Chapter 5 Time-Dependent Conduction

Chapter 5 Time-Dependent Conduction

5.1 The Lumped Capacitance Method This method assumes spatially uniform solid temperature at any instant during the transient process. It is valid if the temperature gradients within the solid are small. Apply energy balance to the control volume of Fig. 5.1 E& out = E& st (5.1) If the energy exchange is through convection, ha s (T T ) = ρvc dt dt Letting θ = T-T, we obtain ρvc ha s dθ dt = θ (5.2)

With the initial condition θ (0)= θ i, it follows θ = T T θ i T i T = exp ha s t ρvc (5.6) The thermal time constant is defined as τ t = ρvc ha s = ( 1 ha s )(ρvc) = R t C t where R t is the resistance to convection heat transfer, C t is the lumped thermal capacitance of the solid. The total energy transfer Q can be obtained by t t Q = 0 qdt = ha s 0 θdt = (ρvc)θ i 1 exp t τ t (5.8)

5.2 Validity of the Lumped Capacitance Method To find the criterion for the validity of the lumped capacitance method, consider the steady-state energy balance (Fig. 5.3) ka L (T s,1 T s,2 ) = ha(t s,2 T ) k--conductivity of the solid T s,1 T s,2 T s,2 T = (L / ka) (1 / ha) = R cond R conv = hl k Bi (5.9) Criterion for the validity of the lumped capacitance method(fig.5.4): Bi = hl c k < 0.1 L c = characteristic length (5.10)

Criterion for the validity of the lumped capacitance method (Fig.5.4): Bi = hl c k < 0.1 L c = characteristic length The dimensionless time, Fourier number, Fo = αt L c 2 Fo, with Bi, characterizes transient conduction problems. EX 5.1 θ θ i = T T T i T = exp( Bi Fo) (5.12) (5.13)

5.3 General Lumped Capacitance Analysis A general situation including convection, radiation, an applied surface heat flux, and internal energy generation (Fig.5.5) can be described as " " " dt qa s s, h + E& g ( qconv + qrad ) As( c, r) = ρvc (5.14) dt " 4 4 dt or qa s s, h + E& g [ ht ( T ) + εσ ( T Tsur )] As( c, r) = ρvc (5.15) dt EXs 5.2, 5.3

5.4 Spatial Effects Consider a representative 1-D heat transfer problem, DE: 2 T IC: BCs: x 2 = 1 α T t T(x,0)= T i T x x=0 = 0 Eqs. 5.26-29 imply k T x x= L = h[t(l,t) T ] T = T(x, t, T i,t, L,k,α,h) However, after nondimensionalization with θ θ θ = T T i T i T 2 θ x* 2 = Fo θ θ (x,0)= 1 x x L θ x* x =0 = 0 t αt L 2 Fo θ x* x =1 = Bi θ (1, t ) (5.26) (5.27) (5.28, 29) (5.30) (5.31-33) (5.34) (5.35-37)

θ = f (x,fo, Bi) (5.38) For a prescribed geometry, the transient temperature distribution is a universal function of x*, Fo, and Bi. That is, the dimensionless solution assumes a prescribed form that does not depend on the particular value of T i, T, L, k, α, or h.. The physical interpretation of Fo: Fo can not only viewed as the dimensionless time, it also provides a measure of the relative effectiveness for a material to conduct and store energy, as 2 3 2 2 ( q / E& ) ~ ( kl ΔT / L) /( ρl cδt / t) = kt / ρcl = αt L = st / Fo

5.5 The Plane Wall with Convection 5.5.1 Exact Solution The problem depicted in Fig. 5.6a has the exact solution as where θ = C n exp( ζ 2 n Fo)cos(ζ n x ) n=1 C n = 4sin ζ n 2ζ n + sin(2ζ n ) (5.39a) (5.39b) and the eigenvalues ζ n are positive roots of the transcendental eq. ζ n tanζ n = Bi (5.39c)

5.5.2 Approximate Solution For Fo > 0.2, the exact infinite series solution can be approximated by the first term of the series: ( ζ 1 < ζ 2 < ζ 3 <, cf. App. B.3) θ C 1 exp( ζ 1 2 Fo) cos(ζ 1 x ) or where θ 0 = C 1 exp( ζ 2 1 Fo) --temperature variation at midplane x* = 0. Eq. 5.40 implies that the time dependence of the temperature at any location within the wall is identical. 5.5.3 Total Energy Transfer Q = ρc[ T( x, t) Ti ] dv; Q [ T( x, t) Ti ] dv 1 sinζ (1 *) ~ 1 1 = = θ dv θ Q T T V V ζ o θ = θ 0 cos(ζ 1 x ) i Q = ρcv[ T T ] o i 1 * o (5.40a) (5.40b)

5.5.4 Additional Considerations The solution is also valid for the problem with insulation on one side (x* = 0) and experiences convective transport on the other side (x* = +1). The foregoing results may be used to determine the transient response of a plane wall to a sudden change in surface temperature (θ* = θ i *, at x* = 1). The process is equivalent to having an infinite convection coefficient, in which case the Biot number is infinite (Bi = ) and the fluid temperature T is replaced by the prescribed surface temperature T s.

5.6 Radial Systems with Convection Infinite cylinder: (Fig. 5.6b) θ = C n exp( ζ 2 n Fo)J 0 (ζ n r ) where n=1 C n = 2 ζ n J 1 (ζ n ) J 0 2 (ζ n ) + J 1 2 (ζ n ) (5.47a) (5.47b) with eigenvalues ζ n being the positive roots of the equation ζ n J 1 (ζ n ) J 0 (ζ n ) = Bi (5.47c) <Home Work> Solve for Eqs. 5.39 and 5.47 using the method of separation of variables. EX 5.4

About Bessel Equation Similar reasoning can be made for the Bessel Equation as for modified Bessel Equation, as both correspond to the cylindrical coordinate. The Bessel Equation of order v is 2 2 d θ 1 v + dθ + (1 ) θ = 0 2 2 dr rdr r The general solution of (3) is θ () r = C J () r + C Y () r 1 ν 2 ν (B3) (B3a) 2.405 5.520 8.654 From www.-efunda.-com/-math

2 Compare d θ 2 + 1 dθ + m θ = 0, with the general solution 2 dr rdr θ () r = C J ( mr) + C Y ( mr) 1 0 2 0 2 d θ 2 and + m θ = 0 2 dx with the general solution θ ( x) = C cosmx + C sinmx 1 2 It can be seen that J 0 (mr) is similar to cosmx in the oscillating behavior, but J 0 (mr) reflects damped amplitude in response to increased r. Their periods are also very similar except for the first half cycle. In fact, for large x 2 Jn( x)~ cos x π x nπ π 2 4 (B4, B4a) (B5) (B5a)

5.7 The Semi-Infinite Solid Three cases of surface conditions: (Fig. 5.7)

Using the similarity method with similarity variable η = x/(4αt) 1/2, the PDE can be transformed into an ODE. Exact solutions can thus be obtained. Case 1 Constant surface temperature: T(0,t)=T s Case 2 Constant surface heat flux: q s = Eq. 5.59 Case 3 Surface Convection: EX 5.6 Eq. 5.60 T( x, t) Ts x = erf Ti T s 2 αt " kt ( s Ti) qs () t = παt q 0 (5.57) (5.58) k T x x=0 = h[t T(0,t)]

A special case: interfacial contact between two semiinfinite solids at different initial temperatures If contact resistance is negligible, there is no temperature jump at the interface. Also, " " qs, A = qs, B Since T s does not change with time (no energy storage), from Eq 5.58 (5.57) ka( Ts TA, i) kb( Ts TB, i) = πα t πα t A B (5.62) (5.58) T = s ( kρc) T + ( kρc) T A A, i B B, i ( kρc) + ( kρc) A B (5.63) With T s determined, the temperature profile can be described by Eq. 5.57.