4510 Lec 7 - Heat Transfer PDF

CENTRAL LUZON STATE UNIVERSITY Science City of Munoz, Nueva Ecija 3120 PHILIPPINES  ABEN 4510 FOOD PROCESS ENGINEERING Melba Domes Denson Department of Agricultural and Biosystems Engineering College of Engineering Central Luzon State...

CENTRAL LUZON STATE UNIVERSITY Science City of Munoz, Nueva Ecija 3120 PHILIPPINES  ABEN 4510 FOOD PROCESS ENGINEERING Melba Domes Denson Department of Agricultural and Biosystems Engineering College of Engineering Central Luzon State University Heat Transfer in Food Processing ✓ Cooking ✓ Baking ✓ Drying ✓ Sterilizing and Pasteurizing ✓ Freezing ✓ Others  Heat Transfer in Food Processing ✓ Heat transfer is a dynamic process in which heat is transferred spontaneously from one body to another cooler body. ✓ The rate of heat transfer depends upon the differences in temperature between the bodies, the greater the difference in temperature, the greater the rate of heat transfer.  Heat Transfer in Food Processing ✓ Temperature difference between the source of heat and the receiver of heat is therefore the driving force in heat transfer. An increase in the temperature difference, increases the driving force and therefore increases the rate of heat transfer. ✓ The heat passing from one body to another travels through some medium which in general offers resistance to the heat flow. Both these factors, the temperature difference and the resistance to heat flow, affect the rate of heat transfer. As with other rate processes, these factors are connected by the general equation: rate of transfer = driving force / resistance Modes of Heat Transfer ✓ Conduction ✓ Radiation ✓ Convection www.scienceabc.com Modes of Heat Transfer: Conduction ✓ In conduction, the molecular energy is directly exchanged, from the hotter to the cooler regions, the molecules with greater energy communicating some of this energy to neighboring molecules with less energy. ✓ An example of conduction is the heat transfer through the solid walls of a refrigerated store. ✓ Fourier equation of heat conduction: dQ/dt = kA dT/dx where dQ/dt is the rate of heat transfer, the quantity of heat energy transferred per unit of time, A is the area of cross-section of the heat flow path, dT/dx is the temperature gradient, that is the rate of change of temperature per unit length of path, and k is the thermal conductivity of the medium. Heat flows from a hotter to a colder body, that is in the direction of the negative temperature gradient. Modes of Heat Transfer: Conduction ✓ Thermal conductivity - is a material property that describes the ability of a material to conduct heat. It is defined as the amount of heat that passes through a unit area of a material per unit time when there is a temperature gradient. ✓ Thermal conductivity does change slightly with temperature, but in many applications, it can be regarded as a constant for a given material. ✓ Simply, it tells us how easily heat can move through a material. Modes of Heat Transfer: Conduction ✓A high thermal conductivity means the material transfers heat efficiently, while a low thermal conductivity means it resists heat flow. ✓Example: Metals like copper and aluminum have high thermal conductivity, making them good conductors of heat. Insulating materials like wood or rubber have low thermal conductivity. Thermal Conductivity ✓ In general, metals have a high thermal conductivity, in the region 50-400 J m-1 s-1 °C-1. ✓ Most foodstuffs contain a high proportion of water and as the thermal conductivity of water is about 0.7 J m-1 s-1°C-1 above 0°C, thermal conductivities of foods are in the range 0.6 - 0.7 J m-1 s-1°C-1. ✓ Ice has a substantially higher thermal conductivity than water, about 2.3 J m-1 s-1°C-1. ✓ The thermal conductivity of frozen foods is, therefore, higher than foods at normal temperatures. Thermal Conductivity ✓ Most dense non-metallic materials have thermal conductivities of 0.5-2 J m-1 s-1°C-1. ✓ Insulating materials, such as those used in walls of cold stores, approximate closely to the conductivity of gases as they are made from non-metallic materials enclosing small bubbles of gas or air. ✓ The conductivity of air is 0.024 J m-1 s-1 °C-1 at 0°C, and insulating materials such as foamed plastics, cork and expanded rubber are in the range 0.03- 0.06 J m-1 s-1 °C-1. Some of the new foamed plastic insulating materials have thermal conductivities as low as 0.026 J m-1 s-1 °C-1. Thermal Conductivity Thermal diffusivity is a material property that measures the rate at which heat spreads through a material. It is defined as the ratio of thermal conductivity to the product of density and specific heat capacity. Heat Conduction through Slab ✓ If a slab of material, as shown in the figure, has two faces at different temperatures T1 and T2, heat will flow from the face at the higher temperature T1 to the other face at the lower temperature T2. ✓ This may be regarded as the basic equation for simple heat conduction. It can be used to calculate the rate of heat transfer through a uniform wall if the temperature difference across it and the thermal conductivity of the wall material are known. Example: ✓ A cork slab 10 cm thick has one face at -12°C and the other face at 21°C. If the mean thermal conductivity of cork in this temperature range is 0.042 J m-1 s-1 °C-1, what is the rate of heat transfer through 1 m2 of wall? Sol’n: Heat Conductances ✓ In tables of properties of insulating materials, heat conductances are sometimes used instead of thermal conductivities. ✓ The heat conductance is the quantity of heat that will pass in unit time, through unit area of a specified thickness of material, under unit temperature difference. For a thickness x of material with a thermal conductivity of k in J m-1 s-1 °C-1, the conductance is k/x = C and the units of conductance are J m-2 s-1 °C-1. Heat conductance = C = k/x ✓ It tells us how efficiently heat is transferred through a specific object made from that material. Heat Conductances in Series ✓ Frequently in heat conduction, heat passes through several consecutive layers of different materials. For example, in a cold store wall, heat might pass through brick, plaster, wood and cork. Heat Conductances in Series ✓ In the steady state, the same quantity of heat per unit time must pass through each layer. q = A1T1k1/x1 = A2T2k2/x2 = A3T3k3/x3 = …….. ✓ If the areas are the same, A1 = A2 = A3 = ….. = A q = AT1k1/x1 = AT2k2/x2 = AT3k3/x3 = …….. So AT1 = q(x1/k1) and AT2 = q(x2/k2) and AT3 = q(x3/k3).….. Heat Conductances in Series (from previous slide) AT1 + AT2 + AT3 + … = q(x1/k1) + q(x2/k2) +q(x3/k3) + … A(T1 + T2 + T3 +..) = q(x1/k1 + x2/k2 +x3/k3 + …) ✓ The sum of the temperature differences over each layer is equal to the difference in temperature of the two outside surfaces of the complete system, i.e. T1 + T2 + T3 + … = T Heat Conductances in Series (from previous slide) and since k1/x1 is equal to the conductance of the material in the first layer, C1, and k2/x2 is equal to the conductance of the material in the second layer C2, x1/k1 + x2/k2 + x3/k3 +... = 1/C1 + 1/C2 + 1/C3 …… = 1/U where U = the overall conductance for the combined layers, in J m- 2 s-1 °C-1 Therefore, AT = q(1/U) and so, q = UAT ✓ U is called the overall heat-transfer coefficient, as it can also include combinations involving the other methods of heat transfer – convection and radiation.  Heat Conductances in Parallel ✓ Heat conductances in parallel have a sandwich construction at right angles to the direction of the heat transfer, but with heat conductances in parallel, the material surfaces are parallel to the direction of heat transfer and to each other. The heat is therefore passing through each material at the same time, instead of through one material and then the next. Heat Conductances in Parallel ✓ An example of heat conductance in parallel is the insulated wall of a refrigerator or an oven, in which the walls are held together by bolts. The bolts are in parallel with the direction of the heat transfer through the wall: they carry most of the heat transferred and thus account for most of the losses.  Example Problem on Heat Conductances in Parallel ✓ The wall of a bakery oven is built of insulating brick 10 cm thick and thermal conductivity 0.22 J m-1 s-1 °C-1. Steel reinforcing members penetrate the brick, and their total area of cross-section represents 1% of the inside wall area of the oven. If the thermal conductivity of the steel is 45 J m-1 s-1 °C-1 calculate (a) the relative proportions of the total heat transferred through the wall by the brick and by the steel and (b) the heat loss for each m2 of oven wall if the inner side of the wall is at 230°C and the outer side is at 25°C. ✓ Applying the eqn. q = ATk/x, we know that T is the same for the bricks and for the steel. Also x, the thickness, is the same.  Solution: a) Consider the loss through an area of 1 m2 of wall (0.99 m2 of brick, and 0.01 m2 of steel). Surface Heat Transfer ✓ Newton found, experimentally, that the rate of cooling of the surface of a solid, immersed in a colder fluid, was proportional to the difference between the temperature of the surface of the solid and the temperature of the cooling fluid. This is known as Newton's Law of Cooling, and it can be expressed by the equation, analogous to: q = hsA(Ta– Ts) where hs is called the surface heat transfer coefficient, Ta is the temperature of the cooling fluid and Ts is the temperature at the surface of the solid.  Surface Heat Transfer (from previous slide) ✓ The surface heat transfer coefficient can be regarded as the conductance of a hypothetical surface film of the cooling medium of thickness xf such that hs = kf /xf, where kf is the thermal conductivity of the cooling medium. ✓ Following on this reasoning, it may be seen that hs can be considered as arising from the presence of another layer, this time at the surface, added to the case of the composite slab considered previously. The heat passes through the surface, then through the various elements of a composite slab and then it may pass through a further surface film. We can at once write the important equation: q = AΔT [(1/hs1) + x1/k1 + x2/k2 +.. + (1/hs2)] = UAΔT Surface Heat Transfer (from previous slide) ✓ where 1/U = (1/hs1) + x1/k1 + x2/k2 +.. + (1/hs2) and hs1, hs2 are the surface coefficients on either side of the composite slab, x1, x2...... are the thicknesses of the layers making up the slab, and k1, k2... are the conductivities of layers of thickness x1,...... The coefficient hs is also known as the convection heat transfer coefficient and values for it will be discussed in detail under the heading of convection. It is useful at this point, however, to appreciate the magnitude of hs under various common conditions and these are shown in the table (next slide). Approximate Range of Surface Heat Transfer Coefficients Example: Heat transfer in jacketed pan ✓ Sugar solution is being heated in a jacketed pan made from stainless steel; 1.6 mm thick. Heat is supplied by condensing steam at 200 kPa gauge in the jacket. The surface transfer coefficients are, for condensing steam and for the sugar solution, 12,000 and 3000 J m-2 s-1 °C-1 respectively, and the thermal conductivity of stainless steel is 21 J m-1 s-1 °C-1. Calculate the quantity of steam being condensed per minute if the transfer surface is 1.4 m2 and the temperature of the sugar solution is 83°C. From steam tables, the saturation temperature of steam at 200 kPa gauge (300 kPa Absolute) = 134°C, and the latent heat = 2164 kJ kg-1. Solution: Heat transfer in jacketed pan : q = AT[(1/hs1) + x1/k1 + x2/k2 +.. + (1/hs2)] = UAT Solution: Heat transfer in jacketed pan Modes of Heat Transfer ✓ Radiation is the transfer of heat energy by electromagnetic waves, which transfer heat from one body to another, in the same way as electromagnetic light waves transfer light energy. ✓ An example of radiant heat transfer is when a foodstuff is passed below a bank of electric resistance heaters that are red-hot. Modes of Heat Transfer: Radiation ✓ The basic formula for radiant-heat transfer is the Stefan- Boltzmann Law q = A T 4 where T is the absolute temperature (measured from the absolute zero of temperature at -273°C, and indicated in Bold type) in degrees Kelvin (K) in the SI system, and  (sigma) is the Stefan-Boltzmann constant = 5.73 x 10-8 J m-2 s-1K-4 The absolute temperatures are calculated by the formula K = (°C + 273).  Modes of Heat Transfer: Radiation ✓ This law gives the radiation emitted by a perfect radiator (a black body as this is called though it could be a red-hot wire in actuality). A black body gives the maximum amount of emitted radiation possible at its particular temperature. Real surfaces at a temperature T do not emit as much energy, but it has been found that many emit a constant fraction of it. For these real bodies, including foods and equipment surfaces, that emit a constant fraction of the radiation from a black body, the equation can be rewritten: n q = A T 4 where  (epsilon) is called the emissivity of the particular body and is a number between 0 and 1. Bodies obeying this equation are called grey bodies. Modes of Heat Transfer: Radiation ✓ Again, grey bodies absorb a fraction of the quantity that a black body would absorb, corresponding this time to their absorptivity  (alpha). For grey bodies it can be shown that  = . The fraction of the incident radiation that is not absorbed is reflected, and thus, there is a further term used, the reflectivity, which is equal to (1 –  ) Modes of Heat Transfer: Radiation ✓ Emissivity is a measure of how efficiently a surface emits thermal radiation compared to a perfect emitter (called a blackbody), which has an emissivity of 1.0. ✓ A surface with high emissivity (close to 1) emits more heat through radiation. ✓ A surface with low emissivity (close to 0) emits very little radiation and reflects most of the heat instead.  Modes of Heat Transfer: Radiation ✓ Emissivities vary with the temperature T and with the wavelength of the radiation emitted. For many purposes, it is sufficient to assume that for: *dull black surfaces (lamp-black or burnt toast, for example), emissivity is approximately 1; *surfaces such as paper/painted metal/wood and including most foods, emissivities are about 0.9; *rough unpolished metal surfaces, emissivities vary from 0.7 to 0.25; *polished metal surfaces, emissivities are about or below 0.05. ✓ These values apply at the low and moderate temperatures which are those encountered in food processing.  Radiation between two bodies ✓ The radiant energy transferred between two surfaces depends upon their temperatures, the geometric arrangement, and their emissivities. ✓ For two parallel surfaces, facing each other and neglecting edge effects, each must intercept the total energy emitted by the other, either absorbing or reflecting it. In this case, the net heat transferred from the hotter to the cooler surface is given by: q = AC(T14- T24 ) where 1/C = 1/1 + 1/2 - 1, 1 is the emissivity of the surface at temperature T1 and 2 is the emissivity of the surface at temperature T2. Radiation to a small body from its surroundings ✓ In the case of a relatively small body in surroundings that are at a uniform temperature, the net heat exchange is given by the equation: q = A(T14- T24 ) where  is the emissivity of the body, T1 is the absolute temperature of the body and T2 is the absolute temperature of the surroundings. Radiation to a small body from its surroundings ✓ In order to be able to compare the various forms of heat transfer, it is necessary to see whether an equation can be written for radiant heat transfer similar to the general heat transfer eqn. This means that for radiant heat transfer: q = hrA(T1 - T2) = hrA T where hr is the radiation heat-transfer coefficient, T1 is the temperature of the body and T2 is the temperature of the surroundings. (The T would normally be the absolute temperature for the radiation, but the absolute temperature difference is equal to the Celsius temperature difference, because 273 is added and subtracted and so T1 - T2 = T Problems to Solve: 1. Calculate the rate of heat transfer through a 3 x 4 m concrete wall. One face of the 0.2-m thick wall is at 22°C and the other face is at 35 °C. The thermal conductivity of the concrete is 1.1 Wm-1 K-1. 2. A tubular heat exchanger will be used to heat tomato paste from 60 °C to 105 °C, prior to holding, cooling, and aseptic filling into drums. The exchanger consists of two concentric tubes. Tomato paste is pumped through the inner tube. Steam condenses at 110 °C in the annular space. Calculate the required heat transfer area. Problem to Solve: Heat Conduction Part 1: In a bakery oven, combustion gases heat one side of a 2.5-cm steel plate at 300°C and the temperature in the oven is 285°C. Assuming steady state conditions and a thermal conductivity for steel of 17 W m-2 °C-1, calculate the rate of heat transfer per m-2 through the plate. Part 2: The internal surface of the oven is at 285°C and air enters the oven at 18°C. Calculate the surface heat transfer coefficient per m-2, using the calculated rate of heat transfer. Modes of Heat Transfer ✓ Convection is the transfer of heat by the movement of groups of molecules in a fluid. The groups of molecules may be moved by either density changes or by forced motion of the fluid. ✓ An example of convection heating is cooking in a jacketed pan: without a stirrer, density changes cause heat transfer by natural convection; with a stirrer, the convection is forced. Heat convection in a jacketed pan www.taizyfoodmachine.com chemico-world.blogspot.com Types of Convection Heat Transfer: Natural and Forced ✓ In natural convection, heat transfer is driven by density differences caused by temperature changes. When a fluid is heated, its temperature increases, which causes it to expand and become less dense. ✓ The less dense, warmer fluid rises because it is buoyant, and cooler, denser fluid sinks to take its place. This creates a continuous flow of fluid, allowing heat to transfer through the fluid.  Natural Convection ✓ It has been found that natural convection rates depend upon the physical constants of the fluid, density , viscosity , thermal conductivity k, specific heat at constant pressure cp and coefficient of thermal expansion  (beta) which for gases = l/T by Charles' Law. Other factors that also affect convection-heat transfer are, some linear dimension of the system, diameter D or length L, a temperature difference term, T, and the gravitational acceleration g since it is density differences acted upon by gravity that create circulation. Heat transfer rates are expressed in terms of a convection heat transfer coefficient hc, which is part of the general surface coefficient hs.  Natural Convection ✓ Experimentally, if has been shown that convection heat transfer can be described in terms of these factors grouped in dimensionless numbers which are known by the names of eminent workers in this field: Nusselt number (Nu) = (hcD/k) Prandtl number (Pr) = (cp/k) Grashof number (Gr) = (D32g  T/2) and in some cases, a length ratio (L/D). Natural Convection ✓ Nusselt number (Nu): This combines the convective heat transfer coefficient and the thermal conductivity of the fluid, simplifying the complex interaction between the fluid and the surface. ✓ Grashof number (Gr): This combines the temperature difference, fluid properties, and characteristic length of the object to represent the buoyancy forces driving convection, relative to the viscous forces that resist the motion. ✓ Prandtl number (Pr): This relates the fluid’s momentum diffusivity to its thermal diffusivity, helping to understand how heat moves within the fluid.  Natural Convection Equation ✓ If we assume that these ratios can be related by a simple power function, we can then write the most general equation for natural convection: (Nu) = K(Pr)k(Gr)m(L/D)n Types of Convection Heat Transfer ✓ In forced convection, external forces (such as a pump or fan) move the fluid. While the external force controls the fluid movement, density still plays a role in the thermal properties of the fluid. ✓ If the fluid is heated and becomes less dense, its flow characteristics change, which may affect the efficiency of heat transfer. Convection Heat Transfer General Equation ✓ For convection, the rate of heat transfer can be described using Newton's Law of Cooling: Q=hA(Ts​−T∞​) where: Q = rate of heat transfer (W, watts) h = convective heat transfer coefficient (W/m²·K) A = surface area through which heat is transferred (m²) Ts​ = temperature of the surface (K or °C) T∞​ = temperature of the surrounding fluid (K or °C) Heat Transfer Applications 1. Heat Exchangers - heat energy is transferred from one body or fluid stream to another ✓ Continuous flow heat exchangers www.rotron.com - One of the fluids is usually passed through pipes or tubes, and the other fluid stream is passed round or across these Heat Transfer Applications Heat exchanger heat transfer: q = UA Tm where: Tm where = (T1 - T2) / ln (T1/ T2) Tm is called the log mean temperature difference Total heat transfer per unit: q = cpG(T1 –T2) G is the rate of mass flow of the fluid that is changing temperature (kg s-1) Sample Problem: Cooling of Milk in a pipe heat exchanger ✓ Milk is flowing into a pipe cooler and passes through a tube of 2.5 cm internal diameter at a rate of 0.4 kg s-1. Its initial temperature is 49°C and it is wished to cool it to 18°C using a stirred bath of constant 10°C water round the pipe. What length of pipe would be required? Assume an overall coefficient of heat transfer from the bath to the milk of 900 J m-2 s-1 °C-1, and that the specific heat of milk is 3890 J kg-1 °C-1. Solution: Heat Transfer Applications ✓ Jacketed Pans - In a jacketed pan, the liquid to be heated is contained in a vessel, which may also be provided with an agitator to keep the liquid on the move across the heat- transfer surface. The source of heat is commonly steam condensing in the vessel jacket. Sample Problem: ✓ Estimate the steam requirement as you start to heat 50 kg of pea soup in a jacketed pan, if the initial temperature of the soup is 18°C and the steam used is at 100 kPa gauge. The pan has a heating surface of 1 m2 and the overall heat transfer coefficient is assumed to be 300 J m-2 s-1 °C-1. Solution: ✓ From steam tables, saturation temperature of steam at 100 kPa gauge = 120°C and latent heat =  = 2202 kJ kg-1. Heat Transfer Applications 2. Thermal Processing ✓ Thermal processing implies the controlled use of heat to increase, or reduce depending on circumstances, the rates of reactions in foods. ✓ A common example is the retorting of canned foods to effect sterilization. The object of sterilization is to destroy all microorganisms, that is, bacteria, yeasts and moulds, in the food material to prevent decomposition of the food, which makes it unattractive or inedible. ✓ Pasteurization is a heat treatment applied to foods, which is less drastic than sterilization, but which is sufficient to inactivate particular disease-producing organisms of importance in a specific foodstuff.  Heat Transfer Applications 3. Refrigeration, Chilling and Freezing ✓ Chilling of foods is a process by which their temperature is reduced to the desired holding temperature just above the freezing point of food, usually in the region of -2 to 2°C. The rates of chilling are governed by the laws of heat transfer. ✓ In the freezing process, added to chilling is the removal of the latent heat of freezing. This latent heat has to be removed from any water that is present. Since the latent heat of freezing of water is 335 kJ kg-1, this represents the most substantial thermal quantity entering into the process. The unit commonly used to measure refrigerating effect is the ton of refrigeration = 3.52 kW. Sample Problem for refrigeration: ✓ It is wished to freeze 15 tonnes of fish per day from an initial temperature of 10°C to a final temperature of -8°C using a stream of cold air. Estimate the maximum capacity of the refrigeration plant required, if it is assumed that the maximum rate of heat extraction from the product is twice the average rate. If the heat-transfer coefficient from the air to the evaporator coils, which form the heat exchanger between the air and the boiling refrigerant, is 22 J m-2 s-1 °C-1, calculate the surface area of evaporator coil required if the logarithmic mean temperature drop across the coil is 12°C. Solution: The specific heat of fish is 3.18 kJ kg-1 °C-1 above freezing and 1.67 kJ kg- 1 °C-1 below freezing, and the latent heat is 276 kJ/kg-1. THANK YOU!