Chapter 2 - Conduction 2012 - Web

Chapter 2- - C HAPTER 2 – C ONDUCTION Consider a wall where the surface temperatures are such that T 1>T 2 q x = flow rate of heat in the x

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: C HAPTER 2 – C ONDUCTION Consider a wall where the surface temperatures are such that T 1 >T 2 q x = flow rate of heat in the x direction (vector!) [units of Watts] q x is dependent on THREE factors: 1. A - surface area perpendicular to the flow of heat (i.e. y x z ), 2. k – the capacity of material to transfer heat across its body (thermal conductivity) 3. L T – temperature gradient across the thickness of the wall, L T 1 T 2 q x L x These three factors are related together through FOURIER’S LAW : dx dT kA q x Q: why the negative sign? Or, in terms of heat flux q” (q” = q/A) dx dT k q x " Although we are only considering 1D heat flow in this case, Fourier’s Law may be expressed in 3 dimensions: dz dT k dy dT j dx dT i k T k q " where is the del operator NOTE: Fourier’s law is typically written individually for each dimension over which heat flow occurs. The driving force of conduction is the temperature gradient dx dT [ o C/m or K/m] Average gradient: L T T run rise dx dT 1 2 here dx dT , thus q x >0 Instantaneous gradient: may be some function of x If T(x)=Px 2 +Qx+R q x = -kA(dT/dx) = -kA(2Px+Q) T 2 T 1 L T 1 T 2 dT dx How can you determine if the temperature gradient is linear? Is there an internal source of heat? Is the material through which heat is flowing homogeneous? o Is k constant? Does the surface area perpendicular to the heat flow change with dx? o Increasing cross-sectional area = lower flux, more uniform temperature as a function of x r q wall T x T q H EAT T RANSFER P ROPERTIES Thermal Conductivity ( k ) : (W/m•K) High k values are good conductors Low k values are better insulators e.g. Material Temp (K) k (W/mK) Material Temp (K) k (W/mK) Brick 300 0.72 Air 300 0.026 Cork 300 0.039 Copper 300 401 Glass 300 1.4 Aluminum 300 237 More values available in Appendix A k is often given as a constant, but it does vary with T– get a value valid for the T range you are looking at. A more accurate relationship is: aT k k o 1 where k o and a are constants. a can be negative or positive depending on the material. For blends of materials A and B, use a weighted average to determine k : B A eff k a ak k 1 ( a = fraction of material A in the blend) For porous media consisting of a mixture of a stationary solid and a fluid, k depends on the geometry of the porous media (size distribution of pores, pore shape, etc.) Minimum k : heat conducts sequentially through a fluid region of length ɛ L and a solid region of length (1- ɛ ) L: ( fluid forms long, random fingers in solid) q Solid k s Pore (fluid) k f L L f s eff k k k 1 1 min , Maximum k : heat can conduct through either a continuous fluid region of cross-section ɛ w or a continuous solid region of cross- section (1- ɛ ) w: (interconnected solid) s f eff k k k...
View Full Document

This note was uploaded on 01/09/2012 for the course CHEM ENG 2A04 taught by Professor Toddhoare during the Spring '10 term at McMaster University.

Page1 / 25

Chapter 2- - C HAPTER 2 – C ONDUCTION Consider a wall where the surface temperatures are such that T 1>T 2 q x = flow rate of heat in the x

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online