = 0.001
So from Fig. 3.8,
ƒ = 0.0053
Therefore the friction loss of energy
= (4ƒv2/2) x (L/D)
= [4ƒv2L/2D]
= [4 x 0.0053 x (1.13)2 x 120]/(2 x 0.15)
= 10.8 J.
TABLE 3.1
RELATIVE ROUGHNESS FACTORS FOR PIPES
Material
Roughness factor (e)
Material
Roughness factor (e)
Riveted steel
0.001- 0.01
Galvanized iron
0.0002
Concrete
0.0003 - 0.003
Asphalted cast
iron
0.001
Wood staves
0.0002 - 0.003
Commercial
steel
0.00005
Cast iron
0.0003
Drawn tubing
Smooth
Friction factors in pipe
TABLE 3.2
FRICTION LOSS FACTORS IN FITTINGS
k
Valves, fully open:
gate
0.13
globe
6.0
angle
3.0
90° standard
0.74
medium sweep
0.5
long radius
0.25
square
1.5
Elbows:
Tee, used as elbow
1.5
Tee, straight through
0.5
Entrance, large tank to pipe:
sharp
0.5
rounded
0.05
Energy loss from bends and discharge
• For the eight right-angled bends, from Table 3.2 we would
expect a loss of 0.74 velocity energies at each, making (8 x
0.74) = 6 in all.
velocity energy = v2/2
= (1.13)2/2
= 0.64 J
• So total loss from bends and discharge energy
= (6 + 1) x 0.64
= 4.5 J
There would be one additional velocity energy loss
because of the unrecovered flow energy discharged into
the reservoir.
Energy to move 1 kg water
• Energy to move 1 kg water against a head of 22 m
of water is
E = Zg
= 22 x 9.81
= 215.8 J.
• Total energy requirement per kg:
Etot = 10.8 + 4.5 + 215.8
= 231.1 J
energy requirement of pump
• and theoretical power requirement
= Energy x volume flow x density
= (Energy/kg) x kgs-1
= 231.1 x 0.02 x 998
= 4613 J s-1.
• Now the head equivalent to the energy requirement
= Etot/g
= 231.1/9.81
= 23.5 m of water,
• and from Fig. 4.4 this would require the 150 mm impeller pump to be
safe, and the pump would probably be fitted with a 5.5 kW motor.
Energy Balance
The Ideal Gas Equation
• Pressure, the force of collision between gas molecules and a surface, is directly
proportional to temperature and the number of molecules per unit volume.
• PV = nRT the ideal gas equation.
• R is the gas constant and has values of 0.08206 L(atm)/(gmole.K); or 8315
N(m)/(kgmole.K) or 1545 ft(lbf)/(lbmole.◦R).
• a fixed quantity of a gas that follows the ideal gas equation undergoes a process
where the volume, temperature, or pressure is allowed to change, the product of
the number of moles n and the gas constant R is a constant
Calculate the quantity of oxygen entering a package in 24 hours if the packaging
material has a surface area of 3000 cm2 and an oxygen permeability 100 cm3/(m2)(24
h) STP (standard temperature and pressure = 0oC and 1 standard atmosphere of
101.325 kPa).
• Jawaban:
Calculate the volume of CO2 in ft3 at 70oF and 1 atm, which
would be produced by vaporization of 1 lb of dry ice.
An empty can was sealed in a room at 80oC and 1 atm pressure. Assuming that only
air is inside the sealed can, what will be the vacuum after the can and contents cool to
20oC?
• Solution:
Calculate the quantity of air in the headspace of a can at 20oC when the vacuum in
the can is 10 in. Hg. Atmospheric pressure is 30 in. Hg. The headspace has a volume
of 16.4 cm3. The headspace is saturated with water vapor
• vapor pressure of water at 20oC = 2336.6 Pa.
• Let Pt be the absolute pressure inside the can.
• Assume there are no dissolved gases in the product at the
time of sealing, therefore the only gases in the headspace
are air and water vapor. The vapor pressure of water at
20◦C and 80◦C are 2.3366 and 47.3601 kPa, respectively.
In the gas mixture in the headspace, air is assumed to
remain at the same quantity in the gaseous phase, while
water condenses on cooling
A gas mixture used for controlled atmosphere storage of vegetables contains 5%
CO2, 5% O2, and 90% N2. The mixture is generated by mixing appropriate quantities
of air and N2 and CO2 gases. 100 m3 of this mixture at 20oC and 1 atm is needed per
hour. Air contains 21% O2 and 79% N2. Calculate the volume at which the component
gases must be metered into the system in m3/h at 20oC and 1 atm.
• All percentages are by volume. No volume changes occur on mixing of ideal
gases. Because volume percent in gases is the same as mole percent, material
balance equations may be made on the basis of volume and volume percentages.
Let X = volume O2, Y = volume CO2, and Z = volume N2, fed into the system per
hour.
• Oxygen balance: 0.21(X) = 100(0.05); X = 23.8 m3
• CO2 balance: Y = 0.05(100); Y = 5 m3
• Total volumetric balance: X + Y + Z = 100
• Z = 100 − 23.8 − 5 = 71.2 m3
Temperature
Pressure
(°C)
20
22
24
26
28
30
40
50
60
70
80
90
100
105
110
115
120
125
130
135
140
150
160
(kPa)
2.34
2.65
2.99
3.36
3.78
4.25
7.38
12.3
19.9
31.2
47.4
70.1
101.35
120.8
143.3
169.1
198.5
232.1
270.1
313.0
361.3
475.8
617.8
Enthalpy
(sat. vap.)
(kJ kg-1)
2538
2542
2545
2549
2553
2556
2574
2592
2610
2627
2644
2660
2676
2684
2692
2699
2706
2714
2721
2727
2734
2747
2758
Latent heat
(kJ kg-1)
2454
2449
2445
2440
2435
2431
2407
2383
2359
2334
2309
2283
2257
2244
2230
2217
2203
2189
2174
2160
2145
2114
2083
Specific
volume
(m3 kg-1)
57.8
51.4
45.9
40.0
36.6
32.9
19.5
12.0
7.67
5.04
3.41
2.36
1.673
1.42
1.21
1.04
0.892
0.771
0.669
0.582
0.509
0.393
0.307
Heat
• Sensible heat is defined as the energy transferred between two bodies at different
temperatures, or the energy present in a body by virtue of its temperature.
• Latent heat is the energy associated with phase transitions, heat of fusion, from
solid to liquid, and heat of vaporization, from liquid to vapor.
• Enthalpy, is an intrinsic property, the absolute value of which cannot be measured
directly.
• However, if a reference state is chosen for all components that enter and leave a
system such that at this state the enthalpy is considered to be zero, then the
change in enthalpy from the reference state to the current state of a component
can be considered as the value of the absolute enthalpy for the system under
consideration.
• The reference temperature (Tref) for determining the enthalpy of water in the steam
tables is 32.018◦F or 0.01◦C.
Specific Heat
• The specific heat (Cp) is the amount of heat that
accompanies a unit change in temperature for a
unit mass.
• The specific heat, which varies with temperature, is
more variable for gases compared with liquids or
solids.
• Most solids and liquids have a constant specific
heat over a fairly wide temperature range.
specific heat
J/(kg K)
Estimation of Cp
• Cavg = 3349M+ 837.36 in J/(kg K) for fat free plant material
• Cavg = 1674.72 F + 837.36 SNF + 4l86.8M in J/(kg K)
• the mass fraction fat (F), mass fraction solids non-fat
(SNF), and mass fraction moisture (M)
• Example: Calculate the heat required to raise the
temperature of a 4.535 kg roast beef containing 15%
protein, 20% fat, and 65% water from 4.44◦C to 65.55◦C
• Solution:
• Cavg = 0.15(837.36) + 0.2(1674.72) + 0.65(4186.8) = 3182
J/(kg K)
• q = 4.535 kg[3182 J/(kg K)] (65.55 − 4.44)K = 0.882 MJ
Specific heat of gas and vapor
• whereCpm is mean specific heat from the reference temperature To
to T1. Tabulated values for the mean specific heat of gases are
based on ambient temperature of 77◦F or 25◦C, as the reference
temperature.
Contoh
• Hitung kebutuhan panas untuk menaikkan suhu
udara pengering pd tekanan 1 atm dari suhu ruang
25oC ke suhu pengeringan 50oC jika tiap menit
dialirkan udara sebanyak 100m3
• q= mCp (50-25)
• m=PVM/RT
• R = 0.08206 m3 atm/kg mole K
PROPERTIES OF SATURATED AND
SUPERHEATED STEAM
•
•
•
•
•
•
Steam and water are the two most used heat transfer mediums in food processing. Water is also a major component of
food products. The steam tables that list the properties of steam are a very useful reference when determining heat
exchange involving a food product and steam or water. At temperatures above the freezing point, water can exist in
either of the following forms.
Saturated Liquid:. Liquid water in equilibrium with its vapor. The total pressure above the liquid must be equal to or be
higher than the vapor pressure. If the total pressure above the liquid exceeds the vapor pressure, some other gas is
present in the atmosphere above the liquid. If the total pressure above a liquid equals the vapor pressure, the liquid is
at the boiling point.
Saturated Vapor: This is also known as saturated steam and is vapor at the boiling temperature of the liquid. Lowering
the temperature of saturated steam at constant pressure by a small increment will cause vapor to condense to liquid.
The phase change is accompanied by a release of heat. If heat is removed from the system, temperature and pressure
will remain constant until all vapor is converted to liquid. Adding heat to the system will change either temperature or
pressure or both.
Vapor-Liquid Mixtures: Steam with less than 100% quality. Temperature and pressure correspond to the boiling point;
therefore, water could exist either as saturated liquid or saturated vapor. Addition of heat will not change temperature
and pressure until all saturated liquid is converted to vapor. Removing heat from the system will also not change
temperature and pressure until all vapor is converted to liquid.
Steam Quality: The percentage of a vapor-liquid mixture that is in the form of saturated vapor.
Superheated Steam: Water vapor at a temperature higher than the boiling point. The number of degrees the
temperature exceeds the boiling temperature is the degrees superheat. Addition of heat to superheated steam could
increase the superheat at constant pressure or change both the pressure and temperature at constant volume.
Removing heat will allow the temperature to drop to the boiling temperature where the temperature will remain constant
until all the vapor has condensed.
Steam table
• The saturated steam table consists of entries under the headings of
temperature, absolute pressure, specific volume, and enthalpy.
Temperature
Pressure
(°C)
20
22
24
26
28
30
40
50
60
70
80
90
100
105
110
115
120
125
130
135
140
150
160
(kPa)
2.34
2.65
2.99
3.36
3.78
4.25
7.38
12.3
19.9
31.2
47.4
70.1
101.35
120.8
143.3
169.1
198.5
232.1
270.1
313.0
361.3
475.8
617.8
Enthalpy
(sat. vap.)
(kJ kg-1)
2538
2542
2545
2549
2553
2556
2574
2592
2610
2627
2644
2660
2676
2684
2692
2699
2706
2714
2721
2727
2734
2747
2758
Latent heat
(kJ kg-1)
2454
2449
2445
2440
2435
2431
2407
2383
2359
2334
2309
2283
2257
2244
2230
2217
2203
2189
2174
2160
2145
2114
2083
Specific
volume
(m3 kg-1)
57.8
51.4
45.9
40.0
36.6
32.9
19.5
12.0
7.67
5.04
3.41
2.36
1.673
1.42
1.21
1.04
0.892
0.771
0.669
0.582
0.509
0.393
0.307
contoh
• Pada tekanan vakum berapa sehingga air mendidih pada suhu 80oC,
nyatakan dalam kPa dan dalam cm Hg
– Lihat tabel uap= 47,4 kPa abs
– Tekanan vakum = 101 - 47,4 = 53,6 kPa
– Tekanan vakum = (53,6/101) x 76 = 40,3 cmHg
• Sterilisasi dilakukan pada suhu 120oC, berapa tekanan yang terbaca
pada manometer yang menggunakan satuan psi?
– Dari tabel pada suhu 120oC tekanan uap = 198,5 kPa, maka tekanan pada
manometer = 198,5 – 101 = 97,5 kPa
– 1 atm = 14,7 psi = 101 kPa
– Tekanan pada manometer = (97,5/101) x 14,7 = 14,2 psig
Freezing Points of Food Products Unmodified
from the Natural State
• the heat to be removed during freezing of a food
product : sensible heat and latent heat.
• determining the amount of heat by calculating the
enthalpy change.
• calculating enthalpy change below the freezing
point (good only for moisture contents between
73% and 94%) is the procedure of Chang and Tao
(1981). In this correlation, it is assumed that all
water is frozen at 227 K (−50◦ F).
Calculate the freezing point and the amount of heat that must
be removed in order to freeze 1 kg of grape juice containing
25% solids from the freezing point to −30◦C.
•
•
•
•
•
•
•
•
Solution:
Y = 0.75.
for juices: Tf = 120.47 + 327.35(0.75) − 176.49(0.75)2 = 266.7 K
Hf = 9792.46 + 405,096(0.75) = 313, 614 J
a = 0.362 + 0.0498(0.02) − 3.465(0.02)2 = 0.3616
b = 27.2 − 129.04(0.1316) − 481.46(0.1316)2 = 1.879
Tr = (−30 + 273 − 227.6)/(266.7 − 227.6) = 0.394
H = 313,614[(0.3616)0.394 + (1 − 0.3616)(0.394)1.879 ]= 79, 457
J/kg
• The enthalpy change from Tf to −30 ◦C is
• ΔH = 313,614 − 79, 457 = 234, 157 J/kg