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Transcript
MHMT9 Momentum Heat Mass Transfer D source Dt Energy balances. FK equations Mechanical, internal energy and enthalpy balance and heat transfer. Fourier´s law of heat conduction. FourierKirchhoff ´s equation. Steady-state heat conduction. Thermal resistance (shape factor). Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010 MHMT9 Power of forces When analysing energy balances I recommend you to imagine a fixed control volume (box) and forces acting on the outer surface ds Power of surface force [W] dF u dF ds u MHMT9 Kinetic energy balance Kinetic energy balance follows directly from the Cauchy’s equation for velocities (kinetic energy is square of velocities, so that it is sufficient to multiply the Cauchy’s equation by velocity vector; scalar product results to scalar energy) Du u u p u u f Dt and this equation can be rearranged (how? see the next slide) to the final form of the kinetic energy transport 1 2 D u 2 ( pu ) p u ( u ) : f u Dt dissipation of reversible work of work done by work done by kinetic energy pressure forces acting at surface expansion viscous forces mechanical energy to heat external forces MHMT9 Kinetic energy balance Proof follows from the previous equation rewritten to the index notation uk uk km p uk ( um ) uk um uk f k t xm xk xk um 1 uk uk u p uk p uk 2 uk p k uk xk xk xk t t 1 uk uk u uk um k u m 2 xm xm km um km u km m xk xk xk um km km um uk ( ) xk 2 xk xm um km km mk xk When considering different forms of energy transport equations it is important to correctly interpret decomposition of energies to reversible and irreversible parts ( u ) ( ) u overall work done by viscous forces reversible change to kinetic energy ( pu ) (p) u overall work done by pressure forces conversion to kinetic energy : irreversible dissipation of viscous forces work to heat p( u ) adiabatic expansion MHMT9 Dissipation of kinetic energy 1 : u : (u (u )T ) : 2 1 (u (u )T ) 2 This identity follows from the stress tensor symmetry Rate of deformation tensor Example: Simple shear flow (flow in a gap between two plates, lubrication) 1 1 ux 1 xy yx ( xu y y ux ) 2 2 y 2 U=ux(H) y : u : xy yx yx xy xy MHMT9 Example: Dissipation Generated heating power in a gap between rotating shaft and casing Rotating shaft at 3820 rpm D=5cm L=5cm U=10 m/s H=0.1 mm y Gap width H=0.1mm, U=10 m/s, oil M9ADS-II at 00C =3.4 Pa.s, =105 1/s, =3.4.105 Pa, = 3.4.1010 W/m3 At contact surface S=0.0079 m2 the dissipated heat is 26.7 kW !!!! MHMT9 Internal energy balance The equation of kinetic energy transport is a direct consequence of Cauchy’s equations and does not bring a new physical information (this is not a new “law”). Such a new law is the equation of internal energy balance which represents the first law of thermodynamics stating, that the internal energy increase is determined by the heat delivery by conduction and by the mechanical work. This statement can be expressed in form of a general transport equation for =uE Pq uE specific internal energy Heat flux by conduction p u reversible expansion (gas cools down when expanded) : dissipated mechanical energy (internal friction ) Q( g ) heat generated by volumetric ohmic or microwave heating DuE q p u : Q ( g ) Dt Internal energy uE is defined as the sum of all energies (thermal energy, energy of phase changes, chemical energy) with the exception of kinetic energy u2/2 and this is the reason why not all mechanical work terms are included in the transport equation and why the reaction heat is not included into the source term Q(g). MHMT9 Internal energy balance Interpretation using the First du = dq - law of thermodynamics p dv DuE q p u : u Q ( g ) Dt Heat transferred by conduction into FE Expansion cools down working fluid This term is zero for incompressible fluid Dissipation of mechanical energy to heat by viscous friction : : u MHMT9 Internal energy balance Remark: Confusion exists due to definition of the internal energy itself. In some books the energy associated with phase changes and chemical reactions is included into the production term Q(g) (see textbook Sestak et al on transport phenomena) and therefore the energy related to intermolecular and molecular forces could not be included into the internal energy. This view reduces the internal energy only to the thermal energy (kinetic energy of random molecular motion). Example: Consider exothermic chemical reaction proceeding inside a closed and thermally insulated vessel. Chemical energy decreases during the reaction (energy of bonds of products is lower than the energy of reactants), no mechanical work is done (constant volume) and heat flux is also zero. Temperature increases. As soon as the internal energy is defined as the sum of chemical and thermal energy, the decrease of chemical energy is compensated by the increased temperature and DuE/Dt=0. MHMT9 Total energy balance The internal energy transport equation is in fact the transport of total energy (this is expression of the law of energy conservation) from which the transport equation for the kinetic energy is subtracted D 1 2 (uE u ) q ( pu ) ( u ) Q ( g ) Dt 2 heat delivered energy comming work done by work done by total energy [J/kg] by conduction pressure viscous forces from ouside (electric heat) The term is potential of conservative external volumetric forces (for example potential energy in gravitational field gh). MHMT9 Enthalpy balance Thermal engineers prefer balancing of steady continuous systems in terms of enthalpies instead of internal energies. The transport equation follows from the equation for internal energy introducing specific enthalpy h=uE+pv (v is specific volume) DuE D (h pv) q p u : Q ( g ) Dt Dt D p Dh p D Dp Dh Dp (h ) p( u ) Dt Dt Dt Dt Dt Dt giving the final form of the enthalpy balance Dh Dp (g) q : Q Dt Dt This enthalpy balance (like all the previous transport equations for different forms of energy) is quite general and holds for compressible/incompressible fluids or solids with variable transport properties (density, heat capacity, …) MHMT9 Fourier Kirchhoff equation Feininger MHMT9 Fourier Kirchhoff equation Primary aim of the energy transport equations is calculation of temperature field given velocities, pressures and boundary conditions. Temperatures can be derived from the calculated enthalpy (or internal energy) using thermodynamic relationship Dh DT v Dp cp (v T ( ) p ) Dt Dt T Dt giving the transport equation for temperature DT cp Dt v Dp T ( ) p T Dt this term is zero for incompressible liquid and reduces to Dp/Dt for ideal gas. q : Q( R ) it is necessary to include also reaction and phase changes enthalpies (and electric heat as previously) The reason why reaction enthalpy and enthalpy of phase changes had to be included into the source term Q(R) is the consequence of limited applicability of thermodynamic relationships between enthalpy and temperature (for example DT/Dt=Dp/Dt=0 during evaporation but Dh/Dt>0). MHMT9 Fourier Kirchhoff equation Diffusive (molecular) heat flux is proportional to the gradient of temperature according to the Fourier’s law q T where is thermal conductivity of material. Fourier Kirchhoff equation for temperature field reads like this DT cp ( T ) : Q ( R ) Dt As soon as thermal conductivity is constant the FK equation is DT cp 2T : Q ( R ) Dt MHMT9 Thermal conductivity Thermal and electrical conductivities are similar: they are large for metals (electron conductivity) and small for organic materials. Temperature diffusivity a is closely related with the thermal conductivity a Memorize some typical values: Material [W/(m.K)] a [m2/s] Aluminium Al 200 80E-6 Carbon steel 50 14E-6 Stainless steel 15 4E-6 Glas 0.8 0.35E-6 Water 0.6 0.14E-6 Polyethylen 0.4 0.16E-6 Air 0.025 20E-6 c p Thermal conductivity of nonmetals and gases increases with temperature (by about 10% at heating by 100K), at liquids and metals usually decreases. MHMT9 Conduction - stationary Let us consider special case: Solid homogeneous body (constant thermal conductivity and without internal heat sources). Fourier Kirchhoff equation for steady state reduces to the Laplace equation for T(x,y,z) 2T 2T 2T 0 T 2 2 2 x y z 2 cylinder sphere 2T 1 T 0 2 (r ) x r r r 1 T 0 2 (r 2 ) r r r The same equation written in cylindrical and spherical coordinate system (assuming axial symmetry) Boundary conditions: at each point of surface must be prescribed either the temperature T or the heat flux (for example q=0 at an insulated surface). Solution of T(x,y,z) can be found for simple geometries in an analytical form (see next slide) or numerically (using finite difference method, finite elements,…) for more complicated geometry. MHMT9 1D conduction (plate) Steady transversal temperature profile in a plate is described by FK equation which reduces to 2T 0 T x 2 2 with a general solution (linear temperature profile) Tf1 T ( x) c1 x c2 Differential equations of the second order require two boundary conditions (one BC in each point on boundary) Tw1 1. BC of the first kind (prescribed temperatures) Tw2 x T ( x) (Tw 2 Tw1 ) Tw1 h these boundary conditions with fixed values are called Dirichlet boundary conditions 2. BC of the second kind (prescribed flux q0 in one point) h x T ( x) q0 x Tw 2 q0 h the boundary conditions of the second kind is called Neumann’s boundary condition 3. BC of the third kind (prescribed heat transfer coefficient) (T f 1 Tw2 ) hT f 1 Tw2 T ( x) x h h the boundary condition of the third kind is called Newton’s or Robin’s boundary condition MHMT9 1D conduction (sphere and cylinder) Steady radial temperature profile in a cylinder and sphere (for fixed temperatures T1 T2 at inner and outer surface) R1 R2 cylinder T r c1 , r r2 T c1 , r Sphere (bubble) T=c1 ln r c2 T=- c1 c2 r T2 T1 T1 ln R2 T2 ln R1 c1 c2 ln R2 / R1 ln R2 / R1 c1 R1 R2 T R T1 R1 (T2 T`1 ) c2 2 2 R2 R1 R2 R1 MHMT9 1D conduction (cylinder-heating power) Knowing temperature profiles it is possible to calculate heat flux q [W/m2] and the heat flow Q [W]. The heat flow can be also specified as a boundary condition (thus the radial temperature profile is determined by one temperature T0 at radius R0 and by the heat flow Q) L For cylinder with thermal conductivity and for specified heat flow Q related to length L, the logarithmic temperature profile can be expressed in the following form T c1 , T=c1 ln r c2 , r T0 =c1 ln R0 c2 r Q R0,T0 Positive value Q>0 represents a line heat source, while negative value heat sink. Q=-2 L r T 2 L c1 r Q R0 T (r ) T0 ln 2L r MHMT9 2D conduction (superposition) Temperature distribution is a solution of a linear partial differential equation (Laplace equation) and therefore is additive. It means that any combination of simple solutions also satisfies the Laplace equation and represents some stationary temperature field. For example any previously discussed solution of potential flows (flow around cylinder, sphere, see chapter 2) represents also some temperature field (streamlines are heat flux lines, and lines of constant velocity potential are isotherms). Therefore the same mathematical techniques (conformal mapping, tables of complex functions w(z) describing dipoles, sources, sinks, circulations,…) are used also for solution of temperature fields electric potential field velocity potential field concentration fields The principle aim is thermal resistance, electrical resistance… Thus it is possible to evaluate for example the effect of particles (spheres) or obstacles (cylinders) to the resistivity of inhomogeneous materials. MHMT9 2D conduction (superposition) As an example we shall analyse superposition of two parallel linear sources/sinks of heat Q Temperature at an arbitrary point (x,y) is the sum y of temperatures emitted by source Q and sink -Q T(x,y) rQ rS x Source Q Sink -Q h h R0 R0 r Q Q T ( x, y ) T0 (ln ln ) T0 ln S 2L rQ rS 2L rQ Lines characterised by constant values k=rS/rQ are isotherms. rS2 ( x h) 2 y 2 2 k 2 Q r ( x h) 2 y 2 This equation describes a circle of radius R and with center at position m 2kh R 2 , k 1 k 2 1 m 2 h k 1 y Resulting temperature field describes T2 for example the following cases: (see also the next slide) y T1 R m x T2 T1 x MHMT9 Thermal resistance Knowing temperature field and thermal conductivity it is possible to calculate heat fluxes and total thermal power Q transferred between two surfaces with different (but constant) temperatures T1 a T2 T T R [K/W] thermal resistance Q T1 T2 1 2 T RT In this way it is possible to express thermal resistance of windows, walls, heat transfer surfaces … T1 T2 T2 S2 R2 S Q S1 R1 h T1 T2 R1 T1 L L h h1 h 2 RT 1 h1 h2 ( ) S 1 2 Serial RT h S11 S 2 2 Parallel RT ln R2 / R1 2L Tube wall RT ln 2h / R 2L Pipe burried under surface MHMT9 EXAM Energy transport MHMT9 What is important (at least for exam) Kinetic energy D 1 2 ( u ) ( pu ) p u ( u ) : f u Dt 2 dissipation of reversible work of work done by work done by kinetic energy pressure forces acting at surface expansion viscous forces mechanical energy to heat external forces Conservation of all forms of energy D 1 (uE u 2 ) q ( pu ) ( u ) Q ( g ) Dt 2 Internal energy (by subtraction kinetic energy, see also 1st law duE=dq-dw) DuE q p u : Q ( g ) Dt Fourier Kirchhoff cp DT 2T : Q ( R ) Dt MHMT9 What is important (at least for exam) Steady state heat conduction (cartesian, cylindrical, spherical coordinates) 2T 2T 2T 0 T 2 2 2 x y z 2 2T 1 T 0 2 (r ) x r r r 1 2 T 0 2 (r ) r r r 1D temperature profiles (cartesian, cylindrical, spherical coordinates) T=c1 ln r c2 T=- c1 c2 r MHMT9 What is important (at least for exam) Thermal resistance Q T1 T2 RT Serially connected plates 1 h1 h2 RT ( ) S 1 2 Cylinder RT ln R2 / R1 2L