**7.1 Hypothesis**

To expand the mathematical model, we look at this simple idea to do the following:

*Assessment of Augmentation Techniques to Intensify Heat Transmission Power DOI: http://dx.doi.org/10.5772/intechopen.101670*


#### **7.2 Modeling equations**

By taking heat balance of hot & cold fluid, we get,

$$Q\_h \mathbf{C} p\_h \ (T\_1 - T\_2) = \mathbf{Q}\_c \mathbf{C} p\_c \ (T\_4 - T\_3) \tag{1}$$

$$Q\_h \mathbf{C} p\_h \left( T\_1 - T\_2 \right) = U A \Delta T\_{lm} \tag{2}$$

$$Q\_h C p\_h \ (T\_1 - T\_2) = \frac{UA \left\{ \left( T\_1 - T\_4 \right) - \left( T\_2 - T\_3 \right) \right\}}{\ln \frac{\left( T\_1 - T\_4 \right)}{\left( T\_2 - T\_3 \right)}} \tag{3}$$

The heat exchanger's mathematical model has been constructed, and it includes a heat balance equation for the two material fluxes Qh and Qc, as well as an expression for transferred heat flow **Table 1**. The overall heat exchange coefficient, U, has a standard expression as the overall HTC, which may be given by Eq. (4), for the heat flow transported in the heat exchanger.

$$U = \frac{1}{\left\{ \begin{pmatrix} \frac{1}{h\_i} \end{pmatrix} \begin{pmatrix} \frac{d\_r}{d\_i} \end{pmatrix} + \begin{pmatrix} \frac{d\_r}{2k} \end{pmatrix} \ln \begin{pmatrix} \frac{d\_r}{d\_i} \end{pmatrix} + \begin{pmatrix} \frac{1}{h\_o} \end{pmatrix} \right\}}\tag{4}$$

#### **7.3 Solving of the mathematical model**

Eq. (1) represents a system of two non-linear equations with two variables having the form,

$$f\_1\left(T\_2, T\_4\right) = \mathbf{0}, \ f\_2\left(T\_2, T\_4\right) = \mathbf{0} \tag{5}$$

$$f\_1 = Q\_h Q p\_h \left(T\_1 - T\_2\right) \\ \quad - Q\_c Q p\_c \left(T\_4 - T\_3\right) \tag{6}$$

$$\begin{aligned} \,\_2f\_2 = \left\{ \mathbf{Q}\_h \mathbf{C} p\_h \, \left( T\_1 - T\_2 \right) \right\} \, \, - \left\{ \frac{UA \, \left\{ \left( T\_1 - T\_4 \right) - \left( T\_2 - T\_3 \right) \right\}}{\ln \frac{\left( T\_1 - T\_4 \right)}{\left( T\_2 - T\_3 \right)}} \right\} \end{aligned} \tag{7}$$

The equation's unknown variables (4), the hot fluid's T2 outlet temperature, and the cold fluid's T4 outlet temperature, are also the heat exchanger's output variables. The expressions of the functions f1 and f2 of the Eq. (5) are defined by the relations (6) and (7).


**Table 1.** *Experimental parameters.* We can get solutions from two methods,


The Jacobean matrix associated with the linear Eq. (5) is given by,

$$f(\mathbf{X}) = \begin{bmatrix} \frac{\partial f\_1}{\partial T\_2} & \frac{\partial f\_1}{\partial T\_4} \\\\ \frac{\partial f\_2}{\partial T\_2} & \frac{\partial f\_2}{\partial T\_4} \end{bmatrix} \tag{8}$$

$$\frac{\partial f\_1}{\partial T\_2} = -Q\_h C p\_h \tag{9}$$

$$\frac{\partial f\_1}{\partial T\_4} = -Q\_c \mathbf{C} p\_c \tag{10}$$

$$\frac{\partial f\_{2}}{\partial T\_{2}} = -Q\_{h} \mathbf{C} p\_{h} - \left\{ \frac{[U \, \mathbf{A} \, \{ \, (T\_{1} - T\_{4}) - (T\_{4} - T\_{3}) \} ] \left[ \frac{(T\_{1} - T\_{4})}{(T\_{2} - T\_{3})} \right]^{2}}{\ln \left[ \frac{(T\_{1} - T\_{4})}{(T\_{2} - T\_{3})} \right]^{2}} \right\} \tag{11}$$

### **7.4 Example a certain amount when using the insert to install a heat transfer system**

Assuming, Temperature T1 = 70 ̊C and T3 = 29 ̊C; μ at 70̊C = 0.0004101 Pa ses; k = 0.62136 W/m K; de = 0.0280 m; di = 0.013 m;

Cp = 4186 J/kg K. Apply Eq. (5)

$$\mathbf{h}\_{\mathrm{i}} = \frac{\mathbf{j}^{\mathrm{H}} \,\mathrm{k}}{\mathrm{d}\_{\mathrm{i}} \left(\mathrm{C}\_{\mathrm{p}} \,\,\frac{\mu}{\mathrm{k}}\right)^{\circ\_{\mathrm{i}}}}$$

$$\mathbf{h}\_{\mathrm{o}} = \frac{\mathbf{j}^{\mathrm{H}} \,\mathrm{k}}{\mathrm{d}\_{\mathrm{e}} \left(\mathrm{C}\_{\mathrm{p}} \,\,\frac{\mu}{\mathrm{k}}\right)^{\circ\_{\mathrm{i}}}}$$

$$\mathbf{N}\_{\mathrm{re}} = \frac{\mathbf{d}\_{\mathrm{i}} \,\mathbf{v} \,\,\mathrm{\rho}}{\mu}$$

$$\mathbf{v} = \frac{\mathbf{Q}}{\mathrm{A}}$$

By using Eqs. (1), (2) and (3) and substituting all above calculated values we get,

$$
\dot{\mathbf{m}} \cdot \mathbf{C\_p} \cdot \Delta \mathbf{T\_{[hot}} = \dot{\mathbf{m}} \cdot \mathbf{C\_p} \cdot \Delta \mathbf{T\_{[cold}}}
$$

T2 = 190.35–4.15 T4

$$\dot{\mathbf{m}} \cdot \mathbf{C\_p} \ (\mathbf{T\_1} - \mathbf{T\_2}) = \mathbf{U} \ \mathbf{A} \Delta \mathbf{T\_{lm}}$$

By substituting value of T2 from Eq. (1)

$$\text{f}(\text{T}\_4) - \frac{(70 - \text{T}\_4) - (161.35 - 4.15 \text{ T}\_4)}{\ln(70 - \text{T}\_4) - \ln(161.35 - 4.15 \text{ T}\_4)} - 66.15 \text{ T}\_4 + 1918.07$$

*Assessment of Augmentation Techniques to Intensify Heat Transmission Power DOI: http://dx.doi.org/10.5772/intechopen.101670*

From trial and error method, T4 = 38 ̊C. From Eq. (1), substituting T4. T2 = 49.25 ̊C.
