Heat transfer due to harmonic variation in the free stream temperatures in a flat plate exposed on both sides

Abstract

The heat transfer through an infinite flat plate is studied when the temperatures of the two free streams surrounding it are varying harmonically with time and out of phase, with a delay period τ_d. The configuration is a simplified model for the heat transfer through the separating wall in the isochoric counter-current heat exchanger. The results show that apart from the τ_d effect, the perturbation parameters depend mainly on the cavity passing frequency f. At the thick plate solution, the combined passing frequency–delay time influences are significant only when the dimensionless frequency is smaller than 10. Within this range τ_d affects seriously not only the temperature perturbation amplitudes (which determine the thermal stresses) but also the heat fluxes and the accumulated energy ones. When f  ≥  10, the plate behaves as two separate semi-infinite slabs. Heat penetration delays greater than one cavity passing period may be possible.

Appendices

Appendix A

The initial thin plate temperature, ${T^{*}(t = 0) \equiv T_{0}^{*}}$ is obtained from Eq. 23 if we take into consideration that, at t =  0, we have T _f1(t) =  T ₁, T _f2(t) =  T ₂ and ${dT_{0}^{*}/ dt = 0}$ . Then, Eq. 23 yields: ${h_{2} (T_{2} - T_{0}^{*}) - h_{1} (T_{0}^{*} - T_{1}) = 0}$ and ${T_{0}^{*}}$ is obtained as:

$$ T_{0}^{*} = \frac{h_{1} T_{1} + h_{2} T_{2}}{h_{1} + h_{2}} = \frac{B_{1} T_{1} + B_{2} T_{2}}{B_{1} + B_{2}} $$

(41)

The non-dimensional thin plate temperature, ${\theta^{*}(\tau) = (T^{*}(t) - T_{0}^{*})/(T_{2} - T_{1})}$ , yields:

$$ T^{*} (t) = T_{0}^{*} + {\left( {T_{2} - T_{1}} \right)}\theta ^{*} (\tau ) $$

(42)

$$ \frac{{\rm d}T^{*} (t)}{{\rm d}t} = {\left( {T_{2} - T_{1}} \right)}\frac{{\rm d}\theta ^{*} (\tau )}{{\rm d}t} = {\left( {T_{2} - T_{1}} \right)}\frac{\alpha }{w^{2}}\frac{{\rm d}\theta ^{*} (\tau )}{{\rm d}\tau} $$

(43)

If we use now Eqs. (41)–(43), and take into consideration equations (1) and (12), we obtain the non-dimensional form of Eq. 23, i.e. Eq. 24.

For the thick plate case, Eq. 5 yields:

$$ T(y,t) = {\left( {T_{2} - T_{1}} \right)}\theta (Y,\tau ) + T_{0} (y) $$

(44)

$$ \frac{\partial T(y,t)}{\partial y} = {\left( {T_{2} - T_{1}} \right)}\frac{\partial \theta (Y,\tau )}{\partial y} + \frac{{\rm d}T_{0} (y)}{{\rm d}y} $$

(45)

At t =  0 it is dT ₀/dy =  constant and from the boundary conditions (3a) and (3b) we obtain:

$$ k\frac{dT_{0}}{dy} = k\frac{T_{0} (w) - T_{0} (0)}{w} = h_{1} {\left[ {T_{0} (0) - T_{1}} \right]} = h_{2} {\left[ {T_{2} - T_{0} (w)} \right]} $$

(46)

The solution of the system of Eq. (46) yields the initial temperature distribution:

$$ T_{0} (y) = \frac{B_{1} T_{1} + B_{2} T_{2} + B_{1} B_{2} T_{1} }{B_{1} + B_{2} + B_{1} B_{2}} + \frac{B_{\rm e}}{1 + B_{\rm e}}{\left( {T_{2} - T_{1}} \right)}\frac{y}{w} = \frac{T_{0}^{*} + B_{\rm e} T_{1}}{1 + B_{\rm e}} + \beta {\left( {T_{2} - T_{1}} \right)}\frac{y}{w} $$

(47)

The B ₁, B ₂, B _e,β are given by equations (12) and (14), while ${T_{0}^{*}}$ is the initial temperature in the thin plate limit, given by Eq. 41. If we use now Eqs. 44–(47) and take into consideration Eqs. 6a, and 6b, we obtain from Eqs. 2, (3a), (3b) and (4), the non-dimensional form of the thick plate equations, i.e. Eqs. (7), (8a), (8b) and (11) respectively.

Appendix B

We have to solve Eq. 7 with boundary conditions given by equations (8a) and (8b). If the equation is solved in the time interval 0  ≤   τ  ≤   τ_d, the initial condition is

$$ \theta (Y,\tau ) = 0,\quad \tau = 0 $$

(48)

while, if the equation is solved in the time interval τ >  τ_d it is:

$$ \theta (Y,\tau ) = F(Y),\quad \tau = \tau _{\rm d} $$

(49)

By separating the time and space variables, it is demonstrated easily that the complete solution of the temperature function θ(Y, τ) is given in the form

$$ \theta (Y,\tau ) = {\sum\limits_{n = 1}^\infty {c_{n} \Psi {\left( {\lambda _{n}, Y} \right)}\bar {\theta}(\lambda _{n}, \tau )}} $$

(50)

where c _n are unknown coefficients to be determined, ${\bar{\theta }(\lambda _{n}, \tau )}$ is the integral transform of θ(Y,τ) with respect to the space variable Y in the range 0 ≤  Y ≤  1, defined as:

$$ \bar{\theta}(\lambda _{n}, \tau ) = \int\limits_{{Y}^{\prime} = 0}^1 {\Psi {\left( {\lambda _{n}, {Y}^{\prime}} \right)}\theta {\left( {{Y}^{\prime},\tau} \right)}}{\rm d}{Y}^{\prime} $$

(51)

while Ψ (λ _n ,Y) has the form

$$ \Psi {\left( {\lambda _{n}, Y} \right)} = \cos \lambda _{n} Y + \frac{B_{1}}{\lambda _{n}}\sin \lambda _{n} Y $$

(52)

In fact, Ψ (λ _n ,Y) is the eigenfunction of the following auxiliary eigenvalue problem

$$ \frac{{\rm d}^{2} \Psi {\left( Y \right)}}{{\rm d}Y^{2}} + \lambda ^{2} \Psi {\left( Y \right)} = 0,\quad 0 < Y < 1 $$

(53)

$$ - \frac{{\rm d}\Psi (Y)}{{\rm d}Y} + B_{1} \Psi (Y) = 0,\quad Y = 0 $$

(54a)

$$ {\kern 1pt} \frac{{\rm d}\Psi (Y)}{{\rm d}Y} + B_{2} \Psi (Y) = 0,\quad Y = 1 $$

(54b)

The coefficients c _n are determined by taking into consideration the orthogonality of the eigenfunctions Ψ (λ _n ,Y) and are given as:

$$ c_{n} = \frac{2\lambda _{n}^{2}}{{{\left( {\lambda _{n}^{2} + B_{1}^{2}} \right)} + {\left( {B_{1} + B_{2}} \right)}\frac{\lambda _{n}^{2} + B_{1} B_{2}}{{\lambda _{n}^{2} + B_{2}^{2}}}}} $$

(55)

By applying now the transform (51) we take the integral transform of Eq. 7

$$ {\int\limits_0^1 {\Psi {\left( {\lambda _{n,} Y} \right)}\frac{{\partial ^{2} \theta {\left( {Y,\tau } \right)}}}{{\partial Y^{2} }}\,} }{\rm d}Y = {\int\limits_0^1 {\Psi {\left( {\lambda _{n,} Y} \right)}\frac{{\partial \theta {\left( {Y,\tau } \right)}}}{{\partial \tau }}\,} }{\rm d}Y = \frac{{{\rm d}\bar{\theta }(\lambda _{n,} \tau )}}{{{\rm d}\tau }} $$

(56)

The term in the first member in Eq. 56 is evaluated by making use of the Green’s theorem and taking into consideration the boundary conditions (8a), (8b) and the corresponding ones for the auxiliary eigenvalue problem, (54a), (54b):

$$ \begin{aligned} {\int\limits_0^1 {\Psi {\left( {\lambda _{n,} Y} \right)}\frac{{\partial ^{2} \theta {\left( {Y,\tau } \right)}}}{{\partial Y^{2} }}\,} }{\rm d}Y = - \lambda _{n}^{2} \bar{\theta }(\lambda _{n,} \tau ) &+ B_{1} \gamma _{1} \Psi {\left( {\lambda _{n,} 0} \right)}{\left[ {1 - \cos \omega (\tau - \tau _{\rm d})} \right]} \\ & + B_{2} \gamma _{2} \Psi {\left( {\lambda _{n,} 1} \right)}{\left( {1 - \cos \omega \,\tau } \right)} \\ \end{aligned} $$

(57)

By substituting Eq. 57 into Eq. 56, the latter it is finally written as:

$$ \frac{{\rm d}\bar{\theta}(\lambda _{n}, \tau )}{{\rm d}\tau} + \lambda _{n}^{2} \bar{\theta}(\lambda _{n}, \tau ) = A(\lambda _{n}, \tau ) $$

(58)

where we have put for brevity

$$ \begin{aligned} A(\lambda _{n}, \tau ) &= B_{1} \gamma _{1} \Psi {\left( {\lambda _{n}, 0} \right)}{\left[ {1 - \cos \omega (\tau - \tau _{\rm d})} \right]} + B_{2} \gamma _{2} \Psi {\left( {\lambda _{n}, 1} \right)}{\left( {1 - \cos \omega \,\tau} \right)} \\& = B_{1} \gamma _{1} {\left[ {1 - \cos \omega (\tau - \tau _{\rm d})} \right]} + B_{2} \gamma _{2} {\left( {\cos \lambda _{n} + \frac{B_{1}}{\lambda _{n}}\sin \lambda _{n}} \right)}{\left( {1 - \cos \omega \,\tau} \right)} \\ \end{aligned} $$

(59)

The application of the integral transform (51) to the initial condition (48) or (49) yields the corresponding initial condition for Eq. 58:

$$ \bar{\theta}(\lambda _{n}, \tau ) = 0 $$

(60)

if the equation is solved in the time interval 0  ≤  τ  ≤  τ_d, and

$$ \left. \bar\theta(\lambda _{n}, \tau ) \right|_{\tau = \tau _{\rm d} } = \int\limits_{{Y}^{\prime} = 0}^1 {\Psi {\left( {\lambda _{n}, {Y}^{\prime}} \right)}F{\left( {{Y}^{\prime}} \right)}}{\rm d}{Y}^{\prime} $$

(61)

if the equation is solved in the time interval τ >  τ_d.

The simplest case is when there is no phase difference between the temperature oscillations of the two fluids. Then τ_d =  0 in Eq. 59 and the solution of Eq. 58 subjected to the transformed initial condition (60) is:

$$ \bar{\theta}(\lambda _{n}, \tau ) = \frac{{B_{1} \gamma _{1} + B_{2} \gamma _{2} {\left( {\cos \lambda _{n} + \frac{B_{1}}{\lambda _{n} }\sin \lambda _{n}} \right)}}}{{\lambda _{n}^{2}}}{\left[ {1 - \frac{{\omega ^{2}}}{{\lambda _{n}^{4} + \omega ^{2}}}{\rm e}^{- \lambda _{n}^{2} \tau} - \frac{\lambda _{n}^{2}}{\sqrt {\lambda _{n}^{4} + \omega ^{2}}}\cos {\left( {\omega {\kern 1pt} \tau - \tan ^{- 1} \frac{\omega}{\lambda _{n}^{2}}} \right)}} \right]} $$

(62)

If we introduce now Eqs. 55, (52) and (62) into the general form of the solution, i.e. Eq. 50 we obtain the following solution

$$ \theta (Y,\tau ) = {\sum\limits_{n = 1}^\infty {{\left( {a_{n} + b_{n}} \right)}\;{\left( {\cos \lambda _{n} Y + \frac{B_{1}}{\lambda _{n}}\sin \lambda _{n} Y} \right)}}}{\left[ {1 - \frac{\omega ^{2} }{{\lambda _{n}^{4} + \omega ^{2}}}{\rm e}^{- \lambda ^{2}_{n} \tau} - \frac{\lambda _{n}^{2}}{\sqrt {\lambda _{n}^{4} + \omega ^{2}}}\cos {\left( {\omega {\kern 1pt} \tau - \tan ^{- 1} \frac{\omega}{{\lambda ^{2}_{n}}}} \right)}} \right]} $$

(63)

where the eigenvalues λ _n are the positive roots of the transcendental equation (33) while the coefficients a _n , b _n are given by Eq. 34.

In the general case, the solution depends on the time interval considered:

• (a) In the time interval [0 ≤ τ ≤  τ_d], the initial condition is given by Eq. 48; so, the solution is given again by Eq. 63 but, the coefficients a _n are set zero, since γ₁ is zero within this time interval.

• (b) In the time interval [τ >  τ_d], the initial condition is given by Eq. 49, where F(Y) =  θ(Y, τ_d) is obtained from Eq. 63 for τ =  τ_d and a _n =  0. Then, the solution of Eq. 58 subjected to the transformed initial condition (61) is:

  $$ \begin{aligned} \bar {\theta}(\lambda _{n}, \tau ) =& \frac{B_{1} \gamma _{1}}{\lambda _{n}^{2}}{\left[ {1 - \frac{\omega ^{2} }{{\lambda _{n}^{4} + \omega ^{2}}}{\rm e}^{- \lambda ^{2}_{n} (\tau - \tau _{\rm d})} - \frac{\lambda _{n}^{2}}{{{\sqrt {\lambda _{n}^{4} + \omega ^{2}}}}}\cos {\left( {\omega {\kern 1pt} \tau - \omega {\kern 1pt} \tau _{\rm d} - \tan ^{- 1} \frac{\omega}{{\lambda _{n}^{2}}}} \right)}} \right]} \\ & + \frac{{B_{2} \gamma _{2} {\left( {\cos \lambda _{n} + \frac{B_{1}}{{\lambda _{n}}}\sin \lambda _{n}} \right)}}}{{\lambda _{n}^{2}}}{\left[ {1 - \frac{\omega ^{2}}{{\lambda _{n}^{4} + \omega ^{2}}}{\rm e}^{- \lambda _{n}^{2} \tau} - \frac{\lambda _{n}^{2}}{{{\sqrt {\lambda _{n}^{4} + \omega ^{2}}}}}\cos {\left( {\omega {\kern 1pt} \tau - \tan ^{{- 1}} \frac{\omega}{{\lambda _{n}^{2}}}} \right)}} \right]} \end{aligned} $$

  (64)

If we introduce now Eqs. 55, 52 and 64 into the general form of the solution, i.e., Eq. 50, we obtain Eq. 32.