Heat Transfer PDF

Heat Transfer Prof. Sabah Mounir November, 7, 2024 Heliopolis University  Outlines  Modes of heat-transfer  Fourier’s law of heat-conduction  Thermal Resistance  Solving heat-transfer problems Modes of Heat-Transfer Basicaly, there are 3 modes...

Heat Transfer Prof. Sabah Mounir November, 7, 2024 Heliopolis University  Outlines  Modes of heat-transfer  Fourier’s law of heat-conduction  Thermal Resistance  Solving heat-transfer problems Modes of Heat-Transfer Basicaly, there are 3 modes of heat-transfer EX: if I give a book to the first student and ask him to transfer the book to the last student In conductive heat-transfer: the first student should pass the book to the second without moving from his place Next: the second one will transfer it to the third again without leaving the place…… and so on till the last student  Similarly in conduction heat-transfer: the molecule transfers heat to another molecule without leaving the place 5 4 3 2 1 Modes of Heat-Transfer EX: I take the same example  another method for transferring the book is that  the first student leaves his place to go directly to the last student and then transfers the book without passing through the other students  Similarly in convective heat-transfer: heat is transferred from one molecule to another molecule and the molecule has to leave its place  and then transfers the heat to the molecule  Heat is transferred by convection currents 5 4 3 2 1 Modes of Heat-Transfer EX: another method for transferring the book is  the student will remain in his place and he is just going to throw  without again passing through the other students  So  in radiative heat-transfer : NO medium is required to transfer heat Modes of Heat-Transfer Basicaly, there are 3 modes of heat-transfer 1- Conduction: mostly in case of solids 2- Convection: between solid and fluid (gas or liquid) or between 2 fluids at least one medium should be a fluid 3-Radiation: heart is transferred in form of electromagnetic waves and  does NOT require any medium e.g. the heat transfers from the sun to the earth  as a result of the movement of the electromagnetic waves  Thermal Properties of Foods 1- Specific heat Specific heat (Cp) is  the quantity of heat that is gained or lost by a unit mass of product to accomplish a unit change in temperature, without a change in state Q is heat gained or lost (kJ), m is mass (kg), Δ T is temperature change in the material (°C), and cp is specific heat (kJ/kg °C).  Specific heat is an essential part of the thermal analysis of food processing or of the equipment used in heating or cooling of foods.  For food materials, this property is a function of the various components that constitute a food  e.g. its moisture content, temperature, & pressure.  Thermal Properties of Foods 1- Specific heat  The specific heat of a food increases as the product moisture content increases  In most food processing applications, we use specific heat at constant pressure, since pressure is generally kept constant except in high-pressure processing  For processes where a change of state takes place, such as freezing or thawing, an apparent specific heat is used.  Apparent specific heat incorporates the heat involved in the change of state in addition to the sensible heat. Latent heat is related to changes in phase between liquids, gases, and solids. Sensible heat is related to changes in temperature of a gas or object with no change in phase.  Thermal Properties of Foods 2- Thermal Conductivity  The thermal conductivity of a food is an important property used in calculations involving rate of heat transfer.  In quantitative terms, this property gives the amount of heat that will be conducted per unit time through a unit thickness of the material if a unit temperature gradient exists across that thickness The unit of thermal conductivity   Most high-moisture foods have thermal conductivity values closer to that of water (0.598 W/m·K at 20°C). On the other hand, the thermal conductivity of dried, porous foods is influenced by the presence of air with its low value.  Thermal Properties of Foods 3- Thermal Diffusivity  Thermal diffusivity, a ratio involving thermal conductivity, density, & specific heat, & is given as,  The unit of thermal diffusivity   Fourier’s law of heat conduction (steady heat conduction) Temperature remains constant  does not change over time  but change from one location to other  It states that  the rate of heat-transfer is directly proportional to the area normal to the direction of heat-flow and temperature gradient T1 > T2 Qα A Q α ΔT T1 Plane-wall (A) Q ΔT = (T1 - T2) Thermal T2 conductivity (K) Temperature gradient = ΔT/Δx (W/m K) Q  J/s  Watt Where, K : is the thermal conductivity of material - ve sign indicates the drop in temperature along the direction of flow Fourier’s law of heat conduction EX. 1: The inner and outer surface of a 5 m and 6 m brick wall of thickness 30 cm and the thermal conductivity 0.69 W/m ̊C, are maintained at 20 ̊C and 5 ̊C. What is the rate of heat transfer through the wall? EX. 2: A house roof with a length of 8 m, width of 6 m, and thickness of 0.25 m. The thermal conductivity of the roof is 0.8 W/m ̊C. The temperature of the inner and outer surface is 15 and 4̊C, respectively. What is the rate of heat transfer through the roof ? Thermal Resistance Concept Current (I) Electrical Resistance Potential (E) (R) Similarly (1) From Equation   (2) Thermal Resistance in parallel  multi-layered wall How to draw a thermal circuit for multilayered wall for conduction heat-transfer T1 > T2 Q R1 R2 R3 Q T1 T2 T1 T2 Q K1 K2 K3 Q Q Q From Fourier’s law of heat – conduction L1 L2 L3 𝑇1 − 𝑇2 𝑄= A1 = A2 = A3 = A 𝐿1 𝐿2 𝐿3 + + 𝐾1 𝐴1 𝐾2 𝐴2 𝐾3 𝐴3 𝑇1 − 𝑇2  𝑇1 − 𝑇2 = 𝑄= 𝑅1 + 𝑅2 + 𝑅3 σ 𝑅𝑡 Thermal Resistance in parallel  composite wall EX. 1: A house window with a height of 0.8 m, width of 1.5 m, composed of two layers of glass with thickness of 4 m, which separated by a layer of air with thickness of 10 m. The thermal conductivity of glass and air is roof is 0.78 and 0.026 W/m ̊C. The temperature of the inner and outer surface is 15 and - 8 ̊C, respectively. What is the rate of heat transfer through the window?  Conductive Heat Transfer through aTubular Pipe Fourier’s law in tubular pipe Consider a long, hollow cylinder of inner radius (ri), outer radius (ro), and length (L), as shown in this figure. Let the inside wall temperature be (Ti) and the outside wall temperature be (To). We want to calculate the rate of heat transfer along the radial direction in this pipe. Assume thermal conductivity of the metal remains constant with temperature Qqr A=2πrL ro Ti L ri r To ri ro  Conductive Heat Transfer through aTubular Pipe Fourier’s law in tubular pipe Thermal Resistance Concept Qqr ro Ti L ri r To ri ro  Ti To Rt  Conductive Heat Transfer through aTubular Pipe Ex: A 2 cm thick steel pipe (thermal conductivity 43 W/m °C with 6 cm inside diameter is being used to convey steam from a boiler to process equipment for a distance of 40 m. The inside pipe surface temperature is 115 °C, and the outside pipe surface temperature is 90 °C. Calculate the total heat loss to the surroundings under steady-state conditions. 90 °C Given Thickness of pipe = 2 cm = 0.02 m Inside diameter = 6 cm = 0.06 m 115 °C Thermal conductivity k = 43 W/(m °C) Length L=40 m 40m Inside temperature Ti = 115°C Outside temperature To= 90°C Approach 115C 90C 0.03m We will determine the thermal resistance 0.05m In the cross-section of the pipe and then 6cm use it to calculate the rate of heat transfer 10cm Conductive Heat Transfer through aTubular Pipe 90 °C 115 °C = 4.727×10-5 °C/W 40m 90C Q =115 - 90 = 528,903 W 115C 0.03m 4.727×10-5 0.05m 6cm 10cm  Conductive Heat Transfer through aTubular Pipe Ex: A 2 cm thick steel pipe (thermal conductivity 43 W/m °C with 6 cm inside diameter is being used to convey steam from a boiler to process equipment for a distance of 40 m. The inside pipe surface temperature is 115 °C, and the outside pipe surface temperature is 90 °C. Calculate the total heat loss to the surroundings under steady-state conditions. 90 °C Given Thickness of pipe = 2 cm = 0.02 m Inside diameter = 6 cm = 0.06 m 115 °C Thermal conductivity k = 43 W/(m °C) Length L= 40 m 40m Inside temperature Ti = 115 °C Outside temperature To= 90 °C Approach 115C 90C 0.03m We will determine the thermal resistance 0.05m In the cross-section of the pipe and then 6cm use it to calculate the rate of heat transfer 10cm Conductive Heat Transfer through aTubular Pipe 90 °C 115 °C = 4.727×10-5 °C/W 40m Q =115 —90 = 528,903W 90C 115C 0.03m 4.727×10-5 0.05m 6cm 10cm  Conductive Heat Transfer through a multilayered Cylindrical Tube r2 T3 B T1 T3 T2 A kB rr11 kA T1 B A Q r1 r3 r2 r3 r2 This figure shows a composite cylindrical tube made of two layers of materials  A and B. An example is a steel pipe covered with a layer of insulating material. The rate of heat transfer in this composite tube can be calculated as follows   Conductive Heat Transfer through a Composite Cylindrical Tube (in Series) Ex 1: A stainless-steel pipe (thermal conductivity 17 W/m °C is being used to convey heated oil. The inside surface temperature is 130 °C. The pipe is 2 cm thick with an inside diameter of 8 cm. The pipe is insulated with 0.04 m thick insulation (thermal conductivity 0.035 W/m °C. The outer insulation temperature is 25°C. Calculate the temperature of the interface between steel and insulation, assume steady-state conditions Given Thickness of pipe 2 cm = 0.02 m Inside diameter 8 cm = 0.08 m k steel = 17 W/m °C Thickness of insulation = 0.04 m k insulation = 0.035 W/m °C Inside pipe surface temperature = 130 °C Outside insulation surface temperature = 25 °C Pipe length 1 m (assumed) Approach  We will first calculate the two thermal resistances, in the pipe and the insulation.  Then we will obtain the rate of heat transfer through the composite layer.  Finally, we will use the thermal resistance of the pipe alone to determine the temperature at the interface between the pipe and insulation. Solution 130 °C 25 °C 25C 130C 6cm 10cm r ri 1. Thermal resistance in the pipe layer 2- The thermal resistance in the insulation layer 3- Using the equation 4- Using the equation Ex: 2 A stainless-steel pipe (thermal conductivity 15 W/m K is being used to transport heated oil at 125°C. The inside temperature of the pipe is 120°C. The pipe has an inside diameter of 5 cm and is 1 cm thick. Insulation is necessary to keep the heat loss from the oil below 25 W/m length of the pipe. Due to space limitations, only 5 cm thick insulation can be provided. The outside surface temperature of the insulation must be above 20°C (the dew point temperature of surrounding air) to avoid condensation of water on the surface of insulation. Calculate the thermal conductivity of insulation that will result in minimum heat loss while avoiding water condensation on its surface.  Given Thermal conductivity of steel = 15 W/m K Inside pipe surface temperature = 120 °C Inside diameter = 0.05 m Pipe thickness = 0.01 m Heat loss permitted in 1 m length of pipe = 25 W Insulation thickness 0.05 m Outside surface temperature > 20°C = 21°C (assumed) 20C 120C 3.5cm 8.5 3.5cm r ri 8.5cm 120 °C 20 °C