Course: Heat Transfer Faculty of Engineering: Program: - - PDF document

HEAT TRANSFER

Course: Heat Transfer Faculty of Engineering: Program: Petrochemical Level: 2 Monday: 8-10 am

• Prof. Yehia Hassan Magdy

HEAT TRANSFER

Convection

Convection is then mode of energy transfer between solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. The faster the fluid motion, the greater the convection heats transfer. As shown in Figure 1.9, Energy is transferred by convection between the solid’s surfaces and the flowing fluid by

• convection. The temperature of the film layer of fluid increases. The rate of conviction heat

transfer is observed to be proportional to the temperature difference, and is conveniently expressed by Newton’s law

( )

fluid surface conv

T T hA T hA Q − =  = (1.17) hA Rconv 1 =

Flowing fluid

Q

Fluid film adjacent to surface Stationary Solid Surface

Figure 1. 9 Heat transfer by convection Where; h: film coefficient of convective heat transfer, W/m2K; A: area of heat transfer parallel to the direction of fluid flow, m2 Ts: solid surface temperature, oC or K

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Tfluid: flowing fluid temperature, oC or K T: temperature difference, K

conv

R thermal resistance for convection, ˚C/W The convection coefficient (h) is not property of the fluid. It is an experimentally determined parameter whose value depends on all variables influencing convection such as the surface geometry, the nature of fluid motion, the properties of the fluid and the bulk fluid velocity. Before the above equation is used to evaluate the convective heat transfer, the fluid flow regime has to be identified; whether it is laminar, transitional or turbulent. Laminar flow of the fluid is encountered at Re <2100. Turbulent flow is normally at Re >4000. Sometimes when Re >2100 the fluid flow regime is considered to be turbulent. If we have a round tube with a liquid flowing in it at a steady state, and if we injected a dye trace with a needle parallel to the axis of the tube, one of two things can happen: a) The dye trace may be smoothly down the tube as a well-defined line, only very slowly becoming thicker which called laminar flow, as shown in Figure 1.10. Figure 1.10 Laminar Flow

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b) The dye trace may flow irregularly down the tube moving back and forth across the diameter of the tube and eventually becoming completely dispersed which called turbulent flow as shown in Figure 1.11. Figure 1.11 Turbulent flow Heat transfer by convection may take the form of either natural or forced convection as shown in Figure 1.12.

• 1. Natural convection:

Natural convection is caused by buoyancy forces due to density differences caused by temperature variations in the fluid. At heating the density change in the boundary layer will cause the fluid to rise and replaced by cooler fluid that also will heat and rise.

• 2. Forced convection:

Forced convection occurs when a fluid flow is induced by an external force, such as a pump, fan or a mixer. Figure 1.12 Forced and Free (Natural) Convection

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In practice, film heat transfer coefficients (h) are generally calculated from empirical equations obtained by correlating experimental data with the aid of dimensional analysis. Example 1.8 A fluid flows through a long pipe 2m and 0.5m diameter. The surface and fluid temperature is 55˚C and 25˚C, respectively. Calculate the convection heat transfer if the coefficient convection is 33W/m².K. Solution: From Newton’s law

( )

fluid surface conv

T T hA T hA Q − =  =

( )

kW W Q T T h DL h Q

conv fluid surface conv

22 . 6 6217 ) 25 55 ( ) 2 ( ) 5 . ( ) 14 . 3 ( 33 ) ( = = −     = −   = 

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Bulk and Film Temperature

The rate of convective heat transfer, q, from a solid surface to a surrounding fluid as shown in Figure 1.13 has given by:

( )

fluid surface conv

T T hA T hA Q − =  = Figure1.13 Convective heat transfer from a solid surface (

s

T ) to surrounding fluid

f

T Where, h = average convective heat transfer coefficient, W/(m2.K) A = surface area for convective heat transfer, m2

s

T = surface temperature of the solid surface, K

f

T = fluid temperature, K

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However, in the case of flow thru ducts or pipes (Figure 1.14), it is more convenient to use the bulk temp.,

s

T instead of free-stream temperature,

f

T . Thus, for the tube flow depicted in Fig. 2, the total energy added can be expressed in terms of bulk temperature difference: q =

( )

fluid S

T T C m −

• 

(1.18)

• r in terms of the heat transfer coefficient :

) ( 2

fluid s

T T dx rh q −   =  Figure 1.14 Total heat transfer expressed in terms of bulk-temperature difference The bulk temperature (Tb) is the mean temperature of the fluid at a given cross-section of the tube. In engineering practice, the bulk temperature is equated to the simple approximate average value: 2

, , f

• utlet

f inlet b

T T T + = (1.19) Which is used to calculate the average heat transfer coefficients.

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Film temperature for fluids flowing thru tubes of ducts is the temperature of the fluid film adjacent to the heating surface. It is the average of the temperature of the heating surface and the free-stream temperature of the fluid (Figure 1.15): 2

f s film

T T T + = (1.20) Figure 1.15 Film and bulk temperature

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DIMENSIONLESS GROUPS

Many of the generalized relationships used in convective heat-transfer calculations have been determined by means of dimensional analysis and empirical considerations. It has been found that certain standard dimensionless groups appear repeatedly in the final equations. Some of the most common dimensionless groups are given below with their names: Reynolds number = Re =   DV Prandtl number = Pr = cpµ k Nusselt number = Nu = hD k Peclet number = Pe = k c DV

p

 = (Re . Pr) Grashof number = Gr =

2 2 3

   t g D  =

2 3

  t g D  Where in the SI system: D: pipe diameter, m V: fluid velocity, m/s : fluid density, kg/m3 µ: fluid dynamic vicosity, N.s/m2 or kg/m.s : fluid kinematic viscosity, m2/s k: fluid thermal conductivity, W/mK h: convective heat transfer coefficient, W/m2.K cp: fluid specific heat capacity, J/kg.K g: acceleration of gravity, m/s2 ß: cubical coefficient of expansion of the fluid = 1

• Tav. , K–1

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where; Tav is the average absolute temperature of the fluid t: temperature difference between surface & fluid, K The magnitudes of the above dimensionless groups are the same provided consistent units are used. The characteristic dimension L, D in Reynolds, Nusselt or Grashof number is defined depending on the situation. For a fluid flowing in a tube, its diameter D is used. For a plate, L is used.

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Calculate the Convection Coefficient Heat Transfer for Natural Convection by using empirical relationships

Natural or free convection heat transfer occurs whenever a body is placed in a fluid at a higher

• r lower temperature than that of the body. As a result of the temperature difference, heat

flows between the fluid and the body and causes a change in the density of the fluid layers in the vicinity of the surface. The difference in density leads to downward flow of the heavier fluid and upward flow of the lighter. If the motion of the fluid is caused solely by differences in density resulting from temperature gradients, without the aid of a pump or a blower for example, the associated heat transfer mechanism is called natural or free convection. Many processing equipment and devices are cooled by free convection. Free convection is the dominant heat flow mechanism from steam radiators, walls of buildings and steam pipes. The fluid velocities in free convection currents are generally low. Fluid motion generated by natural convection may be laminar or turbulent. Many experiments have been performed to establish the functional relationships for different geometric configurations convecting to various fluids. Generally, it is found that Nu = a (GrPr)b (1.21) Where a and b are constants. Laminar and turbulent flow regimes have been observed in natural convection, and transition generally occurs in the range 107 < GrPr < 109 depending on the geometry. Some of the more important formulae obtained for natural convection to or from various geometries as shown in Figure 1.16.

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Figure 1.16 Different geometries and the direction of convective flow for natural convection

• a. Horizontal Cylinders.

Detailed measurements indicate that the convection coefficient varies with angular position round a horizontal cylinder, but for design purposes values given by the following equations are constant over the whole surface area, for cylinders of diameter, d. For laminar flow Nud = 0.525(GrdPr)0.25 (1.22) when 104 < GrdPr < 109 For turbulent flow Nud = 0.129(GrdPr)0.33 (1.23) when 109 < GrdPr < 1012

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Example 1.9 A 2.0 cm-diameter horizontal heater is maintained at a surface temperature of 38oC and submerged in water at 27oC. Calculate the free-convection heat loss per unit length of the heater. The properties of water of at the film temperature interest are: k = 0.623 W/moC, v = 0.7 x 10–6 m2/s, Pr = 5.12 Solution The film temperature is Tƒ = 38 + 27 2 = 32.5oC = 305.5 K ß = 1 305.5 = 3.27 x 10–3K–1 and GrdPr = (

) ( )(

)(

)( )

( )

2 6 3 3

10 7 . 12 . 5 27 38 10 27 . 3 8 . 9 02 .

− −

 −  GrdPr = 29.47 x 106 So, it is laminar flow and by using equation (1.22): Nud = 0.525 (29.47 x 10

6)

0.25

= 38.68 So, Nu = hD k h = (38.68)(0.623) 0.02 = 1204.9 W/m2.oC

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The heat transfer is thus Q L = h d (Ts – Tf) = (1204.9) (3.14) (0.02) (38 – 27) = 832.8 W/m

• b. Vertical Surfaces.

In this case, both vertical flat surfaces and vertical cylinders may be considered using the same correlations of experimental data. The characteristic linear dimension is the length, or height, of the surface, l. With physical properties at the mean film temperature the numerical constants as recommended by McAdams are For laminar, Nul = 0.59(GrlPr)0.25 (1.24) when 104 < GrlPr < 10

9

For turbulent Nul = 0.129(GrlPr)0.33 (1.25) when 109 < GrlPr < 1012 Example 1.10 A large vertical plate 4.0 m high is maintained at 60oC and exposed to atmospheric air at

• 10oC. Calculate the heat transfer if the plate is 10 m wide.

The air properties of interest are thus k = 0.02685 W/mC, v = 16.5 x 10–6 m2/s, Pr = 0.7

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Solution First the film temperature is determined as Tƒ = 60 + 10 2 = 35oC = 308 K ß = 1 308 = 3.25 x 10–3 K–1 and GrlPr = ( ) (

)(

)(

)( )

( )

2 6 3 3

10 5 . 16 7 . 10 60 10 25 . 3 8 . 9 4

− −

 −  GrdPr = 2.618 x 1011 The flow is turbulent so, used equation (1.25) to obtain Nul = 0.129 (2.618 X 1011)0.33 = 756 From, Nu = hD k The heat-transfer coefficient is then h = (

)( )

. 4 02685 . 756 = 5.07 W/m2.oC The heat transfer is Q = h A (Ts – Tf) = (5.07)(4)(10)(60 – 10) = 10149 W = 10.149kW

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• c. Horizontal Flat Surfaces:

Fluid flow is most restricted in the case of horizontal surfaces, and the size of the surface has some bearing on the experimental data. The heat transfer coefficient is likely to be more variable over a smaller flat surface than a large one, when flow effects at the edges become less significant. Further, there will be a difference depending on whether the horizontal surface is above or below the fluid. Similar, though reversed, processes take place for hot surfaces facing upwards (i.e., cold fluid above a hot surface), and cold surfaces facing downwards (i.e., hot fluid below a cold surface). In either case, the fluid is relatively free to move due to buoyancy effects and be replaced by fresh fluid entering at the edges. The following relationships are generally recommended for square or rectangular horizontal surfaces with a mean length of side l: For laminar; Nul = 0.54(GrlPr)0.25 (1.26) When 105 < GrlPr < 108 For turbulent; Nul = 0.14(GrlPr)0.33 (1.27) When GrlPr > 108

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Calculate the Convection Coefficient Heat Transfer for Forces Convection by using empirical relationships

Correlations for a large number of geometrical configurations and conditions are presented in the engineering literature. Equations for some of the more important cases are presented

• below. It should be kept in mind that each of these equations does not represent a new

theoretical principle that must be understood and memorized. It is simply a question of selecting the appropriate equation for the particular case.

• a. Laminar flow in tubes.

An average Nusselt number between entry and distance x from entry is given by Nud = 1.86

( )

x d Pr ) (Re

0.14 w 3 1 3 1 3 1 d

                (1.28) All physical properties are evaluated at the arithmetic mean bulk temperature between entry and x, with the exception of µw which is at the wall temperature, and the equation is valid for heating and cooling in the range 100 <

( ) ( )

x d Pr ) (Re

3 1 3 1 3 1 d

< 10,000.

• b. Turbulent flow in tubes.

For fluids with a Prandtl number near unity, and only moderate temperature differences between the fluid and the wall, (5oC for liquids, 55oC for gases), Dittus and Boelter recommend: Nud = 0.023(Red)0.8 (Pr)n (1.29) where n = 0.4 for heating, and 0.3 for cooling, and Red > 10,000. This is for fully developed flow, i.e., (x/d) > 60, and all fluid properties are at the arithmetic mean bulk temperature.

• c. Turbulent flow along flat plates.

For this type of flow, Chapman recommends: Nux = 0.036 ) , (Re Pr

. x

700 18

8 3 1

− (1.30) This is based on a consideration of laminar flow (for which

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Nux = 0.664

( )

Pr ) (Re

3 1 2 1 x

and turbulent flow after transition at Rex = 400,000, for 10 > Pr > 0.6. Fluid properties are evaluated at the mean film temperature. Example 1.11 Air at 200oC is heated as it flows through a tube with a diameter of 2.54 cm at a velocity

• f 10 m/s. Calculate the heat transfer per unit length of tube if a constant-heat-flux

condition is maintained at the wall and the wall temperature is 20oC above the air temperature, all along the length of the tube. The properties of air at a bulk temperature of 200oC are:  = 1.493 kg/m3 Pr = 0.681 µ = 2.57 x 10–5kg/m . s k = 0.0386 W/m . oC cp = 1.025 kJ/kg . oC Solution First calculate the Reynolds number to determine whether the flow is laminar or turbulent, and then select the appropriate empirical correlation to calculate the heat transfer. Red =   d V . = (1.493)(10)(0.0254) 2.57 x 10–5 = 14,756 so that the flow is turbulent. Thus Eq. (2) is used to calculate the heat-transfer coefficient (note that n = 0.4 for heating). Nud = hd k = 0.023 Re0.8 d Pr0.4 = (0.023)(14,756)0.8(0.681)0.4 = 42.67 h = k d Nud = (0.0386)(42.67) 0.0254 ) = 64.85 W/m

2.

• C

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The heat flow per unit length is then Q L = h d (Tw – Tb) = (64.85)  (0.0254)(20) = 103.5 W/m

THERMAL RESISTANCE NETWORK

One of the common process heat transfer applications consists of heat flow from a hot fluid, through a solid wall, to a cooler fluid on the other side. The solid wall may be a flat surface, but more commonly will be a cylindrical pipe. There may be a scale or dirt deposit that creates a heat transfer resistance on either the inside or outside surface of the pipe. The heat flowing from one fluid to the other must therefore pass through several resistances in series.

• a. Plane Wall

Consider steady one-dimensional heat flow through a plane wall of thickness L and thermal conductivity k that is exposed to convection on both side to fluids at temperatures

i

T and

• T with heat transfer coefficient

i

h and

• h ,respectively, as shown in Figure 1.16.

Under steady conditions;

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Figure 1.16 Thwermal network for convection – conduction plane wall system

( )

        +       +         − = + + =  = A h kA L A h T T Q R R R R R T Q

• i
• i

conve conduction conve total total

1 1 (1.31) ) (

• i

T T UA T UA Q − =  = (1.32) Where U is the overall heat transfer coefficient(

)

K m W

2

. The overall heat transfer coefficient is equal to the inverse of the total thermal resistance.

total

R UA 1 = (1.33) Example 1.12 A flat wall 15 cm thick of thermal conductivity 0.87 W/mK is exposed to air at 30oC on one side where the heat transfer coefficient is 15 W/m2K, and air at 15oC on the opposite side

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where the convection coefficient is 60 W/m2K. Determine the overall heat transfer coefficient and the heat transfer rate per unit area of the wall. Solution The wall area A = 1 m2, for a flat wall, from equation (1.33), the overall heat transfer coefficient is given by: 1 U = 1 hi + x k + 1 ho = 1 15 + 0.15 0.87 + 1 60 = 0.256 m

2K/W

 U = 3.9 W/m

2K

and the rate of heat transfer Q A = UT = (3.9) (30 – 15) = 58.5 W/m

2

• b. Cylinder and Sphere

The thermal resistance for cylinder or spherical shape as shown in Figure 1.17 can be calculated as the following equations.

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Figure 1.17 thermal resistances for cylinder or sphere For Cylinder:

total

R T Q  =

2 , , 1 , conve cyl conduction conve total

R R R R + + =         +                   +         =

• i

total

Lh r Lk r r Lh r R

2 2 1 2 2 1

2 1 2 ln 2 1   

( )

        +                   +         − =

• i
• i

Lh r Lk r r Lh r T T Q

2 2 1 2 2 1

2 1 2 ln 2 1    (1.34) ) (

• i

T T UA Q − =

i i

• total

U A U A R 1 1 = = (1.35)

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Roverall = 1 AoUo = 1 AiUi =

i

U 1

( )

        + +

1 2

1 2 1 h A A L k r r ln A h

• i

a i i

 (1.36) =

• U

1

( )

        + +

1 2

1 2 1 h L k r r ln A h A A

a i i

 (1.37) For Sphere:

2 , , 1 , conve Sph conduction conve total

R R R R + + =         +         − +         =

• i

total

h r k r r r r h r R

2 2 2 1 1 2 2 1

4 1 4 4 1   

( )

        +         − +         − =

• i
• i

h r k r r r r h r T T Q

2 2 2 1 1 2 2 1

4 1 4 4 1    (1.35) ) (

• i

T T UA Q − =

Review Question

• 1. A horizontal large plate with 2m² is maintained at 85˚C and exposed to atmospheric air at

15˚C. Calculate the free-heat transfer if plat is 2m high and the air properties are: k = 0.03 W/m. ˚C v= 12.3 x10-6 m²/s Pr = 1.2

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• 2. Air at 100˚C is heated as it flows through a pipe with a diameter of 3cm at velocity of

15m/s. Calculate the heat transfer if the wall temperature is 120˚C and the pipe length is 1m. The properties of air are: ρ = 0.8741kg/m³ Pr = 2.377 μ = 2.233x10-5kg/m.s k = 3.221x10-5 W/m.˚C.

• 3. A stainless steel pipe is maintained at vertical position with 5cm diameter and 130˚C. It exposed

to atmospheric air at 25˚C. Calculate the heat transfer if the pipe is 10cm length. The air properties are: 697 . Pr / 10 056 . 2 . / 10 003 . 3

2 5 5

=  =  =

− −

s m v K m kW k

• 4. A square cupper plate with 10cm width is exposed to water. The plate surface temperature

is 75˚C and its thermal conductivity is 3.77 W/m.˚C. if the water temperature is 20˚C and the convection coefficient from one side is 25 W/m².˚C and 85 W/m².˚C from the opposite side. Calculate the heat transfer if the plate thickness is 2 cm.

• 5. Air at 100˚C is cooling as it flows through a pipe with a diameter of 3cm at velocity of

15m/s. Calculate the heat transfer if the wall temperature is 120˚C and the pipe length is 1m. The properties of air are: ρ = 0.8741kg/m³ Pr = 2.377

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μ = 2.233x10-5kg/m.s k = 3.221x10-5 W/m.˚C.

• 6. A 2.0 cm diameter vertical heater is maintained at a surface temperature of 38˚C and

submerged in water at 27˚C. Calculate the free-convection heat loss if the pipe long is 2m. The water properties are: k = 0.623 W/m. ˚C v= 0.7x10-6 m²/s Pr = 5.12

THERMAL RADIATION

Radiation is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. Unlike conduction and convection, the transfer of energy by radiation does not require the presence

• f an intervening medium. In fact, energy transfer by radiation is faster (at the speed of the

light) and it suffers no attenuation in a vacuum. In heat transfer studies we are interested in thermal radiation, which is the form of radiation emitted by bodies because of their

• temperature. It differs from other forms of electromagnetic radiation such as x-rays, gamma

rays, microwaves, radio waves and television waves that are not related to temperature. In general, radiation is a volumetric phenomenon, and all solid, liquid and gases emit, absorb,

• r transmit radiation to varying degrees. Transmission of radiation, which can occur in solids

as well as fluids, is an interesting phenomenon because it can occur through a cold non-

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absorbing medium between two other hotter bodies. Thus the surface of the earth receives energy direct by radiation from the sun, even though the atmosphere at high altitude is extremely cold. Similarly, the glass of a green house is colder than the contents and radiant energy does not stop there, it is transmitted to the warmer absorbing surfaces inside. Radiation is also significantly different from conduction and convection in that the temperature level is a controlling factor. In furnaces and combustion chambers, radiation is the predominating mechanism of heat transfer. Since radiation energy exchange depends on the rates at which energy is emitted by one body and absorbed by another, it is necessary to establish definitions relating to these characteristics of surfaces. Further, not all of the energy emitted by one body may necessarily fall on the surface of another due to their geometric arrangement.

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THERMAL RADIATION PROPERTIES OF MATERIALS

When radiant energy strikes a material surface, part of the radiation is reflected, part is absorbed, and part is transmitted, as shown in Figure 1.18. We define the reflectivity (ρ) as the fraction reflected, the absorptivity () as the fraction absorbed, and the transmissivity () as the fraction transmitted. Figure 1.18 Sketch showing effects of incident radiation Thus  +  +  = 1 (1.36) Most solid bodies do not transmit thermal radiation, so that for many applied problems the transmissivity () may be taken as zero. They are termed as thermally opaque. Then,  +  = 1 (1.37) Two types of reflection phenomena may be observed when radiation strikes a surface. If the angle of incidence is equal to the angle of reflection, the reflection is called Specular. On the

• ther hand, when an incident beam is distributed uniformly in all directions after reflection,

the reflection is called diffuse. These two types of reflection are depicted in Figure. 1.19. Note

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that a specular reflection presents a mirror image of the source to the observer. No real surface is either Specular or diffuses. An ordinary mirror is quite Specular for visible light, but would not necessarily be Specular over the entire wave-length range of thermal radiation. Ordinarily, a rough surface exhibits diffuse behavior better than a highly polished surface. Similarly, a polished surface is more Specular than a rough surface. Figure 1.19 (a) Specular (1 = 2) and (b) diffuse reflection

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BLACK BODIES

A blackbody is defined as a perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy that the blackbody. A blackbody absorbs all incident radiation for which  = 1 and  = 0, regardless of wavelength and direction. Also, a blackbody emits radiation energy uniformly in all direction. That is, a blackbody is a diffuse emitter. For real materials the highest values of  are around 0.97. Artificial surfaces may be arranged in practice which is virtually black. Consider Figure. 1.20. The hollow enclosure has an inside surface of high absorptivity. Incident energy passes Figure 1.20. Artificial black-body surface through the small opening and is absorbed on the inside surface. However, some is reflected, but most of this is absorbed on a second incidence. Again, a small fraction is reflected. After a number of such reflections the amount unabsorbed is exceedingly small and very little of the

• riginal incident energy is reflected back out of the opening. The area of the opening may thus

be regarded as black.

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The work of Stefan and Boltzmann led to the law named after them which gives the emission

• f radiant energy from a black body. Thus

4

T A Qb  = is the Stefan-Boltzmann law for black-body radiation. T is the absolute temperature and  is the Stefan-Boltzmann constant and has the value

4 2 8

10 67 . 5 K m W

−

 .

b

Q is the blackbody emissive power, in W. Note that, emission of thermal radiation is proportional to the fourth power of the absolute temperature. So that, the blackbody having the following properties:

• 1. A blackbody absorbs all incident radiation, regardless of wavelength and direction.
• 2. For a prescribed temperature and wavelength, no surface can emit more energy than a

blackbody.

• 3. Although the radiation emitted by a blackbody is a function of wavelength and temperature,

it is independent of direction. That is, the blackbody is a diffuse emitter.

AN EMISSIVITY OF A MATERIAL

The emissivity of a surface represents the ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a blackbody at the same temperature. The emissivity of a surface is denoted by ɛ, and it varies between 1    . Emissivity is a measure of how closely a surface approximates a blackbody, for which ɛ=1. At thermal equilibrium, the emissivity and the absorptivity of a body are equal. A true black body would have an while any real object would have . Emissivity is a dimensionless quantity, so it does not have units. The more reflective a material is, the lower its emissivity. Highly polished sliver has an emissivity of about 0.02.

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GREY BODY

A grey body is defined as a body with constant emissivity over all wavelengths and

• temperatures. Kirchhoff’s law states that the absorptivity and emissivity of a grey body are

equal at any given temperature   = , where is  the total absorptivity and , total emissivity over all wavelengths.

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NET THERMAL RADIATION FROM A GREY BODY

Consider the situation shown in Fig. 1.24, where heat flows from the body at absolute temperature T1 to the surroundings at temperature T2. Due to the fact that the surroundings are black, and the enclosed body is grey with emissivity (1), the net thermal radiation from the grey body will be:

( )

4 2 4 1 1 1

T T A Q − =  (1.38) Figure 1.21 Grey body in black surroundings. The net heat transfer (Q) from a unit surface of a grey body in (W/m2) at temperature T1 to a black enclosure at T2 can be written as: q = hr (T1 – T2) (1.39) Thus hr =       q T1 – T2 (1.40) and (hr) may be looked upon as a radiation transfer coefficient. The above equation for (hr) is also applicable if the surroundings are not black, provided that the body is small and none of its radiation is reflected back to it.

HEAT TRANSFER

Example 1.13 Investigate the possibility of reducing the heat loss from a domestic hot water cylindrical tank by coating it with aluminum paint. The tank is 0.5 m diameter and 1 m high and is situated in a large space effectively forming black surroundings. (Distant non-black surroundings are effectively black because a negligible amount of energy is reradiated to the tank.) The estimation is based on a tank surface temperature of 80

• C and an ambient temperature of 25
• C

and the tank surface is oxidized copper with an emissivity of 0.8. Calculate the amount of heat flow. Solution: The radiation heat flow is:

( )

4 2 4 1 1 1

T T A Q − = 

1

A = the surface area for the cylinder + the top and bottom area of cylinder = DL + (

)

2

2 r 

1

A = (

)( )( ) ( )( )

2

9625 . 1 3925 . 57 . 1 5 . 14 . 3 2 1 5 . 14 . 3 m = + = +

( ) ( )

 

4 4 12

273 25 273 80 9625 . 1 8 . 10 7 . 56 + − +     = 

−

Q

kW Q 68 . =

HEAT TRANSFER

Review Question

• 1. A circular disk with 10cm diameter coating with aluminum paint and situated in a

large space effectively forming black surrounding. The disk surface temperature is 88˚C and an ambient temperature is 20˚C. if the emissivity is 0.55 and the Stefen- boltzmann constant is

4 2 12

. / 10 7 . 56 K m kW

−

 . Calculate the thermal radiation loss from the disk.

• 2. What happen when radiant energy strikes a material surface?
• 3. What is the specular reflection? Give an example.
• 4. What is the diffuse reflection? Give an example.
• 5. Water –cooled, spherical object of diameter 10mm and emissivity 0.9 is maintained at

80˚C when placed in a large vacuum oven whose walls are maintained at 400˚C. Calculate the heat transfer from the oven walls.

• 6. Define the properties emissivity and absorptivity. When are these two properties equal

to each other?

HEAT TRANSFER

EXTENDED SURFACES (FINS)

If heat exchange is occurring between two fluids where one fluid has a very high resistance to heat transfer in comparison to the other, the higher resistance fluid “controls” the rate of heat

• transfer. Such cases occur, for example in the heating of air by steam or in the heating of a

very viscous oil, flowing in laminar flow, by a molten salt mix. The relative magnitude of the heat-transfer coefficient is about 10 for the oil or air, compared with 2000 for the steam or

• salt. This poor, heat-transfer situation will require much transfer surface for a reasonable flow

rate of air or oil. In order to compensate for the high resistance of the oil or air, the heat transfer surface exposed to these fluids may be increased by extension of the surface, as in the addition of fins to the outside of the tube, as illustrated in Figure 1.22. The fins are referred to as an extended surface; they increase the transfer area substantially in a given amount of space. Some automobile radiators are good illustrations of extended-surface heat exchangers. Figure 1.22. Different types of finned surfaces: (a) longitudinal fin of rectangular profile; (b) cylindrical tube equipped with fins of rectangular profile; (c) longitudinal fin of trapezoidal profile; (d) longitudinal fin of

HEAT TRANSFER

parabolic profile; (e) cylindrical tube equipped with radial fin of rectangular profile; (f) cylindrical tube equipped with radial fin

• f

truncated conical profile; (g) cylindrical spine; (h) truncated conical spine; (i) parabolic spine.

Classification of Extended Surfaces.

Figure 1.23. Fin attachment Fins of a number of industrial types are shown in Figure 1.22. Pipes and tubes with longitudinal fins are marketed by several manufacturers and consist of long metal strips or channels attached to the outside of the pipe. The strips are attached either by grooving and peeneding the tube as in Figure (1.23a), or by welding continuously along the base. When channels are attached, they are integrally welded to the tube as in Figure 1.23 b. Longitudinal fins of this type are commonly used in double pipe exchangers when the flow proceeds along the axis of the tube. Longitudinal fins are most commonly employed in problems involving gases and viscous liquids or when the smallness of one of a pair of heat-transfer streams causes streamline flow. Transverse fins are made in a variety of types and are employed primarily for the cooling and heating of gases in cross flow. The helical fins in Figure (1.24a) are classified as transverse fins and are attached in a variety of ways such as by grooving and peeneding, expanding the tube metal itself to form the fin, or welding ribbon to the tube continuously.

HEAT TRANSFER

Figure 1.24 Transverse fins Disc-type fins are also transverse fins and are usually welded to the tube or shrunken to it as shown in Figure (24, b and c). In order to shrink a fin onto a tube a disc, with inside diameter slightly less than the outside diameter of the tube, is heated until its inside diameter exceeds the outside diameter of the tube. It is slipped onto the tube, and upon cooling, the disk shrinks to the tube and forms a bond with it. Another variation of the shrunk-on fin in Figure 1.23c employs a hollow ring in its hub into which a hot metal ring is driven. Other types of transverse fins are known as discontinuous fins, and several shapes such as the star fin are shown in Figure 1.25. Figure 1.25. Discontinuous fins

HEAT TRANSFER