BASIC CONCEPTS. In other words, 0 degrees R = F and 0 degrees K = C (absolute zero).

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BASIC HEAT TRANSFER AND SOME APPLICATIONS IN POLYMER PROCESSING (A version of this was pulished as a oo chapter in Plastics Technician s Toolox, Volume, Pages -33, SPE 00) John Vlachopoulos and David Strutt www.polydynamics.com Heat transfer is a ranch of engineering science which sees to determine the rate of energy transfer etween odies as a result of temperature differences. The concept of rate is the asic difference etween heat transfer and thermodynamics. Thermodynamics deals with systems in equilirium and is concerned with the amount of heat required to change a system from one state to another. Thermodynamics does not answer the question of how fast a change is accomplished. It is the science of heat transfer that deals with this question. BASIC CONCEPTS The terms temperature and heat are understood intuitively. Temperature is the physical property that descries whether a ody is hot or cold. If we touch a hot metal at 0 F (49 C) heat flows from the metal to our hand. If we touch a cold loc of ice heat will flow from our hand to the ice. So, the intuitive concept of temperature is really defined y the heat exchange from one ody to another. Heat is a form of energy that flows from one ody to another as a result of a temperature difference. The two temperature scales used for measurement purposes are the Fahrenheit ( F) and Celsius ( C) scales. These scales were estalished y specifying the numer of increments etween the freezing (3 F, 0 C) and oiling point ( F, 00 C) of water at standard atmospheric pressure. The asolute Celsius scale is called the Kelvin (K) and the asolute Fahrenheit scale is termed Ranine (R). The following conversion relations apply 9 F 3 + C 5 5 C ( F - 3) 9 9 R K 5 R F + 459.69 K C + 73.6 In other words, 0 degrees R -459.69 F and 0 degrees K -73.6 C (asolute zero). Temperature is measured y oserving its effect on some easily oservale property of a measuring device e.g. expansion of mercury in a glass thermometer.

Heat, or energy in general, is usually measured in Btu, cal, cal and Joule (J). Btu will raise lm of water F at 68 F, cal will raise g of water C at 0 C, cal will raise g of water C at 0 C. The definition of Joule (J) comes from the definition of wor done y a force on an oject: (Newton meter N m J) Wor W F S where F is the force and S the distance travelled. Power is the wor done per second and J/s is called Watt (W) Power F S t F V where t is time and V the velocity of travel. If we multiply Watts y time, the result is wor done. The wor done y electricity is usually expressed as ilowatt hours, 000 Watts times 3600 seconds, or 3.6 0 6 Joules. The following conversion relations apply: Btu 055 J cal 48 J Btu 5 cal J W s 3.43 Btu/hr The human metaolism requires aout 500 cal per day (roughly 400 Btu per hour) according to frequently quoted guidelines for daily food consumption. FUNDAMENTAL THERMODYNAMIC LAWS The First Law of Thermodynamics is the principle of conservation of energy, which states that energy cannot e created nor destroyed. A more technical definition is: The increase of internal energy (ΔE) of a given system is equal to heat (Q) asored from the surroundings plus the mechanical wor (W) added. Q + W ΔE If no heat enters or leaves the system the process is referred to as adiaatic thus ΔE W

3 On the other hand, if no wor is done y the surroundings, then ΔE Q Heat and wor are different types of energy. Heat capacity Cp (also called Specific Heat) is the amount of heat required to raise the temperature of a ody y one degree. Here are some typical values: Water (at 0 C, 68 F) cal/g C 48 J/g C Air (approximately) 0.39 cal/g C 000 J/g C Polyethylene (approximately) 0.550 cal/ C 300 J/g C Steel (approximately) 0.08 cal/ C 450 J/g C Bric (approximately) 0.5 cal/ C 900 J/g C For a melting of a solid heat must e added to shae and demolish the crystal structure present. Heat of fusion (ΔHf) is the amount of heat required to melt a crystalline solid without raising its temperature. It is equal in magnitude (ut opposite in sign) to the heat of crystallization. Here are some typical values: Ice ΔHf 333,000 J/g (80 cal/g) This means that to melt g of water we need the same amount of energy as that required to raise g of water y 80 C, and yet with the actual melting, there is no increase in temperature. HDPE ΔHf 50,000 J/g LDPE ΔHf 00,000 J/g For amorphous polymers lie PS, PMMA and PC ΔHf 0, since they do not have any crystal structure. Heat of reaction is the amount of heat involved in a chemical reaction (added or removed). All polymerization reactions (production of polymers) are exothermic, i.e. they involve lieration of heat. The Second Law of Thermodynamics relates to the direction in which energy transfer or conversion may tae place. The result is the increase of Entropy, which is a loss in the availaility of energy for external purposes. Heat will flow from a high temperature to a low temperature. It is impossile to construct a machine or device which will operate continuously y receiving heat from just a single reservoir and producing wor. A heat engine is a device that produces net positive wor as a result of heat transfer from a hightemperature ody to a low-temperature ody. The thermal efficiency of such engines is defined as the ratio of output (energy sought) to the input (energy that costs). The input or the energy that costs is the heat from the high temperature source (indirectly, the cost for the fuel) We define thermal efficiency as

4 W(energy sought) Thermal efficiency Q (energy that costs) The efficiency of heat engines is higher if the heat source has higher temperature. However, all such actual devices have low efficiencies (e.g. comustion engines for cars, no more than 35%). In other words, only aout /3 of the energy in gasoline goes to useful wor (motion of the car). The rest is wasted due to thermodynamic implications. This is the reason why there is so much research on fuel cells nowadays, in which fuels react with oxygen to produce electricity, at higher efficiencies (50 60%). For more information on the thermodynamic laws and their implications the reader is referred to more specialized textoos [-3]. Example Thirty people gather for a coctail party in a asement room that can e assumed completely sealed off and insulated. The room dimensions are 4 ft 8 ft with 8 ft ceiling. Calculate the temperature rise in 30 minutes. Solution We apply the first law of thermodynamics Q + W Δ E Since there is no wor eing added to the air in the room ΔE Q We will assume that each person gives off approximately 500 cal/day (equal to the average metaolic energy consumption for light activity, lie taling and waling around the room) ΔE Q 30 500 75,000 cal/day 35 cal/hour 5 cal/min This amount of energy goes to the air inside the room which is roughly Air volume 4 ft 8 ft 8 ft 5376 ft 3 5.5 m 3 We neglect the volume occupied y the people. The density of air is aout.4 g/m 3. So the total mass of air in the room is The internal energy change will e m.4 5.5 88.79 g.

5 ΔE m Cp ΔT where Cp is the heat capacity of air (0.39 cal/ C). E T m C p 5 cal/min.5c / min.07f / min 88.79 g 0.39 cal/gc So, in 30 minutes the temperature would rise y 30.5 C/min 34.5 C (6. F)! This means that even a cold room would ecome quicly very hot, if the assumption of complete insulation is valid. In reality, there would e considerale heat losses to the surroundings that will slow down the temperature rise. Example In an injection molding machine 0 g of LDPE are molded per hour. The melt temperature entering the mold is 80 C and the mold temperature is maintained at 40 C y a cooling water system. Determine the amount of water required to cool the plastic and eep the mold at 40 C, if the difference in input-output temperatures of the water is not to exceed 5 C. Solution This is a straightforward application of the first law of thermodynamics, that is the principle of conservation of energy. The heat for coming off the solidifying plastic in the mold must e taen away y the water. Qplastic Qwater The heat removed from the plastic is equal to the heat given off as the plastic temperature drops from 80 C to 40 C plus the heat of solidification which is equal to the heat of fusion ut opposite in sign. As the plastic solidifies and the molecules stop moving randomly, heat is lierated. m Q plastic m c p T + m H is amount of material molded per hour, Cp its heat capacity and ΔHf the heat of solidification (00,000 J/g for LDPE which solidifies around 06 C). Q plastic g 0 hr hr 3600s g + 0 hr J 300 (80-40C) gc hr 3600s 00,000 J/g J 789 + 900 s The heat taen up y the water undergoing a 5 C temperature change is f

6 Q water m c p J T m 48 5C m 090 gc Therefore, J J m 090 900 g s m 0.38 g/s 499 g/hr For another example of cooling of a plastic in a mold the reader is referred to the last section of this chapter: Special Heat Transfer Prolems in Plastics Processing. HEAT TRANSFER MODES Temperature differences cause the flow of heat from a high temperature to a low temperature. There are three modes of heat transfer: conduction, convection, and radiation. The asic microscopic mechanism of conduction is the motion of molecules and electrons. It can occur in solids, liquids and gases. In non-metallic solids the transfer of heat energy is due mainly to lattice virations. In metallic solids we have oth lattice virations and random motions of free electrons. Consequently metals are more conductive than non-metals. In gases, we have mainly random motions of molecules. In liquids we have partly random molecular motions and some sort of viration of the liquid lattice structure. Convection is associated with the transport of a mass of liquid or gas. It can e forced i.e. when assisted y a pump or fan, or free (also called natural convection) when the motion of a fluid occurs due to density differences. If there is an electrical heating element at the corner of a room and air is lown onto the element y a fan, this is forced convection. In the asence of a fan the air surrounding the heating element will get hotter, its density will decrease and the air will move upwards causing natural circulation within the room, as the hot air is replenished y colder air, which gets hot and rises again. Radiation involves electromagnetic waves which are emitted y a ody as a result of its temperature. The electromagnetic radiation has a road spectrum from radio waves to x-rays. Between the two extremes a narrow portion of the radiation spectrum is the visile light and a roader one covers the thermal radiation. The earth is heated y sun s radiation. CONDUCTION Consider the solid wall shown in Fig.. Temperature T is higher than temperature T. Heat flows from the high temperature to the low temperature. If A is the area normal to the direction of heat flow, Fourier s Law states that the amount of heat flow is proportional to the area A, the temperature difference T T and inversely proportional to the thicness of the wall.

A Q (T ) The proportionality constant is the thermal conductivity measured in W/m C or Btu/hr ft F (see Tale ). 7 A plastic injected into a mold cavity is cooled y heat conduction through the mold wall. In fact, one of the factors considered in choosing mold materials is their thermal conductivity. Aluminum has roughly 5 times the conductivity of steel and Beryllium copper alloys have aout 3 times higher conductivity than steel. Higher conductivity means faster heat removal (or addition). It is interesting to note the similarity etween Fourier s law of heat conduction and Ohm s law of electricity. We may write: T A E R I Q Fourier's Law Ohm' s Law where E is the voltage (corresponds to T T), R the resistance (corresponds to /A and I the electric current (corresponds to Q). The quantity /A is sometimes referred to as thermal resistance. Now, let us consider a wall composed of three different materials with thermal conductivities, and 3 as shown in Fig.. We will have

8 Q A T - T Q A T - T Q A T - T 3 3 4 3 3 Figure Heat conduction through a composite wall Summing up the aove expressions, we have Q A + A + A T 3 3 4 This equation is similar to the expression of three electrical resistors in series E (R + R + R3) I We can then write a general expression to calculate the rate of heat flow through a composite wall for more than three layers as follows: 3

9 Q T n n + +... A A n A Tale Some typical values of thermal conductivity () W/m C Btu/hr ft F Copper 380 0 Aluminum 04 8 Caron Steel 43 5 Glass 0.78 0.45 Polymer 0. 0.5 Water 0.6 0.347 Air 0.05 0.044 Tale Typical values of convection heat transfer coefficients Mode W/m C Btu/hr ft F AIR, Free Convection 4-8 0.7-5 AIR, Forced Convection 4-570 0.7-00 WATER, Free Convection 84-500 50-65 WATER, Forced Convection 84-7,000 50-3,000 WATER, Boiling 840-57,000 500-0,000 STEAM, Condensing 5680-3,000,000-0,000

0 For the case of composite cylinder (as shown in Fig. 3) the aove equation taes the form Q T 4 r r3 n + n L r L r r4 + 3 L r3 Figure 3 Heat conduction through a composite cylinder. Unsteady heat conduction involves temperature variations with time. For example, if the surface of a ody is suddenly raised to higher temperature how long will it tae for the temperature to penetrate inside? Such prolems require the solution of differential equations. However, many solutions of such prolems for common geometrical shapes appear in the form of charts in specialized texts [4 6]. With the help of Figs. 4 and 5, it is possile to determine how long it will tae for the temperature, at the midplane of plate or axis of a cylinder, to reach a certain value, if the surface temperature is suddenly raised or lowered. On the vertical axes of Figs. 4 and 5, Tm is the (unnown) temperature at the midplane or axis, Ti is the initial temperature of the material (uniform throughout) and T0 the suddenly imposed surface temperature. On the horizontal axes, the quantity t Fo Cp x m is the dimensionless Fourier numer, and t represents time and x the distance from the surface to the center.

Figure 4 Plot for calculating the temperature Tm at the midplane of a plate as a function of time after the two surfaces are suddenly raised to T0 Figure 5 Plot for calculating the temperature Tm at the axis of a cylinder as a function of time after the surface tempearature is suddenly raised to T0

CONVECTION In most heat transfer prolems, we are concerned with solid walls separating liquids or gases from each other. In such cases we usually do not now the temperatures on the wall surfaces, ut rather the temperatures of the ul of fluids on oth sides. Careful experiments supported also y theoretical considerations, have shown that the greatest temperature drop is confined within a thin fluid layer attached to a solid surface, as shown in Fig. 6. Figure 6 Heat transfer through a wall separating two fluids. To explain this oservation, we may assume that a thin film, of thicness δ, adheres to the wall, whereas outside this film all temperature differences vanish as a result of mixing motions. Within the film heat flow taes place y conduction, as in a solid wall. Thus, in general, we may write Q A(T The quantity /δ h is called the heat transfer coefficient and it is an extremely important concept in heat transfer. This simplified model is very useful for practical applications, ecause the calculation of heat transfer can e made in terms of the heat transfer coefficient: w Q h A (T w ) At this point it suffices to say that the heat transfer coefficient depends on the flow conditions and fluid properties. Typical values are given in Tale. In the next section we will present some correlations that can e used for the more accurate determination of this coefficient. For the wall separating the two fluids of Fig. 6, we have )

3 Q h A (T Q A (T Q h A (T w w w ) w ) ) or T T T w w w w Q h A Q A Q h A By summing up T + + Q h A A h A or T + + Q A h h The quantity in the racets is called the total thermal resistance. A more useful concept, however, is the overall heat transfer coefficient, which is defined as follows U h + + h For a composite wall separating two fluids (a and ), we have Then, in general, we have: U h a + + +... + h and T A U Q U A T Q

The significance of the overall heat transfer coefficient is that it permits the calculation of the rate of heat flow y multiplying this quantity y the heat exchange area (perpendicular to the heat flow direction) and the temperature difference. Example Determine the heat loss through an 8-ft y 4-ft glass window of 4 mm thicness. The inside temperature is assumed to e 4 C (75 F) and the outside temperature is 0 C (4 F). The inside heat transfer coefficient is 5 W/m C and the outside aout 0 W/m C (due to moderate wind). The thermal conductivity of window glass is 0.78 W/m C. Solution The overall heat transfer coefficient is 4 U h + + h 5 W / m 0.004 m + + C 0.78 W / m C 0 0. + 0.005 + 0.05 0.55 m C /W W / m C U 3.9 W /m C The rate of heat flow is Q U A (T ) 3.9 W / m C (.44. m ) (4C - (-0 C)) 3.9.98 34 397 W 355 Btu / hr Convection Heat Transfer Coefficient Calculation The most important step in heat convection calculations is the determination of the appropriate heat transfer coefficient. The higher the fluid velocity is, the higher the heat transfer coefficient will e. Numerous correlations have een developed for the calculation of the heat transfer coefficient in terms of dimensionless groups: Nusselt numer is the dimensionless heat transfer coefficient defined as h D Nu

where the heat transfer coefficient h has dimensions of W/m C or Btu/hr m F, the thermal conductivity has dimensions of W/m C or Btu/hr ft F and D is a characteristic length (m or ft), such as the diameter of a pipe in which a fluid is flowing. The Reynolds numer (see also the chapter on Fluid Mechanics), is defined as 5 Re V D where ρ is the density (g/m 3 or lm/ft 3 ), V the fluid velocity (m/s or ft/s), μ the viscosity (Pa s or lfs/ft ) and D a characteristic length (m or ft) such as the diameter of a pipe, in which a fluid is flowing. When the value of the Reynolds numer for pipe flow is less than 00, the flow is streamlined and regular and is called laminar. Aove Re 00, the flow is highly chaotic and irregular and is said to e turulent. The Prandtl numer is defined as C Pr where Cp is the heat capacity (J/g C or Btu/lm F), μ the viscosity (Pa s or lfs/ft ) and the thermal conductivity (W/m C or Btu/hr ft F). p For Laminar Flow in tues, the following correlation applies: For Turulent Flow in tues, we have 0.0668 (D / L) Re Pr Nu 3.66 + + 0.04 [(D / L) Re Pr] Nu 0.03 Re 0.8 Pr n where n 0.4 for heating and n 0.3 for cooling. Numerous other correlations are availale in specialized textoos [4 6]. Generally, for laminar flow Nu depends on the 0.3 power of Re, while for turulent flow on the 0.8 power of Re. To calculate the heat transfer coefficient we plug the various quantities in the appropriate correlation to calculate the Nusselt Numer Nu. Then, from the definition /3 we can calculate h D Nu

6 h Nu The values given in Tale have een calculate from such correlations assuming flow conditions encountered in various practical heat transfer situations (according to reference [4]). Let us now see how much more effective is turulent heat transfer over laminar. Assuming D/L /30, Re 000 and Pr (i.e. water at 00 C), the correlation for laminar flow gives 0.0668 000 30 Nu 3.66 + / 3 + 0.04 000 30 6.37 and h 6.37 D We repeat the calculation for turulent flow assuming Re 3000 the other quantities remaining the same. D Nu 0.8 0.4 0.03 (3000) and h 3.9 D 3.9 We see that while the Reynolds numer (i.e. velocity) was increased y 50% the heat transfer coefficient was increased y 8% i.e. (3.9/6.39) 00 8% 00% 8%. Oviously, to maximize heat transfer rates, we should operate with turulent flow conditions. For example, y increasing the channel diameter and the coolant velocity, the Reynolds numer in the cooling channels of a mold can e increased, eyond the critical value of 00. The resulting turulent flow will e much more effective in removing heat from the mold. Molten polymers are very viscous and the Reynolds numer is in the range 0.000 to 0.0, i.e. polymer melt flows are always laminar. So, for a polymer melt having 0. W/m C, Pr 5 0 6, and flowing through a 8 mm diameter channel the heat transfer coefficient will e roughly 0. h 7 75 W/ m 0.008 C

7 RADIATION Many types of thermoforming machines use radiation to heat the plastic sheet. Metal rod heaters, halogen tues and ceramic plates are used frequently. Heater temperatures can reach 700 C and the corresponding radiative heat transfer coefficients can reach 00 W/m C. Radiation is important wherever very high temperatures are involved. However, radiative effects play a significant role is other less ovious situations. Goose down is nown as the most effective insulation for two reasons: reduction of heat conduction y air pocets trapped y clusters of fiers and entrapment of radiation. The goose down fiers have diameter of few microns which is the wavelength of a significant portion of the infrared radiation that escapes from the human ody and can go through air pocets and farics. Radiation scattering occurs due to equivalence of fier size and wavelength. Synthetic materials have een manufactured that reportedly match the goose down properties oth in reducing conduction and radiation heat loss, again y scattering on micron-sized fiers. The radiant heat transfer (which is a form of electromagnetic radiation), depends on the asolute temperature and the nature of the radiating surface. This is stated y the Stefan-Boltzmann law for a lac ody (perfect radiator) Q σ A T 4 where T asolute temperature of the surface of the ody in K or R, A the surface area (m of ft ) and σ the Stefan-Boltzmann constant (σ 5.669 0 8 W/m K 4 or 0.74 0 8 Btu/hr ft R 4 ). Blac odies are called so, ecause such materials do not reflect any radiation and appear lac to the eye. Thus, a lac ody asors all radiation incident upon it. The "lacness" of a surface to thermal radiation can e quite deceiving insofar as visual oservations are concerned [4]. Some visually lac surfaces are indeed lac to thermal radiation. However, snow and ice appear white and right, ut are essentially "lac" in thermal radiation. When a ody with surface temperature T is placed in a closed environment of temperature T, the net amount of heat transfer depends on the temperature difference, in the form For gray odies, not perfect radiators, we have Q A (T 4 ) Q A (T 4 ) where ε emissivity (ε for lac odies, ε < for gray odies). For lamplac ε 0.96, for oxidized cast iron ε 0.7, and for polished steel ε 0.) If surface A is not completely enclosed y surface A we must introduce an additional factor to account for the relative geometrical orientation of the two radiating surfaces 4 4

8 Q A F- (T 4 ) F is called the shape or view factor or angle factor. During the winter in the northern hemisphere, the temperatures are low ecause the sun's radiation arrives with an unfavorale angle factor. 4 For practical prolems, it is sometimes advisale to define a radiative heat transfer coefficient hr from the following Q h r F- hr T A (T 4 (T ) A 4 ) F The radiative heat transfer coefficient depends strongly on temperature and is less useful as a concept than the convective heat transfer coefficient. However, it is useful for practical prolems involving oth convection and radiation, for which we can write the rate of heat flow as - F (T - + T (T 4 ) (T 4 ) + T ) Q (h + h ) A r w (T w ) where Tw is the temperature of a wall and T the temperature of a surrounding medium which completely encloses surface Aw. Due to the fourth power dependence of radiation heat exchange, this mode is more important in very high temperature applications, as in metallurgical operations. In some polymer processing operations, radiation is less important than convection. For example, in lown film extrusion, the film is liely to emerge from the die lips at 00 C (473 K) or so, and a tangentially impinging cooling air jet is liely to have heat transfer coefficient of the order of 00 W/m C. Let us calculate the radiative heat transfer coefficient assuming F, and T 0 C (93 K) h F r - (T -8 5.669 0 3.44 W / m + T ) W / m C (T + T K 4 ) (473 + 93 ) (473 + 93) This is significantly lower than the convective heat transfer coefficient (00 W/m C) and may e neglected in some calculations. HEAT EXCHANGERS Heat exchangers are devices that transfer heat y convection and conduction etween two fluids which are separated y a wall. The automoile radiator is a heat exchanger in which convection

and conduction tae place ut no radiative heat transfer. Radiative heat exchangers are used in specialized applications in space vehicles and are eyond the scope of most textoos. The simplest type of heat exchanger is the doule pipe system shown in Fig. 7. It involves tue and annular flow. If the flows are in the same direction, the arrangement is called parallel flow and if the flows are in opposite directions, counterflow. Because of the rather small surface areas which are availale for heat transfer, doule pipe heat exchangers are used for low to moderate heat transfer rates. For high rates of heat, other types of heat exchangers are used that provide large surface areas. In this category elong the shell-and-tue heat exchangers (that involve several tue passes) and cross-flow configurations (with interconnected passageways). 9 Figure 7 Doule-pipe heat exchanger (parallel flow) For the design and prediction of performance purposes, the concept of the overall heat transfer coefficient is used, which was descried earlier in this chapter. For the doule pipe arrangement of Fig. 7, we can write again Q U A ΔTm where U is the overall heat transfer coefficient, A the heat transfer area, and ΔTm a suitale average temperature difference etween the entering and exiting fluids. It turns out that this average is the so-called log-mean temperature difference (LMTD) which is defined as T m T - T n [ T / T ] where n represents the natural logarithm having ase e.788, while log has ase 0.

The aove relationships are valid for oth parallel and counterflow heat exchangers. Injection molding machines involving hot oil and cold water are connected counterflow since this arrangement removes roughly 0% more heat from the oil than parallel flow. For other types of heat exchangers (multipass shell-and-tue; cross flow, etc.), a correction factor is needed 0 T m (LMTD for counter flow) F correction factor F is usually etween 0.5 and.0 and values of this factor for various configurations of heat exchangers can e found in specialized textoos [4 6]. Heat exchangers may ecome coated with various deposits present in the flow or corroded. Because of fouling a reduction in heat transfer efficiency is oserved. The overall effect is represented y the fouling factors Definition R f U fouled - U clean taulated values of Rf can e found in specialized textoos [4-6]. For oilers used for long periods of time we may have Rf 0.000 m C/W, while Uclean 3000 W/m C. Thus 0.000 U fouled - 3000 Ufouled 875 W/m C which represents a significant reduction It is important to eep clean channels and clean circulating water or oil in plastics processing machinery, otherwise fouling will result is significant cooling efficiency reduction. SPECIAL HEAT TRANSFER PROBLEMS IN PLASTICS PROCESSING There are two properties of plastics that play a very significant role during their processing: their low thermal conductivity and their high viscosity in the molten state. The thermal conductivity of most plastics is around 0. W/m C (0.5 Btu/hr ft F) which is roughly 00 times smaller than the conductivity of steel and 000 times smaller than the conductivity of copper. Due to low conductivities, cooling of plastics is slow. In fact, in some processes, such as thic profile extrusion, pipe extrusion and film lowing, cooling might e the output rate limiting step. While the mextruder might e capale of pumping out more product, we might not e ale to cool the product fast enough. The viscosity of molten plastics decreases with the rate of shearing, ut it always remains very high, roughly of the order of a million times larger than the viscosity of water. Due to high

viscosity, polymer melts flowing through channels, dies and process equipment tend to raise their temperature y a mechanism of internal friction. Under certain conditions, the frictional heating might result in high temperatures which can cause degradation of the material. Cooling of a Plastic Plate in a Mold Let us assume that a HDPE plate of 0 mm thicness initially at 30 C is cooled in a mold and the mold surface is at 30 C. How long will it tae for the temperature to reach 90 C at the midplane (xm 0/ 5 mm). We will use Fig. 4 for Ti 30 C, T0 30 C, and Tm 90 C. The vertical axis of Fig. 4 is calculated as follows Tm T The corresponding Fourier Numer is i 0 0 90-30 30-30 60 00 0.3 Fo t C p x m 0.58 Using typical thermal properties for HDPE (see Tale 3), we have: 780 0.5 t 0.58 300 0.005 t 04 seconds For computer assisted mold cooling calculations and optimization procedures the reader is referred to a thesis [7]. Temperature Rise Due to Frictional Heating The average temperature rise due to frictional heating (also called viscous dissipation) can e calculated assuming that the process is adiaatic (i.e. there is no heat exchange with the surroundings). This is actually not such a ad assumption for flows through channels and dies. Assuming that the mechanical wor (due to the pressure pushing the molten plastic) is converted into heat, we can easily prove that the temperature rise will e P T C p

Mass So, for a die that has a pressure drop ΔP 4000 psi (7.58 0 6 Pa) and a molten polymer having ρ 780 g/m 3 and Cp 300 J/g C, we get 6 7.58 0 T 5.4C (7.7F) 780 300 This means that the average temperature of the polymer coming out of the die will e 5.4 C (7.7 F) higher than the temperature coming in. Power Requirement of a Single-Screw Extruder This example and the selected thermal properties given in Tale 3 are taen from reference [8]. In the extruder, polymer pellets usually coming in at room temperature (i.e. 0 C) are melted and susequently pumped at the extrusion temperature through the die (i.e. ~00 C, depending on polymer). Most of the energy comes from turning the screw. Of course some energy is supplied y the heating ands around the arrel. In well running extruders usually net energy input occurs in the first section (near the hopper ) and net output in the second section (near the die) i.e. the heat generated y the viscous dissipation is really heating the arrel. The power required y the turning screw is needed to: o Raise the temperature from room temperature to extrusion temperature in the die. o Melt the polymer (heat of fusion). o Pump the molten polymer. The power required y the turning screw is given y the following expression : Po Q C p (T out in) + Q hf + P Q Where: Symol Name Typical Values Throughput (g/hr) Cp Heat Capacity, Average (J/g C) 500-3000 J/g C Tin Usually room temperature C 0 C Tout Extrusion Temperature in the die C 00 C -300 C ρ Density, average (g/m 3) 700-00 g/m 3 Q Volume flow rate (m 3 /hr) ΔP Pressure rise (Pa) 0-50 MPa ΔHf Heat of fusion (J/g) The heat of fusion can vary from zero (for amorphous polymers) to 300,000 J/g (for a very crystalline HDPE). m m Let us calculate the relative contriutions for pressure rise of 30 MPa: Q m 4. g / hr of a typical polymer and

3 Po.4 500 (00-0) +.4 30,000 + 30 0 6.4 760 Po 50,580 J/hr + 4,6 J/hr + 4,437 J/hr 69,69 J/hr Term Relative Significance Contriution (J/hr) Raise temperature to 00C Most Important 50,580 Melt the polymer Somewhat Important 4,6 Pump the molten polymer Insignificant 4,437 Total Power Requirement 69,69 Let us express the final answer in terms of horsepower: Po 69,69 J hr hr 3600 9.34 s W Po 9.34 W hp 5.93 hp 0.746 W So we can size the horsepower of a motor. But we must tae into account its efficiency. If we assume 85% efficiency, then Motor Power 0.85 5.93 hp 30.50 hp.75 W SUMMARY Heat is a form of energy that is transferred from high temperatures to low temperatures y three modes: CONDUCTION, CONVECTION and RADIATION. Conduction through metals e.g. copper, aluminum and eryllium copper alloys is very fast, while through plastics is very slow. The thermal conductivity of aluminum is aout 00 W/m C and of typical plastic aout 0.0 W/m C. This means that heat is conducted one thousand times faster through aluminum than through a plastic. In solid walls the temperature drops linearly from its high value on one side to the low value on the other side. Convection can e natural (free) or forced. Natural convection involves the motion of a gas or liquid due to density differences i.e. hot air is lighter and as it moves upward new air moves near the heating source causing circulation. In forced convection, the movement of a liquid or gas is assisted y a pump or fan. For convective heat transfer calculations it is important to determine the heat transfer coefficient. The heat transfer coefficient increases as the flow velocity increases. Turulent flow is much more effective than laminar flow for transferring of heat. Radiation is the transfer of heat y electromagnetic waves. It is important whenever high temperatures are involved, such as in plastic sheet heating in thermoforming machines. In plastics processing, heat is generated within flowing highly viscous melts due to internal friction (viscous dissipation). Most of the energy necessary to heat and melt plastic pellets in an extruder comes from the motor turning the screw. This means we have conversion of mechanical energy into heat.

It is possile to carry out relatively accurate calculations of temperature rise, drop or distriution and the energy requirements, y applying the principle of conservation of energy and the equations of heat conduction, convection and radiation. For conduction the thermal conductivity of the materials involved must e nown. For convection the ey step is the determination of the heat transfer coefficients, which are usually availale as correlations involving the Reynolds numer of the flow. Radiation calculations require the material emissivities and the relative geometrical orientation of the radiating surfaces. 4 REFERENCES. Holman, J.P., Thermodynamics McGraw-Hill, New Yor (988).. Sonntag, R.E. and C. Borgnae, Introduction to Engineering Thermodynamics, Wiley (000). 3. Smith, J.M. and H.C. Van Ness, Introduction to Chemical Engineering Thermodynamics, McGraw-Hill, New Yor (00). 4. Holman, J.P., Heat Transfer, McGraw-Hill, New Yor (990). 5. Bejan, A., Heat Transfer, Wiley, New Yor (993). 6. Pitts, D.R. and L.E. Sisson, Heat Transfer, (Schaum s Outline Series), McGraw-Hill, New Yor (998). 7. Reynolds, D.W., Optimization of Cooling Circuits in Injection Molds, M.Eng. Thesis, Mechanical Engineering, McMaster University, Hamilton, ON, Canada (000). 8. Vlachopoulos, J. and J.R. Wagner, The SPE Guide on Extrusion Technology and Trouleshooting, Society of Plastics Engineers, Broofield, CT (00). 9. Mar, J.E., Physical Properties of Polymers Handoo, AIP Press, Woodury, NY (996).

5 Polymer Solid Density* ρ (g/cm 3 ) Glass Transition T g Melting Point T m Tale 3 Selected Thermal Properties. Usual Melt Processing Range Melt Density* ρ (g/m 3 ) Thermal Conductivity (W/m C) (Btu/h ft F) Heat Capacity Cp (J/g C) (Btu/l m F) 00-400 0.5-0.57 Heat of Fusion ΔH (J/g) (Btu/l) HDPE 0.94-0.967-30C -0F 30-37C 66-78F 60-40C 30-464F 780 0.5 0.45 0,000-300,000 90-30 LDPE 0.95-0.935-30C 06-C 60-40C 760 0.0 00-400 90,000-40,000-0F 3-34F 30-464F 0.5 0.5-0.57 80-00 LLDPE 0.90-0.95-30C 5C 60-40C 760 0.0 00-400 90,000-40,000-0F 57F 30-464F 0.5 0.5-0.57 80-00 PP 0.890-0.90-0C 65C 80-40C 730 0.8 000-00 0,000-60,000-4F 39F 356-464F 0.0 0.48-0.5 90-0 PVC.30-.58 80C 75C 65-05C 50 0.7 000-700 70,000-90,000 (Rigid) 76F 347F 39-40F 0.0 0.4-0.4 70-80 PS.04-.0 00C amorphous** 80-40C 000 0.5 300-000 amorphous** F 356-464F 0.09 0.3-0.48 PMMA.7-.0 05C amorphous** 80-30C 050 0.9 400-400 amorphous** F 356-446F 0. 0.33-0.57 PET.34-.39 80C 65C 75-90C 60 0.8 800-000 0,000-40,000 76F 509F 57-554F 0.0 0.43-0.48 50-60 ABS.0-.04 05-5C amorphous** 00-90C 990 0.5 300-700 amorphous** -39F 39-554F 0.45 0.3-0.4 Nylon-66.3-.5 90C 65C 75-90C 980 0.0 400-600 90,000-05,000 94F 509F 57-554F 0.5 0.57-0.6 80-88 PC. 40C amorphous** 50-305C 050 0. 300-00 amorphous** 84F 48-58F 0.3 0.3-0.5 * Melt densities have een estimated for roughly the mid-temperature of the processing range. See Mar [9] for expressions in the form ρ A-BT CT. ** Amorphous resin does not possess crystallinity and consequently no melting point or heat of fusion (i.e. heat to rea down crystal structure) can e determined.

6 John Vlachopoulos is Professor of Chemical Engineering and Director of the Centre for Advanced Polymer Processing and Design (CAPPA-D) at McMaster University, Hamilton, Ontario, Canada. He earned his Dipl. Ing. from NTU, Athens, Greece and an M.S. and D.Sc. from Washington University, St. Louis, Mo. He is the author of approximately 00 pulications on various aspects of plastics processing, frequently lectures at conferences and seminars, and serves as consultant to industry worldwide. He estalished and is President of Polydynamics, Inc., a company involved in software research, development and mareting. John was elected an SPE Fellow and received the SPE Education Award in 00.