MECH3002/6002 Heat & Mass Transfer Lecture Notes PDF

SCHOOL OF ENGINEERING Faculty of Science and Engineering MECH3002 Heat and Mass Transfer MECH6002 Advanced Heat and Mass Transfer Applications Professor Yijiao Jiang What have been learned from Lecture 2 Understand thermal conductivity and thermal resistance Concept of thermal resistance...

SCHOOL OF ENGINEERING Faculty of Science and Engineering MECH3002 Heat and Mass Transfer MECH6002 Advanced Heat and Mass Transfer Applications Professor Yijiao Jiang What have been learned from Lecture 2 Understand thermal conductivity and thermal resistance Concept of thermal resistance networks and thermal circuits (Olm’s law) Specify appropriate form of the heat transfer equations through walls, cylinders, and spherical shell Thermal resistance: conduction, convection, radiation Outline: Lecture 3 Conduction with internal heat source: Plane or cylindrical wall (Chapter 2.7: Jack P. Holman, Heat Transfer, 10th Edition) Transient Heat Transfer (Unsteady state conduction) Overall heat transfer coefficient: Composite wall, hollow cylinder Thermal contact resistance Types of heat exchangers Parallel flow and counter flow Fouling factor in a heat exchanger Overall heat transfer coefficient in a heat exchanger Conduction with Internal Heat Source Heat may be generated internally nuclear reactors electrical conductors chemical reactors Conduction with Internal Heat Source One-dimensional system, where the temperature is a function of only one space coordinate – Plane wall with heat source – Cylindrical wall with heat source Plane Wall with Heat Source General conduction heat transfer equation with heat sources: Steady-state one-dimensional heat flow with heat sources: For the boundary conditions, The general solution is Since the temperature must be the same on each side of the wall, then C1=0 At the midplane x=0, 𝑇0 = 𝐶2 The temperature distribution is: One-dimensional conduction or with heat generation or The mid-plane temperature T0 is:  Cylinder with Internal Heat Source If an electric current flows through a wire, the heat generated internally will result in a temperature distribution between the central axis and the surface of the wire. Cylindrical coordinates: Rate of heat generated per unit volume he Steady-state one-dimensional heat flow with heat sources: For the boundary conditions, The heat generated equals heat lost at the surface, Cylinder with Internal Heat Source 𝑑𝑇 So that: −𝑘2𝜋𝑟𝐿 = 𝜋𝑟 2 𝐿𝑄𝐺 𝑑𝑟 𝑑𝑇 𝑄𝐺 𝑟 Hence: = − 𝑑𝑟 2𝑘 𝑄𝐺 𝑟 2 Integrating: T= − +𝐶 4𝑘 T = T0 when r = r0 the radius of wire and hence: 𝑟02 − 𝑟 2 T = 𝑇0 + 𝑄𝐺 4𝑘 or: 𝑄𝐺 𝑟02 𝑟2 T − 𝑇0 = (1 − 2 ) 4𝑘 𝑟0 Cylinder with Internal Heat Source This gives a parabolic distribution of temperature and the maximum temperature will occur at the axis of the wire: Where 𝑄𝐺 𝑟02 T − 𝑇0 = 4𝑘 The arithmetic mean temperature difference: 𝑄𝐺 𝑟02 (T − 𝑇0 )𝑎𝑣 = 8𝑘 Since 𝑄𝐺 𝜋𝑟0 2 is the rate of heat transfer release per unit length of the wire, then putting T1 as the temperature at the centre: 𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑟𝑒𝑙𝑒𝑎𝑠𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑇1 − 𝑇0 = 4π𝑘 Note that the temperature difference is independent of the diameter of the cylinder. Worked Example Solution The rate of heat transfer released per unit length: 0.25×106 𝑊 = 5 × 104 𝑊/𝑚 5𝑚 The temperature difference between the surface and the centre of the uranium element is: 𝑊 4 5 × 10 𝑚 𝑇1 − 𝑇0 = = 121 𝐾 4π𝑘  Transient Heat Transfer In practice the temperatures of heat transfer change with time: - Batch processes - Disturbances from other processes - Automatic controllers - Start up and shut down - Cooling water temperature is affected by weather. Such processes are called transient or unsteady state. Transient Heat Transfer Assume a lumped system - The temperature varies with time but not with position/space - The temperature of the medium changes uniformly. - May be valid if highly conductive, small or well mixed - May not be valid if highly exothermic, low conductivity, poor mixing…..  Energy Balance in a Lumped System An energy balance of the solid body for the time interval dt can be expressed as Note that 𝑚 = ρ𝑉 𝑎𝑛𝑑 𝑑𝑇 = 𝑑 𝑇 − 𝑇∞ since 𝑇∞ is constant, so we get 𝑑 𝑇 − 𝑇∞ ℎ𝐴𝑆 = 𝑑𝑡 𝑇 − 𝑇∞ ρ𝑉𝑐𝑝 Integrate from 𝑡 = 0, 𝑇 = 𝑇𝑖 and At any time t, 𝑇 = 𝑇 𝑡 , so we get 𝑇 𝑡 − 𝑇∞ ℎ𝐴𝑆 𝑙𝑛 =− 𝑡 𝑇𝑖 − 𝑇∞ ρ𝑉𝑐𝑝  Energy Balance in a Lumped System Taking the exponential of both sides and rearranging, give 𝑇 𝑡 − 𝑇∞ ℎ𝐴𝑆 = 𝑒 −𝒃𝑡 where 𝑏= (1/s) 𝑇𝑖 − 𝑇∞ ρ𝑉𝑐𝑝 The rate of convection heat transfer between the solid body and its environment at that time from Newton’s law of cooling, 𝑄ሶ 𝑡 = ℎ𝐴𝑠 𝑇 𝑡 − 𝑇∞ (W) The total amount of heat transfer between the solid body and its environment at that time, 𝑄 = 𝑚𝑐𝑝 𝑇 𝑡 − 𝑇𝑖 (kJ) The maximum heat transfer between the solid body and its environment is, 𝑄𝑚𝑎𝑥 = 𝑚𝑐𝑝 𝑇∞ − 𝑇𝑖 (kJ) Overall Heat-transfer Coefficient Heat flux (heat transfer rate per unit cross-sectional area) is proportional to the temperature difference. The proportionality “constant” is called a heat-transfer coefficient. q = U T Q = U A T where: q = heat flux (W m-2) Q = rate of heat transfer (J/s = W) A = interfacial area through which heat flows (m2) T = temperature difference (driving force) (K) U = overall heat-transfer coefficient (W m-2 K-1) Overall Heat-Transfer Coefficient Q Q Hence, the rate of heat transfer: Q Overall heat-transfer through a plane wall Q Since So the overall heat-transfer coefficient through a plane wall: Where: h1 and h2 are the convection heat transfer coefficient in W/m2·K. Example: Cooling a Tank A kettle contains 700 L of material which needs to be cooled from 55oC to 43oC. Data: Cp (specific heat capacity) = 3500 J/kg  = 1000 kg/m3 U (overall heat-transfer coefficient) = 150 W/(m2K) A (interfacial area) = 6 m2 Cooling water temperature 35 oC Ignore the thermal mass of the tank How long will the tank take to cool? Cooling a Tank dT Q = UAT = mC p dt UA(TJ − T ) = mC p dT dt t =t T =T f dT  UAdt = mC p t =0  T =Ti (TJ − T ) UAt = −mC p (ln (TJ − T f ) − ln (TJ − Ti )) − mC p   TJ − T f  t=  ln   UA   TJ − Ti   where: Q = rate of heat transfer (J/s = W) A = interfacial area through which heat flows (m2) T = temperature difference (driving force) (K) U = overall heat-transfer coefficient (W m-2 K-1) 𝑇𝑗 = the temperature of cooling water; 𝑇𝑖 = the initial temperature of the material; 𝑇𝑓 = the final temperature of the material; Cooling a Tank 𝑚𝐶𝑝 𝑇𝐽 − 𝑇𝑓 𝑡=− 𝑙𝑛 𝑈𝐴 𝑇𝐽 − 𝑇𝑖 ρ𝑉𝐶𝑝 𝑇𝐽 −𝑇𝑓 =− 𝑈𝐴 𝑙𝑛 𝑇𝐽 −𝑇𝑖 1000 𝑘𝑔·𝑚−3 ×700×10−3 𝑚3 ×3500𝐽·𝑘𝑔−1 35℃−43℃ =− 𝑙𝑛 150 𝑊·𝑚−2 ·𝐾−1 35℃−55℃ = 2494.3 s = 42 min Known parameters: Cp (specific heat capacity) = 3500 J/kg  = 1000 kg/m3 U (overall heat-transfer coefficient) = 150W/(m2K) A (interfacial area) = 6 m2 Cooling water temperature 35 oC R Value R value is a measure of thermal resistance used in the building and construction industry. R value is defined as a heat flow per unit area (℃·m2/W): Relation between the overall heat- transfer coefficient and R value: Relationships between the concepts - thermal resistance (R, W m-1 K-1) - the overall heat-transfer coefficient (U) - the R value (℃·m2/W) 𝟏 𝑼= 𝑹𝒕𝒐𝒕𝒂𝒍 𝑨 Example: Heat transfer through a composite wall Solution Q Hollow Cylinder with Convection Boundaries The overall heat transfer: Q The overall heat-transfer coefficient based on either the inside or the outside area of the tube: or Overall Heat Transfer Coefficients for Common Construction Systems Example: Overall Heat-Transfer Coefficient for a Tube Solution Q Q Thermal Contact Resistance Mechanical joining of the two materials According to the energy balance on the two materials, Q or Q Where - the quantity 1/hc is called the thermal contact resistance - hc is called the contact coefficient Temperature profile Thermal Contact Resistance Designating the contact area by Ac and the void area by Av, then the heat flow across the joint: Q Where Lg is the thickness of the void space kf is the thermal conductivity of the fluid which fills the void space Hence, the contact coefficient is Thermal Contact Resistance of Typical Surfaces Example: Contact Resistance Two 3.0-cm-diameter 304 stainless –steel bars with a thermal conductivity of 16.3 W/m·℃ and 10 cm long, have ground surfaces and are exposed to air with a surface roughness of about 1 μm with a thermal contact resistance of 1/hc = 5.28 𝑚2 ·℃/W × 104. If the surfaces are pressed together with a pressure of 50 atm and the two-bar combination is exposed to an overall temperature difference of 100 ℃, calculate the axial heat flow and temperature drop across the contact surface. Solution The contact area 𝐴 = 𝜋𝑟 2 = 𝜋 × 0.0152 = 0.000707 𝑚2 The overall heat flow is subject to three thermal resistances: one conduction resistance for each bar, and the contact resistance. The conduction resistance for each bar: ∆𝑥 0.1 𝑚 𝑅𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = 𝑘𝐴 = = 8.68 ℃/W 16.3 W/m·℃×0.000707 𝑚2 The contact resistance: 1 5.28 × 10−4 𝑚2℃/W 𝑅𝑐𝑜𝑛𝑡𝑎𝑐𝑡 = = = 0.747 ℃/W ℎ𝑐 𝐴 0.000707 𝑚2 The total thermal resistance is therefore: 𝑅𝑡𝑜𝑡 = 2 × 𝑅𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 + 𝑅𝑐𝑜𝑛𝑡𝑎𝑐𝑡 =18.11 ℃/W ∆𝑇 100 ℃ The axial heat flow is: Q= = = 5.52 W 𝑅𝑡𝑜𝑡 18.11℃/𝑊 The temperature drop across the contact is found by taking the ratio of the contact 𝑅 0.0747×100 ℃ resistance to the total thermal resistance: ∆𝑇𝑐 = 𝑅 𝑐 ∆𝑇 = 18.11℃/𝑊 = 4.12 ℃ 𝑡𝑜𝑡 Heat Exchanger What are heat exchangers? – Heat exchangers are devices that allow the exchange of heat between hot and cold fluids. – Often there is no direct contact between the fluids. What are heat exchangers used for? – Heating/cooling feed streams to reactors – Recovering waste heat – Others?  Heat Exchangers double pipe heat exchanger Double pipe heat exchanger schematics Double-pipe heat exchanger Shell and tube heat exchanger Plate heat exchanger the simplest heat exchangers Most frequent type of heat exchanger durable low cost for both design and maintenance in oil refineries and other large Reliable low efficiency chemical processes Not suitable for high high space occupied Match for high pressure applications pressure application There are many types of heat exchangers All have advantages and disadvantages Flow Direction in Double Pipe Heat Exchanger The flow of fluids in a heat exchanger may be: – Counter flow – Parallel flow – A mixture of the two In this course we will only be looking at counter flow and parallel flow Why is flow direction important? Consider the following situations. Water enters a heat exchanger as the cold stream at 20oC. 1. It is heated by a parallel flow of hot oil entering the exchanger at 110oC 2. It is heated by a counter flow of hot oil entering the exchanger at 110oC What might the temperature profiles look like? Parallel Flow in Double-pipe Heat Exchanger Hot Oil THI = 110oC Cold Water TCI = 20oC Both the hot and cold fluids enter the heat exchanger at the same end and move in the same direction Counter Flow in Double-pipe Heat Exchanger Hot Oil THI = 110oC Cold Water TCI = 20oC The hot and cold fluids enter the heat exchanger opposite ends and flow in opposite directions. Shell-and-Tube Heat Exchangers Plate and Frame Heat Exchangers The hot and cold fluids flow in alternate passages, and thus each cold fluid stream is surrounded by two hot fluid streams. Compact Heat Exchangers The ratio of the heat transfer surface area of a heat exchanger to its volume is called the area density β. A heat exchanger with β > 700 m2/m3 is classified as being compact. Gas-to-liquid compact heat exchanger for a residential air-conditioning system Achieve high heat transfer rates between two fluids in a small volume Applications with strict limitations on the weight and volume of heat exchangers Car radiator, residential air conditioning system Cross-flow in Compact Heat Exchangers Two fluids move perpendicular to each other Both fluids in a car radiator are unmixed In the mixed cross-flow, the fluid is free to move in the transverse direction Overall Heat Transfer Coefficient The total thermal resistance is: The rate of heat transfer between the two fluids is: Where U is the overall heat transfer coefficient is: Cancelling ∆T, When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, Fouling Factor Accumulation of deposits on heat transfer surfaces, which induces additional resistance to heat transfer The net effect of these accumulations on heat transfer is represented by a fouling factor, Rf 1 1 𝑅𝑓 = − 𝑈𝑑𝑖𝑟𝑡𝑦 𝑈𝑐𝑙𝑒𝑎𝑛 Considering the effects of fouling on both the inner and the outer surfaces of the tube, the overall heat transfer coefficient becomes:  Work Example Effect of Fouling on the Overall Heat Transfer Coefficient of a Heat Exchanger A double-pipe heat exchanger is constructed of a stainless steel (k = 15.1 W/m·K) inner tube of inner diameter Di = 1.5 cm and outer diameter D0 = 1.9 cm and an outer shell of inner diameter 3.2 cm. The convection heat transfer coefficient is given to be hi = 800 W/m2·K on the inner surface of the tube and h0 = 1200 W/m2·K on the outer surface. For a fouling factor of Rf,i =0.0004 m2·K/W on the inside surface and Rf,0 =0.0001 m2·K/W on the outside surface, determine a) the thermal resistance of the heat exchanger per unit length; b) the overall heat transfer coefficient, Ui and U0 based on the inner and outer surface areas of the tube, respectively. Solution a) The thermal resistance of the heat exchanger per unit length with fouling on both heat transfer surface is: Where Hence, b) the overall heat transfer coefficient, Ui and U0 based on the inner and outer surface areas of the tube, respectively What have been learned from Lecture 3 Conduction with internal heat source (plane or cylindrical) Transient heat transfer conduction (unsteady state conduction) Overall heat transfer coefficient (composite wall or hollow cylinder) Thermal contact resistance Types of heat exchangers Parallel flow and counter flow Fouling fact in a heat exchanger Overall heat transfer coefficient in a heat exchanger Next Lecture Define and apply the log-mean temperature difference When not to use the log-mean temperature difference Correction factors for arbitrary geometries Overall Heat Transfer Equations for Concentric Pipe Heat Exchangers Heat exchanger effectiveness/NTU Method Design of heat exchanger