Heat Conduction (Fourier S Law, Analogous to Fick S Law)

Heat Conduction (Fourier’s Law, analogous to Fick’s Law) ” ” qx = -k | {dT/dx} , x is the direction of heat transfer, qx is the flux, energy/ (area.time) Note the analogy to NA = -D | {dCA/dZ} Convection – energy is transferred by the bulk motion of the fluid ” Newton’s Law of Cooling: q = h(Ts - T∞) Where the heat transfer coefficient “h” is analogous to the mass transfer coefficient. Ts and T∞ are the temperature at the surface and bulk (outer flow) region respectively.

Diagram

Forced convection – external means (fan, pump) is used Natural convection – flow is induced by buoyancy forces Radiation – energy emitted by matter. For the frequently encountered case of radiation exchange between a small surface at Ts and a much larger surface at Tsurr completely surrounding the small surface ” 4 4 -8 ? 4 qrad = Єσ(Ts – Tsurr ) ; T(inK), σ = 5.67 x 10 w/(m K ) Є values are listed in Appendix A The above equation also assumes Є is equal to absorptivity (α)

Radiation (Example Problem) A small oxidized horizontal metal tube with an OD of 0.0254 m and 0.61 m long with a surface temp. of 588K is in a very large furnace enclosure with fire-brick walls and the surrounding air at 1088K. The emissivity of the metal tube is 0.60 at 1088K and 0.46 at 588K. Calculate the heat transfer rate to the tube (in W(J/s)) by radiation. 4 4 qrad = A,Єσ(Ts – Tsurr ) This equation is approximate where the emissivity is taken to be equal to absorptivity and that a single emissivity value at Tsurr is used. π(0.0254)(0.61)(0.6)(5.67 x 10-8)((588)4 – (1088)4) Note: π(0.0254)(0.61) is πDL (0.6) is Є (5.67 x 10-8) is σ(w/(m2K4))

qrad = -2130w

Heat Flux at a Surface

” -k | {dT/dx}|x=0 = qs ||| DIAGRAM Insulated or adiabatic surface: dT/dx}|x=0 = 0 ||| DIAGRAM Convection surface condition -k | {dT/dx}|x=0 = h(T∞ – T(0)) Temperature Distribution (One Dimensional Steady-State Heat Conduction) {d{k | dT/dx}/dx} = 0 (general derivation, p. 63) Alternate derivation Consider a surface perpendicular to x direction of thickness δx ||| DIAGRAM energy in – energy out -k | {dT/dx}|x – (-k | {dT/dx}|x+δx) + energy generation or??? source = energy, accumulation or storage

For no energy generation and accumulation (steady-state) -k | {dT/dx}|x + (-k | {dT/dx}|x+δx) = 0 When δx -> 0, the above equation becomes {d({k | dT/dx})/dx} = 0 Or??? {k | dT/dx} = constant => linear temperature distribution Heat transfer rate = qx = -kA | dT/dx Or??? qx = (kA/L)(Ts,1 – Ts,2) Also from convection surface condition qx = h1A(T∞,1 – Ts,1) h2A(Ts,2 - T∞,1)

DIAGRAM

Define “thermal resistance” Conduction; Rt,cond = (Ts,1 – Ts,2)/qx = L/(kA) Convection; Rt,conv = (Ts - T∞)/q = 1/(hA) Since qx is constant (why?) qx = (T∞,1 – Ts,1) / (1/(h1A)) = (Ts,1 – Ts,2) / (L/(kA)) = (Ts,2 - T∞,2) / (1/(h2A)) also qx = (T∞,1 – T∞,2) / (Rtot) , (overall driving force) / (overall resistance) where Rtot = 1/( h1A) + L/(kA) + 1/( h2A) other variations: [Also note qx” = qx/A] qx = (T∞,1 – Ts,1) / (1/(h1A) + L/(kA)) qx = (Ts,1 - T∞,2) / (L/(kA) + 1/(h2A)) :. qx” = (T∞,1 – Ts,2) / (1/(h1) + L/k) qx” = (Ts,1 - T∞,2) / (L/k + 1/h2)

The Energy(Heat Diffusion) Equation (p.61-65), “Fundamentals of Heat & Mass Transfers” Rectangular coordinates (∂/(∂x))(k(((∂T))/(∂x)))+(∂/(∂y))(k(((∂T))/(∂y)))+(∂/(∂z))( k(((∂T))/(∂x)))+q{with a little dot over it} = PCp | {(∂T)/(∂T)} Cylindrical coordinates (see Fig. 2.9 for coordinate description) (1/r)(∂/(∂r))(kr (∂T)/(∂r)) + (1/rsquared)(2/∂Φ)(k((∂T)/∂Φ)) + (2/∂z)(k((∂T)/∂z)) + q{with a little dot over it} = PCp | {(∂T)/(∂T)} Spherical coordinates (see Fig.2.10) (1/r2)(∂/(∂r))((kr2)((∂T)/(∂r))) + 1/(r2sin2θ)(∂/∂Φ)(k((∂T)/∂Φ)) + (1/r2sinθ)(∂/(∂θ) (ksinθ((∂T)/(∂θ)) + q{with a little dot over it} = PCp | {(∂T)/(∂T)}

When the differential equation is 2nd order in the spatial coordinate (consider x in rectangular coordinates) you will need two boundary (or surface)conditions to solve the problem, i.e. T(o,t) = To ; -k(((∂T)) / (∂x))|x=L = he(T(l,t)-T∞) In Addition he differential equation is first order with respect to time hence you will need one condition (initial condition) for time in order to obtain the solution, i.;e. T(x,o) = To.

The energy equation for the cylindrical geometry: 1/r | {dkr | {dT/dr}/dr} = 0 Diagram {dkr | {dT/dr}/dr} = 0 kr | {dT/dr} = constant r | {dT/dr} = C1 (C1 = constant/k, i.e another constant) {TS1∫TS2} dT = C1 {r1∫r2} dr/r ; TS2 – TS1 = C1ln(r2/r1) C1 = (Ts2 – Ts1)/ln (r2/r1) Or dT/dr = C1/r = (TS2 – TS1) / (rln(r2 / r1)) Fourier’s Law; qr = -kA | {dT/dr} = -k2πrL | {dT/dr} qr = 2πLk((TS2 – TS1) / (ln(r2 / r1))) (Note the heat transfer rate (W, or J/s is constant!)

Or qr = (TS1 – TS2) / Rt, cond {cond is thermal resistance for radial conduction} Rt, cond = (ln(r2 / r1)) / (2πLk) If the fluid temperatures inside and outside the pipe are t∞,1 and t∞,2 , in analogy to the plane wall, one can write: qr = (T∞,1 – T∞,2) / (1/(2πr1Lh1) + (ln(r2 / r1))/(2πkL) + 1/(2πr2Lh2)) Or qr = (T∞,1 – T∞,2) / Rtot Where Rtot = 1/(2πr1Lh1) + (ln(r2 / r1))/(2πkL) + 1/(2πr2Lh2) Often an overall heat transfer coefficient “U” is defined, either based on inner surface area (A1) or outer surface area (A2). Let’s base “U” on inner surface area => Rtot = 1 / (U1A1) 1 / (U1A1) = 1/(U12πr1L) = 1/(2πr1Lh1) + (ln(r2 / r1))/(2πkL) + 1/(2πr2Lh2)

1/U1 = 1/h1 + (r1ln(r2/r1)) / k + r1 / (r2h2) Or U1 = 1 / (1/h1 + (r1/k)ln(r2/r1) + r1/(r2h2)) For the more general case of a composite system (such as one or more layers of insulation can be placed around the pipe) see Equation 3.31 in the text U1 = 1/(1/h1 + (r1/kA)ln(r2/r1) + (r1/kB)ln(r3/r2) + (r1/kC)ln(r4/r3) + (r1/r4)(1/h4))

The next recitation will involve solving a transient heat conduction problem. In transient heat conduction problems, the temperature at any particular location in the solid varies with respect to time. Typically, the temperature of the solid will not be spatially uniform but in many applications, it may be assumed to be so. This assumption simplifies the treatment of the transient heat conduction problem since the problem can now be formulated as an ordinary differential equation where the solid temperature is a function of time rather than the more complicated case of a partial differential equation where the solid temperature is a function of time as well as distance. Here we will first illustrate the criterion for assuming uniform solid temperature, i.e. the lumped capacitance method or the lumped parameter method.

Criterion for the Validity of the Lumped Capacitance method for Transient Conduction The criterion that will be developed here is for the steady-state conduction process but is also applicable for the transient process. DIAGRAM Steady-state conduction through a plane wall of thickness L, consider the surface (boundary) condition -k | {dT/dx} |x=L = h(Ts,2 – T∞) For steady-state conduction dT/dx is linear (k(Ts,1 – Ts,2))/L = h(Ts,2 – T∞) (Ts,1 – Ts,2) / (Ts,2 - T∞) = (hL) / k = Bi For Bi C 0.1, “lumped” method may be used. This criterion can be generalized to any shape by using a characteristic length “Lc” defined as Lc = (volume of the object / surface area of the object) h(Lc/k) < 0.1, for a sphere Lc = (sphere diameter) / 6

Heat Conduction (Transient)

Consider the problem of a long cylindrical object initially at T0 being immersed in a medium (such as H2O or air) at constant temperature of T∞. It was established that if the condition (hLc) / k < 0.1 is satisfied, lumped parameter approach may be used. Otherwise it is necessary to solve the following differential equation PCp | {(∂T)/(∂T)} = (1/r)(∂/(∂r))(kr(∂T)/(∂r)) Or for constant “k” PCp | {(∂T)/(∂T)} = k(1/r)(∂/(∂r))(kr(∂T)/(∂r)) With the following initial & boundary conditions At t=0, T = T0 for all r At r=R, -k((∂T)/(∂r)) = h(T – T∞) for t>0 At r=0, (∂T)/(∂r) = 0 for all t Or // T is finite for all t