Foundation of Technical Education

Third year

College of Technical/ Basrah

32 Lectures

Lectures of Heat Transfer

Heat Transfer Transfer RateTransfer Processes Processes Mechanism Mode Conduction Convection Radiation Radiat ion

Diffusion of energy ener gy due to random random molecular motion Diffusion of energy due to random molecular motion plus bulk motion Energy transfer by electromagnetic waves

Rate of heat transfer (W )

q

=

- kA

dT

dx

q = h A(T s-T ) ) ∞

4

4

) q = σ ε A(T s -T sur )

By Mr. Amjed Ahmed Ali

Syllabus of Heat Transfer (English), (2 hours/ week, Applied 2 hours /week)

Hours

1. Heat Heat transfer by conduction, convection and radiation 2.One-dimensional 2.One-dimensional steady state conduction 3.Systems 3.Systems with conduction-convection 4.Radial 4.Radial systems(cylinder and sphere)

2 2 2 2

4. 5. 6. 7. 8. 9.

Overall heat transfer coefficient Critical thickness of the insulator Heat source systems Extended Surface (Fins) Resistance to heat contact Unsteady state conduction Complete heat capacity system Limited conditions of convection Application and Hessler's diagrams 11. Multi-dimensions systems 12. Principles of heat transfer by convection 13. Boundary layer for laminar and turbulent flow 14. Thermal boundary layer for laminar and turbulent flow

2 2 2 2 2 2 2 2 2 2 1 2 2

16 5.. Experimental Analogy between fluid of friction and heatbytransfer 1 relations heat transfer forced convection inside pipes 17 17.. Flow through cylindrical and spherical bodies 18 18.. Flow through bundle of tubes 19 19.. Heat exchangers Scaling Mean logarithmic difference of temperature NTU method method 20 20.. Heat transfer by radiation 21 21.. Properties of radiation 22 22.. Body in thermal radiation 23 23.. Relation between coefficient and the body 24 24.. Heal exchange between non-black bodies 25 25.. Radiation barriers

2 2 2 2 4 1 1 2 1 2 2 2 2 2

•

•

•

Ch 1: Introdaction

3rd Year College of Technical

Chapter One Introduction Introduction A consider the cooling of a hot steal rod which is place in a cold water Thermodynamics may be used to predict the final equilibrium temperature of the rod-water combination. It will not tell us how long it takes to reach this equilibrium condition. Heat Transfer may be used to predict the temperature of the rod and the water as a function of time.

1.1 Definition:

ﻣﻬﻤﺔ ﺗﻌﺎرﻳﻒ

Heat : is the energy transit as a result of the temperature difference. Heat transfer: is that science which seeks to predict the energy transfer that may take place between material bodes as a result of a temperature difference. difference . Thermodynamics: is the state science of energy, the transformation of energy and the change in the state of matter. (Thermodynamics can be able to determination of heat and work requirements for chemical and

physi phy sica call p roces roc esss and a nd the eq equil uilibr ibrium ium condi con diti tions ons). ). Heat flux : heat transfer flow in the direction per unit area (q”). Steady state: Temperature is very does not very with time (dT/dt) =0. Unsteady state: temperature is depending on time.

1.2 Modes of Heat Transfer

اﻟﺤﺮارة ﻧﻤﺎ اﻧﺘﻘﺎل

The engineering area frequently referred to as thermal science includes thermodynamics and heat transfer. The role of heat transfer is to supplement thermodynamic analyses, which consider only systems in equilibrium, with additional laws that allow prediction of time rates of energy transfer. These supplemental laws are based upon the three fundamental modes of heat transfer conduction, convection, and radiation.

1.3 A Conduction Heat Transfer Conduction may be viewed as the transfer of energy from the more energetic to the less energetic particles of a substance due to interactions between the particles. A temperature gradient within a homogeneous substance results in an energy transfer rate within the medium which can be calculated by Fourier's law dT q = -kA (1.1) 1.1) dx Where q is the heat transfer rate (W (W or J/s) J/s) and k thermal conductivity conductivity ( W/m W/m K K ) is an experimental constant for the medium involved, and it may depend upon other properties, such as temperature and pressure.

dT dx

Is the temperature gradient in the direction normal to the area A.

Mr. Amjed Ahmed Ahmed

1

Ch 1: Introdaction

dT dx

=

∆Τ

3rd Year College of Technical

T 1

Τ2 − Τ1

T1> T2

= ∆ χ χ 2 − χ 1

T 2 q X 2 L

X 1

Figure 1.1 Temperature distribution for steady state conduction. Through pl plate ate wall wal l

The minus sign in Fourier's Law (1.1) is required by the second law of thermodynamics: thermal energy transfer resulting from a thermal gradient must be from a warmer to a colder region. If the temperature profile within the medium is linear Fig. 1.1 it 1.1 it is permissible to replace the temperature gradient (partial derivative) with wit h (1.2) Τ − Τ 2 q = kA 1 L K/W ) which is equal to the The quantity ( L/kA) is equivalent to a thermal resistance Rk ( K/W reciprocal of the conductance. As:

Rk

L

q=

Τ 2 − Τ 1

(1.3)

Rk kA Such linearity always exists in a homogeneous medium of fixed k during during steady steady state heat transfer occurs whenever the temperature at every point within the body, including the surfaces, is independent of time. T1 q T1> T2

T2 Figure 1.2 Associ Ass ociati ation on of con ducti duc tion on he heat at trans tr ansfe ferr wit w ith h dif d iffus fusio ion n of energy due to molecular activity.

If the temperature changes with time

dT dt

, energy is either being stored in or removed

from the body. This storage rate is

q stored = mc p

dT dt

(1.4)

Where m is the mass of substance and Cp is specific heat capacity.

Mr. Amjed Ahmed Ahmed

2

Ch 1: Introdaction

3rd Year College of Technical

1.3.1 Thermal Conductivity The thermal conductivity of a material is a measure of the ability of the material to conduct heat. I Thermal Conductivity of Solids: In general, k for a pure metal decreases with temperature; alloying elements tend to reverse this trend. The thermal conductivity of a metal can usually be represented over a wide range of temperature by

k = k 0(a + bθ + cθ 2 )

(1.5)

Where θ = T − − T ref and k 0 is the conductivity at the reference temperature T ref ref The thermal conductivity of a non homogeneous material is usually markedly dependent upon the apparent bulk density, As a general rule, k for a no homogeneous material increases both with increasing temperature and increasing apparent bulk density II Thermal Conductivity of Liquids: Thermal conductivities of most liquids decrease with increasing temperature. But insensitive to pressure the exception is water, which exhibits increasing k up to about 150°C and decreasing k there after. Water has the highest thermal conductivity of all common liquids except the so-called liquid metals. III Thermal Conductivity of Gases: The thermal conductivity of a gas increases with increasing temperature, but is essentially independent of pressure for pressures close to atmospheric. For high pressure (i.e., pressure of the order of the critical pressure or greater), the effect of pressure may be significant.

Fig(1.3) The mechanism of heat conduction of different phases of a substance.

1.4 A Convection Heat Transfer Whenever a solid body is exposed to a moving fluid having a temperature different from that of the body, energy is carried or convected from or to the body by the fluid If the upstream temperature of the fluid is T ∞ , and the surface temperature of the solid is T s the heat transfer per unit time is given by Newton’s Law of cooling:

q = h A(T - T ) s

∞

(1.6) 2

Where h is Convective Heat transfer coefficient (W/m K ) as the constant of proportionality relating the heat transfer per unit time and area to the overall temperature difference. It is important to keep in mind that the fundamental energy exchange at a solid-fluid boundary is by conduction, and that this energy is then converted away by the fluid flow. The thermal resistance to convection heat transfer Rc, as:

Rc =

1 h A

q = T s - T ∞ Rc Mr. Amjed Ahmed Ahmed

(1.6 ) (1.7) Fig (1.4) Velocity and temperature distribution on flat plate

3

Ch 1: Introdaction

3rd Year College of Technical

1.5 A Radiation Heat Transfer The third mode of heat transmission is due to electromagnetic wave propagation, which can occur in a total vacuum as well as in a medium. Experimental evidence indicates that radiant heat transfer is proportional to the fourth power of the absolute temperature, where as conduction and convection are proportional to a linear temperature difference. The fundamental Stefan-Boltzmann Law Law is

q = σ A A T 4

(1.8)

constant independent of surface, Where T is the absolute temperature, σ is Boltzmann medium, and temperature; its value is 5.6697 X 10-8 W/m2.K 4 ., the thermal emission from many surfaces (gray bodies) can be well represented by 4

4

q = σ ε A(T s -T sur )

(1.9)

Where ε , the emissivity of the surface, ranges (0-1). The ideal emitter or blackbody is one, All other surfaces emit some what less than one. T s and and T T sur The temperature of surface and surroundings respectively. Similarly, Simil arly, The thermal resistance to radiation ra diation heat transfer Rr , as: Τ s − Τ sur (1.11) Rr = 4 4 σε Α( Τ s − Τ sur )

q =

Ts - T sur Rr

(1.12)

Table 1.1 Summary of heat transfer rate processes Rate of heat transfer Mode Transfer Mechanism (W )

Conduction Convection

Radiation

Diffusion of energy due to random molecular motion Diffusion of energy ene rgy due to random molecular motion plus bulk motion Energy transfer by electromagnetic electromag netic waves

q = - kA

dT

dx

Rk = Rc =

q = h A(T s-T ∞ ) ) 4

Thermal Resistance ( K/W K/W )

4

) q = σ ε A(T s -T sur )

Rr =

L kA 1 h Τ s − Τ sur 4

4

σε Α( Τ s − Τ sur )

Figure (1.5) Conduction, Convection and Radiation Heat transfer Modes

The concept of thermal resistance (analogous to electrical resistance) is introduced as an aid to solving conduction heat transfer problems.

Mr. Amjed Ahmed Ahmed

4

Ch 1: Introdaction

3rd Year College of Technical

Example 1.1 Example Calculate the rate of heat transfer by natural convection between a shed roof of area 20 m x 20 m and ambient air, if the roof surface temperature is 27°C , the air temperature 3°C , and the average convection heat transfer coefficient 10 W/m2 K .

Figure 1.7 Schematic Sketch of Shed for Analysis of Roof Roof Temperature. Temperature. Solution Solution Assume that steady state exists and the direction of heat flow is from the air to the roof. The rate of heat transfer by convection from the air to the roof is then given by Eq:

Note we initially assumed that the heat transfer would be from the air to the roof. But Note since the heat flow under this assumption turns out to be a negative quantity the direction of heat flow is actuall actuallyy from the roof roof to to the the air. air. Examplee 1.2 Determine the steady state rate of heat transfer per unit area through a Exampl 4.0cm thick homogeneous slab with its two faces maintained at uniform temperatures of 38I o C and and 21 oC. C. The The thermal conductivity of the material is 0.19 W/m K .

Examplee 1.3 The forced convective heat transfer coefficient for a hot fluid x1 x2 Exampl flowing over a cool surface is 225 W/m2.oC for a particular problem. The fluid temperature C. Determine the heat upstream of the cool surface is 1200C, and the surface is held at 10 0C. transfer rate per unit surface area from the fluid to the surface.

q = h A(Ts-T ∞ ) q/A= 225(120-10)=24750 W/m2 Examplee 1.4 Exampl 1.4 After sunset, radiant energy can be sensed by a person standing near a brick wall. Such walls frequently have surface temperatures around 44 oC , and typical brick emissivity values

are on the order of 0.92. What would be the radiant thermal flux per square foot from a brick wall at this temperature?

Mr. Amjed Ahmed Ahmed

5

Ch 1: Introdaction

3rd Year College of Technical

Example 1.5 1.5 In the summer, parked automobile surfaces frequently average 40-50 40-50 oC . Assuming 45 o

C and surface emissivity of 0.9 0.9,, determine the radiant thermal flux emitted by a car roof

Example 1.6 The air inside an electronics package housing has a temperature of 50°C . A "chip" in this housing has internal thermal power generation (heating) rate of 3 X 10-3 W . This chip is subjected to an air flow resulting in a convective coefficient h of 9 W/m2.oC over its two main surfaces which are 0.5 cm X 1.0 cm. cm. Determine the chip surface temperature neglecting radiation and heat transfer from the edges

Example Exa mple 1.7

Calculate the thermal resistance theand rate0.5 of cm thick, heat transfer a panetemperature of window glass (k = 0.78 W/m K) 1 m high K) 1 , 0.5 m wide, mand wide, cm thick, if thethrough outer-surface is 24°C and and the inner-surface temperature is 24.5°C

24 °C Figure 1.5 heat transfer by conduction through a window pane.

Mr. Amjed Ahmed Ahmed

6

Ch 1: Introdaction

3rd Year College of Technical

Solution Assume that steady state exists and that the temperature is uniform over the inner and outer surfaces. The thermal resistance to conduction R conduction Rk is from Eq

Rk =

L kA

=

0.005m 0.78w / mk × 1m × 0.5m

= 0.0128

K W

The rate of heat loss from the interior to the exterior surface is ∆T 24.5 - 24 q = = = 39.1 W R k 0.0128 Example 1.8 A long, cylindrical electrically heated rod, 2 cm in diameter, is installed in a vacuum furnace as shown in Fig.1.8. The surface of the heating rod has an emissivity of 0.9 and is maintained at 1000 K, while the interior walls of the furnace are black and are at 800 K. Calculate the net rate at which heat is lost from the rod per unit length and the radiation heat transfer coefficient.

Figure 1.8 Schematic Diagram of Vacuum Furnace with Heating Rod Solution Solutio n Assume that steady state has been reached. Moreover, note that since the walls of the furnace completely enclose the heating rod, all the radiant energy emitted by the surface of the

rod is intercepted by the furnace walls. Thus, for a black enclosure, Eq. (1.9) applies and the net heat loss from the rod of surface surface A A1 is

Note that in order for steady state to exist, the heating rod must dissipate electrical energy Note that at the rate of 1893 W and and the rate of heat loss through the furnace walls must equal the rate of electric input to the system, that is, to the rod.

Mr. Amjed Ahmed Ahmed

7

Ch 1: Introdaction

3rd Year College of Technical

Example 1.9 Example An instrument used to study the Ozone depletion near the poles is placed on a large 2-cm-thick duralumin plate. To simplify this analysis the instrument can be thought of as a stainless steel plate 1 cm tall with a 10 cm x 10 cm square base, as shown in Fig. 1.6. The interface roughness of the steel and the duralumin is between 20 and 30 rms (µm) the contact resistance is 0.05 k/w. Four screws at the corners. The top and sides of the instrument are thermally insulated. An integrated circuit placed between the insulation and the upper surface of the stainless steel plate generates heat. If this heat is to be transferred to the lower surface of the duralumin, estimated to be at a temperature of 0°C, determine the maximum allowable dissipation rate from the circuit if its temperature is not to exceed 40°C. 40°C.

Figure 1.6 Schematic Sketch of Instrument for Ozone Measurement. Solution Solutio n Since the top and the sides of the instrument are insulated, all the heat generated by the circuit must flow downward. The thermal circuit will have three resistances the stainless steel, the contact, and the duralumin. Using thermal conductivities k ss = 14.4 W/m K , k M = 164 W/m K the thermal resistances resistances of the metal metal plates are calculated from Equations:

Mr. Amjed Ahmed Ahmed

8

Ch 1:

Introdaction

3rd Year

College Of Technical

1.6 The Energy Balance In this special case the control surface includes no mass or volume and appears as shown in Figure 1.8.Accordingly, the generation and storage terms of the Energy expression, E in 0 in –E out out -E sst t + E g = 0 Consequently, there can be no generation and storage. The conservation requirementt then becomes requiremen E in in –E out = 0 out In Figure 1.8 three heat transfer terms are shown for the control surface. On a unit area basis they are ar e conduction conducti on from the medium medi um to the control surface q" cond from the cond convection from surface to a fluid q" conv conv , and net radiation exchange from the surface to the surroundings q" rad rad . The energy balance then takes the Form and we can express each of the terms according to the appropriate rate equations.

q" cond cond

q" conv conv + q" rad rad

=

1.7 Combined heat transfer systems Summarizes the basic relations for the rate equation of each of the three basic heat transfer mechanisms to aid in setting up the thermal circuits for solving combined heat transfer problems. 1.7.1 Plane Walls in Series

In Fig. 1.15 for a three-layer system, the temperature gradients in the layers are different. The rate of heat conduction through each laye layerr is qk , and from Eq. (1.1) we get

Eliminating the intermediate temperatures T 2 and T 3 in Eq. qk can be expressed in the form

Similarly, for N for N layers in series we have

Mr. Amjed Ahmed Ahmed

9

Ch 1:

Introdaction

3rd Year

College Of Technical

where T 1 is the outer-surface temperature of layer 1 and T N+1 N+1 is the outer-surface temperature of layer N . and ∆T is the overall temperature difference, often called the temperature potential.

Figure 1.9 Conduction Through a Three-Layer System in Series. Example 1.6 Calculate the rate of heat loss from a furnace wall per unit area. The

wall is constructed from an inner layer of 0.5 cm cm thick steel (k (k : 40 W/m K ) and an outer layer of 10 cm cm zirconium brick (k (k = 2.5 W/m K ) as shown in Fig. The inner-surface temperature is 900 K and and the outside surface temperature is 460 K . What is the temperature at the interface?

Figure 1.10 Schematic Diagram of Furnace Wall. Solution Assumptions Assumptio ns: Assume that steady state exists, • neglect effects at the corners and edges of the wall, • the surface temperatures are uniform. • The rate of heat loss per unit area can be calculated from Eq:

The interface temperature T 2 is obtained from

Mr. Amjed Ahmed Ahmed

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Ch 1:

Introdaction

3rd Year

q A

=

T 1

−

T 2

R1

College Of Technical

Solving for T 2 gives

Note that the temperature drop across the steel interior wall is only 1.4 K because the thermal resistance of the wall is small compared to the resistance of the brick. Example 1.7 Example Two large aluminum plates (k = 240 W/m K ), ), each 1 cm thick, with 10 µm surface -4 roughness the contact resistance Ri = 2.75 x 10 m2 K/W. The temperatures at the outside surfaces are 395°C and 405°C . Calculate (a) the heat flux (b) the temperature drop due to the contact resistance. resistance.

Figure 1.11 Schematic Diagram of Interface Between Plates. Solution Soluti on (a) The rate of heat flow per unit area, q'' through the sandwich wall is is

the two resistances is equal to (L/k) = (0.01 m)/(240 W/m K) = 4.17 x 10-5 m2 K/W Hence, the heat flux is

(b) The temperature drop in each section. The fraction of the contact resistance is

Hence 7.67°C of the total temperature drop of 10°C is the result of the contact resistance.

Mr. Amjed Ahmed Ahmed

11

Ch 1:

Introdaction

3rd Year

College Of Technical

1.7.2 Plane Walls in Parallel Conduction can occur in a section with two different materials in parallel between the same potential. Fig. 1.18 1.18 shows a slab with two different materials of areas A A and A B in parallel. parall el. If the temper temperatures atures over the left left and and right right faces faces are uniform uniform at T 1 and T 2, the total rate of heat flow is the sum of the flows through A through A A and and A A B:

Figure 1.12 Heat Conduction Through a Wall Section with Two Paths in Parallel. Note that the total heat transfer area is the sum of A A and and A A B and that the total resistance equals the product of the individual resistances divided by their sum, as in any parallel circuit. A more complex application of the thermal network approach is illustrated in Fig. 1.19, where heat is transferred through a composite structure involving thermal resistances in series and in parallel. For this system the resistance of the middle layer, layer, R R2 becomes becomes and the rate rat e of heat flow is

Where N is Where N is number of layers in series Rn : Thermal resistance of nth layer ∆ T overall o verall : temperature difference across two outer surfaces

Figure 1.13 Conduction Through a Wall Consisting Consist ing of o f Series and Parallel Thermal Thermal Paths. Paths.

Mr. Amjed Ahmed Ahmed

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Ch 1:

Introdaction

3rd Year

College Of Technical

Example 1.8 Example A layer of 2 in thick firebrick (k b = 1.0 Btu/hr ft °F ) is placed between two ¼ in.-thick steel plates (k s = 30 Btu/hr ft °F )).. The faces of the brick adjacent to the plates are rough, having solid-to-solid contact over only 30 % of the total area, with the average height of asperities being L2=1/32 in. in. If the surface temperatures of the steel plates are 200° and and 800°F , respectively. res pectively. the conductivit conductivity y of air ka is 0.02 0.02 Btu/hr Btu/hr ft °F, determine the rate of heat he at flow per unit area.

Figure 1.14 Thermal Circuit for the Parallel-Series Composite Wall. L1 = 1 in.; L2 = 1/32 in.; L 3 = 1/4 in.; T 1 is at the center. Solution Solution The overall unit conductance for half the composite wall is then, from an inspection of the thermal circuit

Since the air is trapped in very small compartments, the effects of convection are small and it will be assumed that heat flows through the air by conduction. At a temperature of 300°F. Then R 5 the thermal resistance of the air trapped between the asperities, is, on the basis of a unit area, equal to

Mr. Amjed Ahmed Ahmed

13

Ch 1:

Introdaction

3rd Year

College Of Technical

The factors 0.3 and 0.7 in R 4 and R 5, respectively, represent the percent of the total area for the two separate heat flow paths. The total thermal resistance for the two paths, R 4 and R 5 in parallel, is

The thermal resistance of half of the solid brick, R l is and the overall unit conductance is

Inspection of the values for the various thermal resistances shows that the steel offers a negligible resistance

Mr. Amjed Ahmed Ahmed

14

Ch 1:

Introdaction

3rd Year

College Of Technical

1.5.2 Convection and Conduction in Series

Figure (1.15) (1.15) shows a situation in which heat is transferred between two fluids separated by a wall, the rate of heat transfer from the hot fluid at temperature T hot to the cold fluid at temperature T cold is

Figure 1.15 Ther Thermal mal Circuit Cir cuit with Conduction Conduction and Convection Convection in Series. Series. Example 1.8 A 0.1 m thick brick wall (k = 0.7 W/m K ) is exposed to a cold wind at 270 K through a convection heat transfer coefficient of 40 W/m2 K. On the other side is air at 330 K, K, 2 with a natural convection heat transfer coefficient of 10 W/m K. Calculate the rate of heat transfer per unit area. Solution Solution The three resistances are the rate of heat transfer per unit area is :

Mr. Amjed Ahmed Ahmed

15

Ch 1:

Introdaction

3rd Year

College Of Technical

1.5.3 Convection and Radiation in Parallel

In many engineering problems a surface loses or receives thermal energy by convection and radiation simultaneously. Figure 1.23 illustrates the co current heat transfer from a surface to its surroundings by convection and radiation. q= qc + qr q = hc A(T 1 - T 2 )+ )+ hr A (T 1 - T 2 ) ) q =( =(h hc+ hr )A (T ( T 1 - T 2 ) ) c where h the, the average convection heatcoefficient transfer coefficient between area A1 and the surroundings air is at T radiation heat transfer 2

The combined heat transfer coefficient is h = hc + hr

Example 1.5 Air at 20C blow over a hot plate 50 x 75 cm cm and thick 2 cm maintained at 250 oC . the convection heat transfer coefficient is 25 W/m2 C. calculate the inside plate temperature if it is mode of carbon steel and that 300 W is lost from the plate surface by radiation. Where thermal conductivity is 43 w/m C. Soluti Sol ution on

qconv = h A(T s-T ∞ ) qconv = 25 (0.5 *0.75) (250 - 20) qconv =2.156 KW qcond = qconv + qrad qcond = 2.156 +0. 3=2.456 kW qcond = kA

Τ 1 − Τ 2

L

2.456 = 43 (0.5 × 0.75) T 1 = 253.05 oC

Mr. Amjed Ahmed Ahmed

Τ1 −

250

0.02

16

Ch 1:

Introdaction

3rd Year

College Of Technical

Example 1.9 Example A 0.5 m diameter m diameter pipe (ε = 0.9) 0.9) carrying steam has a surface temperature of 500 K . The pipe is located located in a room at 300 K, K, and the convectio convection n heat transfer transfer coefficie coefficient nt between between the pipe pipe 2 surface and the air in the room is 20 W/m K . Calculate the combined heat transfer coefficient and the rate of heat loss per meter of pipe length.

Schematic Diagram of Steam Pipe Figure 1.17 Schematic Solution Solu tion

hr = 13.9 W/m2 K The combined heat transfer coefficient is h = hc + hr = 20 + 13.9 = 33.9 W/m2 K and the rate of heat loss per meter is

1.5.4 Overall Heat Transfer Coefficient

We noted previously that a common heat transfer problem is to determine the rate of heat flow between two fluids, gaseous or liquid, separated by a wall. If the wall is plane and heat is trans t ransferr ferred ed only by conve c onvection ction on both bo th sides s ides,, the rate of heat he at transf t ransfer er in terms of the two fluid temperatures tempe ratures is given by:

the rate of heat flow is expressed only in terms of an overall temperature potential and the heat transfer characteristics of individual sections in the heat flow path., the overall transmittance, or the overall coefficient of heat transfer U Writing Eq. (1.29) in terms of an overall coefficient gives

An overall heat transfer coefficient

Mr. Amjed Ahmed Ahmed

can U can

be based on any chosen area

17

Ch 1:

Introdaction

3rd Year

College Of Technical

Example 1.10 In the design of a heat exchanger for aircraft application, the maximum wall temperature in steady state is not to exceed 800 k. For the conditions conditions tabulated below, determine the maximum permissible unit thermal resistance per square meter of the metal wall that separates the hot gas T gh = 1300 K from the cold gas T gas T gc = 300 K. Combined heat transfer coefficient on hot side h 1 = 200 W/m2 K Combined heat transfer coefficient on cold side h3 = 400 W/m2 K

Figure 1.18 Physical System and Thermal Circuit. Solution Solu tion

In the steady state we can write

Solving for R for R2 gives R2 = 0.0025 m2 K/W

Mr. Amjed Ahmed Ahmed

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Ch 1:

Introdaction

3rd Year

College Of Technical

E Ex xample 1.11 The door for an industrial gas furnace is 2 m x 4 m in surface area and is to be insulated 2. to reduce heat loss to no more than 1200 W/m2. The interior surface is a 3/8-in.-thick Inconel 600 sheet 600 sheet ( K= K= 25 W/m K )),, and the outer surface is a l/4 in.-thick sheet of Stainless steel 316. Between these metal sheets a suitable thickness of insulators material is to be placed. The effective gas temperature inside the furnace is 1200°C, and the overall heat transfer coefficient between the gas and the door is U i = 20 W/m2 K . The heat transfer coefficient between the outer surface of the door and the surroundings at 20°C is is hc= 5 W/m2 K . calculate the thickness of insulated should be use

Figure 1.19 Cross section of composite wall of gas furnace door S So olution

The thermal resistance of the two metal sheets are approximately 25 W/m K the the thermal resistance of the two metal sheets are approximately: L1+L2=0.25+0.375= =0.25+0.375=0.625 0.625 in

These resistances are negligible compared to the other three resistances shown in the simplified thermal circuit below;

The temperature drop between the gas and the interior surface of the door at the specified heat flux is: Q=AU ∆T

Hence, the temperature of the Inconel will be about (1200-60)=1140°C . This is acceptable since no appreciable load is applied. The temperature drop at the outer surface is

The insulation thickness for k = 0.27 W/m K is: is: X (1140-240 )

Mr. Amjed Ahmed Ahmed

19

Ch 2:

3rd Year College College

Heat Conduction

of Technical

Chapter Two Heat Conduction 2.1 Introduction A major objective in a conduction analysis is to determine the temperature field in a medium (Temperature Distribution), which represents how temperature varies with position in the medium. knowledge of the temperature distribution: • Determination of thermal stresses, It could be used to ascertain structural integrity through • To determine the optimize thickness of an insulating material • To determine the compatibility of special coatings or adhesives used with the material.

2.2 Conservation of Energy Applying energy conservation to the control volume. At an instant, these include the rate at which thermal and mechanical energy enter E in out ut . through the control surface, in and leave E o Is additional to the rate of change of energy generation E g and stored E sst t . A general form of the energy conservation requirement may then be expressed on rate basis as: E g

E in + E g − E out = E st

2.1

E in in

E st st

E oout ut

2.3 The Conduction Equation of Rectangular Coordinate Coordinate Consider the energy processes that are relevant to this control volume. If there are temperature gradients, conduction heat transfer will occur across each of the control surfaces at the x, y, and z coordinate. The conduction heat rates at the opposite surfaces can then be expressed as a Taylor series expansion where, neglecting higher order terms,

Figure 2.1 Differential control volume, dx dy dz.

q x+ dx = q x + q y + dy = q y +

q z + dz = q z +

Mr. Amjed Ahmed

dq x dx dq y dy dq

z

dz

dx

Slope =

dq x dx

dy

dz

20

Ch 2:

3rd Year College College

Heat Conduction

of Technical

The rate of change of energy generation E g and stored E sst t

= q&dxdydz E g = q&V E st = mC p

dT

= ρ VC p

dT

= ρ (dxdydz )C p

dT

dt dt dt where q& is the rate at which energy is generated per unit volume (W/m 3) and to express conservation of energy using the foregoing rate equation

E in + E g − E out = E st

and, substituting equations, we obtain dq y dq dq dT q x + q y + q z − ( q x + x dx) − (q y + dy ) − ( q z + z dz ) + q& dxdydz = ρ dxdydzC p dz dt dx dy The conduction heat rates may be evaluated from Fourier's law, dT dT = − kdzdy q x = − kA dx dx dT dT = − kdzdx q y = − kA 2.3 dy dy

q z = − kA

dT

= −kdzdx

2.2

dT

dz dz Substituting Equations 2.3 into Equation 2.2 and dividing out the dimensions of the control volume (dx (dx dy dz ), ), we obtain ∂T ∂ ∂T ∂ ∂T ∂ ∂T (k ) + ( k ) + ( k ) + q& = ρ C p ∂t ∂ z ∂ z ∂ y ∂ y ∂ x ∂ x 2.4 It is often possible to work with simplified versions of Heat Equation (k=Const)is ∂ 2T ∂ 2T ∂ 2T q& 1 ∂T

∂ x 2

+

∂ y 2

+

∂ z 2

+

=

α ∂t where α = k/ ρC p (m2/s) is the thermal diffusivity. k

2.5

2.3.1 One Dimension Steady State Conduction A plane wall separates two fluids of different temperatures. Heat transfer occurs by convection from the at T s1, by hot fluid at T ∞ to one surface of the wall at ,1

conduction through the wall, and by convection from the other surface of the wall at T s2 to the cold fluid at T ∞, 2

Figure 2.2 Heat transfer through a plane wall. If the heat transfer one dimensional and under steady-state conditions (there can be no change in the amount of energy storage and generation; hence Heat Equation reduces to ∂ ∂T (k ) = 0 ∂ x ∂ x 2.6 If the thermal conductivity is assumed to be constant (k=Const (k=Const ), ), the equation may be integrated twice to obtain the general solution T(x)=C 1 x+C 2

Mr. Amjed Ahmed

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Ch 2:

3rd Year College College

Heat Conduction

of Technical

To obtain the constants of integration, C 1 and C 2 boundary conditions must be introduced. Applying the conditions B.C.1 x=0 at T=T s1 C 2=T s1 B.C.2 x=L at T=T s2 T − T s1 T s2=C 1 L+C 2 = C 1 L+T s1 C 1 = s 2 L Substituting into the general solution, the Temperature Distribution is then

T(x) = (T S 2 -T S 1 ) x + T s1 L

Linearly equation.

2.7

2.3.2 Contact Resistance The existence of a finite contact resistance is due principally to surface roughness effects. Contact spots are interspersed with gaps that are, in most instances, air filled. Heat transfer is therefore due to conduction across the actual contact area and to conduction and/or radiation across the gaps. The contact resistance may be viewed as two parallel resistances: that due to : (1)the contact spots (2) that due to the gaps (the major contribution to the resistance). The resistance is defined as

Rtc′′ =

T A − T B q x′′

TB TA

Figure 2.3 Temperature drop due to thermal contact resistance.

Mr. Amjed Ahmed

22

Ch 2:

3rd Year College College

Heat Conduction

of Technical

Example 2.1 The temperature distribution across a wall 1 m thick m thick at a certain instant of time is given as 2 (T(x) = a+ bx + cx ) where T is is in degrees Celsius and x and x is in meters, while a = 900° C , b = 2 300°C/m,, and c= -50°C/m . A uniform heat generation q=1000 W/m3, is present in the wall of 300°C/m area 10 m2 having the properties ρ = 1600 kg/m3, k = 40 W/m K , and C p = 4 kJ/kg K . 1. Determine the rate of heat transfer entering (x = 0) and leaving the wall (x = 1 m). 2. Determine the rate of change of energy storage in the wall.

3. Determine the time rate of temperature change at x = 0, 0.25 and 0.25 and 0.5 m. m. Solution Assumptions Assumptio ns:: 1. One-dimensional conduction in the x the x direction. direction. 2. Homogeneous medium with constant properties. 3. Uniform internal heat generation, q (W/m3).

1.

2.

E in + E g − E out = E st

2.1

3. The time rate of change of the temperature at any point in the medium may be determined from the heat equation, Equation 2.15, as

Mr. Amjed Ahmed

23

Ch 2:

3rd Year College College

Heat Conduction

of Technical

Example 2.2 The diagram shows a conical section from pyroceram (k (k = 3.46 W/m K ). ). It is of circular cross section with the diameter D diameter D = ax. ax. The small end is at x at x1 = 50 mm and mm and the large end at x at x2 = 250 mm. The mm. The end temperatures are T 1 = 400 K and and T 2 = 600 K , while the lateral surface is well insulated and a=0.25 a=0.25.. 1. Derive an expression for the temperature distribution T(x) T(x) in symbolic form, assuming one-dimensional conditions.

2. Sketch thethe temperature distribution. 3. Calculate heat rate through the cone. Solution Assumptions Assumptio ns:: 1. Steady-state conditions. 2. One-dimensional conduction in the x direction. 3. No internal heat generation. 4. Constant properties. dT q x = − kA dx With A= With A= л D2 /4= лa2 x2 / 4 and separating variables 4q x dx = −kdT 2 2

a x Integrating from xπ 1 to any x within the, it follows that 4q x

x

∫

dx

π a 2 x1 x 2

T

= −k ∫ dT

(k = const )

T 1

Hence 4q x

1 1 ( − + ) = − k (T − T 1 ) π a x x1 2

and solving for q q x =

π a 2 k (T 1 − T ) 4[((1 / x1 ) − (1 / x))]

or solving for T T ( x) = T 1 −

4q x 2

(−

1

+

1

)

π a k x x1 B.C.2 T=T s2 at x=x2 π a 2 k (T 1 − T 2 ) q x = 4[((1 / x1 ) − (1 / x2 ))] 4q x (T 1 − T 2 ) = 2 π a k [((1 / x1 ) − (1 / x 2 ))] Substituting for q into the expression for T(x), the temperature distribution becomes

⎡ (1 / x) − (1 / x1 )) ⎤ ⎥ ( 1 / ) ( 1 / ) x x − 1 2 ⎦ ⎣

T ( x) = T 1 + (T 1 − T 2 ) ⎢

Substituting numerical values into the foregoing result for the heat transfer rate

Mr. Amjed Ahmed

24

Ch 2:

3rd Year College College

Heat Conduction

of Technical

2.4 The Conduction Equation of Cylindrical Coordinate A common example is the hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures.

Figure 2.4 Hollow cylinder with convective surface conditions.

For a general transient three-dimensional in the cylindrical coordinates T= T(r, φ ,z, t), t), the general form of the conduction equation in cylindrical coordinates becomes 1 ∂ ∂T 1 ∂ 2T ∂ 2T q& 1 ∂T + = + (r ) + 2 2.8 r ∂φ 2 ∂ z 2 k α ∂t r ∂r ∂r If the heat flow in a cylindrical shape is only in the radial direction and for steady-state conditions with no heat generation, the conduction equation reduces to 1 ∂ ∂T ( r ) = 0 r ∂r ∂r Integrating once with respect to radius gives ∂T C 1 ∂T and = r = C 1 ∂r ∂r r A second integration gives T = C 1 ln r + C 2. 2.9 To obtain the constants (C1 and C2), we introduce the following boundary conditions B.C.1 T=T i at r=r i T i = C 1 ln r i+ C 2. B.C.2 T=T o at r=r o T o = C 1 ln r o + C 2. Solving for C1 and C2 and substituting into the general solution, we then obtain T o − T i = C 1 ln r o r i C 1 =

T o − T i ln(r o / r i )

T ( r ) =

C 2 = T o −

T o − T i ln(r o / r i )

T o − T i

r ln( ) + T i r i ln(r o / r i )

2.10

we obtain the following expression for the heat transfer rate C 2π Lk (T i − T o ) dT = −(2π rLk ) 1 = qr = − kA ln(r o / r i ) r dr

qr =

Mr. Amjed Ahmed

(T i − T o ) R

ln r o

R =

ln(r o / r i ) 2π Lk

2.11

2.12

25

Ch 2:

3rd Year College College

Heat Conduction

of Technical

2.4.1 Overall Heat Transfer Coefficient A hot fluid flows through a tube that is covered by an insulating material. The system loses heat to the surrounding air through an average heat transfer coefficient h c,o. the thermal resistance of the two cylinders at the inside of the tube and the outside of the insulation gives the thermal network shown below the physical system T where h∞ hot fluid temperature and T c,∞ the environmental air temperature

the rate of heat flow is 2.13

it is often convenient to define an overall heat transfer coefficient by the equation q = UAo (T hot ) hot -T cold cold The area varies with radial distance. Thus, the numerical value of U will depend on the area selected. Since the outermost diameter is the easiest to measure in practice, Ao= 2 л 2 л r r 3 L is L is usually chosen as the base area. Comparing between above Equations. we see that

Note that UA=U A i i=U o Ao

2.14

Ao = 2π r o L and the overall coefficient becomes 3

2.15

Mr. Amjed Ahmed

26

Ch 2:

3rd Year College College

Heat Conduction

of Technical

Example 2.3 Compare the heat loss from an insulated and an un-insulated copper pipe (k = 400 W/m K) has K) has an internal diameter of 10 cm and cm and an external diameter of 12 cm. cm. Saturated steam 2 flows inside the pipe at 110°C (( hci = 10,000 W/m K ). ). The pipe is located in a space at 30°C and the heat transfer coefficient on its outer surface is estimated to be 15 W/m2 K . The insulation available to reduce heat losses is 5 cm thick and its thermal conductivity is 0.20 W/m K Solution

Figure 2.5 Schematic Diagram and Thermal Circuit for a Hollow Cylinder with Convection Surface Conditions The heat loss per unit length is T s − T ∞ q

L

=

R + R + R 1

Hence we get

2

3

Since R 1 and R 2 are negligibly small compared to R 3 For the un-insulated pipe . q/L = 80/0.177 = 452 W/m For the insulated pipe, we must add a fourth resistance between r 1 and r 3. ln(r i / r o ) ln(11 / 6) R = = = 0.482mK / W 4 2π k 2π (0.2W / mK )

Also, the outer convection resistance changes to Ro =

1

= 0.096mK / W 2π (0.11 × 15) The total thermal resistance per meter length ( RTotal =R4+Ro= 0.578 m K/W) K/W) q/L = 80/0.578 = 138 W/m. Adding insulation will reduce the heat loss from the steam by 70%.

Mr. Amjed Ahmed

27

Ch 2:

3rd Year College College

Heat Conduction

of Technical

Example 2.4 A hot fluid at an average temperature of 200 oC flows through a plastic pipe of 4 cm OD and 3 cm ID. The thermal conductivity of the plastic is 0.5 W/m K, and the heat transfer coefficient at the inside is 300 W/m2 K. The pipe is located in a room at 30°C, and the heat transfer coefficient at the outer surface is 10 W/m 2 K, Calculate the overall heat transfer coefficient and the heat loss per unit length of pipe. Solution

The overall heat transfer coefficient is based on the outside area of the pipe

The heat toss per unit length is

2.4.2 Critical Radius of Insulation Although the conduction resistance increases thearea. addition insulation, convection resistance decreases due to increasing outerwith surface Henceof there may existthe an insulation thickness that minimizes heat loss by maximizing the total resistance to heat transfer. Air T ∞

RTotal =

ln( r / r i )

+

1

2π k 2π rh An optimum insulation thickness would be associated with the value of r that minimized qr or maximized R maximized RTotal . Such a value could be obtained from the requirement that dq = 0 at r=r Critical dr c

=

dr c

1 kr c

− 2π L(T i − T o )((1 / Kr c ) − (1 / hr c )) 2

dq r

−

⎡ ln( r c / r i ) 1 ⎤ + ⎥ ⎢ k rh ⎦ ⎣ 1 2

= 0

r c h

2

r c =

k h

= 0

2.16

For spherical shape:

r c =

Mr. Amjed Ahmed

2k h

28

Ch 2:

Heat Conduction

3rd Year College College

of Technical

Example 2.5 Calculate the total thermal resistance per unit length of tube for a 10 mm diameter tube having the following insulation thicknesses: 0, 2, 5, 10, 20 and 40 mm. The insulation is composed of Cellular Glass ( Glass (k=0.055 k=0.055 w/m K ), ), and the outer surface convection coefficient is 2 5 W/m K. k 0.055 Solution r c = = = 0.011m h 5

Hence r c > r , and heat transfer will increase with the addition of insulation up to a thickness of r cc--r i =(0.011-0.005)=0.006m The thermal resistances corresponding to the prescribed prescribed insulation thicknesses may be calculated and are summarized as follows.

Mr. Amjed Ahmed

29

Ch 2:

3rd Year College College

Heat Conduction

of Technical

2.5 The Conduction Equation of Spherical Coordinate For spherical coordinates, the temperature is a function of the three space coordinates T(r ,θ ,θ , , φ , t). t). The general form of the conduction equation is then 2.17

Figure 2.6 Spherical Spherical Coordinate System

For a hollow sphere with uniform temperatures at the inner and outer surfaces, the temperature distribution without heat generation in the steady state can be obtained by simplifying Eq 2.17. Under these boundary conditions the temperature is only a function of the radius r, and the conduction equation is 1 ∂ 2 ∂T (r )=0 ∂r r 2 ∂r C ∂T r 2 = C 1 ∂T = 21 ∂r ∂r r T (r ) = C 2 −

C 1 r

B.C.1

T=T i

at

r=r i

Ti = C 2 −

C 1

B.C.2

T=T o

at

r=r o

T o = C 2 −

C 1

r i r o

− C 1 = T i − T o 1 1 ( )−( ) r o r i

C 2 = T o + ( T i T o )1 ((1 / r o ) − (1 / r i )) r o

The temperature distribution is T − T 1 1 T (r ) = ( i o )( − ) + T i 1 1 r r o − r i

r o

The rate of heat transfer through the spherical shell is dT dT qr = − kA = − k (4π r 2 ) dr dr may be expressed in the integral form 1

ro

∫

qr dr

4π ri r 2

2.18

To

= − ∫ kdT

A=4πr 2

ﺣﺔ ﻣﺴ

A=πD2

ﻟﻜ ﺮ ة

V=4πr 2/3 ﻟﻜﺮة ﺣﺠﻢ

Ti

Assuming constant k and and qr , we obtain

Mr. Amjed Ahmed

30

Ch 2:

3rd Year College College

Heat Conduction

qr =

4π kr o r i (T i − T o ) r o − r i

R =

(r o − r i ) 4π kr o r i

of Technical

(2.19)(2.20)

Example 2.6 The spherical, thin-walled metallic container is used to store liquid nitrogen at 77 K. The container has a diameter of 0.5 and is covered with an evacuated insulation system composed

of silica powder (k =air0.0017 (k K ). ). latent The insulation is 25 mm thick, its outer surface exposed to ambient at 300W/m K. The heat of vaporization h fg ofand liquid nitrogen is 2 is × 5 2 10 J/kg. If the convection coefficient is 20 W/m K over the outer surface, 1. Determine the rate of liquid b boil-off oil-off of nitrogen per h hour? our? 2. Show expiration of critical radius of insulation? Ans: r c= 2h/k Solution 1. The rate of heat transfer from the ambient air to the nitrogen in the the container can be obtained from the thermal circuit. We can neglect the thermal resistances of the metal wall and between the boiling nitrogen and the inner wall because that heat transfer coefficient is large. Hence

Figure 2.7 Schematic Diagram of Spherical Container

To determine the rate of boil-off we perform an energy balance & h fg = q E in m in=E out out Solving for m gives

&= m

Mr. Amjed Ahmed

q h fg

=

(13.06 J/s)(3600 s/hr)

= 0.235kg / hr

2 x 105 J/kg

31

Ch 2:

3rd Year College College

Heat Conduction

Of Technical

2.6 Heat Generation • • • •

A common thermal energy generation process involves The conversion from electrical to thermal energy in a current-carrying medium E g =I 2 R. The deceleration and absorption of neutrons in the fuel element of a nuclear reactor Exothermic chemical reactions occurring within a medium. Endothermic reactions would, of course, have the inverse effect A conversion from electromagnetic to thermal energy may occur due to the absorption of radiation within the medium. Note: Remember not to confuse energy generation with energy storage.

2.6.1 Plane Wall with Heat Generation

(a) Asymmetrical plane plane wall (b)Symmetrical plane wall (c) Adiabatic surface at midline Figure 2.8 Conduction in a with uniform heat generation Assumptions =Const. • Uniform heat generation per unit volume q =Const. • For constant thermal conductivity k=Const . • One dimension and steady state heat transfer.

The appropriate form of the heat equation, is ∂ 2T ∂ 2T ∂ 2T q& 1 ∂T

∂ x 2

+

∂ y 2

+

∂ z 2

+

k

=

α

∂t

2.5

the equation may be integrated twice to obtain the general solution q & T(x) = - x 2 + C 1 x + C 2 2.21 2k To obtain the constants of integration, C 1 and C 2 boundary conditions must be introduced. q & B.C.1 T=T s1 at x=L T s1 = - L 2 + C 1 L + C 2 2k q & at x= -L T s 2 = − L B.C.2 T=T s2 2 − C 1 L + C 2 2k 2 q & L T − T T + T C 1 = s1 s 2 C 2 = + s1 s 2 2 L 2k 2 In which case the Temperature distribution is 2 2 T ( x ) = q& L (1 − x 2 ) + T s 2 − T s1 x + T s1 + T s 2 2k 2 2 L L

Mr. Amjed Ahmed

2.22

32

Ch 2:

3rd Year College College

Heat Conduction

Of Technical

The Symmetrical Plane Wall when both surfaces are maintained at a common temperature, T s1= T s2= T s. The temperature distribution is given by T(x) =

q& L2 2k

− ( 1

x 2 L2

) + T s

2.23

The maximum temperature (T=T o) exists at the midline ( x=0). x=0). 2 q& L q& L2 T o − T s = 2.24 or T o = 2k + T s 2k which case the temperature distribution, after substitution eq 2.24 into eq 2.23 T ( x ) − T s x 2 2.25 = 1− 2 T o − T s L Consider the surface at x at x = L for L for ( Fig. Fig. 2.8b) 2.8b) or the insulated plane wall ( Fig. Fig. 2.8c). 2.8c). The energy balance given by E g = E out q& V = Ah(T s − T ∞ )

Neglecting radiation radiation

q& AL = Ah(T s − T ∞ ) The surface temperature is

T = T ∞ +

q& L

2.26

h

s

Note :A heat generation cannot be represented by a thermal circuit element Example 2.7 A long electrical heating element made of iron has a cross section of 10 cm x 1.0 cm. It is immersed in a heat transfer oil at 80°C. If heat is generated uniformly at a rate of 10 6 W/m3 by an electric current, determine the heat transfer coefficient necessary to keep the temperature of the heater below 200°C. The thermal conductivity for iron is 64 W/m K.

Solution

T max − T 1 =

q& L2 8k

10 6 × (0.01) 2

=

8 × 64

= 0.2 C o

q& V = Ah(T s − T ∞ ) h=

Mr. Amjed Ahmed

&

L q& A = Ah(T s − T ∞ ) 2

42W / m 2 K q L 2(T s − T ∞ ) =

33

Ch 2:

3rd Year College College

Heat Conduction

Of Technical

Example 2.8 A plane wall is a composite of two materials, A and B. The wall of material A ( k = 75 W/m K ) has uniform heat generation 1.5 X 106 W/m3, and thickness 50 mm. mm. The wall material B has no generation with (k = 150 W/m K) and thickness 20 mm. The inner surface of material A is well insulated, while the outer surface of material B is cooled by a water stream with 30°C and heat transfer coefficient 1000 coefficient 1000 W/m2 K . 1. Sketch the temperature distribution that exists in the composite under steady-state

conditions. 2. Determine the maximum temperature T o of the insulated surface and the temperature of the cooled surface T s.

Solution Assumptions: 1. Steady-state conditions. 2. One-dimensional conduction in x direction. 3. Negligible contact resistance between walls. 4. Inner surface of A adiabatic. 5. Constant properties for materials A and B. q& L A T 2 = T 2.26 ∞ + h

T 2 = 30 +

1.5 × 106 × 0.05

1000

q′′ =

= 105 C o

T 1 − T ∞

′′ + Rconv ′′ Rcond

′′ + Rconv ′′ )q′′ T 1 = T ∞ + ( Rcond ′′ q& = q AL A where the resistances for a unit surface area are Hence T 1 =115oC From Equation 2.24 the temperature at the insulated surface is

T o = T 1 +

q& L2 A 2k A 6

2

T o = 115 + 1.5 × 10 (0.05) = 140o C 2 × 75

Mr. Amjed Ahmed

34

Ch 2:

3rd Year College College

Heat Conduction

Of Technical

2.6.2 Radial Shapes with Heat Generation To determine the temperature distribution in the cylinder, we begin with the appropriate form of the heat equation. For constant thermal conductivity is 1 ∂

(r

∂T q& )+ =0 ∂r k

r ∂r q& 2 ∂T r ∂r = − 2k r + C 1 q& 2 r + C 1 ln r + C 2 T (r ) = − 4k

A Solid Cylinder To obtain the constants (C ( C 1 & C 2), we introduce the following boundary conditions B.C.1 dT/dr=0 at r=0 C 1=0 =0 q& 2 B.C.2 T=T s at r=r o C 2 = r o + T s 4k Solving for C 1 and C 2 and substituting into the general solution, we then obtain T (r ) = −

q& r o

2

(1 −

4k

r 2 r o2

) + T s

The maximum temperature T=T o at r=0 2 q& r o T o = − + T s 4k substitution replace group T ( r ) − T s T o − T s

q& r o

2.28

q& r o

2

4k

= T o − T s

2.29

2

4k

in equation 2.28 2.28 r 2

=1−

r o2

2.30

The energy balance given by E g = E out q&V = Ah(T s − T ∞ )

2

q&π r o L = 2π r o hL(T s − T ∞ )

The surface temperature is

T s = T ∞ +

Mr. Amjed Ahmed

q&r o 2h

2.31

35

Ch 2:

3rd Year College College

Heat Conduction

Of Technical

B For Hollow Cylinder

T (r ) = −

q&

r 2 + C 1 ln r + C 2

4k To obtain the constants (C ( C 1 and C 2), we introduce the following boundary conditions q & 2 B.C.1 T=Ti at r=r i r i + C 1 ln r i + C 2 T i = − 4k q & 2 T=T o at r=r o B.C.2 r o + C 1 ln r o + C 2 T o = − 4k Solving for C 1 and C 2 and substituting into the general solution, we then obtain 2

2

C 1 = (T − T ) + q( r − r ) / 4k ln(r i / r o ) i

C 2 = T o +

&

o

q& 4k

i

o

2

2

r o −

2

(T i − T o ) + q& ( r i − r o ) / 4k ln(r i / r o )

× ln r o

In which case the Temperature distribution is 2

T ( r ) = T o +

2

q& (r i − r o ) 4k

+

ln(r / r o ) ⎡ q& ⎤ 2 2 r r T T ( ) ( ) − + − o i o i ⎥⎦ ln(r o / r i ) ⎢⎣ 4k

2.32

The energy balance given by E g = E out

q& V = Ah (T s − T ∞ ) &

2

2

qπ (r − r ) L = 2π r o hL (T s − T ∞ ) The surface temperature is 2 2 q& (r o − r i ) T s = T ∞ + 2hr o o

i

:ﻣﻬﻤﺔ ﺷﺘﻘﺎﻗﺎت ﻟﺤﻠﻘﻴﺔ ﻟﻤﺠﻮﻓﺔ ﻧﻪ ﻟﻼ ﻄﻮ رة ﻟﺤﺮ ﻟﺘﻮزﻳ درﺟﺔ ﻋﻼﻗﺔ ﺷﺘﻖ ﻣﻌﺰول ﻟﺨﺎرﺟﻲ ﻟﺴﻄﺢ آﺎن ذ ﻟﻤﺠﻮﻓﺔ ﻧ ﻟﺤﻠﻘﻴﺔ ﻮﻄﺳﻼﻟ رة ﻟﺤﺮ درﺟﺔ ﻟﺘﻮزﻳﻊ ﻋﻼﻗﺔ ﺷﺘﻖ ﻟﻠﻜﺮة رة ﻟﺤﺮ درﺟﺔ ﻟﺘﻮزﻳﻊ ﻋﻼﻗﺔ ﺷﺘﻖ

Mr. Amjed Ahmed

2.33

36

Ch 2:

3rd Year College College

Heat Conduction

Of Technical

Example 2.9 A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05 m diameter at the rate of 7.5 x 107 W/m 3. These rods are jacketed by an annulus in which water at an average temperature of 120°C is circulated. The water cools the rods and the average convection heat transfer coefficient is estimated to be 55,000 W/m 2 K. If the thermal conductivity of uranium is 29.5 W/m K, determine the center temperature of the uranium fuel rods.

Figure 2.9 Nuclear Reactor. Solution The rate of heat flow by conduction at the outer surface equals the rate of heat flow by convection from the surface to the water:

T s = T ∞ +

q& r o

T s = 120 +

2h 7 . 5 × 10 7 × 0 . 025 2 × 55000

= 137 o C

The maximum temperature from equation 2.29 2 7 2 q & r 7.5 × 10 × (0.025) T o = o + T s = + 137 = 534o C 4k 4 × 29.5

Mr. Amjed Ahmed

37

Ch 2:

Extended Surfaces

3rd Year

College of Technical

2.7 Heat Transfer In Extended Surfaces

Extended surfaces have wide industrial application as fins attached to the walls of heat transfer equipment in order to increase the rate of heating or cooling cooling q = h As (T s- T ∞ ). Fins come in many shapes and forms, some of which are shown in Fig 2.11.

(a) Bare surface

(b) Finned surface

Figure 2.10 Use of fins to enhance heat transfer from a plane wall.

Figure 2.11 uniform Fin configurations (a) Rectangular Rectangular Fin, (b)& (c)Pin Fin

The selection of fins is made on the basis of thermal performance and cost . the fins is stronger when the fluid is a gas rather than a liquid. The selection of suitable fin geometry requires a compromise among: among: • A cost and weight are available space • Pressure drop of the heat transfer fluid • Heat transfer characteristics of the extended surface.

Figure 2.12 non-uniform Fin configurations

(a) Parabolic

Mr. Amjed Ahmed Ahmed

(b) Triangular

(c) Annular fin

(d) Pin fin.

38

Ch 2:

Extended Surfaces

3rd Year

College of Technical

Consider a pin fin having the shape of a rod whose base is attached to a wall at surface temperature T s. The fin is cooled along its surface by a fluid at temperature T ∞ To derive an equation for temperature distribution, we make a heat balance for a small element of the fin. Heat flows by conduction into the left face of the element, while heat flows out of the element by conduction through the right face and by convection from the surface. Assumptions

1. The The fin fin is has a uniform cross-sectional area conductivity (k = constant) 2. made of a material having uniform 3. The heat transfer coefficient between the fin and the fluid is constant (h=constant). 4. One dimensional steady state condition only. 5. Non heat generation(q=0). 6. Radiation is negligible.

D Figure 2.12 Schematic Diagram of a Pin Fin Protruding from a Wall

E in in = E out out q x = q x+dx +qconv dq q x + dx = q x + x dx dx In symbolic form, this equation becomes dT ( x ) dT ( x ) − kA = − kA + hc dA s (T ( x) − T ∞ ) dx x + dx dx x

2.34

dAs= Pdx Where P is the perimeter of the fin Pdx is the fin surface area between x between x and and x+dx x+dx.. A Cross section area of fin If k and and h are uniform, Eq. Eq. 2.34 simplifies 2.34 simplifies to the form

d 2T ( x ) dx 2

−

hP kA

[T ( x) − T ∞ ] = 0

2.35

It will be convenient to define an excess temperature of the fin above the environment, θ (x) (x) = [T(x) - T ∞ ], ], and transform Eq. 2.35 into the form d 2θ ( x ) − m 2θ = 0 2 Where

dx m2= hP/kA. hP/kA.

Mr. Amjed Ahmed Ahmed

2.36

39

Ch 2:

Extended Surfaces

3rd Year

College of Technical

Last equation is a linear, homogeneous, second-order differential equation whose general solution is of the form θ (x) (x) = C 1 e mx + C 2 e –mx

2.37

To evaluate the constants C1 and C2 it is necessary to specify appropriate boundary conditions. B.C.1

θ (0) (0) = (T s – T ∞ ) θ s = C 1 + C 2

at

x=0

2.38

A second boundary condition depends on the physical physical condition at the end of the fin. we will treat the following Four Cases:

Case1: The fin is very long and the temperature at the end approaches the fluid temperature: θ (( ∞ ) = (T ∞ – T ∞ ) = 0 at x=∞ x=∞ Case2: The end of the fin is insulated: d θ θ ( x ) at = 0 dx Case3: The temperature at the end of the fin is fixed: θ (L) (L) = (T L – T ∞ ) at Case4: The tip loses heat by convection d θ ( x ) − k = hθ ( L ) dx x = L

at

x=L x=L

x=L x=L

Case 2

X

L

X

L

2.13 Representations of Four Boundary Conditions at the Tip of a Fin F i i gure

Mr. Amjed Ahmed Ahmed

40

Ch 2:

Extended Surfaces

3rd Year

College of Technical

Case 1 The second boundary condition is: θ ( (∞ ) = (T ∞ – T ∞ ) = 0 at x = ∞ B.C.2 m ∞ –m ∞ 0= C 1 e + C 2 e B.C.1 θ (0) (0) = (T s – T ∞ ) at x=0 θ s = C 1 + C 2

Differentiating

C 1=0 C 2= θ s

-mx s θ e

θ (x)= (x)=

2.39

∂θ = −mθ s e −mx ∂ x

2.40

Since the heat conducted across the root of the fin must equal the heat transferred by convection from the surface of the rod to the fluid, dT q fin = − kA = h P (T(x) − T ∞ )dx 2.41 dx x =0 The rate of heat flow can be obtained by Two different methods. Method 1. By left term in equation 2.41 substituting Eq. 2.40 for x= for x= 0 0 yields d θ 2.42 q fin = − kA = − kA[− mθ (0 )e (−m )0 ] dx x =0

⎡ h P ⎤ θ s ⎥ kA ⎣⎢ ⎦⎥

q fin = kA[mθ s ] = kA⎢ q fin = h PAk ⋅ θ s

Method 2 . . By right term in equation 2.41 2.41

∫

∞

q fin = hP (T(x) − T ∞ ) dx = hP θ s e −mx dx q fin = h P θ s

e −mx

0

∞

= h PAk ⋅ θ s

m

2.43

0

Case 2 The second boundary condition is : θ s = C 1 + C 2 B.C. 1 B.C.2

dT dx

=0

at

x=L

θ (x) (x) = C 1 e mx + C 2 e –mx

d θ θ ( x ) dx

mC 1e

= mC 1e mL − mC 2 e −mL = 0 x = L

mL

= mC 2 e − mL

C 1 = C 2e −2 mL

Substituting in B.C.1 in B.C.1

θ s = C 2 e −2 mL + C 2

Mr. Amjed Ahmed Ahmed

C 2 = θ s−2 mL 1+ e

41

Ch 2:

Extended Surfaces

3rd Year

C 1 =

θ s 1+ e

− 2 mL

e −2 mL

C 1 =

College of Technical

θ s 1+ e

2 mL

Substituting the above relations for C 1 and C 2 into Eq.(2.37) θ s mx θ s θ ( x) = e e -mx + 2 mL 2 mL 1+ e 1 + e− mL ⎞ ⎛ e -mL ⎞ θ s -mx ⎛ e ⎟+ ⎟ e ⎜ θ ( x ) = e ⎜ mL -mL 2 mL 2 mL − 1+ e ⎜ e ⎠⎟ 1 + e ⎝ ⎜⎝ e ⎠⎟ θ θ s e -mx e mL θ ( x) = -mL s mL e mx e -mL + mL mL − e + e e +e

θ s

mx

⎛ e -m(L-x) e m(L-x) ⎞ ⎟ ⎜ θ ( x ) = θ s ⎜ -mL mL + mL ⎟ − mL e + e ⎠ ⎝ e + e ⎛ e -m(L-x) + e m(L-x) ⎞ ⎛ (e -m(L-x) + e m(L-x) ) / 2 ⎞ ⎟⎟ = θ s ⎜⎜ θ ( x ) = θ s ⎜⎜ mL -mL -mL mL ⎟⎟ + e e ( ) + e / 2 e ⎠ ⎝ ⎠ ⎝ Noting that

Sinh(mL) =

e mL − e − mL 2

Cosh(mL) =

e

mL

+ e − mL 2

The temperature distribution is:

⎛ cosh m( L − x) ⎞ θ ( x) = θ s ⎜ cosh(mL) ⎟ ⎝ ⎠

2.44

The heat loss from the fin can be found by substituting the temperature gradient at the root into Eq.(2.37), we get

d θ ( x) dx

= θ s

d θ ( x) dx

− m sinh m( L − x)

= θ s x =0

cosh( mL )

− m sinh mL cosh mL

=

−θ s m tan mL

d θ q fin = − kA dx x =0

q fin = hPAk ⋅θ s tanh (mL)

Mr. Amjed Ahmed Ahmed

2.45

42

Ch 2:

Extended Surfaces

3rd Year

College of Technical

Case 3 The second boundary condition is : θ s = C 1 + C 2 B.C. 1

C 2 = θ s − C 1

at x=L B.C.2 θ (x)= (x)= θ L Substituting in B.C.2 in B.C.2 θ (x) (x) = C 1 e mx + C 2 e –mx mL θ (L) + C 2 e –mL (L) = C 1 e mL θ (L) +( θ s –C 1 ) e –mL (L) =C 1 e

C 1 =

θ ( L ) − θ s ⋅ e − mL

e mL − e − mL θ ( L ) − θ s ⋅ e − mL

C 2 = θ s − C 2 =

e mL − e − mL θ s ⋅ e mL − θ ( L ) e mL − e − mL

2.37

=

θ s (e mL − e − mL ) − θ ( L ) + θ s ⋅ e − mL e mL − e − mL

Substituting the above relations for C 1 and C 2 into into Eq.(2.37) Eq.(2.37) θ ( L ) − θ s ⋅ e − mL mx θ s ⋅ e mL − θ ( L ) –mx e + θ (x) (x) = e e mL − e − mL e mL − e −mL mx ⎡ − − mx + m ( L− x ) − − m ( L− x ) ⎤ ( L ) s θ (x) (x) = θ s ⎢ (θ / θ )(e e ) e e ⎥ e mL − e −mL ⎥⎦ ⎢⎣ ⎡ θ ( L ) e mx − e − mx e m( L− x ) − e − m( L− x ) ⎤ )+( )⎥ ⎢ ( θ )( 2 2 ⎥ θ (x) (x) = θ s ⎢ s mL − mL e −e ⎢ ⎥ ⎢ ⎥ 2 ⎣ ⎦

The temperature distribution is: ⎡ (θ ( L ) / θ s ) sinh mx + sinh m( L − x) ⎤ θ (x) (x) = θ s ⎢ ⎥ mL sinh ⎣ ⎦

2.46

The heat loss from the fin can be found by substituting the temperature gradient at the root into Eq.(2.37), we get

d θ ( x) mL ) − 0 ⎤ ) cosh2 m( L − x) (sinh = θ s ⎡⎢ (θ ( L ) / θ s )m ⋅ cosh mx + ( − m ⎥ dx x=0 (sinh mL ) ⎣ ⎦

{

d θ ( x) dx

x =0

d θ ( x) dx

x =0

⎡ {(θ / θ )m ⋅ −m cosh mL}(sinh mL )⎤ = θ s ⎢ ( L ) s ⎥ 2 ( ) sinh mL ⎣ ⎦ ⎡ (θ / θ ) − cosh mL ⎤ = mθ s ⎢ ( L ) s ⎥ mL sinh ⎣ ⎦

⎡ − (θ ( L ) / θ s ) + cosh mL ⎤ = − mθ s kA⎢ ⎥ dx x =0 sinh mL ⎣ ⎦ ⎡ cosh mL − (θ ( L ) / θ s ) ⎤ hP θ s kA⎢ ⎥ kA mL sinh ⎣ ⎦

q fin = − kA q fin =

Mr. Amjed Ahmed Ahmed

}

d θ

43

Ch 2:

Extended Surfaces

3rd Year

⎡ cosh mL − (θ ( L ) / θ s ) ⎤ ⎥ mL sinh ⎣ ⎦ M = hPAk ⋅ θ s

q fin = M ⎢

Noting that

College of Technical

2.47

Case 4 The second boundary condition is : B.C. 1

θ s = C 1 + C 2

B.C.2

− k

d θ ( x ) dx

C 2 = θ s − C 1

= hθ ( L) x = L

θ (x) (x) = C 1 e mx + C 2 e –mx θ (L) (L) = C 1 e mL + C 2 e –mL

d θ ( x ) dx

(2.37)

mL = mC − mC 2e −mL 1e x = L

Substituting above equations in B.C.2 in B.C.2 − k (mC 1e mL − mC 2 e − mL ) = h(C1 e mL + C 2 e -mL ) Substituting B.C.2 − k (mC e mL − m(θ − C )e − mL ) = h(C e mL + (θ − C ) e 1

C 1 =

θ s (e

-2mL

s

−e

1 -2mL

1

( h / km))

s

-mL

)

1

+ (h / km )e 2mL − (h / km ) + 1 θ s (e -2mL − ( h / km )e -2mL ) C 1 = 2mL + (h / km )e 2mL + 1 − ( h / km) e -2mL − (h / km )e -2mL ) θ s (e C 1 = 2mL e + (h / km )e 2mL + 1 − ( h / km) e

2mL

C 2 = θ s −

θ s e

-2mL

(1 − ( h / km ))

2mL

+ ( h / km)e

+ 1 − (h / km ) θ s (e 2mL + ( h / km )e 2mL + 1 − ( h / km )) − θ s e -2mL (1 − ( h / km)) C 2 = e 2mL + ( h / km)e 2mL + 1 − ( h / km ) e 2mL + ( h / km)e 2mL + 1 − ( h / km ) − e -2mL + e -2mL ( h / km)) C 2 = θ s e 2mL + ( h / km )e 2mL + 1 − ( h / km) e 2mL − e -2mL + e 2mL ( h / km ) + e -2mL ( h / km ) + 1 − ( h / km)) C 2 = θ s e 2mL + ( h / km )e 2mL + 1 − ( h / km ) e

2mL

Substituting the above relations for C 1 and C 2 into into Eq.(2.37) Eq.(2.37) -2mL -2mL ) θ (e − (h / km)e θ (x) (x) = 2mL s e mx + 2mL e + (h / km)e + 1 − (h / km) θ s

e

2mL

− e -2mL + e e

=

2mL

2mL

( h / km ) + e

+ (h / km )e

-2mL

2mL

( h / km) + 1 − ( h / km ))

+ 1 − (h / km)

( h / km)e m ( L− x ) - ( h / km)e − m ( L− x ) + e m ( L− x ) + e − m ( L − x )

e mL + e −mL + (h / km)e mL − ( h / km)e − mL θ ( x ) θ s The temperature distribution is:

Mr. Amjed Ahmed Ahmed

e –mx

44

Ch 2:

Extended Surfaces

3rd Year

θ ( x) = θ s

( h / km) sinh m( L − x) + coshm( L − x) (h / km) sinh mL + cosh mL

College of Technical

2.48

The heat loss from the fin can be found by substituting the temperature gradient at the root into Eq.(2.37), we get

d θ ( x) dx

= θ s

( − m( h / km) cosh mL − ( −m sinh mL))(cosh mL + ( h / km) sinh mL) − 0

x =0

d θ ( x ) dx

x =0

2 + ((h / km) sinh mL cosh mL) ( h / km) cosh mL − sinh mL ) = −mθ s ( h / km) sinh mL + cosh mL

q fin = − kA

d θ dx x =0

= mθ s kA s

(h / km) cosh mL − sinh mL) (h / km) sinh mL + cosh mL

hP ( h / km) cosh mL − sinh mL) θ s kA kA (h / km) sinh mL + cosh mL (h / km) cosh mL − sinh mL) q fin = M (h / km) sinh mL + cosh mL

q fin =

Noting that that

2.49

M = hPAk ⋅ θ s

Table 1 Temperature distribution and rate of heat transfer for fins

θ = T − T ∞

θ s = θ (0) = T s − T ∞

M =

m2 =

P A

hPAk ⋅ θ s

h P kA

m=

h P kA

: Perimeter of the fin : Cross section area of fin

Mr. Amjed Ahmed Ahmed

45

Ch 2:

Extended Surfaces

3rd Year

College of Technical

2.7.1 Fin Performance

The heat transfer effectiveness of a fin is measured by a parameter called fin effectiveness and the fin efficiency, which is defined as

i

Fin Effectiveness ε . A ratio of the fin heat transfer rate to the heat transfer rate

that would exist without the fin. q fin q fin = ε = f q withou ⋅ fin hA c (T s − T ∞ )

2.52

where Ac is the fin cross-sectional area at the base. the use of fins may rarely be >= 2. justified unless ε >=

ii Fin Efficiency η η f =

q fin qmax

2.53

qmax = hA f (T b − T ∞ ) = hPLθ b

Where A f is the surface area of the fin is A f = 2wLc

[

A f = 2 w L2 + (t / 2)

]

[

A f = 2 r − r π

2 1

Rectangular

2 1/ 2

A f = 2.o 5w L2 + (t / 2) 2 2c

2.54

]

Triangular

2 1/ 2

Parabolic Annular

Where as for a fin of rectangular cross section (length (length L & thickness t ) and an adiabatic end (Case 2) is M tanh mL tanh mL η f = = hPLθ hPL θ b mL 2.55 a corrected fin length of the form L form Lc = L + (t/2). (t/2). tanh mLc or η f = mLc

η f =

tanh h PL2 / kA 2

h PL / kA A fin efficiency for a circular pin fin (Diameter D & Length L) and an adiabatic end (Case 2) is η f =

tanh 4 L2 h / kD 2

2.56

4 L h / kD In Figures 2.14 2.14 and 2.15 2.15 fin efficiencies are plotted as a function of the parameter Lc

3/ 2

( h / kA p )1/ 2

inferred for the straight and the annular fins. Fin efficiencies obtained from the figures may be used to calculate the actual fin heat transfer rate from the expression f f b q f = η f qmax = η hA θ

Mr. Amjed Ahmed Ahmed

2.57

46

Ch 2:

Extended Surfaces

3rd Year

College of Technical

Figure 2.14 Efficiency of straight fins (rectangular, triangular, and parabolic profiles).

Figure 2.15 Efficiency of annular fins of rectangular profile.

Mr. Amjed Ahmed Ahmed

47

Ch 2:

Extended Surfaces

3rd Year

College of Technical

Example 2.10 Consider a copper pin fin 0.25 cm in that protrudes from a wall cm in diameter k k = 396 W/m K that at 95°C into ambient air at 25°C . The heat transfer is mainly by natural convection with a coefficient equal to 10 W/m2 K . Calculate the heat loss, assuming that : (a) the (a) the fin is "infinitely "infinitely long " (b) the fin is 2.5 cm long and the coefficient at the end is the same as around the (b) circumference. circumference.

5 %?

(c) how (c) how long would the fin have to be for the infinitely long solution to be correct within

Solution (a) A (a) A heat loss for the " Infinitely long " fin is

(

( − m )0

q fin = − kA − mθ (0 )e

)=

hPAk θ s

T= 25 C

q= [(10 W/m2 K) K) л л(0.0025 (0.0025 m)(396 W/m K) ( л /4(0.0025 л /4(0.0025 m)2 ]0.5 (95-25)°C q = 0.865 W (b) The (b) The equation for the heat loss from the finite fin is case 4:

q fin =

θ s sinh mL + ( h / mk ) cosh mL = 0.140 W hPAk cosh mL + ( h / mk ) sinh mL

(c) For (c) For the two solutions to be within 5% 5%,, it is necessary that sinh mL + ( h / mk ) cosh mL cosh mL + ( h / mk ) sinh mL

>= 0.95

This condition is satisfied when mL > 1.8 or L > 28.3 cm. cm. Example 2.11 To increase the heat dissipation from a 2.5 cm OD tube, circumferential fins made of aluminum (k (k = 200 W/m K ) are soldered to the outer surface. The fins are 0.1 cm thick and have an outer diameter of 5.5 cm. If the tube temperature is 100°C, the environmental

temperature is 25°C, and the heat transfer coefficient between the fin and the environment is 65 W/m2 K, calculate the rate of heat loss from two fins.

Solution a parameters required to obtain the fin efficiency curve in Fig. 2.15 are

Mr. Amjed Ahmed Ahmed

48

Ch 2:

Extended Surfaces

3rd Year

College of Technical

Example 2.12 The cylinder barrel of a motorcycle is constructed of 2024-T6 aluminum alloy (k = 186 W/m K) and is of height H = 0.15 m and OD = 50 mm. Under typical operating conditions the outer surface of the cylinder is at a temperature of 500 K and is exposed to ambient air at 300 K, with a convection coefficient of 50 W/m2 K. Annular fins of rectangular profile are typically added to increase heat transfer to the surroundings. Assume that five ( N=5 N=5)) such fins, which are of thickness t = 6 mm, length L = 20 mm and mm, length L mm and equally spaced, are added. What

is the increase in heat transfer due to addition of the fins? Solution

Assumptions: 1. Steady-state conditions. 2. One-dimensional radial conduction in fins. 3. Constant properties. 4. No internal heat generation. 5. Negligible radiation exchange with surroundings. 6. Uniform convection coefficient over outer surface (with or without fins). With the fins in place, the heat transfer rate is

q=q f +qb

q f = N η f q max = N η f hA f θ b q f = N η f h 2π r 22c − r 12 (T b − T ∞ ) Heat. transfer from the exposed cylinder surface is

q = hAb (T b − T ∞ ) Hence

Ab = ( H − Nt ) 2π r 1

)

q = N η f h 2π r 22c − r 12 (T b − T ∞ ) + h ( H − Nt ) 2π r (T b − T ∞ )

The fin efficiency may be obtained from Figure 2.19 with

Hence q = 5 (100.22) + 188.5 = 690 W Without the fins, the heat transfer rate is q f = hAwo ( T b − T ∞ ) Awo = H ( 2π r 1 ) Hence qwo = 50 W/m2 K (0.15 x л x л x x 0.025) m2 (200 K) = 236 W

Mr. Amjed Ahmed Ahmed

49

49

Ch 3: Unsteady State Conduction

3rd Year College of Technical

Chapter Three Unsteady State Conduction 3.1 Introduction To determine the time dependence of the temperature distribution within a solid during a transient process,. One such approach may be used under conditions for which temperature gradients within the solid are small. It is termed the lumped capacitance method. method.

3.2 The Lumped Capacitance Method The lumped capacitance method is the assumption that the temperature of the solid is spatially uniform at any instant during the transient process.(The process.(The temperature gradients within the solid are negligible). negligible ). From Fourier's law, heat conduction in the absence of a temperature gradient implies the existence of infinite thermal conductivity

Figure 3.1 Cooling of a hot metal forging .( Rcond << Rconv )

Applying energy conservation to the control volume. the energy terms E in + E g − E out = E st

− E out = E st

∞ ) = ρ VC p − hA s (T − T Assume Assume

θ = (T − T ∞ )

− hA sθ = ρ VC p

dT dt d θ / dt = dT / dt

d θ

dt Separating variables and integrating equation, we then obtain t ρ VC p θ d θ − dt = hA θ s θ i 0

∫

t =

∫

ρ VC p hA s

ln

θ i θ

or

t =

ρ Lc C p h

ln

T i − T ∞ T − T ∞

3.1

This equation used to determine the time required for the solid to reach some temperature θ θ i

= exp ( −

hA s ρVC ρVC p

)t

or

T − T ∞ T i − T ∞

= exp ( −

h ρ Lc C p

)t

3.2

This Equation used to compute the temperature reached by the solid at some time Where θ i = (T i − T ∞ ) and exponent group is Mr. Amjed Ahmed Ahmed

50

Ch 3: Unsteady State Conduction

hA s VC p ρ ρVC

t =

h

t =

ρC ρC p Lc

3rd Year College of Technical

hL k k L k t )( 2 ) = Biα t c = ( c )( = BiFo C p k ρC ρ ρC ρC ρC p Lc k Lc ρC p Lc h

3.3

Where Lc is the characteristic length as the ratio of the solid's volume to surface area Lc=V/A s . s . Lc = L/2 for a plane wall of thickness 2L 2L.. Lc = r/2 for a long cylinder (end ( end edge are negligible) negligible) Lc = r/3 r/3 for a sphere Lc = r o-r i for a long annular cylinder(end cylinder(end edge are negligible). negligible). k α is termed the Fourier number It is a dimensionless time and substituting F O = C p ρ ρC

equation 3.3 3.3 into into 3.2 3.2,, we obtain θ i θ

=

T − T ∞ T i − T ∞

= exp( − BiFo)

3.4

The difference between the solid and fluid temperatures must decay exponentially to zero as approaches infinity time. The quantity ρVC p /hA s may be interpreted as a thermal time constant. as 1 τ t = ( )( ρ VC p ) = Rt C t 3.5 hA s

where R where Rt is the resistance to convection heat transfer C t t is the lumped thermal capacitance of the solid. Any increase in R t t or Ct , will cause a solid to respond more slowly to changes in its thermal environment and will increase the time required to reach thermal equilibrium (θ (θ = = 0). 0).

Figure 3.2 Transient temperature response of lumped capacitance solids

3.2.1 Energy Transfer between a Solid and Surrounding To determine the total energy transfer Q occurring up to some time t t

t

t

∫

∫

∫

0

0

0

s )dt Q = qdt = hA s (T − T = hA s θ dt Substitution equation 3.2 t

∫

Q = hA s θ i exp ( − 0

hA s ρVC ρVC p

)t

Q = ρ VC pθ i (1 − exp ( −

hA s ρ ρVC VC p

Q = ρ VC pθ i (1 − exp(− 1 t )) τ

Mr. Amjed Ahmed Ahmed

dt

)t ) 3.6

51

Ch 3: Unsteady State Conduction

3rd Year College of Technical

3.2.2 A dimensionless group Biot number : Applying energy balance to the surface under steady state E in in=E out out kA hA(T s 2 − T (T s1 − T s 2 ) ∞ ) = L T s1 − T s 2 L / kA hL = = T − T k 1 / hA Where

hL k

s 2

∞

is dimensionless group Biot number (Bi). hL R Bi = c = cond k Rconv

Figure 3.3 Transient temperature distribution for different Biot No. in a plane wall cooled by convection.

Applicability of Lumped Capacity Analysis When confronted with transient conduction problems, the very first thing that one should do is calculate the Biot number. If the following condition is satisfied hLc Bi = ≤ 0.1 k the error associated with using the lumped capacitance method is small.

Mr. Amjed Ahmed Ahmed

52

Ch 3: Unsteady State Conduction

3rd Year College of Technical

Example 3.1 A thermocouple junction, which may be approximated as a sphere, is to be used for temperature measurement in a gas stream. The convection coefficient between the junction 2 surface and the gas is known to be h = 400 W/m K , and the junction properties are k = 20 W/m K, C p = 400 J/kg K, and ρ and ρ = = 8500 kg/m3. Determine the junction diameter needed for the thermocouple to have a time constant of 1 s. If the junction is at 25°C and is placed in a gas stream that is at 200°C, how long will it take for the junction to reach 199°C?

Solution

Assumptions: 1. Temperature of junction is uniform at any instant. 2. Radiation exchange with the surroundings is negligible. 3. Losses by conduction through the leads are negligible. 4. Constant properties. 5. Using the lumped capacitance method. A s = л D2 and V = л = л D3 /6 for for a sphere D 3 1 1 ρπ τ t = ( hA )( ρ VC p ) = hDπ 2 6 C p s

D =

6hτ ρ C p

−4

= 7.07 × 10 m = 0.71mm hLc

400 × 0.000353

= 0.000235 k 3 × 20 the lumped capacitance method may be used to an excellent approximation. Lc=r/3

Bi =

=

2. The time required for the junction to reach T = 199°C is ρ Lc C p T i − T ∞ t = ln h T − T ∞ 4 8500 × 7.06 × 10 − × 400 25 − 200 ln t = 6 × 400 199 − 200 = 5.2 s

Mr. Amjed Ahmed Ahmed

53

Ch 3: Unsteady State Conduction

3rd Year College of Technical

3.3 Transient Heat flow in a Semi-Infinite Solid Solid If a thermal change is suddenly imposed at this surface, a one-dimensional temperature wave will be propagated by conduction within the solid. The appropriate equation is ∂ 2T 1 ∂T = 0 ≤ x ≤ L ∂ x 2 α ∂t To solve this equation we must specify two boundary conditions and the initial temperature distribution. For the initial condition we shall specify that the temperature inside the solid is uniform at Ti, that is, B.C.1 T(x, 0) = Ti. Assumptions: 1. One Dimensional 2. Extended body to infantry

Figure 3.4 Schematic Diagram and Nomenclature for Transient Conduction in a Semi-Infinite Solid.

Closed-form solutions have been obtained for Three Cases of changes in surface conditions, instantaneously applied at t = = 0: These three cases are Case 1 Change in surface temperature: a sudden change in surface temperature T (0, t ) = T s T ( x, t ) − T s T i − T s

⎛ x ⎞ = erf ⎜ m ⎟ ⎝ 2 α t ⎠

3.8

3.9 Case 2 Constant surface heat flux: a sudden application of a specified heat flux q'' s =q'' o as, for example, exposing the surface to radiation

3.10 Case 3. Surface convection a sudden exposure of the surface to a fluid at a different temperature through a uniform and constant heat transfer coefficient h 3.11 3.12

the specific temperature histories computed from Eq. (3.12) are plotted in next Fig.

Mr. Amjed Ahmed Ahmed

54

Ch 3: Unsteady State Conduction

3rd Year College of Technical

Figure 3.5 Dimensionless Transient Temperatures

B.C. 1 B.C. 2

Figure 2 6 Transient Transient Temperature Distributions in a Semi-Infinite Solid

where erf is the Gaussian error function, which is encountered frequently in engineering and is defined as

3.13 Values of this function are tabulated in the appendix. The complementary error function, erfc(w), is defined as erfc(w)=1-erf(w)

Mr. Amjed Ahmed Ahmed

55

Ch 3: Unsteady State Conduction

3rd Year College of Technical

Table 3.1 The Error Function x

erf(x)

x

erf(x)

x

erf(x)

0.00

0.00000

0.76

0.71754

1.52

0.96841

0.02

0.02256

0.78

0.73001

1.54

0.97059

0.04

0.04511

0.80

0.74210

1.56

0.97263

0.06

0.06762

0.82

0.75381

1.58

0.97455

0.08

0.09008

0.84

0.76514

1.60

0.97635

0.10

0.11246

0.86

0.77610

1.62

0.97804

0.12

0.13476

0.88

0.78669

1.64

0.97962

0.14

0.15695

0.90

0.79691

1.66

0.98110

0.16

0.17901

0.92

0.80677

1.68

0.98249

0.18

0.20094

0.94

0.81627

1.70

0.98379

0.20

0.22270

0.96

0.82542

1.72

0.98500

0.22

0.24430

0.98

0.83423

1.74

0.98613

0.24

0.26570

1.00

0.84270

1.76

0.98719

0.26

0.28690

1.02

0.85084

1.78

0.98817

0.28

0.30788

1.04

0.85865

1.80

0.98909

0.30

0.32863

1.06

0.86614

1.82

0.98994

0.32

0.34913

1.08

0.87333

1.84

0.99074

0.34

0.36936

1.10

0.88020

1.86

0.99147

0.36

0.38933

1.12

0.88679

1.88

0.99216

0.38

0.40901

1.14

0.89308

1.90

0.99279

0.40

0.42839

1.16

0.89910

1.92

0.99338

0.42

0.44749

1.18

0.90484

1.94

0.99392

0.44

0.46622

1.20

0.91031

1.96

0.99443

0.46

0.48466

1.22

0.91553

1.98

0.99489

0.48

0.50275

1.24

0.92050

2.00

0.99532

0.50

0.52050

1.26

0.92524

2.10

0.997020

0.52

0.53790

1.28

0.92973

2.20

0.998137

0.54

0.55494

1.30

0.93401

2.30

0.998857

0.56

0.57162

1.32

0.93806

2.40

0.999311

0.58

0.58792

1.34

0.94191

2.50

0.999593

0.60

0.60386

1.36

0.94556

2.60

0.999764

0.62

0.61941

1.38

0.94902

2.70

0.999866

0.64

0.63459

1.40

0.95228

2.80

0.999925

0.66

0.64938

1.42

0.95538

2.90

0.999959

0.68

0.66378

1.44

0.95830

3.00

0.999978

0.70

0.67780

1.46

0.96105

3.20

0.999994

0.72

0.69143

1.48

0.96365

3.40

0.999998

0.74

0.70468

1.50

0.96610

3.60

1.000000

Mr. Amjed Ahmed Ahmed

Mr. Amjed Ahmed Ahmed

56

Ch 3: Unsteady State Conduction

3rd Year College of Technical

Example 3.2 Estimate the minimum depth xm at which one must place a water main below the surface to avoid freezing. The soil is initially at a uniform temperature of 20°C . Assume that under the worst conditions anticipated it is subjected to a surface temperature of -15°C for for a period of 60 days. Use the following properties for soil (300 K ): ): 3 ρ = 2050 kg/m k = 0.52 W/m K

C p= 1840 J/kg-6K 2 α =0.138 x 10 m /s Solution To simplify the problem assume that 1. Conduction is one-dimensional 2. The soil is a semi-infinite medium 3. The soil has uniform and constant properties.

The prescribed conditions correspond to those of Case 1, the temperature distribution in the soil is T ( x, t ) − T s ⎛ x ⎞ = erf ⎜ m ⎟ T i − T s ⎝ 2 α t ⎠ 0 − (−15)

⎛ x ⎞ = 0.43 = erf ⎜ m ⎟ 20 − (−15) ⎜ 2 α t ⎠⎟ ⎝ From Table 43 we find by interpolation that when xm / 2 α t = 0.4 to satisfy the above relation. Thus

xm = 0.4 × 2 α t = 0.68m Another Solution: To use Fig. 2.35, first calculate T ( x, t ) − T s 0 − 20 = 0.57 = − 15 − 20 T ∞ − T s

and

h α t / k = ∞

Then enter the curve Fig.(3.5) obtain xm / 2 α t = 0.4, the same result as above .

Mr. Amjed Ahmed Ahmed

5

Ch 3: Unsteady State Condiction

3rd Year College of Technical

3.4 Heisler Charts For Transient Heat Conduction The temperature distribution and the heat flow have been calculated and the results are available in the form of charts. we shall illustrate the application of some of these charts to typical problems of transient heat conduction in solids (One-Dimensional) having a Bi > 0.1. Three simple geometries for which results have been prepared in graphic form are: 1. An infinite plate of width 2L (see Fig. 3.7) 3.7) a) Calculate T(0, t) from t) from Figure 3.7 -a :(Midplate temperature vs time for an infinite plate) b) After that, calculate surface temperature T(x, t) from t) from Figure 3.7 --b b c) Calculate total heat transfer Q at any time from Figure 3.7 -c , note that: Qo≡ρ ≡ρC C V(T i-T ∞ )= )= ρ ρC C V θ θi 2. An infinitely long cylinder of radius r o (see Fig. 3.8) 3. A sphere of radius r o (see Fig. 3.9) One boundary condition for all three geometries are similar requires that the temperature gradient at the midplane of the plate, the axis of the cylinder, and the center of the sphere be equal to zero. Physically, this corresponds to no heat flow at these locations. The other boundary condition requires that the heat conducted to or from the surface be transferred by convection to or from a fluid at temperature through a uniform and constant heat transfer coefficient hA(T s

T ) = kA dT dx

− ∞

Applicability of the Heisler Charts The calculations for the Heisler charts were performed by truncating the infinite infinite series solutions for the problems into a few terms. This restricts the applicable the charts to values of the Fourier number greater than 0.2. 0.2.

F O

Mr. Amjed Ahmed Ahmed

k α

=

ρ ρC C p

>

0.2

58

Ch 3: Unsteady State Condiction

(a)

3rd Year College of Technical

(b)

(c)

Figure 3.7 Dimensionless Dimensionless Transient Temperatures and Heat Flow in an Infinite Plate of Width 2L

Mr. Amjed Ahmed Ahmed

Mr. Amjed Ahmed Ahmed

59

Ch 3: Unsteady State Condiction

3rd Year College of Technical

(a)

(b)

(c)

Heat Flow for a Long Cylinder. Cylinder. Figure 3.8 Dimensionless Transient Temperatures and Heat

Mr. Amjed Ahmed Ahmed

60

Ch 3: Unsteady State Condiction

3rd Year College of Technical

(a)

(b)

(c)

Figure 3.9 Dimensionless transient temperatures and heat flow for a sphere.

Mr. Amjed Ahmed Ahmed

61

Ch 3: Unsteady State Condiction

3rd Year College of Technical

Example 3.3 In a fabrication process, steel components are formed hot and then quenched in water. Consider a 2.0 m m long, 0.2 m m diameter steel cylinder (k = 40 W/m K , α = 1.0 x10-5 m2 /s), /s), initially at 400°C , that is suddenly quenched in water at 50°C. If the heat transfer coefficient is 200 W/m2 K, calculate the following 20 min after immersion: 1. the center temperature 2. the surface temperature

3. the heat transferred to the water during the initial 20 min Solution Since the cylinder has a length 10 times the diameter, we can neglect end effects. we calculate first the Biot the Biot number hr 200 × 0.1 Bi = o = = 0.5 > 0.1 40 k 1. we cannot use the lumped-capacitance method. To use the chart solution we calculate the appropriate dimensionless parameters: α t and Bi2 F O = (0.52 )(1.2) = 0.3 Fo = 2 = 1.2 r o

The dimensionless centerline temperature for 1/Bi = 2.0 and Fo = 1.2 from Fig. 2.38(a) is T ( 0 , t ) − T T (0, t ) − 50 = 0 . 35 T i − T 400 − 50 = 0.35 ∞

∞

T(0,t) = 172.5 C 2. The surface temperature at r/r o =1.0 and t = 1200 s is obtained from Fig. 3.8(b) 3.8(b) in terms of the centerline temperature: T ( r o , t ) − T = 0.8 T (0, t ) − T ∞

∞

T ( r o , t ) − 50

= 0.8 172.5 − 50 and the surface temperature after 20 min is: is:

T(r o , t) = 148°C

3. The initial amount of internal energy stored in the cylinder per unit length is Qi = C pπ r o 2 (T i − T ) = ( k α / F o )π r o 2 (T i − T ) = 4.4 × 107 W / m Then the amount of heat transferred from the steel rod to the water can be obtained from Fig. 3.8(c). Since Q(t)/Qi = = 0.61 0.61 2m × 4.4 × 10 7 W ⋅ s / m Q(t ) = 0.61 × = 14.9 kW ⋅ hr 3600hr ∞

∞

Mr. Amjed Ahmed Ahmed

62

Ch 3: Unsteady State Condiction

3rd Year College of Technical

Example 3.4 Example 3.4 A large concrete wall 50 cm thick cm thick is initially at 60°C . One side of the wall is insulated. The other side is suddenly exposed to hot combustion gases at 900°C through through a heat transfer 2 coefficient of 25 W/m K . Determine (a) the (a) the time required for the insulated surface to reach 600°C. (b) the (b) the temperature distribution in the wall at that instant (c) the (c) the heat transferred during the process.

The following average physical properties are given: k = 1.25 W/m K , C p=837 J/kg K , ρ ρ = = 500 kg/m3 ,

α =0.30 x 10-5 m2 /s

Solution (a). that the wall thickness is equal to L since L since the insulated surface corresponds to the center plane of a slab of thickness 2L 2L when when both surfaces experience a thermal change. The temperature ratio for the insulated face at the time sought is

T s (t ) − T

∞

T s (0) − T

=

∞

Bi=10

,

1

=

x =0

0.1

Bi From Fig. 2.37(a) 2.37(a) we we find that

600 − 900 60 − 900 and

=

0.357 Fo =

α t L2

=

0.7

2

×

t = 0.7 0.55 = 58333 s = 16.2hr 0.3 × 10 (b). The temperature distribution in the wall 16 hr after after the transient was initiated can be obtained from Fig. from Fig. 2.37(b) for 2.37(b) for various values of x/L of x/L,, as shown below: −

Assume of positions

From the above dimensionless data we can obtain the temperature distribution as a function of distance from the insulated surface:

Mr. Amjed Ahmed Ahmed

63

Ch 3: Unsteady State Condiction

3rd Year College of Technical

Teme ratue ratue Distrbution 900

861

850 777

800 750

708

700

651 612

) x ( T

600 650 600 550 500

0.5

0.4

0.3

0.2 x

0.1

0

(c). The heat transferred to the wall per square meter of surface area during the transient can be obtained from Fig. from Fig. 3.7(c). 3.7(c). for Bi = 10 and Bi2 Fo = 70 is Q(t)/Qi=0.70. 8 2 Q(t) = C p ρ L(T i - T )837 × 500 × 0.5 × (-840 ) = -1.758 × 10 J/m ∞

The minus sign indicates that the heat was transferred into the wall and the internal energy increased during the process.

Mr. Amjed Ahmed Ahmed

64

Ch 3: Unsteady State Condiction

3rd Year College of Technical

Start

h, k, ρ , T ∞ , T s C , V, A s

Input

hLc Calculate Bi

No

Calculate F O P r o c e s s i n g

No

=

k

k and α = ρ C p

Bi<0.1

Yes

k α

=

ρC ρC p Yes

F o>0.2

The Lumped Capacitance Method

The Infinite Body

Sem-Infinite Bod

Heisler Charts Charts t =

h=Const Eq. 3.11 Eq. 3.12

T(0,t)=T s Eq. 3.8 Eq. 3.9

ρ Lc C p h

T − T

∞

T i − T

ln

T i − T

∞

T − T

∞

=

exp ( −

∞

h ρ Lc C p

)t

Q = ρ VC pθ i (1 − exp( −t / τ )) qo=Const Eq. 3.10

Output

A sphere Fig. 2.39

T,

A long A plate Cylinder wall Fig. 2.37 Fig. 2.38

t,

Q

End

Figure 3.10 Flow Chart for the solution of Unsteady state conduction problem. problem.