TEM – Lecture 2 Basic concepts of heat transfer: Heat Conduction. Relation to Thermodynamics.

Summary of Last Lecture The Course Organization The importance of Energy and Mass Transfer at global and at the Engineering scale. The close linkage between Mass and Energy Transfer Heat Conduction and Radiation as the only energy transport processes not associated to mass transfer. This is why the course will start with heat conduction.

Heat Transfer Processes Why doesn’t a fluid support a horizontal temperature gradient at rest?

Temperature vs Heat Temperature is a measure of the kinetic energy of a molecule. It is Energy per molecule. Heat is the Energy of an ensemble of molecules (usually expressed per unit of volume or per unit of mass). Heat and Temperature are related by the Heat capacity: Heat capacity depends on the number of molecules per unit of volume and on the molecular mass. The energy necessary to change the temperature of a body of mass M is: To store a big heat quantity I need a large Mass and a large “c” (or both…). What is the material with higher “c” on earth?

Heat Conduction Process Fourier Law of heat conduction

Heat Conduction: Thermal conductivity (The Fourier Law) Units:

NOTE: The result should be 375mm NOTE: The result should be 375mm. In fact the heat flux required for the concrete wall is 0.8 of the composite wall and not the way around!!!

Thermal Conductivity of different materials Why are thermal and electrical conductivity well correlated? http://www.thermopedia.com/pt/content/1187/

Why do metals usually touch “cold” while other materials can touch “hot”? http://www.thermopedia.com/pt/content/1187/

Conductivity vs Diffusivity High conductivity means that it is easy to move a large amount of heat, to a long distance, i.e. if the conductivity of a wall material is high, heat can cross it easily. 𝑞 𝑥 " =−𝑘 𝑑𝑇 𝑑𝑥 𝑘 𝜌𝑐 = 𝑚 2 𝑠 −1 Conductivity is the product of density (mass per volume) by the heat capacity (heat per mass) that gives heat per unit of volume, by a velocity and by a distance.

Diffusivity α = 𝑘 𝜌𝑐 = 𝑚 2 𝑠 −1 α = 𝑘 𝜌𝑐 = 𝑚 2 𝑠 −1 In Fluid Mechanics we have seen that these units are also the units of the kinematic viscosity and the units of mass diffusivity. In fact these are the units of diffusivity of any property (heat, mass, momentum, …). Latter in this course we will see that we need diffusivity because we can not quantify the transport associated to each elementary particle. In fluids, when we have movement diffusivity increases a lot as a consequence of the difference between the velocity of elementary particles and the macroscopic definition of velocity.

Intermediate Summary Conduction is in fact heat diffusion due to the movement of elementary particles (molecules, atoms, electrons) not represented by velocity. These particles exchange kinetic energy and consequently they exchange heat. The heat flux is quantified by the Fourier Law. Heat conductivity is a property of each material and has to be measured. Values for common materials can be found in Table A3, page 989 of the recommended text book.

Energy Conservation The variation of the energy stored in the control volume must equal: the amount of energy that enters the control volume, minus the amount of energy that leaves the control volume, plus the amount of energy that is generated within the control volume. Or, per unit of time:

Internal + Mechanical Energy in a open system heat flux got through the solid wall Work done by a machine inside the volume work done by pressure (pressure energy) Flow work done by pressure (pressure energy) Flow velocity => kinetic energy) Flow velocity => kinetic energy) Under steady state: Internal energy) Internal energy)

If there is no work done inside the volume and changes of mechanical energy (pressure, kinetic and potential) can be neglected then one can write: This equation holds for perfect gas and for liquids without phase change and negligible changes of mechanical energy. In many systems the mechanical energy can change a lot and variations have to be accounted. In some of those cases mechanical energy changes are more important than internal energy changes and the Bernoulli Equation can be used. m 𝑝 𝜌 + 𝑉 2 2 +gz 1 - m 𝑝 𝜌 + 𝑉 2 2 +gz 2 -W=0 When both mechanical and internal energy components are changing, equations 1.12d has to be used.

In this problem we have water being vaporised and vapor is heated up to 600ºC and being elevated by 10 meters and the pressure is changing by 3 bar. In this case every form of energy is changing and thus one has to use the generic equation for stationary conditions (1.12d)

(Resolution of problem 1.36) Equation to use: We know: The mass flow rate: (1.5 kg/s) The pressure and temperature at the entrance / exit and The height difference between entrance and exit (10m) The pipe diameter (0.1m) => Area=3.16*0.1*0.1/4=0.0079m2 There is no work produced/supplied. Using thermodynamics tables one can get the internal energy/enthalpy at the entrance and exit, as well as the specific volume/density. At the entrance (T=110ºC) the internal energy of the water is (see table in next slide): 461kJ/kg and the velocity is 0.19m/s (density 1000kg/m3) and the kinetic energy is negligible (0.018m2/s2) At the exit the specific volume can be obtained by interpolation (0.5858m3/kg) as well as the enthalpy (3701kJ/kg). The velocity at exit is 106m/s. The kinetic and potential energy cost only 5725/11615*100%=50% of the total energy. 1.5*(461+10*105/1000+0+0)-1.5*(3700+106*106/2+9.8*10)+q=0 q = 11615kJ/s = 11.6 kW

Note: The temperature between parenthesis is the saturation temperature.

Summary In thermodynamics we have learned about energy conservation (1st law) and we have applied it to open systems. Then, heat flux was a statement. In this course its calculation is the major issue. Mechanical energy change (conservation) was a major issue in Fluid Mechanics and changes of internal energy were usually neglected. That allowed the Bernoulli equation to be used. In this course changes of internal energy are the main concern. In real cases one has to verify if any of those extreme simplifications is allowed. If not, the problem has to be solved into its full complexity.