Theoretical Entropy Generation Analysis for Forced Convection Flow Around a Horizontal Cylinder

The steady-state external flow around the horizontal cylinder and entropy generation rate is controlled in volume as shown in Figure 1.

1.png

Figure 1. External flow convection heat transfer. (a) Shape geometry. (b) Control volume

Mass balance [5]:

$\dot{m}_{in }=\dot{m}_{out}=\dot{m}$     (1)

Heat balance [5]:

$\dot{m}_{in} h_{in}-\dot{m}_{out} h_{out}+\iint q^{\prime \prime} d A=0$        (2)

Entropy generation balance [5]:

$\dot{S}_{gen}=\dot{m}_{out} S_{out}-\dot{m}_{in} s_{ in }-\iint \frac{q^{\prime \prime} d A}{T_s}$        (3)

The Gibbs equation is given as [4]:

$d h=T d S+V d P$             (4)

Applying the Gibbs Eq. (4) for the control volume in Figure 1 with Eq. (2) gives:

$\begin{aligned} \dot{m}_{in} h_{in}-\dot{m}_{out} h_{out } &=T_{\infty}\left(\dot{m}_{\text {out }} s_{out}-\dot{m}_{in} s_{in}\right)  +\frac{m_{in}}{\rho_{\infty}}\left(P_{out}-P_{in}\right)\end{aligned}$         (5)

Following the studies [5-7], it is assumed that the average quantities of the control volume are equal to the free stream quantities. Combining Eq. (1) with Eq. (5) and substituting into Eq. (3) yields:

$\dot{S}_{\text {gen }}=\iint q^{\prime \prime}\left(\frac{1}{T_{\infty}}-\frac{1}{T_s}\right) d A-\dot{m}_{\text {in }} \frac{P_{\text {out }}-P_{\text {in }}}{\rho_{\infty} T_{\infty}}$        (6)

The flow rate and the drag force may be given as:

$\dot{m}=A_{cross} \rho_{\infty} U_{\infty}$ and $F=A_{cross}\left(P_{in}-P_{out}\right)$           (7)

Eq. (7) substituting in Eq. (6) to find:

$\dot{S}_{g e n}=\iint q^{\prime \prime}\left(\frac{1}{T_{\infty}}-\frac{1}{T_s}\right) d A+\frac{U_{\infty} F}{T_{\infty}}$           (8)

Assuming $\left|T_s-T_{\infty}\right| \ll T_{\infty}$ or $T_s$, and neglecting the higher order, Eq. (8) by Taylor series becomes:

$\dot{S}_{g e n}=\frac{1}{T_{\infty}^2} \iint q^{\prime \prime} \cdot\left(T_s-T_{\infty}\right) \cdot d A+\frac{U_{\infty} \cdot F}{T_{\infty}}$                 (9)

On the right, the first term shows the irreversibility due to forced convection heat transfer and the second term represents the irreversibility due to friction:

The drag force per unit length, F/l [6, 9]:

$\begin{aligned} F / l=\int_0^\pi \frac{1}{2} \rho_{\infty} U_{\infty}^2 C_{D, x} r d \theta =\frac{1}{2} \rho_{\infty} U_{\infty}^2 \frac{D}{2} \int_0^\pi C_{D . x} d \theta\end{aligned}$            (10)

The temperature difference due to heat fluxes can equal to:

$\frac{q^{\prime \prime}}{\alpha}=T_s-T_{\infty}$             (11)

Using Eqs. (10) and (11) and substituting into Eq. (9):

$\frac{\dot{S}_{g e n}}{l}=\frac{q^{\prime 2}}{T_{\infty}^2} 2 \frac{D}{2} \int_0^\pi \frac{d \theta}{\alpha}+\frac{U_{\infty}^3 \rho_{\infty}}{2 T_{\infty}} \frac{D}{2} 2 \int_0^\pi C_{D, x} d \theta$           (12)

where, $\alpha=a \operatorname{Re}_D^m \operatorname{Pr}^{\mathrm{n}} \frac{\mathrm{k}_{\infty}}{\mathrm{D}} ;$ and $\int_0^\pi C_{D, x} d \theta=C_f$ following [4, 9].

$C_D=\mathrm{b} \operatorname{Re}_D^{-y}$        (13)

Substituting Eqs. (11) and (13) into (12) yields:

$\begin{gathered}\frac{\dot{S}_{g e n}}{l}=\frac{q^{\prime \prime 2}}{T_{\infty}^2} \frac{D^2 \operatorname{Re}_D^{-m} \operatorname{Pr}^{-n}}{a k_{\infty}} \int_0^\pi d \theta+\frac{U_{\infty}^3 \rho_{\infty} D}{2 T_{\infty}} b R e_D^{-y} \\ \frac{\dot{S}_{g e n}}{l}=\frac{q^{\prime \prime 2}}{T_{\infty}^2} \frac{D^2 \operatorname{Re}_D^{-m} \operatorname{Pr}^{-n}}{a k_{\infty}} \pi+\frac{U_{\infty}^3 \rho_{\infty} D}{2 T_{\infty}} b R e_D^{-y}\end{gathered}$                   (14)

where, $q^{\prime}=\frac{q}{l}=q^{\prime \prime} \pi D \rightarrow q^{\prime \prime}=\frac{q^{\prime}}{\pi D} \rightarrow q^{\prime \prime 2}=\frac{q^{\prime 2}}{\pi^2 D^2}$, and $R e_D=$ $\frac{\rho U D}{\mu} U=\frac{R e_D \mu}{\rho D}$.

These definitions will be used into Eq. (14):

$\frac{\dot{S}_{g e n}}{l}=\frac{1}{a \pi} \frac{q^{\prime 2}}{T_{\infty}^2} \frac{P r^{-n}}{k_{\infty}} R e_D^{-m}+\frac{b}{2} \frac{U_{\infty}^2 \mu_{\infty}}{T_{\infty}} R e_D^{1-y}$              (15)

The general form for external flow irreversibility (entropy generation) around a cylinder is represented in Eq. (15), and the values of n, a, and m are shown in Table 1 depending on the references [10, 11]:

Table 1. Constant for Eq. (15)

Drag coefficients are a function of Re_D. The values needed for that are obtained by fitting curves to the Re_D range as in the Nu equations above. The map from reference [12], Figure 2, is used to define the constants b and y in Eq. (13) by curve fitting and the standard errors and correlations coefficients is shown in Table 2.

2.png

Figure 2. The relation between the Re and C_D (drag force coefficient) for force flow around cylinder [12]

Table 2. Drag coefficient correlations

New form of Eq. (15) will be as follow:

0.4<Re_D<4:

$\frac{\dot{S}_{g e n}}{l}=0.322 \frac{q^{\prime 2}}{T_{\infty}^2} \frac{P r^{-\frac{1}{3}}}{k} R e_D^{-0.33}+5.085 \frac{\mu_{\infty} U_{\infty}^2}{T_{\infty}} R e_D^{0.216}$            (16)

4<Re_D<40:

$\begin{aligned} \frac{\dot{S}_{g e n}}{l}= & 0.349 \frac{q^{\prime 2}}{T_{\infty}^2} \frac{P r^{-\frac{1}{3}}}{k} \operatorname{Re}_D^{-0.385}+ 3.317 \frac{\mu_{\infty} U_{\infty}^2}{T_{\infty}} \operatorname{Re}_D^{0.698}\end{aligned}$           (17)

40<ReD<4000:

$\frac{\dot{S}_{g e n}}{l}=0.466 \frac{q^{\prime 2}}{T_{\infty}^2} \frac{P r^{-\frac{1}{3}}}{k} R e_D^{-0.466}+2.3405 \frac{\mu_{\infty} U_{\infty}^2}{T_{\infty}} R e_D^{0.8}$              (18)

4000<Re_D<40000:

$\frac{\dot{S}_{g e n}}{l}=1.649 \frac{q^{\prime 2}}{T_{\infty}^2} \frac{P r^{-\frac{1}{3}}}{k} R e_D^{-0.618}+0.55 \frac{\mu_{\infty} U_{\infty}^2}{T_{\infty}} R e_D$                  (19)

The optimum Reynolds number (Re_D,opt) can be found in the Eqs. (16)-(19) can be found by diverting these equations with respect to Reynolds number as follows:

0.4<Re_D<4:

$R e_{\text {Dopt }}=0.139 \beta^{\frac{1}{0.546}}$         (20)

4<Re_D<40:

$R e_{\text {Dopt }}=0.722 \beta^{\frac{1}{1.083}}$            (21)

40<Re_D<4000:

$R e_{\text {Dopt }}=1.824 \beta^{\frac{1}{1.266}}$               (22)

4000<Re_D<40000:

$R e_{\text {Dopt }}=14.64 \beta^{\frac{1}{1.618}}$             (23)

The duty parameter β in Eqs. (20)-(23) is found. The duty parameter is defined by the ratio between the irreversibility due to convection and the irreversibility due to pressure drop [5-7] and is represented in Figure 3. By finding the duty factor the optimum Re_D,opt can be get from the figure which is helpful to go to fond the most important parameter that is the entropy generation number Ns :

$\beta=\frac{q^{\prime^2}}{U_{\infty}^2\,\, k_{\infty}\,\, \mu_{\infty}\,\, T_{\infty}\,\, P r^n}$         (24)

3.png

Figure 3. The relation between Re_D,opt and Duty factor [5, 6]

The irreversibility distribution ratio ∅ defines: fluid friction irreversibility due to pressure drop in relation to force convection and heat transfer irreversibility [5, 6].

Bejan number Be: which is the alternative distribution parameter. This parameter is defined by the ratio of the irreversibility due to force convection heat transfer to the summation of the irreversibility due to force convection heat transfer and friction due to pressure drop [7]:

$B e=\frac{1}{1+\varnothing}$              (25)

where, Be=1 is the limit at which heat transfer irreversibility dominates, Be=0 is the opposite limit at which friction irreversibility is dominated by fluid effects, and Be=1/2 is the case in which the heat transfer and fluid friction entropy generation rates are equal.

Substitute Re_D,opt: Eq. (20) in Eq. (16), Eq. (21) in Eq. (17), Eq. (22) in Eq. (18), finally Eq. (23) in Eq. (19) to find new entropy generation number, Ns:

0.4<ReD<4:

$N_{s 1}=\frac{\dot{S}_{gen}}{\dot{S}_{gen.min}}=\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{-0.33}+\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{0.216}$              (26)

4<Re_D<40:

$N_{s 2}=\frac{\dot{S}_{g e n}}{\dot{S}_{gen.min}}=\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{-0.385}+\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{0.698}$             (27)

40<Re_D<4000:

$N_{s 3}=\frac{\dot{S}_{g e n}}{\dot{S}_{gen.min}}=\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{-0.466}+\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{0.8}$           (28)

400<Re_D<4000:

$N_{s 4}=\frac{\dot{s}_{g e n}}{\dot{s}_{\text {gen.min }}}=\left(\frac{R e_D}{R e_{\text {Dopt}}}\,\, \right)^{-0.618}+\left(\frac{R e_D}{R e_{\text {Dopt}}}\,\, \right)$                (29)

As the above forms result in Ns=2 where heat transfer and pressure drop occur simultaneously, the entropy generation number must be Ns=1. This combines force convection with heat transfer irreversibility and friction drop due to pressure drop irreversibility. Now let's consider Eqs. (6) and (7) how to represent the irreversibility of heat transfer due to forced convection and the irreversibility of pressure drop due to friction as percentages. Pareto theory is the key, and Eqs. (26)-(29) will be:

0.4<Re_D<4:

$N_{s 1}=\frac{\dot{S}_{gen }}{\dot{S}_{gen.min}}=\frac{2}{3}\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{-0.33}+\frac{1}{3}\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{0.216}$               (30)

4<Re_D<40:

$N_{s 2}=\frac{\dot{S}_{\text {gen }}}{\dot{S}_{\text {gen.min }}}=\frac{2}{3}\left(\frac{R e_D}{R e_{\text {Dopt}}}\,\, \right)^{-0.385}+\frac{1}{3}\left(\frac{R e_D}{R e_{\text {Dopt}}}\,\, \right)^{0.698}$                (31)

0<Re_D<4000:

$N_{s 3}=\frac{\dot{S}_{gen}}{\dot{S}_{gen.min}}=\frac{2}{3}\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{-0.466}+\frac{1}{3}\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{0.8}$                 (32)

4000<Re_D<40000:

$N_{S 4}=\frac{\dot{S}_{gen}}{\dot{S}_{gen.min }}=\frac{2}{3}\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)^{-0.618}+\frac{1}{3}\left(\frac{R e_D}{R e_{Dopt\,\,}}\right)$               (33)

The irreversibility of force convection heat transfer and friction flow for a horizontal cylinder is studied and classified into four groups. The results are:

• The entropy generation for forced convection around a horizontal heated cylinder is represented in four Reynolds number ranges.

• The relation shape is the same for all ranges but different in values which mean it is important to find the entropy generation number for each range separately.

• The optimum entropy generation number for all Reynolds number ranges via Re_D / Re_D,opt is done at Ns=1 and Re_D /Re_D,opt=1. in this point the entropy generation is lowest and the design of any heat exchangers must be near this point.

• The thermal system optimum point is found at Be, opt =0.667 where the lowest irreversibility of heat transfer and pressure drop is reached when Ns =1, and Re_D/ Re_D,opt =1, and ∅opt = 0.5.

• At low Reynolds range, Reynolds number for heat transfer is not correct. It is actually the irreversibility of natural convection and radiation that is absent.