**4. Simultaneous heat and mass transfer in cooling and dehumidifying coils**

In the cooling coil, the coolant fluid "chilled water or refrigerant" flows inside the tubes and the air passes across the tube bundle. Since the coolant fluid temperature is less than the dew point temperature to ensure the dehumidification process there is possibility of heat and moisture transfer between them. The directions of heat and moisture transfer depend upon the temperature and vapor pressure differences between air and wetted surface. As a result, the direction of the total heat transfer rate, which is a sum of sensible heat transfer and latent heat transfers. The concept of enthalpy potential [4] is very useful in quantifying the total heat transfer in these processes and its direction.

The sensible (QS) and latent (QL) heat transfer rates are given by:

$$\mathbf{Q\_S = h\_o A\_S \ (t\_i - t\_a)}$$

$$\mathbf{Q\_L = h\_{mass} A\_S \ (W\_i - W\_a)} \ \mathbf{h\_{fg}}$$

the total heat transfer QT is given by:

$$\mathbf{Q\_{T} = Q\_{S} + Q\_{L} = h\_{o} A\_{S} \ (t\_{i} - t\_{a}) + h\_{mass} \ A\_{s} \ (W\_{i} - W\_{a}) \ h\_{\tilde{y}}}$$

Where:

370 Heat Exchangers – Basics Design Applications

There are three standard plate fin patterns that are usually used in the cooling coil: flatplate, wavy-plate, and star-plate fin patterns, as shown in Figure 5. They are made of Aluminum, copper, and stainless steel or carbon steel. The fins are permanently attached to the tubes by expansion of each tube. Full fin collars allow for both precise fin spacing and maximum fin-to-tube contact. The flat-plate fin type has no corrugation, which results in the lowest possible air friction drop and lowest fan horsepower demands while the wavy-plate fin corrugation across the fin provides the maximum heat transfer for a given surface area, and is the standard fin configuration used. The star-plate fin pattern corrugation around the tubes provides lower air friction. This pattern is used when lower air friction is desired

(a) (b) (c)

Fig. 5. (a) Wavy-plate fin; (b) Star-plate fin; (c) Flat-plate fin (*Aerofin heat transfer products).*

**4. Simultaneous heat and mass transfer in cooling and dehumidifying coils** In the cooling coil, the coolant fluid "chilled water or refrigerant" flows inside the tubes and the air passes across the tube bundle. Since the coolant fluid temperature is less than the dew point temperature to ensure the dehumidification process there is possibility of heat

Fig. 4. Full circuit and half circuit four row coils with 4-tube face.

without a large decrease in heat transfer capacity.

**3.1 Fin patterns**

t a = dry-bulb temperature of air, oC t i= temperature of water/wetted surface, oC Wa = humidity ratio of air, kg/kg Wi= humidity ratio of saturated air at ti, kg/kg ho = convective heat transfer coefficient, W/m2.oC hmass = convective mass transfer coefficient, kg/m2 hfg = latent heat of vaporization, J/kg

Since the transport mechanism that controls the convective heat transfer between air and water also controls the moisture transfer between air and water, there exists a relation between heat and mass transfer coefficients, hC and hD as discussed in an earlier chapter. It has been shown that for air-water vapor mixtures,

$$\mathbf{H}\_{\text{mass}} \approx \mathbf{h}\_o / \mathbf{c}\_{\text{pm}} \text{ or } \mathbf{h}\_o / \mathbf{h}\_{\text{mass}}. \mathbf{c}\_{\text{pm}} = \text{Lewis number} \approx 1.0 \,\text{s}$$

Where cpm is the humid air specific heat ≈ 1.0216 kJ/kg.K. Hence the total heat transfer is given by:

$$\mathbf{Q\_{\overline{t}} = \mathbf{Q\_{\overline{s}}} + \mathbf{Q\_{\overline{t}}} = \mathbf{h\_{0}}} \text{ A\_{\overline{s}} (t\_{\overline{i}} - t\_{\overline{a}}) + \mathbf{h\_{\text{mass}}} \text{ A\_{\overline{s}} (W\_{\overline{i}} - W\_{\mathbf{a}}) } \mathbf{h\_{\overline{t}}} = \left( \mathbf{h\_{0}} \, \mathbf{A\_{\overline{s}}} / \mathbf{C\_{pm}} \right) \left[ \left( \mathbf{t\_{\overline{i}} - t\_{\overline{a}}} \right) + \left( W\_{\overline{i}} - W\_{\mathbf{a}} \right) \, \mathbf{h\_{\overline{\overline{s}}}} \right]$$

by manipulating the term in the parenthesis of RHS, it can be shown that:

$$\mathbf{Q\_{T} = Q\_{S} + Q\_{L} = (h\_{o} \ A\_{S} / c\_{\text{pm}}) [ (h\_{i} - h\_{a}) ]}$$

The air heat transfer coefficient, ho has been computed from the experimental correlations derived in [3]. The heat transfer parameter is written as Stanton number, *St* times Prandtl number, *Pr* to the 2/3 power. It is given as a function of Reynolds number, *Re* where the function was established through curve-fitting of a set of the experimental data as follow:

$$St \times Pr^{(2/3)} = 0.1123 \times Re^{-0.261}$$

Where these three dimensionless parameters are defined as:

$$\text{St} = \frac{\{\text{A}\_{\text{min}} \ge \text{h}\_{\text{o}}\}}{\left(\text{m}\_{\text{a}} \times \text{c}\_{\text{pm}}\right)}, \text{Pr} = \frac{\{\mu\_{\text{a}} \times \text{c}\_{\text{pm}}\}}{\text{k}\_{\text{a}}}, \text{and } \text{Re} = \frac{\{\text{m}\_{\text{a}} \times \text{d}\_{\text{o}}\}}{\{\text{A}\_{\text{min}} \ge \mu\_{\text{a}}\}}$$




$$
\Delta \mathbf{Q}\_{\rm cl} = \mathbf{m}\_{\rm a} (\mathbf{ha}\_{\rm l} - \mathbf{ha}\_{\rm l+1}) \tag{1}
$$

$$
\Delta \mathbf{Q}\_{\rm cl} = \frac{\eta\_{\rm s}}{c\_{\rm pm}} \mathbf{h}\_{\rm o} \Delta \mathbf{A}\_{\rm o} (\mathbf{h}\_{\rm mil} - \mathbf{h} \mathbf{s}\_{\rm ml}) \tag{2}
$$

$$
\Delta \mathbf{Q}\_{\rm cl} = \mathbf{m}\_{\rm W} \mathbf{C} \mathbf{p}\_{\rm W} (\mathbf{T} \mathbf{w}\_{\rm l+1} - \mathbf{T} \mathbf{w}\_{\rm l}) \tag{3}
$$

$$
\Delta \mathbf{Q}\_{\rm cl} = \mathbf{h}\_{\rm l} \Delta \mathbf{A}\_{\rm l} (\mathbf{T} \mathbf{s}\_{\rm ml} - \mathbf{T} \mathbf{w}\_{\rm ml}) \tag{4}
$$

$$\mathbf{h}\_{\rm ml} = \frac{(\mathbf{h}\_{\rm l} + \mathbf{h}\_{\rm l+1})}{2} \qquad , \mathbf{h}\_{\rm l+1} = 2\mathbf{h}\_{\rm ml} - \mathbf{h}\_{\rm l} \tag{5}$$

$$\mathbf{T} \mathbf{w}\_{\rm ml} = \frac{(\mathbf{T} \mathbf{w}\_{\rm l} + \mathbf{T} \mathbf{w}\_{\rm l+1})}{2} \quad , \mathbf{T} \mathbf{w}\_{\rm l+1} = 2 \mathbf{T} \mathbf{w}\_{\rm ml} - \mathbf{T} \mathbf{w}\_{\rm l} \tag{6}$$

$$
\Delta \mathbf{Q}\_{\rm cl} = 2 \mathbf{m}\_{\rm a} (\mathbf{h} \mathbf{a}\_{\rm l} - \mathbf{h} \mathbf{a}\_{\rm ml}) \tag{7}
$$

$$
\Delta \mathbf{Q}\_{\rm cl} = 2 \mathbf{m}\_{\rm w} \mathbf{C} p\_{\rm w} (\mathbf{T} \mathbf{w}\_{\rm mi} - \mathbf{T} \mathbf{w}\_{\rm l}) \tag{8}
$$

$$
\Delta \mathbf{Q}\_{\rm cl} = \frac{\eta\_{\rm s} \,\mathrm{h}\_{\rm o} \Delta \mathrm{A}\_{\rm o} / c\_{\rm pm}}{1 + \Delta \mathrm{NTU}\_{\rm o}/2} \ast \left( \mathrm{ha}\_{\rm l} - \mathrm{h} \mathrm{s}\_{\rm mi} \right) \tag{9}
$$

$$
\Delta \mathbf{Q}\_{\rm cl} = \frac{\mathbf{h}\_{\rm l} \Delta \mathbf{A}\_{\rm l}}{\mathbf{1} + \Delta \mathbf{N} \mathbf{T} \mathbf{U}\_{\rm l} / 2} \ast \left( \mathbf{T} \mathbf{s}\_{\rm mi} - \mathbf{T} \mathbf{w}\_{\rm l} \right) \tag{10}
$$

$$\frac{\mathbf{h}\_{\rm h} - \mathbf{h}\_{\rm ml}}{\mathbf{T} \mathbf{s}\_{\rm ml} - \mathbf{T} \mathbf{w}\_{\rm l}} = \mathbf{R} \tag{11}$$

$$R = \begin{bmatrix} \frac{\mathbf{h}\_{\text{l}} \mathbf{c}\_{\text{pm}}}{\mathbf{h}\_{\text{o}} \eta\_{\text{s}}} \left( \frac{\Delta \mathbf{A}\_{\text{l}}}{\Delta \mathbf{A}\_{\text{o}}} \right) \end{bmatrix} \ast \begin{bmatrix} \left( 1 + \frac{\Delta \text{NTU}\_{\text{o}}}{2} \right) \\ \left( 1 + \frac{\Delta \text{NTU}\_{\text{l}}}{2} \right) \end{bmatrix} \tag{12}$$

$$\Delta \text{NTU}\_{\text{o}} = \frac{\eta\_{\text{s}} \text{ h}\_{\text{o}} \Delta \text{A}\_{\text{o}}}{\text{m}\_{\text{a}} \text{c}\_{\text{pm}}} \; , \qquad \Delta \text{NTU}\_{\text{i}} = \frac{\text{h}\_{\text{l}} \Delta \text{A}\_{\text{l}}}{\text{m}\_{\text{W}} \text{C}\_{\text{pw}}}$$

$$\mathbf{h}\mathbf{s}\_{\rm ml} = \mathbf{h}\_{\rm a} + \mathbf{c}\_{\rm p} \text{ (Ts}\_{\rm ml}\text{-T}\_{\rm a}\text{)}\tag{13}$$

$$\rm hs\_{ml} = 10.76 + 1.4 \,\rm Ts\_{ml} + 0.046 \,\rm Ts\_{ml}^2 \tag{14}$$

$$\text{ha}\_{\text{l}}-\text{hs}\_{\text{ml}} = \text{R} \text{(Ts}\_{\text{ml}}-\text{Tw}\_{\text{l}}\text{)}$$

$$\text{ha}\_{\text{l}}-\left\{10.76+1.4\*\text{Ts}\_{\text{ml}}+0.046\*\text{Ts}\_{\text{mi}}^{2}\right\} = \text{R} \text{(Ts}\_{\text{ml}}-\text{Tw}\_{\text{l}}\text{)}\;\prime$$

$$0.046\text{ }\text{Ts}\_{\text{ml}}^{2}+\text{(R}+1.4)\*\text{Ts}\_{\text{ml}}-\left(\text{ha}\_{\text{l}}+\text{R}\*\text{Tw}\_{\text{l}}-10.76\right) = 0$$

$$\mathbf{a} \mathbf{T} \mathbf{s}\_{\rm mi}^2 + \mathbf{b} \mathbf{T} \mathbf{s}\_{\rm mi} - \mathbf{c} = \mathbf{0}$$

$$\mathrm{Ts}\_{\mathrm{m}} = \frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2 - 4\mathrm{ac}}}{\mathrm{za}} \tag{15}$$

$$\Delta \mathbf{Q}\_{\rm cl} = \frac{\mathbf{h}\_{\rm l} \Delta \mathbf{A}\_{\rm l}}{\mathbf{1} + \Delta \mathbf{T} \mathbf{U}\_{\rm l}/2} \ast \left( \mathbf{T} \mathbf{s}\_{\rm ml} - \mathbf{T} \mathbf{w}\_{\rm l} \right), \quad \mathbf{h}\_{\rm l+1} = \mathbf{h}\_{\rm l} - \frac{\Delta \mathbf{Q}\_{\rm cl}}{\mathbf{m}\_{\rm a}} \quad , \mathbf{T} \mathbf{w}\_{\rm l+1} = \mathbf{T} \mathbf{w}\_{\rm l} - \frac{\Delta \mathbf{Q}\_{\rm cl}}{\mathbf{m}\_{\rm w} \mathbf{C} \mathbf{p}\_{\rm w}}$$

$$\Delta \mathbf{Q}\_{\rm cl} = \mathbf{m}\_{\rm a} \mathbf{c}\_{\rm pa} (\mathbf{T} \mathbf{a}\_{\rm l} - \mathbf{T} \mathbf{a}\_{\rm l+1}) \tag{16}$$

$$
\Delta \mathbf{Q}\_{\rm cl} = \eta\_{\rm s} \mathbf{h}\_{\rm o} \Delta \mathbf{A}\_{\rm o} (\mathbf{T} \mathbf{a}\_{\rm ml} - \mathbf{T} \mathbf{s}\_{\rm ml}) \tag{17}
$$

$$\mathbf{Q}\_{\rm cl} = \eta\_{\rm s} \mathbf{h}\_{\rm o} \Delta \mathbf{A}\_{\rm o} \left( \frac{\mathbf{T} \mathbf{a}\_{\rm l+1} + \mathbf{T} \mathbf{a}\_{\rm l}}{2} - \mathbf{T} \mathbf{s}\_{\rm ml} \right) \tag{18}$$

$$\mathrm{Ta}\_{\mathrm{l}+1} = \begin{bmatrix} \left( 1 \frac{\Delta \mathrm{NTU}\_{\mathrm{o}}}{\mathrm{z}} \right) \\ \left( 1 + \frac{\Delta \mathrm{NTU}\_{\mathrm{o}}}{\mathrm{z}} \right) \end{bmatrix} \times \mathrm{Ta}\_{\mathrm{l}} + \begin{bmatrix} \left( \Delta \mathrm{NTU}\_{\mathrm{o}} \right) \\ \left( 1 + \frac{\Delta \mathrm{NTU}\_{\mathrm{o}}}{\mathrm{z}} \right) \end{bmatrix} \times \mathrm{Ts}\_{\mathrm{mi}} \tag{19}$$

$$\mathbf{W}\mathbf{a}\_{l+1} = \begin{bmatrix} \frac{\left(\mathbf{h}\mathbf{a}\_{l+1} - \mathbf{c}\_{\mathrm{pa}} \times \mathbf{T}\mathbf{a}\_{l+1}\right)}{\left(250\mathbf{1} + \mathbf{1}.6 \times \mathbf{T}\mathbf{a}\_{l+1}\right)} \end{bmatrix} \tag{20}$$

$$\text{Ts}\_{\text{ml}} = \frac{-(\text{R} + 1.4) + \sqrt{(\text{R} + 1.4)^2 + 0.104 \times (\text{ha}\_l + \text{R} \times \text{Tw}\_l - 10.76)}}{0.092} \tag{21}$$

$$\mathbf{T} \mathbf{w}\_{\mathbf{l}+\mathbf{1}} = \mathbf{T} \mathbf{w}\_{\mathbf{l}} - \frac{\omega\_{\mathbf{C}\mathbf{l}}}{\mathbf{m}\_{\mathbf{w}} \mathbf{C} \mathbf{p}\_{\mathbf{w}}} \tag{22}$$

$$\mathbf{ha\_{l+1}} = \mathbf{ha\_l} - \frac{\mathbf{a\_{Cl}}}{\mathbf{m\_a}} \tag{23}$$

$$\mathrm{Ta}\_{\mathrm{l}+1} = \begin{bmatrix} \left( 1 \frac{\mathrm{\Delta NTU\_{o}}}{\mathrm{z}} \right) \\ \left( 1 + \frac{\mathrm{\Delta NTU\_{o}}}{\mathrm{z}} \right) \end{bmatrix} \times \mathrm{Ta}\_{\mathrm{l}} + \begin{bmatrix} \left( \mathrm{\Delta NTU\_{o}} \right) \\ \left( 1 + \frac{\mathrm{\Delta NTU\_{o}}}{\mathrm{z}} \right) \end{bmatrix} \times \mathrm{Ts}\_{\mathrm{mi}} \tag{24}$$

$$\mathbf{W}\mathbf{a}\_{\mathbf{l}+1} = \begin{bmatrix} \frac{\left(\mathbf{h}\mathbf{a}\_{\mathbf{l}+1} - \mathbf{c}\_{\mathbf{p}\mathbf{a}} \times \mathbf{T}\mathbf{a}\_{\mathbf{l}+1}\right)}{\left(250\mathbf{1} + \mathbf{1}.6 \times \mathbf{T}\mathbf{a}\_{\mathbf{l}+1}\right)} \end{bmatrix} \tag{25}$$

$$R = \begin{bmatrix} \frac{\mathbf{h}\_{\text{l}} \mathbf{c}\_{\text{pa}}}{\mathbf{h}\_{\text{o}} \eta\_{\text{s}}} \left( \frac{\Delta \mathbf{A}\_{\text{l}}}{\Delta \mathbf{A}\_{\text{o}}} \right) \end{bmatrix} \* \begin{bmatrix} \left( 1 + \frac{\Delta \mathbf{T} \mathbf{U}\_{\text{o}}}{2} \right) \\ \left( 1 + \frac{\Delta \mathbf{T} \mathbf{U}\_{\text{l}}}{2} \right) \end{bmatrix} \tag{26}$$


$$\mathbf{m\_a} = \frac{\mathbf{q\_{\mathcal{L}}}}{(\mathbf{ha\_{\text{in}}} - \mathbf{ha\_{out}})} \qquad \qquad \mathbf{kg/s}$$

$$\mathbf{m\_w} = \frac{\mathbf{q\_c}}{\mathbf{c\_{Pw}(T\_{Wout} - T\_{W\_{\ln}})}} \quad \text{kg/s}$$

$$\Delta\mathbf{A\_o} = \left[\frac{(\mathbb{g} \times \mathbf{S\_L} \times \mathbf{m\_a})}{(\rho\_{\text{a}} \times \mathbf{V\_{face}})}\right] \text{ m}^2$$


$$\begin{aligned} m\_a &= \frac{Q\_\mathcal{C}}{(h a\_{\text{in}} - h a\_{\text{out}})} = \frac{60}{(54 - 33)} = 2.857 \text{ kg/s} \\\ m\_\text{w} &= \frac{Q\_\mathcal{C}}{C p\_\text{w} (T \nu\_{\text{out}} - T \nu\_{\text{in}})} = \frac{60}{4.14 \ast 5} = 2.90 \text{ kg/s} \end{aligned}$$

$$
\Delta A\_o = \frac{\beta \times \mathbf{S}\_\mathbf{L} \times \mathbf{m}\_a}{\rho\_\text{a} \times \mathbf{v}\_{\text{face}}} = \frac{1060 \times 0.02616 \times 2.857}{1.16 \times 2.8} = 24.39 \text{ m}^2
$$

$$
\Delta NTU\_o = \frac{\eta\_s \ h\_o \Delta A\_o}{m\_a c p m} = \frac{0.85 \times 60 \times 24.39}{2.857 \times 1001} = 0.435
$$

$$
\Delta A\_l = \left(\frac{A\_l}{A\_o}\right) \times \Delta A\_o = \frac{24.39}{23} = 1.06 \text{ m}^2
$$

$$
\Delta \text{NTU}\_l = \frac{h\_l \Delta A\_l}{m\_w c p\_w} = \frac{4000 \times 1.06}{2.9 \times 4114} = 0.355
$$

$$
\Delta = \left[\frac{h\_l c\_{\text{pa}}}{h\_l c\_{\text{pa}}} \left(\frac{\Delta A\_l}{\Delta}\right)\right]\_\text{+} \left[\left(1 + \frac{\Delta \text{NTU}\_o}{2}\right)\right] = 3.5725 \text{ Kf/kg K}
$$

$$TS\_{m1} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \times (h a\_1 + R \ast T \mathbf{w}\_1 - 10.76)}}{0.092} = 14.65 \text{ °C}$$

$$\Delta Q\_{c1} = \frac{h\_l \Delta A\_l}{1 + \Delta \text{NTU}\_l / Z} \times (\text{Ts}\_{m1} - \text{Tw}\_1) = 13.15 \text{ kW}$$

$$Tw\_2 = Tw\_1 - \frac{\Delta Q\_{c1}}{m\_w C p\_w} = 11 - \frac{13.15}{2.9 \ast 4.14} = 9.9 \text{ °C}$$

$$ha\_2 = ha\_1 - \frac{\Delta Q\_{c1}}{m\_a} = 49.4 \text{ kJ/kg}$$

$$\text{Ta}\_{i+1} = \left[\frac{\left(1 - \frac{\Delta \text{NTU}\_0}{2}\right)}{\left(1 + \frac{\Delta \text{NTU}\_0}{2}\right)}\right] \times \text{Ta}\_1 + \left[\frac{\left(\Delta \text{NTU}\_0\right)}{\left(1 + \frac{\Delta \text{NTU}\_0}{2}\right)}\right] \times \text{Ts}\_{\text{m1}} = 21.87 \text{ °C}$$

$$\text{Wa}\_2 = \frac{ha\_2 - CpaTa\_2}{2501 + 1.9 \ast Ta\_2} = \frac{49.4 - 1(21.87)}{2501 + 1.8(21.87)} = 0.01083 \text{ kg} \text{/kg} \text{s}$$

$$T s\_{m2} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \times (ha\_2 + R \ast T w\_2 - 10.76)}}{0.092} = 13.28 \text{ eV}$$

$$\Delta Q\_{c2} = \frac{h\_1 \Delta \mathbf{A}\_1}{1 + \Delta \text{NTU}\_1/2} \times (T s\_{m2} - T w\_2) = 12.17 \text{ kW}$$

$$T w\_3 = T w\_2 - \frac{\Delta q\_{c2}}{m\_w c p\_w} = 9.9 - \frac{12.17}{2.9 \ast 4.14} = 8.89 \text{ °C}$$

$$ha\_3 = ha\_2 - \frac{\Delta q\_{c2}}{m\_a} = 45.14 \text{ kJ/kg}$$

$$Ta\_3 = \begin{bmatrix} \left(1 \frac{\Delta \text{NTU}\_0}{\text{2}}\right) \\ \left(1 + \frac{\Delta \text{NTU}\_0}{\text{2}}\right) \end{bmatrix} \times Ta\_2 + \begin{bmatrix} \left(\frac{\Delta \text{NTU}\_0}{\text{1} + \frac{\Delta \text{NTU}\_0}{\text{2}}}\right) \times Ts\_{m2} = 18.74 \text{ °C} $$
 
$$Wa\_3 = \frac{ha\_3 - CapTa\_3}{2501 + 1.8 \times Ta\_3} = \frac{45.14 - 1(18.74)}{2501 + 1.8(18.74)} = 0.0104 \text{ kg} / \text{kg} \text{ } $$

$$Ts\_{m3} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.164 \cdot (ha\_3 + R \cdot Tw\_3 - 10.76)}}{0.092} = 12.0 \text{ °C}$$

$$\Delta Q\_{c3} = \frac{\text{h}\_1 \Delta \text{h}\_1}{1 + \Delta \text{NTU}\_1/2} \times (Ts\_{m3} - Tw\_3) = 11.2 \text{ kW}$$

$$Tw\_4 = Tw\_3 - \frac{\Delta Q\_{c3}}{m\_w c p\_w} = 8.89 - \frac{11.2}{2.9 \times 4.14} = 7.96 \text{ °C}$$

$$ha\_4 = ha\_3 - \frac{\Delta Q\_{c3}}{m\_a} = 41.22 \text{ kJ/kg}$$

$$Ta\_4 = \left[\frac{\left(1 - \frac{\Delta \text{NTU}\_0}{2}\right)}{\left(1 + \frac{\Delta \text{NTU}\_0}{2}\right)}\right] \times Ta\_3 + \left[\frac{\left(\Delta \text{NTU}\_0\right)}{\left(1 + \frac{\Delta \text{NTU}\_0}{2}\right)}\right] \times Ts\_{m3} = 16.27 \text{ °C}$$

$$\text{Wa}\_4 = \frac{ha\_4 - CpaTa\_4}{2501 + 1.87Ta\_4} = \frac{41.22 - 1(16.27)}{2501 + 1.8(16.27)} = 0.00986 \text{ kg}\_{\text{v}}/\text{kg}\_{\text{s}}$$

$$Ts\_{m4} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.194 \times (ha\_4 + R \ast Tw\_4 - 10.76)}}{0.092} = 10.8 \text{ \textdegree C}$$

$$\Delta Q\_{C4} = \frac{h\_1 \Delta A\_1}{1 + \Delta \text{NTU}\_1 / 2} \times (Ts\_{m4} - Tw\_4) = 10.22 \text{ kW}$$

$$Tw\_5 = Tw\_4 - \frac{\Delta Q\_{c4}}{m\_w C p\_w} = 7.96 - \frac{10.22}{2.9 \text{ \textdegree A}} = 7.11 \text{ \textdegree C}$$

$$ha\_5 = ha\_4 - \frac{\Delta Q\_{c4}}{m\_a} = 37.64 \text{ kJ/kg}$$

$$Ta\_5 = \left[\frac{\left(1 - \frac{\Delta \text{NTU}\_0}{2}\right)}{\left(1 + \frac{\Delta \text{NTU}\_0}{2}\right)}\right] \times Ta\_4 + \left[\frac{\left(\Delta \text{NTU}\_0\right)}{\left(1 + \frac{\Delta \text{NTU}\_0}{2}\right)}\right] \times Ts\_{m4} = 14.27 \text{ \textdegree C}$$

$$\text{Wa}\_5 = \frac{ha\_{5-\text{cpu}} \cdot a\_5}{2501 + 1.8 \text{ }Ta\_5} = \frac{37.64 - 1(14.27)}{2501 + 1.8(14.27)} = 0.00925 \text{ kg} \text{\textdegree C} \text{g/kg}$$

$$T s\_{m5} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.164 \times (ha\_5 + R \ast T \mathbf{w}\_5 - 10.76)}}{0.092} = 9.68 \text{ °C}$$

$$\Delta Q\_{c5} = \frac{h\_i \Delta A\_i}{1 + \Delta N \text{TU}\_1/2} \times (T s\_{m5} - T \mathbf{w}\_5) = 9.25 \text{ kW}$$

$$T \mathbf{w}\_6 = T \mathbf{w}\_5 - \frac{\Delta Q\_{c5}}{m\_w c p\_w} = 7.11 - \frac{9.25}{2.9 \ast 4.14} = 6.34 \text{ °C}$$

$$ha\_6 = ha\_5 - \frac{\Delta\_{65}}{m\_a} = 34.40 \text{ kJ/kg}$$

$$Ta\_6 = \begin{bmatrix} \left(1 - \frac{\Delta \text{NTU}\_o}{2}\right) \\ \left(1 + \frac{\Delta \text{NTU}\_o}{2}\right) \end{bmatrix} \times Ta\_5 + \begin{bmatrix} \left(\frac{\Delta \text{NTU}\_o}{\left(1 + \frac{\Delta \text{NTU}\_o}{2}\right)}\right) \times Ts\_{m5} = 12.59 \text{ °C} \end{bmatrix}$$

$$\text{Wa}\_6 = \frac{ha\_6 - CapTa\_6}{2501 + 1.8 \text{ °Ta}\_6} = \frac{34.40 - 1(12.59)}{2501 + 1.8(12.59)} = 0.00864 \text{ kg/kg}$$

$$TS\_{m6} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.164 \times (Ra\_6 + R \cdot Tw\_6 - 10.76)}}{0.092} = 8.645 \text{ °C}$$

$$\Delta Q\_{c6} = \frac{h\_1 \Delta h\_1}{1 + \Delta \text{NTU}\_1/2} \times (Ts\_{m6} - Tw\_6) = 8.3 \text{ kW}$$

$$Tw\_7 = Tw\_6 - \frac{\Delta Q\_{c4}}{m\_w c p\_w} = 6.34 - \frac{8.3}{2.9 \times 4.14} = 5.65 \text{ °C}$$

$$ha\_7 = ha\_6 - \frac{\Delta q\_{c6}}{m\_a} = 31.50 \text{ kJ/kg}$$

$$Ta\_7 = \left[\frac{\left(1 - \frac{\Delta W\_{c4}}{2}\right)}{\left(1 + \frac{\Delta \text{NTU}\_2}{2}\right)}\right] \times Ta\_6 + \left\{\left[\frac{(\Delta \text{NTU}\_0)}{\left(1 + \frac{\Delta \text{NTU}\_0}{2}\right)}\right] \times Ts\_{m6} = 11.14 \text{ °C}$$

$$\text{Wa}\_7 = \frac{h\_7 - CpaTa\_7}{2501 + 1.987a\_7} = \frac{31.50 - 1(11.14)}{2501 + 1.8(11.14)} = 0.0081 \text{ kg} \text{ }/\text{ kg}$$

$$Q\_c = \sum\_{l=1}^{N\_r} \Delta Q\_{cl} = (13.15 + 12.17 + 11.12 + 10.22 + 9.25 + 83)$$


$$m\_{\le} = \frac{N\_t}{N\_p} \rho\_{\le} \left(\frac{\pi}{4} d\_l^2\right) V\_{\le}$$

$$N\_{\rm t} = \frac{4N\_p m\_a}{\pi \rho\_w d\_l^2 V\_w} = 120 \text{ tubes}$$

$$N\_t = N\_r \* N\_c$$

$$N\_c = \frac{120}{\epsilon} = 20$$

$$A\_o = \sum\_{i=1}^{N\_r} \Delta A\_o = N\_r \* \Delta A\_o = 6 \* 24.39 = 146.34 \text{ m}^2\text{s}$$

$$\mathbf{A}\_{\mathbf{i}} = 6.363 \text{ m}^2 \,= N\_{\mathbf{f}} (\pi d\_l L)$$

$$L = Length \text{ of } the \text{ } coil \text{ } = \frac{A\_l}{\pi N\_t d\_l} = 1.4 \text{ } m$$

$$Q\_c = m\_a(ha\_1 - ha\_2) \tag{1}$$

$$Q\_c = \frac{\eta\_s}{c\_p} h\_o A\_o (ha\_m - h s\_m) \tag{2}$$

$$Q\_c = m\_w C p\_w (T w\_2 - T w\_1) \tag{3}$$

$$Q\_c = h\_l A\_l (T \text{s}\_m - T \text{w}\_m) \tag{4}$$

$$\mathbf{R} = \frac{\mathbf{h}\mathbf{a}\_1 - \mathbf{h}\mathbf{s}\_1}{\mathbf{T}\mathbf{s}\_1 - \mathbf{T}\mathbf{w}\_1} = \frac{\mathbf{h}\mathbf{a}\_2 - \mathbf{h}\mathbf{s}\_2}{\mathbf{T}\mathbf{s}\_2 - \mathbf{T}\mathbf{w}\_2} \tag{5}$$

$$TS\_1 = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \ast (ha\_1 + R \ast T\nu\_1 - 10.76)}}{0.092} \tag{6}$$

$$TS\_2 = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \ast (ha\_2 + R \ast T\overline{w\_2 - 10.76})}}{0.092} \tag{7}$$

$$R = \begin{bmatrix} \frac{c\_{pa} \ h\_l}{\eta\_{\bar{s}} h\_o} \left(\frac{A\_l}{A\_o}\right) \end{bmatrix} \tag{8}$$

$$
\Delta T \omega\_m = \left( T s\_m - T \omega\_m \right) = \frac{[(Ts\_1 - T\omega\_1) - (Ts\_2 - T\omega\_2)]}{\ln \left[ \frac{(Ts\_1 - T\omega\_1)}{(Ts\_2 - T\omega\_2)} \right]} \tag{9}
$$

$$A\_l = \frac{q\_c}{h\_l \* \Delta T\_m} \tag{10}$$

$$A\_o = \left(\frac{A\_o}{A\_l}\right) A\_l \tag{11}$$

$$\text{DHL} = \beta \mathcal{A}\_o \tag{12}$$

$$m\_{\rm w} = \frac{N\_t}{N\_p} \rho\_{\rm w} \left(\frac{\pi}{4} d\_l^2\right) V\_{\rm w}$$

$$N\_t = \frac{4N\_p m\_a}{\pi \rho\_{\rm w} d\_l^2 V\_{\rm w}}\tag{13}$$

$$L = \frac{A\_l}{N\_t \pi d\_l} \tag{14}$$

$$A\_{face} = HL = \frac{m\_a}{\rho\_a V\_{faca}}\tag{15}$$

$$H = \left(\frac{m\_a}{\rho a V\_{facc}}\right) \* \left(\frac{N\_t \pi d\_l}{A\_l}\right) \tag{16}$$

$$N\_r = \frac{w}{s\_T} \tag{17}$$

$$D = N\_r \ast \mathcal{S}\_L \tag{18}$$

$$Q\_s = m\_a C p\_a (T a\_1 - T a\_2) \tag{19}$$

$$Q\_s = \eta\_s h\_o A\_o \left[ \frac{Ta\_1 + Ta\_2}{2} - \frac{Ts\_1 + Ts\_2}{2} \right] \tag{20}$$

$$Ta\_2 = \begin{bmatrix} \left(\frac{1 - \frac{\Delta \text{NTU}\_0}{2}}{2}\right) \\ \left(1 + \frac{\Delta \text{NTU}\_0}{2}\right) \end{bmatrix} \times Ta\_1 + \begin{bmatrix} \left(\frac{\Delta \text{NTU}\_0}{2}\right) \\ \left(1 + \frac{\Delta \text{NTU}\_0}{2}\right) \end{bmatrix} \times \begin{pmatrix} \frac{\text{Ts}\_1 + \text{Ts}\_2}{2} \end{pmatrix} \tag{21}$$

$$\begin{aligned} \text{2.} \quad m\_a &= \frac{Q\_\text{C}}{\frac{(h a\_{\text{in}} - h a\_{\text{out}})}{Q\_\text{C}}} = \frac{60}{54 - 33} = 2.857 \text{ kg/s} \\\text{2.} \quad m\_W &= \frac{Q\_\text{C}}{Cp\_\text{W}(T\nu\_{\text{out}} - T\nu\_{\text{in}})} = \frac{60}{4.14 \ast 5} = 2.90 \quad \text{kg/s} \end{aligned}$$

$$R = \begin{bmatrix} \frac{h\_l C p\_a}{\eta\_s h\_o} \left(\frac{A\_l}{A\_o}\right) \end{bmatrix} \implies 3.41 \text{ kJ/kg.K}$$

$$Ts\_1 = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \ast (ha\_1 + R \ast T w\_1 - 10.76)}}{0.092}$$

$$Ts\_1 = 14.71 < T\_{d, point} = 15 \qquad \text{[Coil surface is wet]}$$

$$Ts\_2 = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \ast (ha\_2 + R \ast T w\_2 - 10.76)}}{0.092}$$

$$\Delta T \mathbf{w}\_m = (T\mathbf{s}\_m - T\mathbf{w}\_m) = \frac{[(T\mathbf{s}\_1 - T\mathbf{w}\_1) - (T\mathbf{s}\_2 - T\mathbf{w}\_2)]}{\ln\left[\frac{(T\mathbf{s}\_1 - T\mathbf{w}\_1)}{(T\mathbf{s}\_2 - T\mathbf{w}\_2)}\right]}$$

$$\begin{aligned} A\_l &= \frac{Q\_c}{h\_l \* \Delta T\_m} = 5.95 \text{ m}^2 \\\\ A\_o &= \left(\frac{A\_o}{}\right) A\_l = 136.85 \text{ m} \end{aligned}$$

$$N\_{\rm t} = \frac{4N\_{\rm p}m\_{\rm w}}{\pi \rho\_{\rm w} d\_{\rm t}^2 V\_{\rm w}} = 12o \, tubes$$

$$L = \frac{A\_l}{N\_t \pi d\_l} = 1.30 \text{ m}$$

$$W = \left(\frac{m\_a}{\rho\_a V\_{facs}}\right) \* \left(\frac{N\_t \pi d\_l}{A\_l}\right) = \ 0.88 \* 0.766 = 0.674 \text{ m}^3$$

$$N\_r = \frac{N\_t}{N\_c} = \frac{N\_t \* S\_n}{W} = 5.65 \approx 6 \text{ rows}$$

$$D = N\_r \ast \mathcal{S}\_L = 0.157 \quad \text{m}$$

$$Ta\_2 = \frac{\left[\left(1 - \frac{\Delta \text{NTU}\_0}{\sharp}\right)\right]}{\left(1 + \frac{\Delta \text{NTU}\_0}{\sharp}\right)} \times Ta\_1 + \left[\frac{\left(\Delta \text{NTU}\_0\right)}{\left(1 + \frac{\Delta \text{NTU}\_0}{\sharp}\right)}\right] \times \left(\frac{Ts\_1 + Ts\_2}{2}\right) = 10.95 \text{ }^\circ\text{C}$$

$$Wa\_2 = \frac{ha\_2 - CapTa\_2}{2501 + 1.8 \text{ }^\circ\text{Ta}\_2} = 0.00874 \text{ kg}\_\text{V} / \text{kg}\_\text{s}$$

$$Q\_L Q\_S = 60 - \ 2.857 \ast \text{(26 - 10.95)} = 17 \text{ kW} \text{kW}$$

$$CSHF = \frac{60 - 17}{60} = 0.717$$



$$\begin{aligned} m\_a &= \frac{Q\_c}{(ha\_{ln} - ha\_{out})} = \frac{60}{(46 - 30.6)} = 3.53 \text{ kg/s} \\ m\_W &= \frac{Q\_c}{Cp\_W(Tw\_{out} - Tw\_{in})} = \frac{60}{4.14 \times 5} = 2.90 \text{ kg/s} \end{aligned}$$

$$
\Delta A\_o = \frac{\beta \times \mathbf{S\_L} \times \mathbf{m\_a}}{\rho\_\text{a} \times \mathbf{v\_{face}}} = \frac{1060 \times 0.0261 \times 3.53}{1.16 \times 2.6} = 29.95 \text{ m}^2$$

$$
\Delta NTU\_o = \frac{\eta\_s \ h\_o \Delta A\_o}{m\_a c\_p m} = \frac{0.85 \times 60 \times 29.95}{3.53 \times 1001} = 0.432$$

$$
\Delta A\_l = \left(\frac{A\_l}{A\_o}\right) \times \Delta A\_o = \frac{29.95}{23} = 1.3 \text{ m}^2$$

$$
\Delta \text{NTU}\_l = \frac{h\_l \Delta A\_l}{m\_w c\_p m\_w} = \frac{4000 \times 1.3}{2.9 \times 4114} = 0.435$$

$$
\text{F}\_{\text{D}} \quad \left[\text{h}\_l \text{c}\_{\text{p\_B}} \left(\text{AA}\_l\right)\right] \quad \left[\left(1 + \frac{\text{ANTU\_0}}{2}\right)\right] \quad 5.9 \text{ mV} \ (1 + \text{K})$$

$$TS\_{m1} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \ast (ha\_1 + R \ast T\omega\_1 - 10.76)}}{0.092} = 13.5 \text{ °C}$$

$$\mathrm{Ts}\_{m1} = \frac{(\mathrm{Ta}\_1 + \mathrm{R} \, ^\circ \mathrm{Tw}\_1)}{(\mathrm{R} + 1)} = 13.6 \, ^\circ \mathrm{C}$$

$$\Delta Q\_{c1} = \frac{\mathrm{h}\_{\mathrm{l}} \Delta \mathrm{A}\_{\mathrm{l}}}{(1 - \frac{\Delta \mathrm{NTU}\_{\mathrm{l}}}{2})} \times (\mathrm{Ts}\_{\mathrm{m1}} - \mathrm{Tw}\_1) = 17.27 \, ^\circ \mathrm{kW}$$

$$T\mathbf{w}\_2 = T\mathbf{w}\_1 - \frac{\Delta Q\_{c1}}{m\_\text{w}Cp\_\text{w}} = 11 - \frac{14.5}{2.9 \ast 4.14} = 9.8 \text{ °C}$$

$$\text{Ta}\_2 = \text{Ta}\_1 - \frac{\Delta Q\_{c1}}{m\_\text{w} \text{Сра}} = 22.2 \text{ °C}$$

$$\text{Wa}\_3 = \text{Wa}\_1 = 0.0081 \text{ kg}\_\text{w} / \text{kg}\_\text{s}$$

$$T\mathbf{s}\_{m2} = \frac{(T\mathbf{a}\_2 + \mathbf{R} \ast T\mathbf{w}\_2)}{(R+1)} = 11.81 \text{ °C}$$
 
$$\mathbf{h}\_i \Delta \mathbf{A}\_i \qquad \dots \qquad \dots \qquad \dots \qquad \dots$$

$$
\Delta Q\_{c2} = \frac{\text{m}\_{\text{l}} \Delta \text{m}\_{\text{l}}}{(1 - \frac{\Delta \text{NTU}\_{\text{l}}}{2})} \times (T \text{s}\_{m2} - T \text{w}\_2) = 13.35 \text{ kW}
$$

$$\mathrm{Tw\_3} = \mathrm{Tw\_2} - \frac{\Delta Q\_{c2}}{m\_w c p\_w} = 9.8 - \frac{11.2}{2.9 \ast 4.14} = 8.86 \text{ °C}$$

$$\mathrm{Ta\_3} = \mathrm{Ta\_2} - \frac{\Delta Q\_{c1}}{m\_a \mathrm{xCpa}} = 18.5 \text{ °C}$$

$$\mathrm{Wa\_3} = \mathrm{Wa\_2} = 0.0081 \ \mathrm{kg\_v/kg\_a}$$

$$Ts\_{m3} = \frac{(Ta\_3 + R \ast Tw\_3)}{(R+1)} = 10.42 \text{ °C}$$

$$
\Delta Q\_{c3} = \frac{\text{h}\_{\text{l}} \Delta \text{A}\_{\text{l}}}{(1 - \frac{\Delta \text{NTU}\_{\text{l}}}{Z})} \times (T s\_{m3} - T w\_3) = 10.36 \text{ kW}
$$

$$
T w\_4 = T w\_3 - \frac{\Delta q\_{c3}}{m\_w c p\_w} = 8.86 - \frac{8.7}{2.9 \ast 4.14} = 8.13 \text{ °C}
$$

$$
\text{Ta}\_4 = \text{Ta}\_3 - \frac{\Delta q\_{c1}}{m\_a \text{xCpa}} = 15.6 \text{ °C}
$$

$$
\text{T}\_{\text{max}}
$$

$$T s\_{m4} = \frac{(Ta\_4 + R \ast T\omega\_4)}{(R+1)} = 9.34 \text{ } ^\circ \text{C}$$

$$
\Delta Q\_{c4} = \frac{\text{Ti}\_1 \Delta \text{H}\_1}{(1 - \frac{\Delta \text{NTU}\_1}{2})} \times (T\_{5m4} - T\_{W4}) = 8 \text{ kW}
$$

$$
T \text{w}\_5 = T \text{w}\_4 - \frac{\Delta Q\_{c4}}{m\_w C p\_w} = 8.13 - \frac{6.75}{2.9 \times 4.14} = 7.57 \text{ °C}
$$

$$
\text{Ta}\_5 = \text{Ta}\_4 - \frac{\Delta Q\_{c4}}{m\_a \text{xCpa}} = 13.3 \text{ °C}
$$

$$
\text{Wa}\_5 = \text{Wa}\_4 = 0.0081 \text{ kg}\_7/\text{kg}\_4
$$

$$ha\_5 = 1.001 \text{xTa5} + \text{Was}^\*(2501 + 1.8^\circ \text{Tas}) = 33.76 \text{ kJ/kg}$$

$$T s\_{m5} = \frac{-(R + 1.4) + \sqrt{(R + 1.4)^2 + 0.184 \ast (ha\_5 + R \ast T \mathbf{w}\_5 - 10.76)}}{0.092} = 9 \text{ °C}$$

$$\Delta Q\_{c5} = \frac{\text{h}\_{\text{l}} \Delta \text{A}\_{\text{l}}}{(1 - \frac{\Delta \text{NTU}\_{\text{l}}}{2})} \times (T s\_{m5} - T \mathbf{w}\_5) = 9.5 \text{ kW}$$

$$T \mathbf{w}\_6 = T \mathbf{w}\_5 - \frac{\Delta Q\_{c5}}{m\_w c p\_w} = 7.57 - \frac{7.97}{2.9 \ast 4.14} = 7 \text{ °C}$$

$$ha\_6 = ha\_5 - \frac{\Delta q\_{cs}}{m\_a} = 31.1 \text{ kJ/kg}$$

$$Ta\_6 = \left[\frac{\left(1 - \frac{\Delta \text{NTU}\_o}{2}\right)}{\left(1 + \frac{\Delta \text{NTU}\_o}{2}\right)}\right] \times Ta\_5 + \left\{\frac{(\Delta \text{NTU}\_o)}{\left(1 + \frac{\Delta \text{NTU}\_o}{2}\right)}\right\} \times Ts\_{m5} = 11.7 \text{ °C}$$

$$\text{Wa}\_6 = \frac{ha\_o - \text{CpaTa}\_6}{2501 + 1.6 \text{ °Ta}\_6} = \frac{31.1 - 1(11.7)}{2501 + 1.6(11.7)} = 0.0077 \text{ kg} \text{/kg} \text{°A}$$

$$Ts\_{m6} = \frac{-(R+1.4) + \sqrt{(R+1.4)^2 + 0.184 \cdot (ha\_a + R \cdot Tw\_a - 10.76)}}{0.092} = 8.1 \text{ °C}$$

$$\Delta Q\_{c6} = \frac{h\_1 \Delta \mathbf{l}\_1}{(1 - \frac{\Delta \mathbf{NT} \mathbf{U}\_1}{2})} \times (Ts\_{m6} - Tw\_6) = 7.31 \text{ kW}$$

$$Tw\_7 = Tw\_6 - \frac{\Delta Q\_{ca}}{m\_w c p\_w} = 7 - \frac{6.37}{2.9 \ast 4.14} = 6.4 \text{ °C}$$

$$ha\_7 = ha\_6 - \frac{\Delta Q\_{c5}}{m\_a} = 30 \text{ kJ/kg}$$

$$Ta\_7 = \left[\frac{\left(1 - \frac{\Delta \mathbf{NT} \mathbf{U}\_0}{2}\right)}{\left(1 + \frac{\Delta \mathbf{NT} \mathbf{U}\_0}{2}\right)}\right] \times Ta\_6 + \left\{\left(\frac{\left(\Delta \mathbf{NT} \mathbf{U}\_0\right)}{\left(1 + \frac{\Delta \mathbf{NT} \mathbf{U}\_0}{2}\right)}\right) \times Ts\_{m6} = 10.3 \text{ °C}\right\}$$

$$\text{Wa}\_7 = \frac{ha\_7 - CpaTa\_7}{2501 + 1.8 \ast Ta\_7} = \frac{30 - 1(10.3)}{2501 + 1.8(10.3)} = 0.0076 \text{ kg}\_{\text{v}}/\text{kg}\_{\text{s}}$$

$$CSHF = \frac{Q\_S}{Q\_c} = \frac{m\_a C p (T a\_1 - T a\_7)}{65.8} = 0.885$$


$$m\_{\le} = \frac{N\_t}{N\_p} \rho\_{\le} \left(\frac{\pi}{4} d\_l^2\right) V\_{\le}$$

$$N\_{\rm t} = \frac{4N\_p m\_a}{\pi \rho\_w d\_l^2 V\_w} = 120 \text{ tubes}$$

$$N\_t = N\_r \ast N\_c$$

$$N\_c = \frac{120}{6} = 20$$

$$A\_o = \sum\_{l=1}^{N\_r} \Delta A\_o = N\_r \ast \Delta A\_o = 6 \ast 29.95 = 179.7 \text{ m}^2$$

$$\text{Ta}\_{\text{out}} = 10.3 \text{ } \text{°C}$$

$$\begin{aligned} \text{2. } \quad m\_{a} &= \frac{\eta\_{c}}{(h\_{w} - h\_{a}\eta\_{w})} = \frac{\alpha}{(49 - 36.6)} = 3.53 \text{ kg/s} \\ \text{3. } \quad m\_{w} &= \frac{\eta\_{c}}{c p\_{w}(\gamma w\_{w\text{or}} - \gamma w\_{w})} = \frac{\alpha}{4.14 \text{ s}} = 2.90 \text{ kg/s} \end{aligned}$$

$$R = \left[\frac{h\_{l}c p\_{a}}{\eta\_{l}h\_{o}}\left(\frac{A\_{l}}{A\_{o}}\right)\right] = 3.41 \text{ K/kg.K}$$

$$Ts\_{1} = \frac{-(R + 1.4) + \sqrt{(R + 1.4)^{2} + 0.184 \ast (ha\_{1} + R \ast Tw\_{1} - 10.76)}}{0.092}$$

$$Ts\_{1} = 13.74 > T\_{d, point} = 10 \qquad \text{[Coil surface is dry]}$$

$$Ts\_{1} = \frac{(Ta\_{1} + R \ast Tw\_{1})}{(R + 1)} = 14.62 \text{ } ^\circ \text{C}$$

$$Ts\_{2} = \frac{(Ta\_{2} + R \ast Tw\_{2})}{(R + 1)} = 7.02$$

$$
\Delta T \mathbf{w}\_m = (T\mathbf{s}\_m - T\mathbf{w}\_m) = \frac{| (T\mathbf{s}\_1 - T\mathbf{w}\_1) - (T\mathbf{s}\_2 - T\mathbf{w}\_2) |}{\ln \left| \frac{(T\mathbf{s}\_1 - T\mathbf{w}\_1)}{(T\mathbf{s}\_2 - T\mathbf{w}\_2)} \right|}
$$

$$
\Delta \mathbf{w}\_m = \mathbf{n} \cdot \mathbf{n} \cdot \mathbf{w}
$$

$$A\_{l} = \frac{q\_{c}}{h\_{l} \* \Delta T\_{m}} = 7.31 \text{ m}^{2}$$

$$A\_{o} = \left(\frac{A\_{o}}{A\_{l}}\right) A\_{l} = 168.3 \text{ m}^{2}$$

$$N\_t = \frac{4N\_p m\_\text{w}}{\pi \rho\_\text{w} d\_l^2 V\_\text{w}} = 12o \text{ tubes}$$

$$L = \frac{A\_l}{N\_t \pi d\_l} = 1.62 \text{ m}$$

$$H = \left(\frac{m\_a}{\rho\_a V\_{facc}}\right) \* \left(\frac{N\_t \pi d\_l}{A\_l}\right) = 0.683 \text{ m}.$$

$$N\_r = \frac{N\_t}{N\_c} = \frac{N\_t \* S\_t}{H} = 5.5 \approx 6 \text{ rows}$$

$$D = N\_r \* S\_L = 0.157 \,\mathrm{m}$$

$$NTU\_o = \frac{h\_o A\_o \eta\_s}{m\_a Cp\_a} = 2.43$$

$$Ta\_2 = \left[\frac{\left(1 - \frac{\text{NTU}\_0}{2}\right)}{\left(1 + \frac{\text{NTU}\_0}{2}\right)}\right] \times Ta\_1 + \left[\frac{\left(\text{NTU}\_0\right)}{\left(1 + \frac{\text{NTU}\_0}{2}\right)}\right] \times \left(\frac{\left(\text{TS}\_1 + \text{TS}\_2\right)}{2}\right) = 9.2 \, ^\circ \text{C}$$

$$Wa\_2 = \frac{ha\_2 - CapTa\_2}{2501 + 1.8 \, ^\circ Ta\_2} = 0.008 \text{ kg} / \text{kg}$$

$$Q\_L Q\_S = \text{m}\_a \text{ (Wa}\_2\text{- Wa}\_1\text{)} \ge 2501 \text{ =} 0.88 \text{ kW}$$

$$CSHF = \frac{60 - 0.99}{60} = 0.98$$





$$\begin{array}{l} \text{s}\_{\text{a}} = \frac{\text{Qc}}{(\text{ha}\_{\text{in}} - \text{ha}\_{\text{out}})} = \frac{60}{(54 - 33)} = 2.857 \text{ kg/s} \\\\ \text{T}\_{\text{b}} = \frac{\text{R}\_{\text{c}} \text{Cp}\_{\text{a}} \left( \frac{\text{A}\_{1}}{\text{A}\_{0}} \right)}{\text{T}\_{\text{b}} \text{s}\_{\text{b}}} = 1.7 \text{ K} / \text{kg.K} \\\\ \text{Ts}\_{1} = \frac{-(\text{R} + 1.4) + \sqrt{(\text{R} + 1.4)^{2} + 0.184 \ast (\text{ha}\_{1} + \text{R} \ast Tev})}{0.092} \\\\ \text{Ts}\_{1} = 14.23 < \text{T}\_{\text{d}, \text{point}} = 15 \qquad \text{[Coil surface is wet]} \\\\ \text{T}\_{\text{S}\_{7}} = \frac{-(\text{R} + 1.4) + \sqrt{(\text{R} + 1.4)^{2} + 0.184 \ast (\text{ha}\_{2} + \text{R} \ast Tev})}{0.092} \end{array}$$

 $\Delta \text{TeV}\_{\text{m}} = (\text{Ts}\_{\text{m}} - \text{TeV}\_{\text{m}}) = \frac{[(\text{Ts}\_{1} - \text{TeV}) - (\text{Ts}\_{2} - \text{TeV})]}{\ln[\frac{(\text{Ts}\_{1} - \text{TeV})}{(\text{Ts}\_{2} - \text{TeV})}]}$ 
$$\Delta \text{TeV}\_{\text{m}} = 5.33 \quad \text{°C}$$

$$\mathbf{A}\_{\mathrm{l}} = \frac{\mathbf{q}\_{\mathrm{c}}}{\mathrm{h}\_{\mathrm{l}} \* \Delta \mathrm{TeV}\_{\mathrm{m}}} = 5.63 \text{ m}^2$$

$$\mathbf{A}\_{\mathrm{o}} = \left(\frac{\mathrm{A}\_{\mathrm{o}}}{\mathrm{A}\_{\mathrm{l}}}\right) \mathrm{A}\_{\mathrm{l}} = 129.5 \text{ m}^2$$

 $Ta\_{2} = \left[\frac{\left(1 - \frac{\Delta \text{NTU}\_{0}}{2}\right)}{\left(1 + \frac{\Delta \text{NTU}\_{0}}{2}\right)}\right] \times Ta\_{1} + \left[\frac{\left(\Delta \text{NTU}\_{0}\right)}{\left(1 + \frac{\Delta \text{NTU}\_{0}}{2}\right)}\right] \times \left(\frac{Ts\_{1} + Ts\_{2}}{2}\right) = 10.75 \text{ }^{\circ}\text{C}$  
$$\Delta \text{NTU}\_{0} = \frac{\eta\_{\text{s}} \,\text{h}\_{0} \text{A}\_{0}}{\text{m}\_{\text{a}} \text{Cp}\_{\text{a}}} = 2.30$$

$$\text{Wa}\_2 = \frac{\text{na}\_{2-\text{Cp}\_3 \ast \text{Ta}\_2}}{2501 + 1.0 \ast \text{Ta}\_2} = 0.00823 \text{ Kgy } / \text{ kga}$$

$$\mathbf{Q\_{C}} = \mathbf{m\_{a}}(\mathbf{ha\_{1}} - \mathbf{ha\_{2}}) = 64.28 \text{ kW}$$

$$\mathbf{Q\_{L}} = \mathbf{Q\_{C}} - \mathbf{Q\_{S}} = 64.28 - 2.857 \ast \{26 - 10.72\} = 20.62 \text{ kW}$$

$$\text{CSFF} = \frac{64.28 - 20.6}{64.28} = 0.68$$

$$\mathbf{m\_r = \frac{Q\_0}{x \cdot \mathbf{h\_{\oplus}}} = 0.33 \text{ kg/s}} \quad \text{[Assume inlet dryness fraction, } x = 0.9]$$

$$\mathbf{N\_t = \frac{4N\_\mathrm{p^m}}{\pi \rho\_\mathrm{gd}^2 V\_\mathrm{g}}} \approx 96 \,\mathrm{tube}$$

**Part 4** 

**Plate Heat Exchangers** 

**The Length of the Tube (Coil), L**

$$\mathcal{L} = \frac{\mathbf{A\_4}}{\mathbf{N\_t} \pi \mathbf{d\_4}} = \mathbf{1.54 \text{ m}}$$

**Height of the Coil, H** 

$$\text{Air face area, A}\_{\text{face}} = \frac{\text{m}\_{\text{a}}}{\rho\_{\text{a}} \text{V}\_{\text{face}}} = 0.88 \text{ m} \text{2}$$

$$\text{A}\_{\text{face}} = H \text{L}$$

$$H = \frac{\text{A}\_{\text{face}}}{\text{L}} = 0.57 \text{ m} \text{2}$$

**Number of Rows, Nr**

$$\mathbf{N\_r = \frac{N\_t}{N\_c} = \frac{N\_t \* S\_t}{W}} = 5.35 \approx 6 \text{ rows}$$

**Depth of the Coil, D:** 

$$D = N\_r \ast \mathbb{S}\_L = 0.157 \quad \text{m}$$

#### **9. Conclusion**

In this chapter, simulation of the cooling coil using a discrete technique "row-by-row method" has been presented. The main advantage of this method is to trace the air and coil surface temperature locally. In addition, this method gives more accurate results for the cooling coil design or simulation compared with those given by ordinary method such as log mean enthalpy method. Step-by-step procedure has been introduced and worked examples are presented. The deviation between the two methods "numerical discrete method and treating the coil as a single zone" is around of 12%.

#### **10. Nomenclature**


#### **11. References**


**Part 4** 

**Plate Heat Exchangers** 

394 Heat Exchangers – Basics Design Applications

A���� = �L

= 1.54 m

�������

= 0.57 m2

� <sup>=</sup> 5.35 ≈ 6 rows

= 0.88 m2

L = �� �����

Air face area, A���� <sup>=</sup> ��

� = ����� �

<sup>=</sup> �����

 � = �� � �� = 0.157 m

In this chapter, simulation of the cooling coil using a discrete technique "row-by-row method" has been presented. The main advantage of this method is to trace the air and coil surface temperature locally. In addition, this method gives more accurate results for the cooling coil design or simulation compared with those given by ordinary method such as log mean enthalpy method. Step-by-step procedure has been introduced and worked examples are presented. The deviation between the two methods "numerical discrete

 N� <sup>=</sup> �� ��

method and treating the coil as a single zone" is around of 12%.

[1] ASHRAE Systems and Equipment Handbook (SI), 2000, Chapter 21

[3] ASHRAE Fundamental Handbook (SI), 2001, Chapter 6

[2] Wibert Stoecker, and Jerold Jones. "Refrigeration & air-conditioning",2nd, Ed., 1982,

[4] Kays, W.M., and London A.L. Compact Heat Exchangers, 3rd edition, McGraw-Hill, New

[5] Threlkeld, J.L. Thermal Environment Engineering, Prentice-Hall Inc., New Work, NY.

**The Length of the Tube (Coil), L**

**Height of the Coil, H** 

**Number of Rows, Nr**

**Depth of the Coil, D:** 

**9. Conclusion** 

**10. Nomenclature** 

**11. References** 

A = surface area, m2 Cp = specific heat, kJ/kg. C

NTU = number of transfer unit Q = heat transfer, W T = temperature , oC W = humidity ratio, kgv/kga

McGraw-Hill

York.1984

1970.

h = heat transfer coefficient, W/m2. C hmass = mass transfer coefficient, kg/m2.S

**15** 

*Oman* 

Muthuraman Subbiah *Higher College of Technology,* 

**The Characteristics of Brazed Plate Heat** 

**Exchangers with Different Chevron Angles** 

Plate heat exchangers (PHEs) were introduced in the 1930s and were almost exclusively used as liquid/liquid heat exchangers in the food industries because of their ease of cleaning. Over the years, the development of the PHE has generally continued towards larger capacity, as well as higher working temperature and pressure. Recently, a gasket sealing was replaced by a brazed material, and each thermal plate was formed with a series of corrugations (herringbone or chevron). These greatly increased the pressure and the

The corrugated pattern on the thermal plate induces a highly turbulent fluid flow. The high turbulence in the PHE leads to an enhanced heat transfer, to a low fouling rate, and to a reduced heat transfer area. Therefore, PHEs can be used as alternatives to shell-and-tube heat exchangers. Due to ozone depletion, the refrigerant R22 is being replaced by R410A (a binary mixture of R32 and R125, mass fraction 50 %/50 %). R410A approximates an azeotropic behavior since it can be regarded as a pure substance because of the negligible temperature gliding. The heat transfer and the pressure drop characteristics in PHEs are related to the hydraulic diameter, the increased heat transfer area, the number of the flow channels, and the profile of the corrugation waviness, such as the inclination angle, the corrugation amplitude, and the corrugation wavelength. These geometric factors influence the separation, the boundary layer, and the vortex or swirl flow generation. However, earlier experimental and numerical works were restricted to a single-phase flow. Since the advent of a Brazed PHE (BPHE) in the 1990s, studies of the condensation and/or evaporation heat transfer have focused on their applications in refrigerating and air conditioning systems, but only a few studies have been done. Much work is needed to understand the features of the two-phase flow in the BPHEs with alternative refrigerants. Xiaoyang *et al*., [1] experimented with the two-phase flow distribution in stacked PHEs at both vertical upward and downward flow orientations. They indicated that non-uniform distributions were found and that the flow distribution was strongly affected by the total inlet flow rate, the vapor quality, the flow channel orientation, and the geometry of the inlet port Holger [2].Theoretically predicted the performance of chevron-type PHEs under singlephase conditions and recommended the correlations for the friction factors and heat transfer coefficients as functions of the corrugation chevron angles. Lee *et al*., [3] investigated the characteristics of the evaporation heat transfer and pressure drop in BPHEs with R404A and

**1. Introduction** 

temperature capabilities.
