Basic Concepts PDF
Basic Concepts PDF
Basic Concepts PDF
of Heat Flow 1
1.1. What is Heat Transfer ? 1.2. Modes of Heat Transfer. 1.3. Physical Mechanism of Modes of Heat Transfer—Conduction
—Convection—Radiation. 1.4. Laws of Heat Transfer—Law of conservation of mass : Continuity equation—Newton’s second
law of motion—Laws of thermodynamics—Fourier law of heat conduction—Newton’s law of cooling—The Stefan Boltzmann
law of thermal radiation. 1.5. Combined Convective and Radiation Heat Transfer—Equation of state. 1.6. Thermal
Conductivity—Variation in thermal conductivity—Determination of thermal conductivity—Variable thermal conductivity.
1.7. Isotropic Material and Anisotropic Material. 1.8. Insulation Materials—Superinsulators—Selection of insulating
materials—The R-Value of insulation—Economic thickness of insulation. 1.9. Thermal Diffusivity. 1.10. Heat Transfer in
Boiling and Condensation. 1.11. Mass Transfer. 1.12. Summary—Review Questions—Problems—Multiple Choice Questions.
1
2 ENGINEERING HEAT AND MASS TRANSFER
state. In case of steady state heat transfer, the moving molecules from a region of lower temperature,
temperature at any location on the system does not vary the heat energy transfer takes place between them. The
with time. The temperature is function of space low energy molecules absorb energy and thus their
coordinates only, but it is independent of time. temperature is increased and the temperature of high
Mathematically, for rectangular coordinate system ; energy molecules is lowered.
T = f(x, y, z) ...(1.3) The conduction heat transfer in liquids and gases
During steady state conditions, the heat transfer occurs due to collisions and diffusion of molecules during
rate is constant and there is no change of internal energy their random motion. However, the nature is much more
of the system. For example, the heat transfer in coolers, complex.
heat exchangers, heat transfer from large furnaces, etc. The temperature gradient is the potential for heat
In unsteady state heat transfer, the temperature conduction. If a body in any phase exists a temperature
varies with time as well as position. The temperature is gradient, will definitely have the conduction heat
a function of time and space coordinates. transfer.
Mathematically, for rectangular coordinates ; 1.3.2. Convection
T = f(x, y, z, t) ...(1.4)
The convection is a mode of heat transfer in which the
During unsteady state or transient heat transfer,
energy is transported by moving fluid particles. The
rate of heat transfer varies with time due to change in
convection heat transfer comprises two mechanisms.
internal energy of the system. Most of the actual heat
First is transfer of energy due to random molecular
transfer processes are unsteady in nature , but some of
motion (diffusion) and second is the energy transfer by
them are considered in steady state to simplify them.
bulk or macroscopic motion of the fluid (advection). The
For example, heat transfer from hot coffee left in a room,
molecules of fluid are moving collectively or as aggre-
cooling and heating process, etc. are transient processes.
gates thus carry energy from high temperature region
The heat transfer may be one, two or three to low temperature region. Therefore, the faster the fluid
directional, depends upon the configuration of the motion, the greater the convection heat transfer.
system considered.
Convection heat transfer may be classified
according to nature of fluid flow.
1.2. MODES OF HEAT TRANSFER If the fluid motion is artifically induced by a
pump, fan or a blower, that forces the fluid over a surface
When the temperature gradient exists in a medium,
to flow as shown in Fig. 1.1(a), the heat transfer is said
which may be solid, liquid, or gas, heat transfer occurred
to be by the forced convection.
is called conduction. In contrast, the convection refers
to heat transfer that will occur between a surface and a
moving medium, when they are at different tempera-
tures. The third mode of heat transfer is thermal
radiation. All surfaces at finite temperature emit energy
in the form of electromagnetic waves. The thermal
Heated
radiation can also occur in absence of any medium. plate Tw
Q Fan
1.3. PHYSICAL MECHANISM OF MODES OF
HEAT TRANSFER Air at T¥
1.3.1. Conduction Fig. 1.1. (a) Forced convection of air (Tw > T∞)
The conduction occurs usually in the stationary If the fluid motion is set-up by buoyancy effects,
mediums. It is the mode of heat transfer in which energy resulting from density difference caused by temperature
exchange takes place from a region of high temperature difference in the fluid as shown in Fig. 1.1(b), the heat
to that of low temperature by direct molecular transfer is said to be by the free or natural convection.
interactions and by the drift of electrons. The heat For example, a hot plate vertically suspended in stag-
conduction may be viewed as the transfer of energy from nant cool air, causes a motion in air layer adjacent to
more energetic molecules to adjacent less energetic the plate surface because of temperature difference in
molecules of a substance. When a fast moving molecules air gives rise to density gradient which in turn sets-up
from a region of high temperature collide with slower the air motion.
CONCEPTS AND MECHANISMS OF HEAT FLOW 3
Q
dx = – kdT
A
T2 = 1100 K
L = 15 cm
T
z z
Fluid flow condition h (W/m2.K)
Q r2 dr T2
Air (1 bar, free convection) 6 – 30
Hence =–k dT
2πL r1 r T1
Air (1 bar, forced convection) 10 – 200
Q FG IJ = – k(T
r
ln 2
Water (free convection) 500 – 1000
or
2 πL H K
r1 2 – T1) Water (forced convection)
Vapourisation of water
600
2500
–
–
8000
1,00,000
Q ln(r2 /r1 )
or k= Condensation of steam 4000 – 25,000
2πL (T1 − T2 )
Substituting numerical values, Example 1.3. Hot air at 150°C flows over a flat plate
maintained at 50°C. The forced convection heat transfer
(1.8 W) × ln(0.125/0.0025)
k= coefficient is 75 W/m2.K. Calculate the heat gain rate by
2π × (0.3 m) × (200°C – 150°C) the plate through an area of 2 m2.
= 0.075 W/m.K. Ans. Solution
1.4.5. Newton’s Law of Cooling Given : Flow of hot air over a flat plate
It is the fundamental law for heat convection and it T∞ = 150°C, Ts = 50°C
states that the rate of heat transfer is directly h = 75 W/m2.K, A = 2 m2.
proportional to the temperature difference between a To find : Heat transfer rate by air to plate.
surface and a fluid, or mathematically Assumptions :
Q (i) Steady state conditions,
∝ (Ts – T∞) (ii) Constant properties,
A
6 ENGINEERING HEAT AND MASS TRANSFER
(iii) Heat is transferred by forced convection only. (ii) It is also the net heat flux conducted through
Analysis : According to Newton’s law of cooling the wall, therefore for plane wall
T¥ = 150°C Q k(Ts, o − Ts, i )
q= =
h = 75 W/m .K
2 A L
Ts = 50°C (0.10 W/m.K ) × (16° C − Ts , i )
or (50 W/m2) =
(0.03 m )
Fig. 1.6. Flow over a flat plate
Q = hAs(T∞ – Ts) (50 W/m 2 ) × (0.03 m )
or Ts, i = 16°C –
= (75 W/m2.K) × (2 m2) × (150 – 50) (°C or K) (0.10 W/m.K )
= 15 × 103 W = 15 kW. Ans. = 1°C. Ans.
Example 1.4. A refrigerator stands in a room, where Example 1.5. A hot plate is exposed to an environment
air temperature is 21°C. The surface temperature on the at 100°C. The temperature profile of the environment
outside of the refrigerator is 16°C. The sides are 30 mm fluid is given as T(°C) = 60 + 40 y + 0.1 y2. The thermal
thick and has an equivalent thermal conductivity of conductivity of the plate material is 40 W/m.K. Calculate
0.10 W/m.K. The heat transfer coefficient on the outside the heat transfer coefficient.
is 10 W/m2.K. Assume one dimensional conduction Solution
through the sides, calculate the net heat flow rate and
the inside surface temperature of the refrigerator. Given : A hot plate exposed to an environment
T = (60 + 40 y + 0.1 y2)°C
Solution
k = 40 W/m.K
Given : Heat transfer from a refrigerator wall.
T∞ = 100°C
T∞ = 21°C, Ts, o = 16°C,
L = 30 mm = 0.03 m,
k = 0.10 W/m.K, h = 10 W/m2.K. Q conv.
FG q IJ
1/4 1.5.1. Equation of State
or Ts =
H σK It is the relation between the properties of an ideal gas.
The perfect gas law is used in convection heat transfer,
= M
L 950 W/m 2 OP 1/ 4
which is :
N 5.67 × 10 W/m .K Q
–8 2 4 p
ρ
= RT ...(1.21)
= 359.78 K
where, p = gas pressure in kN/m2,
The equilibrium temperature of black surface will
ρ = gas density in kg/m3,
be,
R = specific gas constant in kJ/kg.K,
Ts = 86.78°C. Ans.
(= 0.287 kJ/kg.K for air)
Example 1.7. A black body at 30°C is heated to 100°C. T = absolute temperature of gas in K.
Calculate the increase in its emissive power.
1.6. THERMAL CONDUCTIVITY
Solution
Given : A black body emission The thermal conductivity is the property of materials
T1 = 30 + 273 = 303 K, and is defined as the ability of the materials to conduct
the heat through it. By inspection of equation (1.9),
T2 = 100 + 273 = 373 K.
thermal conductivity can be interpreted as the rate of
To find : The increase in emissive power.
heat transfer through a unit thickness of material per
Assumptions : unit area per unit temperature difference. The thermal
1. Stefan Boltzmann’s constant, conductivity of a material is a measure of how fast heat
σ = 5.67 × 10–8 W/m2.K4. will flow in that material. The large value of thermal
2. No heat loss by conduction and convection. conductivity indicates that the material is a good heat
Analysis : The radiant heat flux or emissive power conductor and low value indicates that the material is a
for a black surface can be expressed as ; poor heat conductor or an insulator.
Eb = σTs4 The thermal conductivity is measured in watts per
Hence the increase in emissive power of a black metre per degree Celsius or Watt per metre per kelvin,
body can be calculated as ; when heat flow rate is expressed in watts. The thermal
Eb2 – Eb1 = σ(T24 – T14) conductivity of a substance is highest in solid phase and
lowest in gaseous phase. Fig. 1.10 shows typical range of
or Eb2 – Eb1 = (5.67 × 10–8 W/m2.K4)
thermal conductivity of various materials at 20°C.
× [(373 K)4 – (303 K)4]
2 The value of thermal conductivity depends upon
= 619.62 W/m . Ans. the manner in which energy is transferred. The pure
Example 1.8. The surface temperature of a central metals allow faster transmission of heat energy through
heating radiator is 60°C. What is the net black body the vibrations of the crystal lattices. Therefore, a metal
radiation heat transfer unit surface area between the in pure state has the maximum thermal conductivity
radiator and its surroundings at 20° ? and is a good conductor of heat. The thermal conductivity
Take σ = 5.67 × 10–8 W/m2.K4 decreases with increasing amount of impurities in the
metals. Most non-metals are poor conductor of heat
Solution
transfer, therefore, have low values of thermal
Given : Central heating radiator as black body conductivity and are called the thermal insulators.
with In gases, the faster the molecules move, the faster,
Ts = 60°C = 333 K, T∞ = 20°C = 293 K. they will transport energy. Therefore, the thermal
To find : Radiation heat transfer conductivity of gases depends on the square root of
Analysis : The black body radiation heat transfer absolute temperature. The thermal conductivities of
rate per unit area between radiator surface and its some typical gases are given in Table 1.2.
surroundings is expressed as ; The physical mechanism of heat conduction in
Q liquids is also same as in gases, however, the mechanism
q= = σ (Ts4 – T∞4) is slightly complex due to close spacing of molecules and
A
= (5.67 × 10–8 W/m2.K4) molecular attraction force exerts a strong influence on
× [(333 K)4 – (293 K)4] energy exchange in collision process. The thermal con-
ductivity of liquids usually lies between those of solids
= 5.67 × 10 × 4.9263 × 109
–8
and gases. The value of thermal conductivity for some
= 279.32 W/m2. Ans. standard liquids is also given in Table 1.2.
CONCEPTS AND MECHANISMS OF HEAT FLOW 9
Non-metallic
crystals
Diamond
1000 Pure
Graphite metals
Silver Metal
Silicon Copper
Carbide alloys
100 Aluminium
alloys Non-
Iron metals
Thermal conductivity, k (W/m. K)
Bronze Oxides
Steel
Quartz
10 Manganese
Nichrome Liquids
Rock
Mercury
Food
1 Water Insulators
Fibres
Rubber
Oils Gases
Wood
0.1 H2
He
Foams Air
CO2
0.01
Fig. 1.10. Typical range of thermal conductivity of various materials at room temperature
TABLE 1.2. Typical values of thermal conductivities at 20°C
Material Thermal conductivity, k (W/m.K)
Metals :
Diamond 2300
Silver 429
Copper (pure) 401
Gold 317
Aluminium (pure) 237
Iron (pure) 73
Carbon steel, 1% C 43
Non-metallic solids :
Window glass 0.780
Brick 0.720
Asbestos 0.149
Cork 0.045
Glass wool 0.038
Liquids :
Water 0.556
Ethylene glycol 0.249
Ammonia 0.54
Gases :
Helium 0.152
Air 0.024
Steam 0.0206
Carbon dioxide 0.0146
10 ENGINEERING HEAT AND MASS TRANSFER
10 Me
He
Aluminium
oxide
5 0.2
Pyroceram
Fused
2 quartz
1
100 300 500 1000 2000 4000 0.1 )
Steam (1atm
Temperature (K)
CO2
Fig. 1.11. Effect of temperature on thermal Air
Freon-12
conductivity of selected solids
0 200 400 600 800 1000
The thermal conductivity of mercury increases Temperature (K)
with increase in temperature is an exceptional case of Fig. 1.13. Effect of temperature on thermal conductivity of
metals. selected gases at normal pressure
CONCEPTS AND MECHANISMS OF HEAT FLOW 11
Gases. For the gases, the molecules are in α may be negative or positive as shown in Fig. 1.15,
continuous random motion. As temperature increases, depending upon whether thermal conductivity increases
velocities of molecules become higher than in some lower or decreases with rise in temperature. The constant α
temperature region. The molecules move from high is usually positive for non-metals and insulating
temperature region to a region of low temperature and materials and negative for most of the metals.
give up its energy through collisions to lower energy
molecules. Thus the thermal conductivity of gases
T(x)
increases with increase in temperature and it is pro- T1
a=0
portional to square root of the absolute temperature. It
a<0
is also affected by change in pressure and humidity. Q
a>0
1.6.2. Determination of Thermal Conductivity
The thermal conductivity, k can be defined by Fourier T2
law, equation (1.8)
(Q/A) x
k=– ...(1.22)
(dT/dx) L
This equation is used for determination of thermal Fig. 1.15. Variation of thermal conductivity
with temperature
conductivity of a material. A layer of solid material of
thickness L and area A is heated from one side by an With a variable thermal conductivity, Fourier law
electric resistance heater as shown in Fig. 1.14. If the of heat conduction through a plane wall can be expressed
outer surface of heater is perfectly insulated, then all the as ;
heat generated by resistance heater will be transferred
Q dT dT
through the exposed layer of material. When steady state =–k = – k0(1 + αT) ...(1.25)
condition is reached, the temperature of two surfaces of A dx dx
material T1 and T2 are measured and thermal Q
conductivity of material is determined by relation or dx = – k0(1 + αT) dT
A
Integrating both sides within the boundary
Sample
Insulation conditions :
z z
material
T1
Q L T2
dx = – k0 (1 + αT) dT
Resistance T2 A 0 T1
heater
Q=W
or
Q
(L – 0) = – k0 T + α
T2 LM OP T2
A 2 N Q T1
A k0 A LM(T − T1 ) + α
(T2 2 − T12 ) OP
or Q=–
MN PQ
L
2
L 2
Rearranging
Fig. 1.14. Experimental set-up for determination of thermal
conductivity k0 A(T1 − T2 ) α LM OP
k=
QL
...(1.23)
Q=
L
1 + (T1 + T2 )
2 N Q
A (T1 − T2 )
km A(T1 − T2 )
1.6.3. Variable Thermal Conductivity or Q=
L
For many materials, the thermal conductivity can be
where, km = k0 1 + LM α OP ...(1.26)
approximated as a linear function of temperature over
limited range and it is expressed as ; N 2
(T1 + T2 )
Q
The quantity km represents the mean value of
k = k0(1 + αT) ...(1.24)
thermal conductivity evaluated at arithmetic mean
where, k0 is the value of thermal conductivity at some
reference temperature and α is an empirical constant, T1 + T2 FG IJ
dT
whose value depends on material. The value of constant
temperature
2 H
. The term
dxK in eqn. (1.25)
12 ENGINEERING HEAT AND MASS TRANSFER
represents the slope of the temperature profile for the Analysis : According to Fourier law of heat
conducting medium. Mathematically, the slope is : conduction
dT
= −
LM Q OP ...(1.27)
Q
=q=–k
dT
dx N
k0 (1 + αT)A Q A dx
dT
(i) When constant α = 0, the equation (1.24) or q = – k0(1 + bT + cT2)
dx
reduces to k = k0 and thermal conductivity does not or qdx = – k0(1 + bT + cT2) dT
change with temperature. The slope of curve is constant
and the temperature profile is linear.
(ii) When constant α > 0, the slope or temperature
profile follows a positive curved line along the material
thickness. Therefore, the thermal conductivity increases T1
with increase in temperature and vise versa. k = k0(1 + bT + cT )
2
T1
(1 + bT + cT2) dT
ro = 10 cm
T1 = 1350°C T1
k(T) p/2 cm
Q = 50
Sector of circle L
T1
(1 + 0.0007 T) dT
kelvin scale for temperature.
Analysis : The area of cylindrical piece :
Q LM
T2 F I OP T2 πro2 π(0.1 m) 2
or
A
(L – 0) = – 0.838 × T + 0.0007
MN
2 GH JK PQ A=
4
=
4
= 7.854 × 10–3 m2
T1
Fourier law,
Q 0.838 RS
× (T2 − T1 ) +
0.0007
× (T2 2 − T12 )
UV Q dT
or
A
=–
L T 2 W A
=−k
dx
Using numerical values or Q dx = – 111.63 × 7.854 × 10–3
Q 0.838 × (1 – 10–4 T) dT
=– × [(50 – 1350) + 0.00035
A 0.25 Integrating both sides within boundary conditions
z z
× (502 – 13502)]
L T2
= 6492.82 W. Ans. Q dx = – 0.8767 × (1 – 1 × 10–4 T) dT
0 T1
Example 1.11. A metal piece, 50 cm long is in the form
of a sector of a circle of radius 10 cm and includes an
L
Q L = – 0.8767 × MT − 1 × 10 G
F T I OP 2
T2
H 2 JK PQ
angle of π/2. The thermal conductivity of the metal piece or
−4
varies as ; MN T1
k = k0(1 + αT)
where, k0 = 111.63 W/m.K and α = – 1 × 10–4 W/m.K2.
L
= – 0.8767 × M(T − T ) − 1 × 10
−4
with air. The air is trapped inside small cavities of contents. Fig. 1.21 shows the ranges of effective thermal
solids. The same effect can also be produced by filling conductivity for evacuated and non evacuated insulat-
the space across which heat flow is to be minimised with ions.
small solid particles and trapping the air between them.
A gas has very poor thermal conductivity. Commercial Conduction
insulators are ceramics (e.g., insulating bricks), rock- Convection Radiation
wool, gypsum and polymeric (expanded polyurethane,
expanded polystyrene etc.) materials. These materials
are highly porous, or have a high void volume filled with
an inert gas.
The heat transfer through an insulation is by
conduction through the solid material, and by
conduction and convection through the air space as well
as by radiation as shown in Fig. 1.20. Such insulation
materials are characterised by an apparent thermal
conductivity keff.. It is an effective value that accounts
for all mechanisms of heat transfer and it should not
change with temperature, pressure and moisture Fig. 1.20. Heat transfer through an insulating material
Evacuated Non-evacuated
Powders, fibres,
foams, cork, etc.
Powders, fibres,
and foams
Opacified powders
and fibres
Multilayer
insulations
–5 –4 –3 –2 –1
10 10 10 10 10 1.0
Effective thermal conductivity keff (W/m.K)
Insulating materials are classified into three cellular materials are polyurethane, expanded polysty-
categories : rene, cellular glass, and cellular silica, etc.
1. Fibrous. The fibrous insulations are obtained 3. Granular. Granular insulations consist of
by mixing small particles or flakes of low density small flakes or particles of inorganic materials, bonded
materials with air. The material is poured into small into some common shapes or used as powders. For
gaps as loose fill or formed into boards or blankets. example, asbestos, perlite powder, diatomaceous silica
Fibrous materials have very high porosity. Mineral wool and vermiculite.
is a common fibrous material for application at
1.8.1. Superinsulators
temperature below 700°C and fibrous glass is used for
temperature below 200°C. For temperature range 700°C The insulators with extremely low apparent thermal
to 1700°C, the refractory fibre, such as alumina or silica conductivity (about one thousand of that of air), called
is used. superinsulators and are obtained by using layers of
2. Cellular. Cellular insulations are closed or highly reflective sheets separated by glass fibres in an
open cell materials that are usually in the form of evacuated space. Like a thermos bottle in which the
extended flexible or rigid boards. These can easily be space between the two surfaces is evacuated to
formed or sprayed in the place to achieve desired suppress conduction and convection and inner surface
geometrical shapes. These materials have low density, is coated with reflective layer to prevent the radiation
low heat capacity, and good compressive strength. Some heat transfer, the heat transfer between two surfaces
16 ENGINEERING HEAT AND MASS TRANSFER
can also be reduced by placing highly reflective sheets. superinsulators are used in space applications and
The radiation heat transfer is inversely proportional to cryogenics.
the number of such reflective sheets placed between the
1.8.2. Selection of Insulating Materials
surfaces. Very effective insulations are obtained by
using closely packed layers of highly reflective thin The selection and design of a suitable insulation depends
metal sheets such as aluminium foils separated by upon following factors :
fibres made of insulating materials like glass. Further, 1. Thermal conductivity,
the space between the layers is evacuated to form 2. Density of material,
a very strong vacuum to eliminate the conduction 3. Upperlimit of operating temperature,
and convection heat transfer through the air space. 4. Structural rigidity,
The resulting materials have an apparent thermal 5. Degradation rate,
conductivity below 2 × 10–5 W/m.K, which is one 6. Chemical stability,
thousand times less than the conductivity of air or 7. Cost, i.e., economic thickness of insulation.
any common insulating material. These specially The range of thermal conductivities for common
built insulators are called superinsulators. The temperature insulating materials is given in Table 1.3.
TABLE 1.3. Some insulating materials with their range
of operating temperatures
Insulating materials Max. operating Density Thermal conductivity
temperature (K) (kg/m3) (W/m.K)
Asbestos fibre 420 190–300 0.078–0.098
Cellular glass 700 145 0.046–0.079
Diatomaceous 1145 345 0.92–0.104
Alumina silica fibre 1530 48 0.071–0.15
Magnesia 85% 590 185 0.051–0.061
Polystyrene 350 16–56 0.023–0.040
Mineral fibres 922–1255 430–560 0.071–0.137
1.8.3. The R-Value of Insulation new and more effective materials are developed and use
For building materials, the effectiveness of insulation of insulation is considerably increased. The walls and
is characterised by a term called R-value. The R-value roofs of our house are also applied with some insulation
is the thermal resistance of material for a unit area and like plaster of paris, etc., to minimise the heat transfer
it is the ratio of thickness and effective thermal rate with its surroundings.
conductivity of the material. That is : When an insulation layer is put on a heating
system, the heat loss from the system reduces. The cost
Thickness L of insulation material adds the system cost, while cost
R-value = =
Effective thermal conductivity k eff of reduction in heat loss reduces the operating cost.
...(1.28) Therefore, the cost of insulation material is subsidised
by saving in energy cost. Insulation pays for itself from
The R-value is generally given in English unit
the energy it saves.
h.ft2. °F/Btu.
The insulating material has certain period of
For example, R-value of 6″ thick glass fibre service. Over the service life of the insulation material,
insulation (keff = 0.025 Btu/h.ft2.°F) is designated as R-20 the thickness of insulation at which the sum of cost of
insulation by builders, i.e., insulation and cost of heat loss is minimum as shown in
6″ × (1/12) ft Fig. 1.22, is referred to as economic thickness of
L
R-value = = insulation. If the thickness of insulation is more than
keff 0.025 Btu/h.ft 2 . ° F
the economic thickness, the cost of insulation will not
= 20 h.ft2. °F/Btu. compensate the energy it saves. Thickness of insulation
is controlled by density of material. As density of
1.8.4. Economic Thickness of Insulation material increases, the required thickness of insulation
The energy crisis of 1970s had a tremendous impact on decreases for same thermal effect. If the layer after layer
energy awareness and energy conservation. Since then, of insulation is applied over a surface, the heat transfer
CONCEPTS AND MECHANISMS OF HEAT FLOW 17
reduces gradually. The inner layer saves more heat than Solution
outer one. A limiting thickness of insulation balances Given : A furnace wall exposed to convection
the cost of energy saved and cost of insulation itself. environment on one side.
This particular critical layer is called optimum
insulation thickness. k = 1.35 W/m.K
The optimum thickness of insulation can be L = 200 mm = 0.2 m
obtained by plotting a graph of value of heat loss and T1 = 1400°C
cost of insulation against the thickness of insulation.
T∞ = 40°C
h = 7.85 + 0.08 (∆T) (W/m2.K)
t
os
nc
T1
Combined cost
tio
ula
Co
Cost
st o Ins
fh
ea h = f(T)
t lo
ss Q
T¥
T2
Economic thickness
Insulation thickness L
Fig. 1.22. Economic thickness of insulation
Fig. 1.23. Schematic of a furnace wall
To find : Heat flux.
1.9. THERMAL DIFFUSIVITY
Analysis : Steady state heat transfer rate per unit
Thermal diffusivity is an important thermophysical area
property. It is the ratio of thermal conductivity k of the k(T1 − T2 )
medium to heat capacity ρC. It is denoted by α, and q= = h(T2 – T∞)
measured in m2/s. L
k 1.35 × (1400 − T2 )
α= ...(1.29) or = [7.85 + 0.08 (T2 – 40)] × (T2 – 40)
ρC 0.2
The thermal conductivity k indicates how well a or 9450 – 6.75 T2 = 7.85 T2 – 314 + 0.08 (T2 – 40)2
material can conduct heat and the heat capacity ρC
= 7.85 T2 – 314 + 0.08
represents how much energy a material can store per
unit volume. Therefore, the thermal diffusivity of a × (T22 – 80 T2 + 1600)
material is viewed as the ratio of the heat conducted
= 7.85 T2 – 314 + 0.08 T22
through the material to the heat stored per unit volume.
In other words, the thermal diffusivity of a material is – 6.4 T2 + 128
associated with the propagation of heat energy into the Rearranging, 0.08 T22 + 8.2 T2 – 9636 = 0
medium during the change of temperature with time.
The higher the thermal diffusivity, faster the propaga- − 8.2 + (8.2) 2 + 4 × 0.08 × 9636
tion of heat into the medium. or T2 =
2 × 0.08
Example 1.13. The inside temperature of a furnace wall
(k = 1.35 W/m.K), 200 mm thick, is 1400°C. The heat = 299.57°C
transfer coefficient at the outside surface is a function of The heat flow rate per unit area
temperature difference and is given by k
h = 7.85 + 0.08 ∆T (W/m2.K) q= (T1 – T2)
L
where ∆T is the temperature difference between outside 1.35
wall surface and surroundings. Determine the rate of = × (1400 – 299.57)
0.2
heat transfer per unit area, if the surrounding tempera-
= 7427.88 W. Ans.
ture is 40°C.
18 ENGINEERING HEAT AND MASS TRANSFER
Example 1.14. An uninsulated steam pipe is passed Example 1.15. A horizontal plate (k = 30 W/m.K)
through a room in which air and walls are at 25°C. The 600 mm × 900 mm × 30 mm is maintained at 300°C.
outer diameter of the pipe is 50 mm and surface The air at 30°C flows over the plate. If the convection
temperature and emissivity are 500 K and 0.8, coefficient of air over the plate is 22 W/m2.K and 250 W
respectively. If the free convection heat transfer heat is lost from the plate by radiation. Calculate the
coefficient is 15 W/m2.K, what is the rate of heat loss bottom surface temperature of the plate. (P.U., Dec. 2008)
from the surface per unit length of pipe ? Solution
Solution Given : A horizontal plate as shown in Fig. 1.25.
Given : An uninsulated pipe exposed to room air k = 30 W/m.K, A = 600 mm × 900 mm
T∞ = Tw = 25°C = 298 K Ts = 300°C, T∞ = 30°C
Ts = 500 K, D = 50 mm = 0.05 m h = 22 W/m2.K, L = 30 mm = 0.03 m
ε = 0.8, h = 15 W/m2.K. Qrad = 250 W.
0.05 m L
Qrad = 250 W
T¥ = 30°C
Air
Qconv
2
h = 22 W/m .K
2
h = 15 W/m .K e = 0.8
T¥ = 25°C
Ts = 300°C
Fig. 1.24. Schematic of example 1.14
L = 30 mm k = 30 W/m·K
To find : Heat loss per unit length of pipe.
Ti
Assumptions : Qcond
1. Steady state conditions.
Fig. 1.25. Schematic of horizontal plate, conducting,
2. Heat loss by radiation and convection only. convecting and radiating heat
3. Stefan Boltzmann constant, σ = 5.67 × 10–8 To find : Temperature of bottom surface of the plate.
W/m .K4.
2
Assumptions :
4. Constant properties. 1. Steady state conditions.
Analysis : 2. One dimensional heat conduction in the plate.
(i) Heat loss from the pipe by convection is given 3. Constant properties.
by
Analysis : The surface area of the plate
Qconv = hAs(Ts – T∞) = h(πDL)(Ts – T∞)
A = 600 mm × 900 mm
Q conv
or = 15 × (π × 0.05) × (500 – 298) = 5.40 × 105 mm2 = 0.54 m2
L
= 476 W/m Making the energy balance for the plate :
(ii) Heat loss per unit length of pipe by radiation Rate of heat conduction = Rate of heat convec-
is given by tion + Rate of heat radiation
Q rad or Qcond = Qconv + Qrad
= σε(πD)(Ts4 – T∞4) kA (Ti − Ts )
L or = hA(Ts – T∞) + 250
= 5.67 × 10–8 × 0.8 × (π × 0.05) L
× (5004 – 2984) Using numerical values :
= 389.13 W/m 30 × 0.54 × (Ti − 300)
= 22 × 0.54
Total heat loss from pipe surface per unit length : 0.03
× (300 – 30) + 250
Q Q conv Q rad
= + 0.03
L L L or Ti – 300 = × (3207.6 + 250)
30 × 0.54
= 476 + 389.13
or Ti = 300 + 6.40 = 306.4°C. Ans.
= 865.13 W/m. Ans.
CONCEPTS AND MECHANISMS OF HEAT FLOW 19
Example 1.16. A black metal plate (k = 25 W/m.K) at at 30°C. The heat transfer coefficient between plate
300°C is exposed to surrounding air at 30°C. It convects surface and air is 20 W/m2.K. The emissivity of the plate
and radiates heat to surroundings. If the convection surface is 0.8. Calculate.
coefficient is 25 W/m2.K, what is the temperature gradient (i) Rate of heat loss by convection.
in the plate ? (ii) Rate of heat loss by radiation.
Solution (iii) Combined convection and radiation heat
Given : An iron plate convects and radiates heat transfer coefficient. (P.U., May 2009)
to surroundings. Solution
k = 25 W/m.K Given : A thin metal plate surface exposed to
Ts = 300°C = 573 K convection and radiation environment.
T∞ = 30°C = 303 K A = 5 m × 3 m = 15 m3
h = 25 W/m2.K. Ts = 300°C = 573 K
Air h = 20 W/m2 . K,
Qrad
2
T∞ = 30°C = 303 K ε = 0.8
h = 25 W/m .K Qconv
To find :
T¥ = 30°C (i) Rate of heat transfer by convection.
Ts = 300°C (ii) Radiation heat transfer rate.
k = 25 W/m.K
(iii) Combined heat transfer coefficient.
Assumption : The Stefan Boltzmann’s constant
Fig. 1.26. Schematic of iron plate
σ = 5.67 × 10–8 W/m2.K4.
To find : Temperature gradient in the plate.
Analysis :
Assumptions :
1. Steady state conditions. (i) Convection heat transfer rate from the plate
surface
2. Black metal plate is black body for radiation.
Qconv = hA(Ts – T∞)
3. Stefan Boltzmann constant, σ = 5.67 × 10–8
W/m .K4.
2 = 20 × 15 × (300 – 30)
Analysis : The energy balance for the metal plate = 81,000 W. Ans.
is given as ; (ii) Radiation heat transfer rate from the plate
Heat conducted through the plate Qrad = εσA(Ts4 – T4∞ )
= Heat convection from the surface
= 0.8 × 5.67 × 10–8 × 15
+ Heat radiated from the surface
× (5734 – 3034)
i.e., Qcond = Qconv + Qrad
= 67,612 W. Ans.
dT
or – kA = hA(Ts – T∞) + σA(Ts4 – T∞4) (iii) Combined convection and radiation heat
dx
transfer coefficient
Using numerical values
Total heat transfer rate by convection and
dT radiation
– 25 × = 25 × (573 – 303) + 5.67 × 10–8
dx
Q = Qconv + Qrad
× (5734 – 3034)
= 81,000 + 67,612
= 25 × 270 + 5.67 × 10 × 9.937 × 1010
–8
dT
Conduction Exchange of energy due to q(W/m2) = – k Thermal conductivity
direct molecular interactions dx k(W/m.K)
Convection Diffusion of energy due to q(W/m2) = h(Ts – T∞) Heat transfer coefficient
random molecular motion h(W/m2.K)
plus energy due to bulk
motion (advection)
Radiation Energy transfer by q(W/m2) = ε σ(Ts4 – Tsur4 ) Emissivity ε, or radiation
electromagnetic waves or Q(W) = hr A(Ts – Tsur) heat transfer coeff. hr
The thermal insulation is a material or 17. What is the R-value of an insulation ? How is it
combination of materials, which is mainly used to determined ?
minimise the heat flow to or from a system. 18. How does the R-value of an insulation differ from
Thermal diffusivity is the ratio of thermal its thermal resistance ?
conductivity k of the medium and heat capacity ρC. It
19. What is the physical significance of thermal
is denoted by α, and measured in m2/s, i.e., diffusivity ?
k
α=
ρC PROBLEMS