Highlights of Sixth Session

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• Highlights of Sixth Session
• Filtration equipment
• Mixing
• Different types of mixer
• Solid solid mixing
• Liquid mixing
• Power requirement for Liquid mixing

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• INTRODUCTION TO HEAT TRANSFER
• Heat transfer As the name suggests is the
  transfer of heat from hotter body to colder body.
  There are three modes by which heat transfer
  occurs namely
• 1) Conduction
• 2) Convection
• 3) Radiation

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CONDUCTION In
conduction, heat can be conducted through solids,
liquids and gases. The heat is conducted by the
transfer of the energy of motion between adjacent
molecules. In a gas the hotter molecules which
have greater energy and motions impart energy to
the adjacent molecules at lower energy levels.
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CONVECTION Convection is the transfer of heat
from one point to another within a fluid, gas, or
liquid by the mixing of one portion of the fluid
with another. In natural convection, the motion
of the fluid is entirely the result of
differences in density resulting from temperature
differences in forced convection, the motion is
produced by mechanical means. When the forced
velocity is relatively low, it should be
realized that free-convection factors, such as
density and temperature difference, may have an
important influence.
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RADIATION Radiation
is the transfer of heat from one body to
another, not in contact with it, by means of wave
motion through space.
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HEAT-TRANSFER APPLICATIONS      
Cooling of milk in a pipe heat exchanger
    Water chilling in a counter flow heat
exchanger     Steam required to heat pea soup
in jacketed pan     Time to heat pea soup in a
jacketed pan     Time/Temperature in a can
during sterilization     Pasteurization of
milk     Freezing of fish     Rate of boiling
of refrigerant.     Operation of a compressor
in a refrigeration system     Chilling of fresh
apples     Freezing of a slab of meat
    Freezing time of a carton of meat
controllable factors
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FOURIERS LAW

• Fouriers law is the fundamental differential
  equation for heat transfer by conduction
• q/A -k(dt / dx)
• where q (quantity of heat per unit time) is the
  rate of flow of heat,
• A is the area at right angles to the direction in
  which the heat flows,

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-dt /dx is the rate of change of temperature with
the distance in the direction of the flow of
heat, i.e., the temperature gradient. The
factor k is called the thermal conductivity it
is a characteristic property of the material
through which the heat is flowing and varies with
temperature.
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• Heat Loss Through
• An Insulated wall
• Calculate the heat loss per m2 of surface area
  for an
• insulating wall composed of 25.4 mm thick fiber
• insulating board whose thermal conductivity is
• 0.048w/mK), where the inside temperature is
  352.7K
• and the outside temperature is 297.1K?

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• Solution
• Thickness of the wall x2-x10.0254m
• Thermal conductivity of insulating board (k)
  0.048w/ m K
• Delta T(temperature difference) 352.7-297.1
  55.6K
• q / A k (T2-T1) / (x2-x1)
• 105.1w/m2

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Radial conduction through a hollow cylinder.
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• q / A -k dt / dr
  Equation 1
• Cross sectional area normal to the heat flow
  2prL
• Substituting this in equation 1 we get
• q / 2 p r L -k dt / dr
  Equation 2
• By integrating and putting limits from r1 to
  r2 and T1 to T2 this we get
• (q / 2pL) ? dr/r -k ? dT
• q 2 p k L/ ln (r2 / r1) (T1-T2)

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In the steady state, the same quantity of heat
per unit time must pass through each layer. q
A1 ?T1k1/x1 A2 ? T2k2/x2 A3?T3k3/x3 if
the areas are the same,                         
              A1 A2 A3 A q A?T1k1/x1 A
? T2k2/x2 A?T3k3/x3 So A ? T1 q(x1/k1) and
A ? T2 q(x2/k2) and A ? T3 q(x3/k3) A ?T1
A ?T2 A ?T 3 q(x1/k1) q(x2/k2)
q(x3/k3) A(? T1 ? T2 ? T3 ) q(x1/k1
x2/k2 x3/k3 )
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The sum of the temperature differences over each
layer is equal to the difference in temperature
of the two outside surfaces of the complete
system, i.e. ? T1
? T2 ? T3 ? T q A ? T / (x1/k1 x2/k2
x3/k3 )
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• Problem
• A cold storage room is constructed of an inner
  layer of 12.7 mm of pine (thermal conductivity
  k0.151w/mK), a middle layer of 101.6 mm of cork
  board (k0.0433w/ m K), and an outer layer of
  76.2 mm of concrete(0.762w/ m K). The wall
  surface temperature is 255.4 K inside the cold
  room and 297.1 K at the outer surface of the
  concrete. Calculate the heat loss in watts per m2
  and the temperature at the interface between wood
  and cork board

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• Solution
• T1255.4K,T4297.1K,
• Take Pine as material A (ka 0.151w/m K,?xa
  0.0127 m)
• Take cork as material B (kb 0.0433 w/m K, ?xb
  0.1016 m)
• Take concrete as material C (kc 0.762w/m, ?xc
  0.0762m)

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• Ra ?xa/kaA 0.0127/0.151(1) 0.0841
  K/W
• Rb ?xb/kbA 0.1016/0.0433(1) 2.346 K/W
• Rc ?xc/kcA 0.0762/0.762(1) 0.1K/W
• q (T1-T4) / (Ra RbRc)
• -41.7/2.530 -16.48 W
• Substituting the known values and solving we get
• -16.48 (255.4-T2)/0.0841
• T2 256.79K

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Problem

• The wall of a bakery oven is built of insulating
  brick 10 cm thick and thermal conductivity 0.22 J
  m-1 s-1 C-1. Steel reinforcing members penetrate
  the brick, and their total area of cross-section
  represents 1 of the inside wall area of the
  oven. If the thermal conductivity of the steel is
  45 J m-1 s-1 C-1 calculate
• (a) the relative proportions of the total
  heat transferred through the wall by the brick
  and by the steel and
• (b) the heat loss for each m2 of oven wall if
  the inner side of the wall is at 230C and the
  outer side is at 25C

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(a) Consider the loss through an area of 1 m2 of
wall (0.99 m2 of brick, and 0.01 m2 of steel)For
brick qb AbDT kb/x

• Consider the loss through an area of 1 m2 of
  wall (0.99 m2 of brick, and 0.01 m2 of steel)For
  brick qb Ab ?T kb/x
• 0.99(230-25)0.22/0.1 446j/s
• For steel qs As ?T ks/x
• 0.01(230-25) 45/0.10 923 j /s
• Therefore qb /qs 0.48

For steel qs AsDT ks/x
Therefore qb /qs 0.48
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(b) Total heat loss      q (qb qs ) per m2 of
wall        446 923         1369 J
s-1Therefore percentage of heat carried by
steel        (923/1369) x 100         67
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• Reference
• Unit Operations in food processing
• R L Earle with M D Earle
• Transport Processes And Unit Operations (third
  edition)
• Christie j.Geankoplis