2.1.2. Heat Transfer through Channel Walls

Let us have two meshes representing the tube and the shell sides of a heat exchanger and focus on an arbitrary pair of adjacent control volumes representing a portion of the tube side (i.e., a tube segment) and the enclosing portion of the shell side (see Figure 5). Let . *m*t, *c*p,t, *T*t,1, and *T*t,2 denote the tube side mass flow rate, mean specific heat capacity at constant pressure, and inlet and outlet temperatures and . *m*s, *c*p,s, *T*s,1, and *T*s,2 the corresponding shell-side quantities.

**Figure 5.** A pair of adjacent control volumes representing a portion of the tube side and the enclosing portion of the shell side.

Should, e.g., the hot fluid be placed in the shell side, then the overall heat balance could be written as .

$$
\dot{m}\_{\rm t}c\_{\rm p,t}(T\_{\rm t,2} - T\_{\rm t,1}) = \dot{m}\_{\rm s}c\_{\rm p,s}(T\_{\rm s,1} - T\_{\rm s,2}).\tag{3}
$$

Let us for a moment assume that the temperature of the fluid in the shell-side control volume is constant and is equal to the shell-side inlet temperature, *T*s,1. Let us also assume that the tube-side inlet temperature, *T*t,1, is known. Additionally, let *L* denote the length of the tube-side mesh edge and *U* the cumulative overall heat transfer coefficient. The heat flux for a small portion of this edge having the length d*x* can then be expressed as

$$\mathrm{d}\dot{Q} = \dot{m}\_{\mathrm{f}}c\_{\mathrm{p},\mathrm{t}}(T\_{\mathrm{t}}(\mathbf{x}+\mathbf{dx}) - T\_{\mathrm{t}}(\mathbf{x})) = \mathrm{d}\mathrm{l}\frac{\mathrm{dx}}{L}(T\_{\mathrm{s},1} - T\_{\mathrm{t}}(\mathbf{x})).\tag{4}$$

This can be modified, rearranged, and written in integral form,

$$\int\_{T\_{\rm t,1}}^{T\_{\rm t}(\mathbf{x})} \frac{1}{T\_{\rm s,1} - T\_{\rm t}(\mathbf{x})} \mathrm{d}T\_{\rm t}(\mathbf{x}) = \int\_{0}^{\mathbf{x}} \frac{\mathcal{U}}{\dot{m}\_{\rm t} c\_{\rm p,1} L} \mathrm{d}x,\tag{5}$$

which yields

$$T\_{\mathbf{t}}(\mathbf{x}) = T\_{\mathbf{s},1} + \left(T\_{\mathbf{t},1} - T\_{\mathbf{s},1}\right) \exp\left(-\frac{\mathcal{U}}{\dot{m}\_{\mathbf{t}}c\_{\mathbf{p},\mathbf{t}}} \frac{\mathbf{x}}{L}\right), \quad \mathbf{x} \in [0, L]. \tag{6}$$

From Equation (6), we immediately see that the temperature at the end of the edge can be obtained using

$$T\_{\rm t,2} = T\_{\rm s,1} + (T\_{\rm t,1} - T\_{\rm s,1}) \exp\left(-\frac{\mathcal{U}}{\dot{m}\_{\rm t}c\_{\rm P,t}}\right). \tag{7}$$

*Energies* **2020**, *13*, 1664

Equation (7) must be linearized for it to be used in a matrix solver. This is done in a straightforward manner by taking

$$\mathbb{C}\_{\mathcal{U}} = \exp\left(-\frac{\mathcal{U}}{\dot{m}\_{\text{f}}c\_{\text{p},\text{t}}}\right) = \text{const.},\tag{8}$$

where the necessary cumulative overall heat transfer coefficient, *U*, is computed from the tube-side and shell-side heat transfer coefficients. These, in turn, are calculated using equations from literature (e.g., [28] in the case of plain tubes or [29] if the tubes are finned) depending on the actual bundle geometry and flow conditions. Additional information regarding validation of the commonly used empirical equations for estimation of heat transfer coefficient in the case of plain and serrated fins can be found, for instance, in [30]. One could also use the correlations from [31], which have been obtained via the machine learning technique. If U-shaped or helical fins of various types are employed, then the two-part article by Hofmann and Heimo [32,33] can be recommended to the reader. The final, linearized equation for a single edge, therefore, is

$$\mathbb{C}\_{L}T\_{\mathbf{t},1} - T\_{\mathbf{t},2} = \left(\mathbb{C}\_{L} - 1\right)T\_{\mathbf{s},1}.\tag{9}$$

Considering the shell-side outlet temperature, for cross-flow with *T*s,1 = const. on the entire inlet face of the control volume, we have

$$\mathrm{d}\dot{Q} = \mathrm{l}\varGamma \frac{\mathrm{d}\mathbf{x}}{L}(T\_{\mathrm{s},1} - T\_{\mathrm{t}}(\mathbf{x})) = \dot{m}\_{\mathrm{s}}c\_{\mathrm{p},\mathrm{s}}\frac{\mathrm{d}\mathbf{x}}{L}(T\_{\mathrm{s},1} - T\_{\mathrm{s},2}(\mathbf{x})).\tag{10}$$

Just as before, this can be modified and rearranged to yield

$$T\_{\rm s,2}(\mathbf{x}) = T\_{\rm s,1} + \frac{\mathcal{U}}{\dot{m}\_{\rm s}c\_{\rm P,s}} (T\_{\rm s,1} - T\_{\rm t,1}) \exp\left(-\frac{\mathcal{U}}{\dot{m}\_{\rm f}c\_{\rm P,t}} \frac{\mathbf{x}}{L}\right), \ \mathbf{x} \in [0, L]. \tag{11}$$

The mean shell-side outlet temperature then is

$$T\_{\rm s,2} = \frac{1}{L} \int\_0^L T\_{\rm s,2}(\mathbf{x}) \mathbf{dx} = T\_{\rm s,1} + \frac{\dot{m}\_{\rm t,1}}{\dot{m}\_{\rm t} c\_{\rm p,s}} \Big( \exp\Big(-\frac{\mathcal{U}}{\dot{m}\_{\rm t} c\_{\rm p,t}}\Big) - 1\Big) (T\_{\rm s,1} - T\_{\rm t,1}) \tag{12}$$

which corresponds to the shell-side outlet temperature obtained using the respective set of linear equations.

One could also simplify the model even further by using a one-dimensional shell-side mesh (i.e., a mesh such that each cross-section of the shell along the general flow direction is spanned by just one cell). With a row of *n* tubes being present in a specific shell-side cell, Equation (3) would simply become

$$\sum\_{i=1}^{n} \dot{m}\_{\mathbf{t},i} c\_{\mathbf{p},\mathbf{t},i} (T\_{\mathbf{t},i,2} - T\_{\mathbf{t},i,1}) = \dot{m}\_{\mathbf{s}} c\_{\mathbf{p},\mathbf{s}} (T\_{\mathbf{s},1} - T\_{\mathbf{s},2}) \,. \tag{13}$$

while Equation (7), still necessary for each of the *n* tubes, would remain almost identical:

$$T\_{\rm t,i,2} = T\_{\rm s,1} + (T\_{\rm t,i,1} - T\_{\rm s,1}) \exp\left(-\frac{\mathcal{U}\_{\rm i}}{\dot{m}\_{\rm t,i}c\_{\rm p,t,i}}\right) \quad i = 1, \ 2, \ldots, n. \tag{14}$$

There also is a special case of no heat transfer, which can be treated similarly. The necessary equation can easily be obtained by setting the heat transfer coefficient in Equation (14) to zero, which results in the equation being reduced to the equality between the temperatures in the end nodes of an edge. This is important because the number of linear equations describing heat transfer is always constant no matter if heat transfer occurs or not.
