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Question 11.6: A heat exchanger is made up of two identical pipes separated...

A heat exchanger is made up of two identical pipes separated by an impermeable diathermal wall of section area A, thickness h and thermal conductivity κ. In both pipes, a liquid flows at uniform velocities ν_1 = ν_1 \hat{x} and ν_2 = −ν_2 \hat{x} , with ν1 > 0 and ν_2 > 0 , where x is the unit vector that is parallel to the liquid flow in pipe 1. The temperature T_1 of the liquid in pipe 1 is larger than the temperature T_2 of the liquid in pipe 2, i.e. T_1 > T_2 .
Thus, there is a heat current density j_Q = j_Q \hat{y} , with j_Q > 0 going across the wall separating the pipes, where \hat{y} is the unit vector orthogonal to the wall and oriented positively from pipe 1 to pipe 2. There is no liquid current density across the wall, i.e. j_C = 0 . Heat conductivity is considered negligible in the direction of the flow and yet large enough to ensure a homogeneous temperature across any section of both pipes. Consider that the heat exchanger has reached a stationary state.

a) Show that the temperature profiles in the fluids are given by the differential equations,

∂_x T_1 = -\frac{κ}{h \ell c_1 ν_1}( T_1 − T_2 ) .

∂_x T_2 = -\frac{κ}{h \ell c_2 ν_2}( T_1 − T_2 ) .

where c_1 . and c_2 are the specific heat densities of liquids 1 and 2, and κ is the thermal conductivity of the diathermal wall and \ell is a characteristic length for the thermal transfer.
b) Show that the convective heat current density j = c_1 ν_1 T_1 + c_2 ν_2 T_2 . is homogeneous.
c) Determine the temperature difference ΔT (x) = T_1 (x) − T_2 (x) ..
d) Determine the temperature profiles T_1 (x) and T_2 (x) .
e) Show that on a distance that is short enough, i.e. x/d \ll 1 ,

T_1 (x) = \frac{j + c_2 ν_2 ΔT (0)}{c_1 ν_1 + c_2 ν_2} - \frac{κ ΔT (0)}{h\ell c_1 ν_1}x

T_2 (x) = \frac{j - c_1 ν_1 ΔT (0)}{c_1 ν_1 + c_2 ν_2} + \frac{κ ΔT (0)}{h\ell c_2 ν_2}x.

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a) According to relation (10.101), since there is no liquid current density across the wall, i.e. j_C = 0 , the heat current density j_Q across the wall is equal to the internal energy current density j_{u2}, which is the opposite of the internal energy current density j_{u1} due to energy conservation,

− κ ∇^2 T = −c_e ∂_t T = 0                              (11.101)

j_Q = −j_{u1} = j_{u2}

Since the flow of liquid in both pipes is uniform, there is no fluid expansion, i.e. ∇· ν_1 = ∇· ν_2 = 0 . Moreover, since there is no mechanical constraint on the liquids, i.e. π_{u1} = π_{u2} = 0 , the internal energy continuity equations (10.43) for the liquid in pipes 1 and 2 can be recast as,

\dot{u} + (∇· ν) u +∇· j_u = π_u .                             (10.43)

\dot{u}_1 = −∇· j_{u1} = ∇· j_Q .

\dot{u}_2 = −∇· j_{u2} = – ∇· j_Q .

The internal energy densities u_1 and u_2 are written in terms of the temperatures T_1 and T_2 of the liquids as,

u_1 = c_1 T_1         and        u_2 = c_2 T_2

In a stationary state the time derivatives of the internal energy densities vanish, i.e. ∂_t u_1 = ∂_t u_2 = 0. Thus, by applying relation (10.18) to the internal energy densities u_1 and u_2 in the particular case of a uniform liquid flow at velocities ν_1 =ν_1 \hat{x} and ν_2 = −ν_2 \hat{x}, we find,

\dot{f} (x, t) = ∂_t f (x, t) + (ν (x, t) ·∇) f (x, t)          (10.18)

\dot{u}_1 = ∂_t u_1 + ν_1 ·∇u_1 = ν_1 ∂_x u_1 = c_1 ν_1 ∂_x T_1

 

\dot{u}_2 =∂_t u_2 + ν_2 ·∇u_2 = −ν_2 ∂_x u_2 = −c_2 ν_2 ∂_x T_2

According to equation (10.104) for an infinitesimal section of wall of volume dV, of infinitesimal cross section dA and of thickness h, the thermal power is given by,

P_Q = – \int_{S}{dS .j_Q} = -\int_{V}{dV(\nabla .j_Q )}       (10.104)

 

P_Q = −dV ∇· j_Q = −h dA ∇· j_Q

In view of the discrete Fourier law (11.21), the divergence of the heat current density j_Q is written as,

P_Q = \kappa \frac{A}{\ell } \Bigl(T^+-T^-\Bigr)       (11.21)

Thus, the spatial derivatives of the tem ∇· j_Q = − \frac{P_Q}{h dA} = – \frac{κ}{h \ell}( T_1 − T_2 )

Thus, the spatial derivatives of the temperatures are given by,

∂_x T_1 = – \frac{κ}{h \ell c_1 ν_1}( T_1 − T_2 )

 

∂_x T_2 = – \frac{κ}{h \ell c_2 ν_2}( T_1 − T_2 )

b) In view of the spatial derivatives of the temperatures,

∂_x j = ∂_x (c_1 ν1 T_1 + c_2 v_2 T2) = c_1 ν_1 ∂_x T_1 + c_2 ν_2 ∂_x T_2 = 0

which implies that the convective heat current density j is homogeneous.

c) The difference between the spatial derivatives of the temperatures can be written as,

∂_x (T_1 − T_2) = −\frac{1}{d}(T_1 − T_2)

where the decay length d is given by,

\frac{1}{d}=\frac{κ}{h \ell c_1 ν_1}+\frac{κ}{h \ell c_2 ν_2}=\frac{k}{h\ell} (\frac{c_1 ν_1+c_2 ν_2}{c_1 ν_1c_2 ν_2} )

Thus, the temperature difference ΔT (x) = T_1 (x) − T_2 (x) decays exponentially

ΔT (x) = ΔT (0) exp(-\frac{x}{d} )

d) The convective heat current density j can be recast as,

j = c_1 ν_1 T_1 (x) + c_2 ν_2 \biggl(T_1(x) – \Delta T(0) \exp\Bigl(-\frac{x}{d}\Bigr) \biggr)

 

j = c_1 ν_1 \biggl(T_2(x) – \Delta T(0) \exp\Bigl(-\frac{x}{d}\Bigr) \biggr)+ c_2 ν_2 T_2 (x)

Thus, the temperature profiles are given by,

T_1 (x) = \frac{1}{c_1 ν_1 + c_2 ν_2} \biggl(j+ c_2 ν_2\Delta T(0) \exp\Bigl(-\frac{x}{d}\Bigr) \biggr)

 

T_2 (x) = \frac{1}{c_1 ν_1 + c_2 ν_2} \biggl(j- c_1 ν_1\Delta T(0) \exp\Bigl(-\frac{x}{d}\Bigr) \biggr)

e) When the heat transfer occurs on a distance that is short enough, i.e. to first-order in x/d, the temperature profiles reduce to,

T_1 (x) = \frac{j + c_2 ν_2 ΔT (0)}{c_1 ν_1 + c_2 ν_2} – \frac{κ ΔT (0)}{h \ell c_1 ν_1} x

 

T_2 (x) = \frac{j – c_1 ν_1 ΔT (0)}{c_1 ν_1 + c_2 ν_2} + \frac{κ ΔT (0)}{h \ell c_2 ν_2} x .

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