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Heat Transfer Revision – Examples Heat transfer: energy transport because of a temperature difference. Thermal energy is transferred from one region to another. Heat transport is the same phenomena like mass transfer, momentum transfer and electrical conduction. Similar rate equations, where flux is proportional to a potential difference. Potential difference temperature difference Three modes of heat transfer: - Thermal Conduction - Thermal Convection - Thermal Radiation Example 1 Given is a solid brick wall, with the following data Material: brick, fired clay, density: 1760 kg/m3, conductivity (k): 0.8 W/m K Surface (A): 5 m x 3 m = 15 m2 Interior surface temperature (t1): 20 °C Exterior surface temperature (t2): 0 °C Thickness of the wall (L): 25 cm = 0.25 m Calculate: a) the heat flow through the wall b) the thermal resistance Phenomenon: Thermal conduction The mechanism: energy is transported between parts of continuum by the transfer of kinetic energy between particles or groups of particles at the atomic level. Purely thermal conduction: in solid opaque bodies (opaque: not permeable for radiation) the thermal conduction is the significant heat transfer mechanism because the material doesn’t flow and there is no radiation. In flowing fluids, thermal conduction dominates in the region very close to the boundary layer, where – the flow is laminar – the flow parallel to the surface – there is no eddy motion Steady-state conduction Steady-state conditions: t = F(t) = constant For steady-state heat conduction, in one dimension, the Fourier-law is q (kA) q k A dt/dx dt dx – heat flow rate, W – thermal conductivity or heat conduction coefficient, W/m·K – cross sectional area normal to flow, m2 – temperature gradient, K/m The equation states that the heat flow rate (q) in the x direction is directly proportional to the temperature gradient dt/dx and the cross sectional area A normal to the heat flow. The proportionality factor is the thermal conductivity k. The minus sign indicates that the heat flow is positive in the direction of decreasing temperature. One-layer flat wall t2 t1 t1 t2 Q k A kA , W x2 x1 x2 x1 Or R – thermal resistance, K/W x 2 x1 L R , k k m2 K W If A and k are constant (e.g. at a wall) the equation (t1 t 2 ) qkA L L t1 t2 A k – – – – – wall thickness temperature at x = 0 temperature at x = L surface of the wall thermal conductivity and t qA , W R where L R , k m2 K W Example 1 Given is a solid brick wall, with the following data: Material: brick, fired clay, density: 1760 kg/m3, conductivity: from table k = 0.8 W/m K Surface (A): 5 m x 3 m = 15 m2 Interior surface temperature (t1): 20 °C Exterior surface temperature (t2): 0 °C Thickness of the wall (L): 25 cm = 0.25 m Calculate a) the heat flow through the wall b) the thermal resistance (t1 t 2 ) qkA L L R , k m2 K W t qA , W R Example 2 Given is an insulated concrete wall Layers: 1 – sand and gravel concrete, 2400 kg/m3, k1 = 1.5 W/m K, L1 = 16 cm 2 – expanded polystyrene, 20 kg/m3, k2 = 0.036 W/m K, L2 = 6 cm 3 – sand and gravel concrete, 2400 kg/m3, k3 = 1.5 W/m K, L3 = 8 cm Surface (A): 5 m x 3 m = 15 m2 Interior surface temperature (t1): 20 °C Exterior surface temperature (t4): 0 °C Calculate: the heat flow through the wall and the temperature on the surface of the layers Multi-layer flat wall t 2 t3 t3 t 4 t1 t 2 q A· A· A· , W R1 R2 R3 If the number of layers is n: t1 t n 1 q A· n Ri i 1 t1 t 4 q A· , W R1 R2 R3 L1 R1 , k1 L2 R2 k2 L3 R3 k3 m2 K W Example 3 Given are a room and a radiator. Surface temperature: ts = 85 °C Parameters of ambient air: tair = 30 °C, 1 bar, dry air Natural convection, length in direction of flow: 0,5 m Phenomenon: Thermal convection Energy transfer is involved by fluid movement and molecular conduction. Heat transfer means energy transfer • from liquids and gases to the surface of a body or a wall, or • from the surface of a body to the liquid. If in the flow the Reynolds number is large enough, three different flow regions exist: At the wall: laminar sublayer (boundary layer) thermal conduction Outside the laminar layer: buffer layer eddy mixing and conduction Beyond the buffer layer: turbulent region eddy mixing Convection is divided into two categories: Free convection, natural convection: the flow is generated by nonhomogeneous densities caused temperature difference. Forced convection: the flow is produced by external sources (pump, fan). The heat flow rate is described by the Newton’s formula: q hc A (t s t f ) hc A t Where q – heat flow rate, W hc – convectional heat transfer coefficient, W/m2K A – surface of the wall, m2 ts – temperature of the surface, K or °C tf – temperature of the fluid, K or °C The heat transfer coefficient can be calculated with dimensionless numbers, from similarity theory. Generally Nu = F (Pr; Gr; Re) It means, Nusselt number is a function of the product of Prandtl number, Grashof number and Reynolds number. Dimensionless Numbers Nu hc L k hc – heat transfer coefficient, W/m2·K L – characteristic dimension, m k – thermal conductivity of the fluid, W/m·K Dimensionless Numbers In the equation of Nusselt number Pr cp – absolute viscosity, kg/m·s or Ns/m2 cp – specific heat, J/kg·K k k – thermal conductivity of the fluid, W/m·K L3 2 g t L – characteristic dimension, m Gr 2 – density, kg/m3 – coefficient of thermal expansion, 1/K g Re vL – gravitational acceleration, m/s2 t – temperature difference, K or °C – absolute viscosity, kg/m·s or Ns/m2 v – velocity, m/s L – characteristic dimension, m D, diameter – kinematic viscosity = /, m2/s Practical cases of convection Natural or free convection: effect of temperature difference. General relationship: Nu hc L C (Gr Pr) n k n k L g t hc C L 2 f 3 2 c p k n f Open spaces: hc = F(t; direction of heat flow) Direction of heat flow: Ceiling down Wall horizontal Floor up t = ts – tair > 0 hc up >hc hor > hc down Some values for hc heat transfer coefficient Air, gas 4 … 10 W/m2 K Water, liquid 100 … 500 W/m2 K Forced convection: fluid stream derived from outer force Nu = F(Pr, Re) = F(v) Some values for hc heat transfer coefficient Air, gas Water, liquid 10 … 50 W/m2 K 500 … 5000 W/m2 K Example 3 Given are a room and a radiator. Surface temperature: ts = 85 °C Parameters of ambient air: tair = 30 °C, 1 bar, dry air Natural convection, length in direction of flow: 0,5 m Solution: h L Nu c C (Gr Pr) n k n k L3 2 g t hc C L av2 f cp k n f Properties of dry air at tair = 30 °C 30 = 1,1270 kg/m3 density, coefficient of thermal expansion, 30 = 3,255·10-3 1/K = 9,806 m/s2 gravitational acceleration, g absolute viscosity, 30 = 1,8682 ·10-5 Ns/m2 specific heat, cp30 = 1013 kJ/kg·K thermal conductivity of the fluid, k30 = 0.0258 W/m·K t = 85 – 30 = 55 °C Temperature difference Average temperature tav = (85 + 30)/2 = 115/2 ~ 60 °C absolute viscosity at 60 °C 60 = 1,9907 ·10-5 Ns/m2 ASHRAE Fundamentals 2005, Chapter 3, Page 3.17 Table 10 A special problem of convection: Convection in closed spaces: hollows, air layer between surfaces e.g. window construction t1 > t2 A1 = A2 Complex process of heat transfer: conduction convection radiation Equivalent heat conduction coefficient: ke Equivalent conduction resistance of air layer: Requi L kequi From table, e.g. ASHRAE Fundamentals 2001, Chapter 25, Table 3 Thermal Resistances of Plane Air spaces, m2 K/W Phenomenon: Thermal radiation The radiation energy transfer is through energy-carrying electromagnetic waves that are emitted by atoms and molecules due to change in their energy content. It means: does not depend on an intermediate material. The rate of thermal energy emitted by a surface depends on its quantity and its absolute temperature. A black surface absorbs all incident radiation. The total energy emitted per unit time and unit area is given by the Stefan-Boltzman law: Eblack T 4 where – Stefan-Boltzmann constant: 5.670·10-8 W/m2 K4 For nonblack surfaces E Eblack T 4 where – hemispherical emittance or emissivity. is a function of the material, condition of its surface. Lambert’s cosine law: Lambert's cosine law is the statement that the total power observed from a "Lambertian" surface is directly proportional to the cosine of the angle Φ made by the observer's line of sight and the line normal to the surface. E En cos Utilising the Lambert’s law the total energy radiated to the hemisphere is: ETotal En The heat flow rate between the surfaces Given are two surfaces: First let’s examine two general plane elements! The energy flux which leaves dA1 element and which is absorbed by dA2 element is quadraticly small: d2E1 = E1n · cos 1 · dA1 ·dW1 From this, the radiation absorbed by dA2 is: d2E1-2 = 2 ·d2E1 The heat exchange through radiation between the two plane elements is: d2q1-2 = d2E1-2 – d2E2-1 Introducing the angle factor, which is affected by geometrical parameters, the equation is: q A1 1 2 T T F12 4 1 4 2 A – surface m2 – hemispherical emittance or emissivity – Stefan-Boltzmann constant: 5.670·10-8 W/m2 K4 F12 – angle factor, shape factor, Definition of Angle Factor The fraction of all radiant energy leaving a surface i that is directly incident on surface k is the angle factor Fik (also known as view factor, shape factor, and configuration factor). Example 4: Calculation of a U-value Given is a threelayer wall The wall construction is the following: Material 1 – Brick, fired clay Density Conductivity Thickness , kg/m3 k, W/m,K L, m 1760 0.8 0.25 2 – Mineral fibreboard 260 0.049 0.06 3 – Wood wallboard, pine 660 0.16 0.02 Phenomenon: Overall Heat Transfer Given is a multilayer wall. The number of layers is 3. From inside air to outside air there is a complex process: Inside convection In the wall construction conduction Outside also convection The heat flow rate with overall heat transfer coefficient: q U A (t i t o ) Example 4 Given is a multilayer wall, where the wall construction data are the following: Material 1 – Brick, fired clay Density Conductivity Thickness , kg/m3 k, W/m,K L, m 1760 0.8 0.25 2 – Mineral fibreboard 260 0.049 0.06 3 – Wood wallboard, pine 660 0.16 0.02 Heat transfer coeff./surface resistance on the laminar air layer ASHRAE F.25.2 EN 12831 Still air (inside) Horizontal surface (upward) 9.26/0.11* 10.0/0.1 Horizontal surface (downward) 6.13/0.16* 5.9/0.17 Vertical surface 8.29/0.12* 7.7/0.13 * non reflective surfaces Moving air (outside) Any position 34/0.03* 25/0.04 * wind velocity 6,7 m/s (24 km/h) Calculations 1. Overall heat transfer coefficient: With overall thermal resistances: U U 1 1 n Li 1 hi j 1 ki ho 1 n Ris R j Ros j 1 2. The heat flow, if Inside temperature: 20 °C Outside temperature: 4 °C Area of the surface: 15 m2 q U A (ti to ) 3. The temperatures of the layers 1 1 tis ti q A hi 1 L1 t1 tis q A k1 1 1 tos q to A ho