Heat Transfer Main ref: Chemical Engineering, Volume 1: Fluid Flow, Heat Transfer and Mass Transfer, Coulson and Richardson, 6th ed, Chapter 9 Chapter 9 Thermodynamics and Heat Transfer • Heat: The form of energy that can be transferred from one system to another as a result of temperature difference. • Thermodynamics is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. • Heat Transfer deals with the determination of the rates of such energy transfers as well as variation of temperature. • The transfer of energy as heat is always from the highertemperature medium to the lower-temperature one. • Heat transfer stops when the two mediums reach the same temperature. • Heat can be transferred in three different modes: conduction, convection, radiation 3 Heat Transfer in Chemical Processes • In most chemical processes, heat is given out or absorbed • Fluids must often be heated or cooled, e.g. distillation units, reaction vessels • Important that heat is transferred at suitable rate • It can also be important to minimize heat loss e.g. from a pipe • Control of flow of heat important in chemical engineering Conduction, Convection and Thermal Radiation Conduction refers to the transport of energy in a medium due to a temperature gradient. Solid: Transfer of vibrational energy between molecules. Liquid: Also transfer of kinetic energy between molecules. Metals: Also via movement of free electrons (hence metals have high thermal conductivity). Convection refers to heat transfer between a surface and a fluid (at rest or in motion), or between two fluids. This process involves mixing of fluid elements – not governed by temperature difference alone. Two types: Natural convection: Due to density difference, e.g. when a pool is heated from below Forced convection: Eddy movement in a fluid, e.g. when a pipe is heated from the outside Thermal Radiation All materials radiate thermal energy in the form of electromagnetic waves. When radiation falls on a second body, it is reflected, transmitted or absorbed – if absorbed, the temperature (heat) of the second body increases. Individual and Overall Heat Transfer Coefficients Consider heat transfer between flames heating a pipe and the liquid inside: Radiation from flames, conduction through the pipe wall, and convection to/through the liquid. Q = UADT Q = heat transferred per unit time (J/s) A = area available for flow of heat (m2) DT = temperature difference between flame and liquid in pipe (K) U = overall heat transfer coefficient (W/(m2 K)) UUdepends (heaton: transfer coeff) is the constant of between the “rate of heat •proportionality Mechanism of heat transfer •transfer Fluid dynamics (Q)”of fluid and “the driving force for the heat • Properties of wall of pipe (thermal conductivity) and its thickness (DT) multiplied byvary thewith area”. •transfer U not necessarily a constant – may temperature and DT (eqn above not necessarily a linear relationship) Heat Transfer by Conduction Consider heat transfer through a wall of thickness x: Rate of heat flow Q over the area A and a small distance dx: The negative sign indicates temperature gradient is in the opposite direction of the flow of heat. k = thermal conductivity of the material (W/(m K)) If wall thickness is x, and boundary temperatures are T1 and T2: Overall heat transfer coefficient (U) is here equal to k/x Thermal Conductivities (room temperature) Think of it as transfer of kinetic/vibrational energy between adjacent atoms/molecules – if they are far away from one another, as in gases, very low thermal conductivity. The exact opposite case is a metal, where atoms are organized close to one another in the solid state. • • • • Metals: very high Non-metallic solids: lower Non-metallic liquids: even lower Gases: very low Thermal Diffusivity How fast does heat diffuse through a material? cp Specific heat, J/kg · °C: Heat capacity per unit mass c p Heat capacity (storage), J/m3·°C: Heat capacity per unit volume Thermal diffusivity, m2/s: Represents how fast heat diffuses through a material A material that has a high thermal conductivity or a low heat capacity will obviously have a large thermal diffusivity. The larger the thermal diffusivity, the faster the propagation of heat through the medium. A small value of thermal diffusivity means that heat is mostly absorbed by the material and a small amount of heat is conducted further. Heat Transfer by Convection Heat transfer by convection is a result of movement of fluid on a macroscopic scale in the form of eddies and circulating currents. Natural Convection: Currents arise from the heating process itself. Consider heating of a beaker containing a liquid using a heat source under the beaker. The density of liquid near the bottom decreases as the temperature increases, and it thus rises, being replaced by colder liquid from above. Forced Convection: Currents are provided by an agitator or turbulent flow in a pipe. If convective heat transfer is taking place from a heated surface to a fluid, there are no circulating currents in the immediate vicinity of the surface – a film of fluid (free of turbulence) covers the surface. Heat transfer by thermal conduction occurs through this film. More agitation Thinner film Higher rate of heat transfer. If resistance mainly lies in this film: The film thickness is typically unknown, and the eqn is written in terms of h, the heat transfer coefficient of the film: EXAMPLE 1 A warm object of initial temperature 60 C is cooling in air of temperature 22 C. The temperature of the object is measured as a function of time. Given the data below, what is the value of the heat transfer coefficient? Air temperature Newton’s Law of Cooling: The rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings provided the temperature difference is small and the nature of radiating surface remains same. Consider a warm object cooling in air: y (t ) Tt Tair dy d Tt Tair hTt Tair hy dt dt y y0 e ht Tt Tair T0 Tair e DT e ht DT0 DT ht ln DT0 ht Because the rate of change in temperature difference, i.e. differential with respect to time, is proportional to the temperature difference as stated in Newton’s Law of Cooling (constant of proportionality = h) First order kinetics! Same type of eqn as in reaction kinetics. A plot of ln(DT/DT0) vs t should give a straight line with slope –h. EXAMPLE 1 A warm object of initial temperature 60 C is cooling in air of temperature 22 C. The temperature is measured as a function of time. What is the value of the heat transfer coefficient? Approximately linear – follows Newton’s Law of Cooling. h = 0.0607 Jmin-1K-1 = 1.01 x 10-3 W K-1 Note: Here h also incorporates the area term. EXAMPLE 2 The Cost of Heat Loss through a Roof The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete with thermal conductivity k = 0.8 W/m °C . The temperatures of the inner and the outer surfaces of the roof one night are measured to be 25°C and 0°C, respectively, for a period of 10 hours. (a) What is the rate of heat loss through the roof? (b) What is the cost of that heat loss if the cost of electricity is $0.2/kWh. The kilowatt hour (symbol kWh) is a derived unit of energy equal to 3.6 megajoules. If the energy is being transmitted at a constant rate over a period of time, the total energy in kilowatt hours is the power in kilowatts multiplied by the time in hours. The kilowatt hour is commonly used as a billing unit for energy delivered to consumers by electric utilities. SOLUTION (a) Heat transfer through the roof is by conduction and the area of the roof is: A = 6 m × 8 m = 48 m2 The steady rate of heat transfer through the roof is determined to be Q = kA(T1-T2)/x = (0.8)(48)(25-0)/0.25 = 3840 W = 3.84 kW (b) The amount of heat lost through the roof during 10 h and its cost: Q x Dt =(3.84 kW)(10 h) = 38.4 kWh Cost of that heat loss (i.e. per 10 h) = = (Amount of energy) x (Unit cost of energy) = (38.4 kWh)($0.2/kWh) = $7.68 Cost/month = (cost/day)×(30 days/month)= $7.68×30 = $230.4 EXAMPLE 3 Estimate the heat loss per square meter of surface through a brick wall 0.5 m thick when the inner surface is at 400 K and the outer surface is at 300 K. The thermal conductivity of the brick may be taken as 0.7 W/mK. SOLUTION EXAMPLE 3 Estimate the heat loss per square meter of surface through a brick wall 0.5 m thick when the inner surface is at 400 K and the outside surface is at 300 K. The thermal conductivity of the brick may be taken as 0.7 W/mK. Note that here the area of the wall is set as 1, thus giving us the answer per square meter. Thermal resistances (x/k) in series Chemical Engineering, Volume 1: Fluid Flow, Heat Transfer and Mass Transfer, Coulson and Richardson, 6th ed, p. 390 EXAMPLE 4 EXAMPLE 4 Temperature drop is proportional to thermal resistance Material Balances Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed, Chapter 2 Chemical Plant Extremely complex !! • In industry, chemical processes typically involve several steps • An industrial process is normally treated as a series of unit operations – one unit operation refers to a basic step involving a chemical transformation (chemical reaction) or physical transformation (e.g. filtration, crystallization) • Materials are transferred between the different steps (unit operations) • Essential to know amount/composition/condition of material entering/leaving process unit • We need means of representing complex processes (qualitatively and quantitatively) General material balance equation qin and qout denote flow rates into and out of the process unit. Methane flows in and out of the unit. If qin ≠ qout Possible explanations: 1. Methane is leaking 2. Methane is consumed or generated (chemical reaction) 3. Methane is accumulating in the unit 4. The measurement(s) of qin and/or qout are wrong Input – Output + Generation – Consumption = Accumulation This balance can refer to total mass of material or to any molecular or atomic species involved Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed; p. 85 Material balances on reactive systems • Chemical reaction(s) occur, i.e. compounds are consumed/generated • Must include generation and consumption terms in balance equation • Stoichiometric equation imposes constraints on the relative amounts of compounds (e.g. if A → B, cannot start with 1 mol of A and end up with 2 mol of B) If no accumulation: Input + Generation = Output + Consumption Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed; p. 118 Limiting reagents Consider the chemical reaction: 2SO2 + O2 → 2SO3 The reactants are in stoichiometric proportions if the molar ratio equals that of the balanced reaction equation, i.e. 2 mol SO2 react with 1 mol O2. If reactants are fed into a reactor in stoichiometric proportions, and if the reaction proceeds to completion, all reactants are consumed. But, consider you have 2 mol SO2 and only 0.8 mol O2 – in this case reaction will stop once all O2 is consumed, and 0.4 mol SO2 (= 20% of 2) remain unreacted. In this case, O2 is the limiting reagent. Chemical reactions take time to proceed – it is often not practical to design processes such that complete conversion of the limiting reactant is reached. Instead, the output stream contains both reactants and products, and the separated reactant is then recycled back into the reactor. Consider shape of conversion-time plot (slope decreases with time). Limiting reagents Consider the chemical reaction: 2A + 3B → 2C We conduct this reaction in batch, loading the reactor with 3.21 moles of A and 4.03 moles of B. The reaction is run to maximum conversion. 1. Which is the limiting reagent? 2. What is the number of moles of C generated? 3. How much more of the limiting reagent would have to be added in order for there to be no reactants remaining at full conversion? Limiting reagents 2A + 3B → 2C We conduct this reaction in batch, loading the reactor with 3.21 moles of A and 4.03 moles of B. The reaction is run to maximum conversion. 1. Which is the limiting reagent? For all of A to react, we need 3/2 molecules of B for each A. 3.21 moles x 3/2 = 4.815 moles We only have 4.03 moles of B B is the limiting reagent Limiting reagents Consider the chemical reaction: 2A + 3B → 2C We conduct this reaction in batch, loading the reactor with 3.21 moles of A and 4.03 moles of B. The reaction is run to maximum conversion. 2. What is the number of moles of C generated? • Given that B is the limiting reagent, the reaction will stop when all B is gone. • 1 mole of B gives 2/3 mole of C based on the stoichiometric equation. • The number of moles of C obtained will equal the initial number of moles of B multiplied by 2/3, i.e. 2/3 x 4.03 = 2.69 moles Limiting reagents Consider the chemical reaction: 2A + 3B → 2C We conduct this reaction in batch, loading the reactor with 3.21 moles of A and 4.03 moles of B. The reaction is run to maximum conversion. 3. How much more of the limiting reagent would have to be added in order for there to be no reactants remaining at full conversion? For all of A to react, we need 3/2 molecules of B for each A. 3.21 moles x 3/2 = 4.815 moles We only have 4.03 moles of B We need to add 4.815 – 4.03 = 0.785 moles of B Example of a material balance involving chemical reaction Production of acrylonitrile Acrylonitrile Reactor Mol per 1 mol air Composition per 1 mol total Information provided: • Feed contains 10% propylene, 12% ammonia, and 78% air (all mol%) • 30% conversion of the limiting reagent is achieved • 100 mol (total) feed as basis (this is what is fed to reactor) Questions: • Which is the limiting reagent? • By how much are the other two reagents present in excess? • Determine molar amounts of all product gases. Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed; p. 123 Reactor Composition per 1 mol Limiting reagent: (NH3/C3H6) = 1.2 (O2/C3H6) = 1.64 Stoichiometric ratio = 1 Stoichiometric ratio = 1.5 O2 being fed: 0.78 x 100 x 0.21 = 16.4 mol (mol fraction O2 in air = 0.21) C3H6 being fed: 0.100 x 100 = 10 mol Both NH3 and O2 are fed in more than stoichiometric ratio relative to C3H6 – this is the limiting reagent. Excess reagents: % excess NH3 = (12 – 10)/10 = 0.20 = 20% % excess O2 = (16.4 – 15)/15 = 0.093 = 9.3% Reactor Composition per 1 mol Molar amounts of product gases: Conversion of C3H6 = 30% 0.7 x 10 = 7 mol C3H6 From balanced reaction equation (3 mol of C3H6 have reacted) NH3: 12 – 3 = 9 mol O2: 16.4 – (3 x 1.5) = 11.9 mol C3H3N: 3 mol N2: 61.6 mol (not involved in reaction) H2O: 3 x 3 = 9 mol Input - Consumption Input - Consumption Generation Input = Output Generation The same principles apply to more complex systems The task is typically to determine unknown process parameters for a system based on the information given. The way forward is to set up equations based on what is known in the form of material balances, then solve these equations for the unknown(s). Note that to solve for X unknown parameters, you need at least X independent equations (material balances). The more complex case below illustrates the process for production of non-alcoholic beer. From CEIC1000 Lecture Notes Multiphase systems • Many industrial processes do not involve any chemical reactions. • Many process units are simply mixers that blend streams of different compositions. • Anther common type of process is separation, e.g. isolate A from mixture of A and B. • Separation processes typically include two phases: gas/liquid, solid/liquid, gas/solid, immiscible liquids. • Much easier to separate two different phases than two compounds in the same phase. Examples of separation processes in two-phase systems: • Distillation - difference in boiling point between two liquids • Crystallization - difference in melting point • Liquid extraction - contact a liquid mixture with a solvent that is miscible with one component but immiscible with the other, then separate the two phases Example of liquid extraction We wish to separate a mixture of benzene and methanol (miscible liquids). Water and methanol are miscible, but water and benzene are not. If water is added to a mixture of benzene and methanol, most of the methanol will end up in the aqueous phase. Once the mixture has settled (into two immiscible layers), separation is straightforward. Benzene Methanol Add water Benzene Water + Methanol