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Department of Metallurgical EngineeringHeat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani

Heat transfer1. IntroductionHeat transfer which is defined as the transmission of energy from one region to another as a result of temperature gradient takes place by the following three modes:1.Conduction.2.Convection.3.Radiation.

1-1 CONDUCTION HEAT TRANSFER:- When a temperature gradient exists in a body, experience has shown that there is anenergy transfer from the high-temperature region to the low-temperature region. We say that the energy is transferred by conduction and that the heat-transfer rate per unit area is proportional to the normal temperature gradient:qA

～ ∂ T∂ x When the proportionality constant is inserted,

qx

A=−kA ∂T

∂ x this equation is called Fourier’s law of heat conductionWhere (qx) is the heat-transfer rate(W) [∂T/∂x]is the temperature gradient in the direction of the heat flow.(oC/m) The positive constant [ k] is called the thermal conductivity of the material(W/m.oC)The minus sign is inserted so that the second principle of thermodynamics will be satisfiedqx

A=−k ∂T

∂ x since all properties are constant ,then the equation can

be written as; qx

A=−k dT

dx the integration equation for x from x1 to

x2 and T from T1 to T2 qx∫

x1

x2

dx =∫T 1

T 2

kAdT

qx =−KA ΔTΔx

[ W ]

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Sketch showingdirection of heat flow


1-2 THERMAL CONDUCTIVITY:-

From Fourier’s law ,we have

k= q

A. dxdT

The value of k = 1 when q = 1, A = 1 and dTdx

=1

Now k= q

I. dx

dT (unit of k :W × 1

m2× m

k (or °C )=W /mK .

or W /m °C )

Thus, the thermal conductivity of a material is defined as follows : "The amount of energy conducted though a body of unit area, and unit thickness in unit time when the difference in temperature between the faces causing heat flow is unit temperature difference ". It follows from equation that materials with high thermal conductivities are good conductors of heat, where as materials with low thermal conductive are good thermal insulator. Conduction of heat occurs most readily in pure metals, less so in alloys, and much less readily in non-metals. The very low thermal conductivities of certain thermal insulators e.g., cork is due to their porosity, the air trapped within the material acting as an insulator. Thermal conductivity (a property of material) depends essentially upon the following factors:(i) Material structure.(ii) Moisture content.(iii) Density of the material.(iv) Pressure and temperature (operating conditions). (Table1.1gives the thermal conductivities of various materials at 0ºC.)

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Table 1.1 Thermal conductivity of various materials at 0°C

MaterialThermal

Conductivity(W/m K)

MaterialThermal

Conductivity(W/m K)

Gass Solids : MetalsHydrogen 0.175 Silver , pure 410Helium 0.141 Copper, pure 385Air 0.024 Aluminum, pure 202Water vapour (saturated) 0.0206 Nickel, pure 93Carbon dioxide 0.0146 Iron, pure 73(Thermal conductivity of helium and hydrogen are much higher than other gases, because their molecules have small mass and higher mean travel velocity)

Carbon steel, 1%CLead pureChrome-nickel-steel (18%Cr, 8% Ni)

4335

16.3

LiquidsNon-metalsQuartz, parallel to axis 41.6

Mercury 8.21 magnesite 4.15Water* 0.556 marble 2.08 to 2.94Ammonia 0.54 sandstone 1.83Lubricating oil Glass, window 0.78SAE 40 0.147 Maple or Oak 0.17Freon 12 0.073 Saw dust 0.059

Glass wool 0.038

Note:- water has its maximum thermal conductivity (k= 0.68 W/m K) at about 150ºC.

Example 1 CONDUCTION THROUGH COPPER PLATE. One face of a copper plate 3 cm thick is maintained at 400℃, and the other face is maintained at 100℃. How much hear is transferred through the plate?Solution. From Appendix A the thermal conductivity for copper is 370 W/M·℃ at 250℃. From Fourier’s lawqA

=−k dTdx

Integration givesqA = -k

ΔTΔx =

−(370 )(100−400)3×10 = 3.7 MW/m2　　[1.173¿106

Btu/h·ft2]

Example 2 The following data relate to an oven.

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10-2


Thickness of side wall of the oven = 82.5 mm Thermal conductivity of wall insulation = 0.044 W/m°CTemperature on inside of the wall = 17 5° C Energy dissipated by the electrical coil within the oven = 40.5 WDetermine the area of wall surface, perpendicular to heat flow, so that temperature on the other side of the wall does not exceed 75°C.Solution. x = 82.5 mm = 0.0825 m; k = 0.044 W/m°C; T1 = 175°C; T2 = 75°C; q = 40.5WArea of the wall surface, A Assuming one- dimensional steady state heat conduction,Rate of electrical energy dissipation in the oven. = rate of heat transfer (conduction) across the wall

i.e. q=−kA dT

dx=−kA

(T 2−T 1 )x

=kA (T1−T 2)

x

or 40 . 5=0 .044 A (175−75 )

0 . 0825

or A=40 . 5×0 . 0825

0 . 044 (175−75 )=0. 759

Example3A plane wall has thickness L and its two surfaces are maintained at temperatures T1, and T2. If the thermal conductivity of the material varies with the temperature and is given by k=k0 (1+α T ) , derive an expression for steady state heat transfer rate.

Solution :-

¿¿1-3 CONVECTION HEAT TRANSFERIt is well known that a hot plate of metal will cool faster when

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placed in front of a fan than when exposed to still air. We say that the heat is convected away, and we call the process convection heat transfer. Consider the heated plate shown in Figure. The temperature of the plate isTw, and the temperature of the fluid is T∞. The velocity of the flow will appear as shown, being reduced to zero at the plate as a result of viscous action. Since the velocity of the fluid layer at the wall will be zero .The temperature gradient is dependent on the rate at which the fluid carries the heat away; a high velocity produces a large temperature gradient, and so on. Newton’s law of cooling:q = hA(Tw - T∞)

Here the heat-transfer rate is related to the overall temperature difference between the wall and fluid and the surface area A. The quantity h[W/m2.0C] is called the convection heat-transfer coefficient. Example 3 CONVECTION CALCULATION. Air at 20℃ blows over a hot plate 50 by 75 cm maintained at 250℃. The convection heat-transfer coefficient is 25 W/m2·℃. Calculate the heat transfer.Solution. From Newton’s law of coolingq = hA(Tw-T∞)= (25)(0.50)(0.75)(250-20)= 2.156 kW　　[7356 Btu/h]

Example 4 A wire 1.5 mm in diameter and 150 mm long is submerged in water at atmospheric pressure. An electric current is passed through the wire and is increased until the water boils at 100°C. Under the condition if convective heat transfer coefficient is 4500 W/m2°C find how much electric power must be supplied to the wire to maintain the wire surface at 120°C?Solution:-Diameter of the wire,d = 1.5 mm = 0.0015 m Length of the wire, L = 150 mm = 0.15 m .∴ Surface area of the wire (exposed to heat transfer)

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A=π d L=π×0. 0015×0 .15=7 . 068×10−4 m2 Wire surface temperature, Ts = 120°C Water temperature, Tf = 100°CConvective heat transfer coefficient, h = 4500 W/m2 °C Electric power to be supplied:Electric power which must be supplied = total convection loss (q): ∴ q=hA (T s−T f )=4500×7 . 068×10−4 (120−100 )=63 . 6 W1-4 RADIATION HEAT TRANSFERIn contrast to the mechanisms of conduction and convection, where energy transfer through a material medium is involved, heat may also be transferred through regions where a perfect vacuum exists. The mechanism in this case is electromagnetic radiation. We shall limit our discussion to electromagnetic radiation which is propagated as a result of a temperature difference; this is called thermal radiation. Thermodynamic considerations show that an ideal thermal radiator, or blackbody, will emit energy at a rate proportional to the fourth power of the absolute temperature of the body and directly proportional to its surface area. Thus[qemitted = σ A T4](σ) Stefan-Boltzmann constant with the value of [5.669*10-

8W/m2·K4].q netexchange

A ¿ σ（T14 – T2

4） , q = FεFGσA（T14 – T2

4）where Fε is the emissivity function and FG is the geometric “view factor” function.

Example 5 RADIATION HEAT TRANSFER. Two infinite black plates at 800 and 300℃ exchange heat by radiation. Calculate the heat transfer per unit area.

Solution. q/A = σ(T14 – T2

4) = (5.669¿10-8)(10734-5734)= 69.03 kW/m3　　

[21,884 Btu/h·ft2]Example 5 MULTIMODE HEAT TRANSFER. Assuming

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that the plate in Ex. 3 is made of carbon steel (1%) 2 cm thick and that 300 W is lost from the plate surface by radiation, calculate the inside plate temperature.

Solution. The heat conducted through the plate must be equal to the sum of convection and radiation heat losses:qcond = qconv + qrad

-kAΔTΔx = 2.156 + 0.3 = 2.456 kW

ΔT =

(−2456)(0 . 02)(0 .5 )(0 . 75)(43 )

= -3.05℃　　[-5.49℉]where the value of k is taken from Table 1-1. The inside plate temperature is therefore

Ti = 250 + 3.05 = 253.05 ℃Example 7 A carbon steel plate (thermal conductivity = 45 W/m°C) 600mm ¿ 900mm ¿ 25 mm is maintained at 310°C. Air at 15°C blows over the hot plate. If convection heat transfer coefficient is 22W/m2°C and 250 W is lost from the plate surface by radiation, calculate .the inside plate temperature.Solution. Area of the plate exposed to heat transfer,A = 600 mm ¿ 900 mm = 0.6 ¿ 0.9 = 0.54 m2

Thickness of the plate, L = 25 mm = 0.025 mSurface temperature of the plate Ts = 310°CTemperature of air (fluid), Tf = 15°CConvective heat transfer coefficient, h = 22 W/m2°CHeat lost from the plate surface by radiation, qrad = 250WThermal conductivity, k = 45 W/m °CInside plate temperature, Ti . In this case the heat conducted through the plate is removed from the plate surface by a combination of convection and radiation.

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Tf =15°C

qconv.=hA(Ts -Tf)

qrad


Heat conducted through the plate = convection heat losses + radiation heat losses. or qcond .=qconv .+qrad .

−kA dT

dx=hA (T s−T f )+F σ A (T

s4−T

f 4 )

or −45×054×

(T s−T i)L

=22×0 .54 (310−15 )+250( given)

or −45×0 . 54×

(310−T i)0. 025

=22×0 .54×295+250

or T i=

3754 .6972

+310=313 . 86 °C

PROBLEMS

1-A glass panel (k = 0.78 W/mK) 1.5 m ¿ 2.5 m is 16 mm thick. If its inside and outside surface temperatures are 30°C and 5°C respectively, calculate the heat loss by conduction through the panel.

2-A truncated cone 25 cm high is made of aluminum (k = 204 W/mK). The cross-sectional area at the top and bottom are 50 cm2 and 200 cm2 respectively. The lower surface is maintained at 500°C and the upper surface is at 95°C. The lateral surface is insulated. Assuming one-dimensional flow of heat, calculate the rate of heat transfer.

3-A conical section has its circular cross-section with diameter D = 0.25x. The small end is at x = 50 mm and the large end is at x = 250 mm. The end temperatures are respectively 400K and 600K. The lateral surface is well insulated. Derive an expression for the temperature distribution assuming one-dimensional conditions and calculate the rate of heat flow through the cone.

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