Unit Operations & Heat Transfer
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UNIT OPERATIONS & HEAT TRANSFER CHE 315 By ADEWUYI Adewale Chemical Sciences Redeemer’s University This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License. HEAT TRANSFER Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. Heat transfer is classified into various mechanisms, such as: Thermal conduction Thermal convection Thermal radiation, and Transfer of energy by phase changes. Conduction So we can say: Another property that characterize the flow of heat which is related to thermal conductivity is the thermal diffusivity which is given as: α = k/ρ C. Where α is thermal diffusivity, ρ (kg/m3) is density and C is specific heat capacity (J/kgK) Conductance Writing U = k/ΔX where U is the conductance, in W/(m2 K). Fourier's law can also be stated as: ΔQ/Δt = UA (-ΔT) The reciprocal of conductance is resistance, R, given by: R = 1/U = Δx\k = A(-ΔT)/ ΔQ/Δt Convection Convection uses the motion of fluids to transfer heat. In a typical convective heat transfer, a hot surface heats the surrounding fluid, which is then carried away by fluid movement such as wind. The warm fluid is replaced by cooler fluid, which can draw more heat away from the surface. Since the heated fluid is constantly replaced by cooler fluid, the rate of heat transfer is enhanced. Natural convection(or free convection) refers to a case where the fluid movement is created by the warm fluid itself. The density of fluid decrease as it is heated; thus, hot fluids are lighter than cool fluids. Warm fluid surrounding a hot object rises, and is replaced by cooler fluid. The result is a circulation of air above the warm surface, as shown in Figure below: Forced convection uses external means of producing fluid movement. Forced convection is what makes a windy, winter day feel much colder than a calm day with same temperature. The heat loss from your body is increased due to the constant replenishment of cold air by the wind. Natural wind and fans are the two most common sources of forced convection. Convection coefficient, h, is the measure of how effectively a fluid transfers heat by convection. It is measured in W/m²K, and is determined by factors such as the fluid density, viscosity, and velocity. Wind blowing at 5 mph has a lower h than wind at the same temperature blowing at 30 mph. The basic relationship for heat transfer by convection is: A fluid flows over a plane surface 1 m by 1 m. The surface temperature is 50oC, the fluid temperature is 20oC and the convective heat transfer coefficient is 2000 W/m2oC. The convective heat transfer between the hotter surface and the colder air can be calculated as: q = [2000 W/(m2oC)] [(1 m) (1 m)] [(50 oC) - (20 oC)] = 60000 (W) = 60 (kW) For convection we use the convection heat transfer coefficient hc, W/(m2K). A different approach is to define h through the Nusselt number Nu, which is the ratio between the convective and the conductive heat transfer: Where: − Nu = Nusselt number − hc= convective heat transfer coefficient − k = thermal conductivity, W/mK − L = characteristic length, m The convection heat transfer coefficient is then defined as following: The Nusselt number depends on the geometrical shape of the heat sink and on the air flow. For natural convection on flat isothermal plate the formula of Na is given in table 1. Radiation Heat transfer through radiation takes place in form of electromagnetic waves mainly in the infrared region. Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. Radiation heat transfer can be described by reference to the 'black body'. The Black Body The black body is defined as a body that absorbs all radiation that falls on its surface. Actual black bodies don't exist in nature - though its characteristics are approximated by a hole in a box filled with highly absorptive material. The emission spectrum of such a black body was first fully described by Max Planck. A black body is a hypothetical body that completely absorbs all wavelengths of thermal radiation incident on it. Such bodies do not reflect light, and therefore appear black if their temperatures are low enough so as not to be self-luminous. All black bodies heated to a given temperature emit thermal radiation. The radiation energy per unit time from a black body is proportional to the fourth power of the absolute temperature and can be expressed with Stefan-Boltzmann Law as q = σ T4 A (1) where q = heat transfer per unit time (W) σ = 5.6703 10-8 (W/m2K4) - The Stefan-Boltzmann Constant T = absolute temperature in Kelvin (K) A = area of the emitting body (m2) The Stefan-Boltzmann Constant in Imperial Units -8 2 4 σ = 5.6703 10 (W/m K ) = 1.714 10-9 ( Btu/(h ft2 oR4) ) = 1.19 10-11 ( Btu/(h in2 oR4) ) Example - Heat Radiation from the surface of the Sun If the surface temperature of the sun is 5800 K and if we assume that the sun can be regarded as a black body the radiation energy per unit area can be expressed by modifying (1) to q / A = σ T4 = (5.6703 10-8 W/m2K4) (5800 K)4 = 6.42 107 (W/m2) Gray Bodies and Emissivity Coefficients For objects other than ideal black bodies ('gray bodies') the Stefan-Boltzmann Law can be expressed as q = ε σ T4 A (2) where ε = emissivity coefficient of the object (one - 1 - for a black body) For the gray body the incident radiation (also called irradiation) is partly reflected, absorbed or transmitted. The emissivity coefficient is in the range 0 < ε < 1, depending on the type of material and the temperature of the surface. oxidized Iron at 390 oF (199 oC) > ε = 0.64 polished Copper at 100 oF (38 oC) > ε = 0.03 emissivity coefficients for some common materials Net Radiation Loss Rate If a hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as 4 4 q = ε σ (Th - Tc ) Ac (3) where Th = hot body absolute temperature (K) Tc = cold surroundings absolute temperature (K) 2 Ac = area of the object (m ) Example problems I have a system in space that is generating a lot of waste heat which I need to get rid of by radiation into deep, cold space. The power I need to dump is 1.0×10³ W. My external radiator has a surface area of 1.0 m by 2.0 m, has useful emission from only one side, and has an emissivity of 0.99 (pretty good absorber). What is the equilibrium temperature of my radiator in °C? Solution: Since I am given the amount of radiation that I need to emit, I can determine Qemitted without having to worry about Qincoming. From the equation Qemitted = P/A = 1.0×10³ W/(1.0 m × 2.0 m) = 500 W/m² 4 Qemitted = P/A = ꜫsT so that -8 4 T = [Qemitted/ꜫs] = {500 W/m²/[0.99 × 5.67×10 W/(m²·K )]} = 307.2 K We were asked for °C, so convert from K to °C: T = 307.2 - 273.15 = 34°C Question 2 You work part time as a contractor building houses (to supplement your teaching income). Your client asks you what difference there is (if any) between using a light color roof and a dark color roof in the south where the summer is very hot (we all know that). The homeowner wants to use very dark shingles with an emissivity of 0.90. The roof has a 6.0"-thick fiberglass insulation and the attic temperature is always maintained at 80.0°F (26.7°C= 299.8 K). Thermal conduction through the roof to the attic and radiative loss are the main sources of heat transfer. Determine the equilibrium temperature in the day time for the shingles if the solar power is 1.0 kW/m². Solution: Energy absorption rate = 1.0 kW/m² × 0.90 = 9.0 kW/m². Heat loss through conduction = (0.040 W/m·K) / (6.0×0.0254 m) DT = 0.26 (T- 300) W/m² Heat loss through radiation = sT4 = 0.90 × 5.67×10-8 W/(m²·K4) × T4 = 5.6×10-8 T4 W/(m²·K4) Balancing heat absorption and loss yields: 900 = 0.26T - 78 + 5.6×10-8 T4 5.6×10-8 T4 + 0.26T - 978 = 0 (NOW WHAT DO WE DO????) Boiling Vapor pressure If we enclose a liquid and its vapor in a system, such as that illustrated in Fig. below and allow equilibrium conditions to be reached we find that for each temperature, Tsat at which we maintain the system there corresponds a pressure, Pvap, at which the two phases can co-exist. We will refer to this temperature as the saturation temperature and the pressure as the vapor pressure. Boiling, or formation of vapor bubbles within the liquid phase, usually occurs when the vapor pressure is greater than the surrounding ambient pressure. This can be caused by either raising the temperature of the liquid above the saturation temperature or lowering the ambient pressure below the vapor pressure. If we measure Lsat and Pvap we can construct a vapor-pressure curve such as the one shown in Fig. below for water. On the basis of simplifying assumptions which are generally valid away from the critical point one can derive the Clausius-Clapeyron equation which relates the vapor pressure to the temperature by the expression.