Energy Balance
1
• Most food processes require input of energy in one form or another
• Energy takes many forms such as heat, kinetic energy, chemical
energy, potential energy but, because of inter-conversions, it is not
always easy to isolate separate constituents of energy balances.
• However, under some circumstances certain aspects predominate. In
many heat balances, other forms of energy are insignificant.
• In the food industries, large proportion of energy leaving or entering
the system is in the form of heat
2
• In some chemical situations, mechanical energy is insignificant and in
some mechanical energy situations, as in the flow of fluids in pipes,
the frictional losses appear as heat but the details of the heating need
not be considered.
• Energy balances can be calculated on the basis of external energy used
per kilogram of product, or raw material processed, or on dry solids,
or some key component.
• In the SI system there is only one energy unit, the joule. However,
kilocalories are still used by some nutritionists, and British thermal
units (Btu) in some heat-balance work.
3
• The most common important energy form is heat energy and the
conservation of this can be illustrated by considering operations
such as heating and drying.
• In these, enthalpy (total heat) is conserved and as with the mass
balances so enthalpy balances can be written round the various
items of equipment, or process stages, or round the whole plant, and
it is assumed that no appreciable heat is converted to other forms of
energy such as work.
4
Heat Balances
Heat to surrounding
Heat out in product.
Heat from electricity
Heat from fuel combustion
Heat out in waste
Heat stored
Heat from mechanical sources
Heat in raw materials
Heat in from surrounding
5
Heat Balance Cont’d
• Sensible heat………………..
• Latent heat …………………
• First law of thermodynamics; ΔE=Q-W
6
Heat Transfer Theory
• Heat transfer: Dynamic process where heat is transferred
spontaneously from one source to another cooler source
• The rate of heat transfer depends upon temperature
difference between the sources.
• Increase in temperature difference increases driving force
in heat transfer
• Resistance affects rate of heat transfer
• Rate of transfer = driving force/resistance
7
Heat Transfer Cont’d
• Rate of heat transfer= ΔT / R
• Where R= heat flow resistance of medium
• Unsteady state heat transfer : Temperatures may change during
processing and therefore the rate of heat transfer will change.
• Steady state heat transfer: Temperature doesn’t change during
processing.
8
Heat Transfer Cont’d
Heat transfer can be transferred by:
• Conduction
• Convection
• Radiation
9
• Conduction: The molecular energy is directly exchanged from the
hotter to the cooler regions, the molecules with greater energy
communicating some of this energy to neighbouring molecules with
less energy.
• Convection: The transfer of heat by the movement of groups of
molecules in a fluid. The groups of molecules may be moved by
either density changes or by forced motion of the fluid.
• Radiation: The transfer of heat energy by electromagnetic waves,
which transfer heat from one body to another.
10
Heat Conduction
• Molecular energy is directly exchanged, from the hotter to the cooler
regions.
• Heat transfer of conduction in and fluids follows a basic equation
known as Fourier’s law which states that heat transfer rate per unit
area is proportional to the normal temperature gradient.
• Heat flows from a hotter to a colder body, that is in the direction of
the negative temperature gradient.
• Thermal conductivity does change slightly with temperature, but in
many applications it can be regarded as a constant for a given
material.
11
Heat Conduction Cont’d
• Generally, metals have a high thermal conductivity (50-400 J m-1 s-1
oC-1).
• Most foodstuffs contain a high proportion of water. Water has a
thermal conductivity of 0.7 J m-1 s-1 oC-1 above 0oC.
• Thermal conductivities of foods are in the range 0.6-0.7 J m-1 s-1 oC-1.
• Ice has a substantially higher thermal conductivity than water, about
2.3 Jm-1 s-1 oC-1. the thermal conductivity of frozen foods is therefore
higher than foods at normal temperatures.
12
Heat Conduction Cont’d
• Rate of heat transfer =driving force × conductance, where driving
force is temperature gradient
𝑑𝑄
𝑑𝑇
= −k A
𝑑𝑡
𝑑𝑥
where,
dQ/dt or q = the rate of heat transfer by conduction (W)
A = the area of cross-section of heat flow path (m2)
dT/dx = the temperature gradient
k = constant = thermal conductivity of the medium (W/m °C)
13
Heat Conduction Cont’d
• X =length (m)
• T= Temperature (oC)
Conduction through a slab;
•
𝑑𝑄
𝑘𝐴∆𝑇
=𝑑𝑡
∆𝑋
•
𝑑𝑄
𝑑𝑡
= q (constant)
• q= •
𝑑𝑇
𝑑𝑥
=
𝑘𝐴 ∆𝑇
−
∆𝑋
𝑘𝐴 ∆𝑇
∆𝑥
=
𝑇1−𝑇2
;
𝑋
where x is thickness of the slab
14
• 𝑞=−
𝑘𝐴 (𝑇1−𝑇2)
𝑋
OR,
q= -
kA ∆T k
= A ∆T
x
x
15
Example
A cork slab 10 cm thick has one face at -12°C and the other
face at 21 °C. If the mean thermal conductivity of cork in this
temperature range is 0.042 J m-1 s-1 oC-1, what is the rate of
heat transfer through 1 m2 of wall?
16
Solution
T1 = 21 oC
T2 = -12 oC
DT = 33 oC
A=1m2
K = 0.042 J m-1 s-1 oC-1
X= 0.1 m
0.042
q=
× 1 × 33
0.1
q=13.9 J s-1
17
Heat Conductances
• Heat conductance is the quantity of heat that will pass in unit time,
through unit area of a specified thickness of material, under unit
temperature difference.
•C=
𝑘
𝑥
where,
• C= heat conductance; x = thickness; k= thermal conductivity (Jm-1 s-1
oC-1)
• Heat passes through several consecutive layers of different materials
frequently in heat conduction.
18
q=
𝐴∆𝑇1𝑘1
𝑥1
=
𝐴∆𝑇2𝑘2
𝑥2
=
𝐴∆𝑇3𝑘3
𝑥3
Heat conductances in series
At steady state,
q=
𝑇1−𝑇2
𝑘1 𝐴1
= k2
∆𝑋
𝑇2−𝑇3
A2
=
∆𝑋
(𝑇3−𝑇4)
k3 A3
∆𝑋
19
Heat Conductance conti…
If the areas are the same,
A1 =A2 =A3 =………=A
∆𝑥1
𝑞( )
𝑘1𝐴1
= 𝑇1 − 𝑇2
∆𝑥2
𝑞(
)
𝑘2 𝐴2
(1)
(2)
=𝑇2 − 𝑇3
∆𝑥3
𝑞( )=
𝑘3𝐴3
(3)
𝑇3 − 𝑇4
(1) +(2) + (3)
∆𝑥1
𝑞(
𝑘1 𝐴1
𝑞=
+
∆𝑋2
𝑘2 𝐴2
+
𝑇1−𝑇4
∆𝑋3
)
𝐾3𝐴3
∆𝑋1 ∆𝑋2 ∆𝑋3
+
+
𝑘1 𝐴1 𝐾2𝐴2 𝐾3𝐴3
=
=𝑇1 − 𝑇4
𝑇1−𝑇4
𝑅1+𝑅2+𝑅3
20
Heat Conductance conti…
𝑘1
= conductance
𝑥1
𝑘2
=
𝑥2
𝑋1
𝐾1
of he material in the first layer , C1
conductance of the material in the second layer , C2
+
𝑋2
𝐾2
+
𝑋3
+
𝐾3
…..
=
1
𝐶1
+
1
1
+
𝐶2
𝐶3
1
=
𝑈
• Where U= overall conductance for combined layers (J m-2 s-1 oC-1)
Therefore,
So,
1
𝑈
𝐴∆𝑇 = 𝑞( )
𝑞 = 𝑈𝐴∆𝑇
21
Example
A cold store has a wall comprising of 11cm of brick on the
outside, then 7.5cm of concrete and then 10cm of cork . The
mean temperature within the store is maintained at -18°𝐶 and
the mean temperature outside of the surface of the wall is
18°𝐶. Calculate the rate of heat transfer through the wall. The
appropriate thermal conductivities are for the brick, concrete
and cork respectively 0.69, 0.76, 0.043 Jm-1s-1 ℃. Determine
also the temperature at the interfaces between the concrete and
cork layers, and the brick and concrete layers.
22
Solution
For brick
𝑥1
𝑘1
=
0.11
0.69
= 0.16
For concrete 𝑥2 = 0.075 = 0.10
𝑘2
For cork
But 1/U
𝑥3
𝑘3
𝑥1
=
𝑘1
0.76
=
+
0.10
=
0.043
𝑥2
𝑘2
+
2.33
𝑥3
𝑘3
=0.16 + 0.10 + 2.33
=2.59
23
U= 0.38 Jm−2s−1 oC-1
∆T= 18−(−18) = 36 oC
A= 1 m2
q = UA ∆ T
=0.38 × 1 × 36
=13.7 Js−1
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24
A3∆T3k3
q=
𝑥3
13.7 = 1 × ∆T3 × 1/2.33
∆T3 = 32 oC
∆t3 = difference between temp. of cork /concrete surface (Tc) and temp. of
cork surface inside the cold store.
Tc− (−18) = 32
Tc= 14 oC
If ∆T1= difference between the temp. of the brick/ concrete surface (Tb), and
the temp. of the external air. Then,
13.7 = 1 × ∆T1 × 1/0.16
= 6.25 ∆T1
25
18− Tb = T1
=13.7/ 6.25 = 2.2
Tb = 15.8 oC
18/05/2022
26
Heat Conductance in Parallel
• In heat conductances in parallel, the surfaces of the material are
parallel to the direction of heat transfer and to each other.
• Heat passes through each material at the same time.
qa=
𝐴𝑎∆𝑇𝑘𝑎
𝑥
qb=
𝐴𝑏∆𝑇𝑘𝑏
𝑥
qc=
𝐴𝑐∆𝑇𝑘𝑐
𝑥
27
Example
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
(b) the heat loss for each m2 of oven wall if the inner side of
the wall is at 230 °C and the outer side is at 25 °C.
28
Solution
(a) Considering the loss through an area of 1 m2 of wall (0.99 m2 of
brick, and 0.01 m2 of steel)
A ∆T kb
For brick; qb = b
x
0.99 230−25 0.22
=
= 446 J s-1
0.10
For steel; qs =
As ∆T ks
x
0.01 230−25
=
0.10
45
= 923 J s-1
Therefore, qb/ qs = 0.48
29
(b) Total heat loss
q= (qb + qs) per m2 of wall
q= 446 + 923
q= 1369 J s−1
Percentage of heat carried by steel
=
923
1369
× 100
= 67%
30
Hollow Cylinder
• Most process materials are transported through hollow cylinder as
pipes that may or may not be insulated.
• Let the internal or inner radius be r where the temperature is T1 and
outer radius r2 with temperature T2. The length of the pipe segment is
L and heat flows radially from the inner surface to the outside.
𝑞
𝑑𝑇
= −𝑘
𝐴
𝑑𝑟
• For a cylinder
A = 2𝜋𝑟𝐿
•
𝑞
2𝜋𝑟𝐿
=
18/05/2022
𝑑𝑇
−𝑘
𝑑𝑟
31
=
𝑞
= 2𝜋𝑙
ln
=
𝑞
2𝜋𝐿
=
𝑞
2𝜋𝐿
= -k (T2 –T1)
ln 𝑟2 − ln 𝑟1 = -k (T2 –T1)
𝑟
𝑟1
ln 2= k (T1 –T2)
q= k.
q=(
𝑟21
𝑟
2𝜋𝑙
ln
𝑟2
𝑟1
(T1 –T2)
𝑇1 −𝑇2
)
𝑅
𝑟
R=
ln 𝑟2
1
2𝜋𝑙𝑘
32
Multilayer Cylinder
33
At steady state;
𝑇 −𝑇2
q= (𝑟 −𝑟 1)
2
1
𝑘𝐴
where
AA𝑙𝑚 =
𝑇 −𝑇3
AA
= (𝑟 −𝑟 2)
3
2
A
𝐾𝐵
𝑙𝑚
𝑇 −𝑇4
= (𝑟 −𝑟 3)
4
𝐵𝑙𝑚
3
𝑘𝐶𝐴𝐶𝑙𝑚
𝐴2−𝐴1
ln 𝐴2 𝐴
1
A𝐵𝑙𝑚 =
𝐴3−𝐴2
ln(𝐴3
𝐴2
𝐴𝐶𝑙𝑚 =
)
𝐴4−𝐴3
ln(𝐴4 𝐴3)
34
𝑇1−𝑇4
q= (𝛾2−𝛾1) (𝛾3−𝛾2) (𝛾4−𝛾3)
+
+
𝑘 𝐴 𝑙𝑚
𝑘 AA
𝐾 A
𝐴
=
𝑙𝑚
𝐵
𝐵𝑙𝑚
𝐶
𝐶
𝑇1−𝑇4
𝑅𝐴+𝑅𝐵+𝑅𝐶
35
Example
A thick walled cylindrical tubing material, having an inside
radius of 5 mm and an outside radius of 20 mm is being used
as a cooling coil in a bath. Iced waters is flowing rapidly
inside it, and the inside wall temperature is 274.9 K. The
outside wall temperature is 297.1 K. A total of 14.65 W must
be removed from the bath by the cooling coil, how many
meters of tubing are needed. [k = 0.151 W/mk].
36
Solution
Find the heat flow rate for 1m length of tubing
r1 = 5 mm = 0.005m
r2= 20 mm= 0.02m
A1 = 2𝜋𝑟1𝐿 = 2𝜋 0.005 1.0
= 0.0314 m2
A2 = 2𝜋𝑟2𝐿 = 2𝜋(0.02)(1.0)
= 0.1257 m2
q= 𝑘𝐴𝑙𝑚
𝐴𝑙𝑚 =
𝑇1−𝑇2
𝑟2−𝑟1
0.1257−0.0314
01.257
ln( 0.0314)
37
𝐴𝑙𝑚 =
0.0943
1.387
= 0.0688
q= 0.151 × 0.0688
274.9−297.1
0.02−0.005
q= -15.2 W
(-ve means the heat is from r2 to r1)
1.52
1m
14.65
14.65
15.2
1
15.2
× 14.65𝑚
= 0.964m
L= 0.964m
38
Trial Question
A hollow cylinder has the inner radius, r1, and outer radius,
r0. Inner and outside temperatures T1 and T0 respectively,
made of material whose thermal conductivity is k for the
length l. Derive the expression for the conducted heat loss
in Watts. If r1 = 7.5cm, r0 = 4.5cm. k = 52 W/mk, where
heat conducted out from the cylinder of length 1m is 465 W.
If the inside temperature is 400 oC determine the outside
temperature.
39
Example
A thick walled tube of stainless steel (A) having a
k=21.63 W/mk, with dimensions of 0.0254m ID and
0.058m OD is covered with a 0.0254m layer of (B),
insulation of k=0.2423 W/mK The inside wall
temperature of the steel is 811K while the outside
temperature of the insulation is 310.8 K, for 0.30m length
pipe, calculate the heat loss and also the temperature at
the interface between the metal and the insulation.
18/05/2022
40
Solution
T1  811k, T2  interface temprature, T3  310.8k
18/05/2022
41
0.0254
r1=
2
=0.0127m
0.0508
r2=
2
= 0.02254m
r3 = r2 + thicknesss of insulator= 0.20254 + 0.20254
= 0.0508m
For l = 0.305m
A1=2𝜋𝑟1𝑙= 2𝜋 0.0127 (0.305)
=0.0243𝑚2
A2= 2𝜋𝑟2𝑙 = 2𝜋 0.0254 (0.305)
=0.0487𝑚2
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42
A3 = 2𝜋𝑟3𝑙 = 2𝜋 0.0508 (0.305)
= 0.0974𝑚2
A A/M =
𝐴2−𝐴1
𝐴
ln𝐴2
1
=
0.0487−0.0243
0.0487
ln0.0243
= 0.0351m2
A B/M =
𝐴3−𝐴2
ln
𝐴3
𝐴2
=
0.0974−0.0487
0.0974
ln0.0487
= 0.073m2
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43
𝑅𝐴 =
𝑟2−𝑟1
𝑘𝐴 AA𝑙𝑚
𝑅𝐵 =
𝑟3−𝑟2
𝑘B AB𝑙𝑚
=
0.0127
(0.2423)(0.0703)
= 0.01673K/W
=
0.0254
(0.2423)(0.0703)
=1.491K/W
The heat transfer rate
q=
q=
𝑇1−𝑇3
𝑅𝐴+𝑅𝐵
=
811−310.8
=331.7W
0.01673+1.491
𝑇1−𝑇2
𝑅𝐴
T2= 𝑞 𝑅𝐴 - T1
T2= 805.5K
18/05/2022
44
Hollow sphere
For a hollow sphere, A = 4𝜋𝑟2
From the Fourier’s law
𝑞
𝑑𝑇
= −𝑘
𝐴
𝑑𝑟
𝑞
𝑑𝑇
= −𝐾
2
4𝜋𝑟
𝑑𝑟
𝑞 𝑟2 𝑑𝑟
=
4𝜋 𝑟1 𝑟2
𝑞
4𝜋
−
𝑞
1
4𝜋 𝑟1
1
𝑟
-k
𝑇2
𝑑𝑇
𝑇1
=k(T1-T2)
1
−
𝑟2
=k (T1-T2)
4𝜋𝑘(𝑇1 − 𝑇2)
𝑞=
1 1
−
𝑟1 𝑟2
45
(𝑇1 − 𝑇2)
= 𝑘(4𝜋𝑟1𝑟2)
(𝑟2 − 𝑟1)
=
𝑇1−𝑇2
𝑟2−𝑟1
4𝜋𝑟1𝑟2𝑘
46
Trial Question
A hollow sphere contains 100W electrical filament and
the following data are known. The internal radius equals
24cm, external =32cm, k=4.16 W/mk. Assuming steady
state, determine the internal temperature if the outside
temperature is 22oC.
47
Trial Question
A hollow steel sphere contain 100W electrical filament, and the
following data re known:
Internal radius = 24cm
External radius = 32cm
Inner surface film confident of heat = 30 W/m2K
Outer surface film coefficient of heat = 10 W/m2-K
Neglecting the thermal resistance of the metal and assuming steady
conditions. Determine the internal temperature if the outside
temperature is 22oC.
Neglecting the thermal resistance of the metal and assuming steady
conditions.
Determine the internal temperature if the outside temperature is
22℃
18/05/2022
48
Convective Heat Transfer Coefficient
The rate of heat transfers from a fluid to a solid surface or vice versa is
usually
expressed as
q = h A(Tw – Tr)
q = heat transfer rate
A= Area of heat transfer (m2)
TW = Temperature of solid wall surface
Tr = Average of bulk temperature of the fluid (k)
h = convective heat transfer coefficient (W/m2k)
The coefficient h is a function of many variable including system
geometry, fluid properties, and flow velocity and temperature
differences. It is often predicted by empirical relations. h is also
called the film coefficient
18/05/2022
49
𝑞 = ℎ𝐴(𝑇𝑊 − 𝑇𝑓)
=
q=
(𝑇𝑊−𝑇𝑓)
1
ℎ𝐴
𝑇1−𝑇4
1 ∆𝑋 1
+ +
ℎ𝑖𝐴 𝑘𝐴 ℎ𝑜𝐴
50
Trial Question
A flat composite plate is made of two layers of aluminum and
steel of 5cm and 2cm thick respectively. The aluminum side is in
contact with a fluid at 200oC. The heat transfer coefficient of
liquid film being 14 W/m2K. The cold surface of steel side is in
contact with liquid at 25oC. The heat transfer coefficient being 29
W/m2K
Determine
(i) heat transfer rate from the hot liquid to the cold liquid,
through a surface of interface temperatures
(ii) Thermal conductivities of Aluminum and steel are 205 and 45
(W/mK) respectively
18/05/2022
51
SOLUTION
(i)
Roverall =
=
1
14𝑥10
q=
1
ℎ𝑖𝐴
+
∆𝑋𝐴
𝐾𝐴𝐴
+
∆𝑋𝐵
𝐾𝐵𝐴
0.05
205𝑥10
+
0.02
45𝑥10
+
𝑇𝑖 −𝑇0
𝑅𝑜𝑣𝑒𝑟𝑎𝑙𝑙
18/05/2022
=
200−25
0.01066
+
1
ℎ𝑜𝐴
+
1
29𝑥10
=0.01066k/W
= 16416.5 =
200−𝑇1
1
14𝑥10
52
(16416.5 x
1
)
14𝑋10
=200-T1
T1= ? (solve)
(ii) q=
𝑇𝑖 −𝑇1
1
ℎ𝐴𝑖
=
𝑇1−𝑇2
∆𝑋𝐴
𝐾𝐴𝐴𝐴
=
𝑇2−𝑇3
∆XB/KBAB
𝑇 −𝑇0
= 13
ℎ0𝐴0
calculate for T2 and T3 temperature
18/05/2022
53
Example
A cold storage room is constructed of an inner layer of
12.7mm of pine, a middle layer of 101.6mm of cork board
and an outer layer of 76.2mm of concrete. The wall
surface temperature is (255.4K inside the cold room and
297.1 K at the surface of the concrete. Calculate the heat
loss in W for 1 m2 and the temperature at the interface
between the wood and cork board.
Thermal conductivities :pine – 0.151 W/mk
cork – 0.0433 W/mk
concrete - 0.762 W/mk
18/05/2022
54
SOLUTION
∆𝑋𝐴 = 0.0127𝑚
∆𝑋𝐵 = 0001016𝑚
∆𝑋𝐶 = 0.0762𝑚
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∆𝑋𝐴
RA=
𝑘𝐴𝐴
=
0.0127
(0.151)(1.0)
∆𝑋𝐵
RB=
𝑘𝐵𝐴
=
0.1016
(0.0433)(1.0)
∆𝑋𝐶
𝑘 𝐶𝐴
=
0.0762
(0.762)(1.0)
RC=
q=
𝑇1−𝑇4
𝑅𝐴+𝑅𝐵+𝑅𝐶
=
=0.0841K/W
= 2.346K/W
= 0.100K/W
255.4−297.1
0.0841+2.346+0.100
=
−41.7
2.530
= -16.49W
(-ve means heat flow is from outside). To calculate the
temperature T2 at the interface between the pine and cork:
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q=
𝑇1−𝑇2
𝑅𝐴
-16.48=
255.4−𝑇2
0.0841
T2= 256.79K
Calculate T3
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EXAMPLE
A cold room has one of the walls 5m x 2.5m made of bricks
12cm thick insulated externally by cork slabbing 8cm thick.
Cork is protected externally by 2.5cm wood. Estimate the heat
infiltration through the wall in 24 hours if the interior of the
cold room is maintained at a temperature of 0 oC and the
outside temperature is 20 oC. Thermal conductivities for brick,
cork and wood are 0.93, 0.044 and 0.175 W/m.k respectively.
What will be the interface temperatures?
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Solution
RA =
∆𝑋𝐴
𝐾𝐴𝐴
A = 5 X 25 = 12.5m2
KA = 0.93
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KB = 0.044
KC = 0.175W/mk
59
∆𝑋𝐴 = 0.12𝑚
∆𝑋𝐵 = 0.08𝑚
∆𝑋𝐶 = 0.025𝑚
0.12
RA =
X 12.5 = 0.01K/W
0.93
0.08
RB =
X 12.5 = 0145K/W
0.044
0.025
RC=
X
0.175
12.5=0.01143K/W
RTOTAL = 0.01 + 0.145 + 0.01143 = 0.16643 K/W
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𝑇1−𝑇2
q=
𝑅𝑇𝑂𝑇𝐴𝐿
20−0
=
=
0.16643
q
120.17W
120.17 𝐽 3600 𝑠
=
.
𝑠
ℎ
𝐽
𝑠
𝑠
ℎ
= 120.17 .3600
𝐽
= 120.17𝑥 3600
ℎ
If
1 hr = 120.17 x 3600 J
24 hrs =
120.17 𝑥 3600 𝑥24 𝐽
24ℎ𝑟𝑠
103820. 𝑘𝐽
=
24ℎ𝑟𝑠
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Heat transfer through each layer is the same
𝑇1−𝑇2
=
𝑅𝐶
=
𝑇3−𝑇4
𝑅𝐴
20−𝑡2
0.01143
q= 120.17 =
=
t2= 20 – (120.17)(0.01143)
= 18.63 oC
Also, 120.17 =
𝑡3−𝑡4
0.01
=0
t3 = 1.2 oC (t4 - 0)
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Question
A composite insulating wall has three layers of material held together
by 3cm diameter aluminium rivet per 0.1m2 surface. The layer of
materials composite are 10cm thick brick with hot surface at 200oC,
1cm thick timber with cold surface at 10oC. These two layers are
interposed by a third layer of insulating wall 25cm thick. The
conductivities of walls are;
K(brick) = 0.95 w/m.k
k (wood) = 0.175 w/mk
k(insulation) = 0.115w/mk
k(aluminium) =200w/mk
If the heat transfer is assumed to take place only in one direction
normal to the layers, determine the percentage increase in heat transfer
rate due to rivets.
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Trial question
A flat composite plate made of two layers of aluminium and steel of
5cm and 2cm thickness respectively. The thermal conductivities are
205 and 45 W/mk for aluminium and steel respectively. Aluminium
side is in contact with hot liquid at 200oC. The heat transfer
coefficient of liquid film being 14 W/mk . The cold surface of steel
side is in contact with liquid at 25 oC. The heat transfer coefficient
of liquid film being 29 W/m oK.
Determine the :
(i) heat transfer rate from hot liquid to cold liquid through a surface
of 10m2
(ii) overall heat transfer coefficient
(iii) interface temperatures.
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Solution
ti
t1
t2
t3
to
q
𝑅 𝑜𝑣𝑒𝑟𝑎𝑙𝑙
𝑅𝑜𝑣𝑒𝑟𝑎𝑙𝑙
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∆𝑋1 ∆𝑋2
1
=
+
+
+
ℎ𝑖𝐴 𝐾1𝐴 𝐾2𝐴 ℎ𝑂𝐴
1
0.05
0.02
1
=
+
+
+
14 𝑥10 205 𝑥10 45 𝑥10 29 𝑥10
65
𝑡𝑖−𝑡𝑜
𝑅𝑜𝑣𝑒𝑟𝑎𝑙𝑙
(i) q=
=
200−25
0.01066
= 16416.5𝑊
= 16.4165𝑘𝑊
ii. Q = 𝑈𝐴∆𝑇
iii. U =
𝑄
𝐴∆𝑇
Or U =
=
16416.5
10 𝑋 175
1
1 ∆𝑋1 ∆𝑋2 1 = 0.1066
+
+ 𝑘 +ℎ
ℎ𝑖 𝑘1
2
𝑜
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1
= 9.38 W/m2
= 9.38 𝑊/𝑂𝐾
66