Question 14.6: Estimate the heat, mass, and friction coefficients for a fou......

The correlations of Figs. 14-14 and 14-15 will be used with the parameters JP, FP, J(s), and F(s) computed from Eqs. 14-46 through 14-60. The mass velocity is

∆P_{0i}  =  K_{i}  \frac{G^{2}_{c}}{2ρ_{i}g_{c}}                                (14-46)

\dot{q}=h _dAη_{ms}Δ i_m                               (14-60)

G_{c}  =  \frac{60  (600)  14.5  (144)}{0.54  (53.35)  540}  =  4900  lbm/(hr – ft²)

The Reynolds number based on tube diameter is then

\frac{G_{c}  D}{μ}  =  \frac{4900  (0.525/12)}{0.044}  =  4870

where data from Fig. 14-12 and Table A-4a are used. To compute the parameter A/A_{t} ,assume a four-row coil, 1 ft in length, with 10 tubes in the face. This coil has a volume of

V  =  \frac{10(1.25)  (12)  4  (1.083)}{1728 }  =  0.376  ft^{2}

and a total of 40 tubes. The total outside surface area of the tubes is

A_{t}  =  40π,  DL  =  40π  (0.525/12)(1)  =  5.498  ft^{2}

From Fig. 14-12, A/V = α = 238 ft²/ft³; then

\frac{A}{A_{t}}  =  \frac{238(0.376)}{5.498}  =  16.28

Using Eq. 14-39,

JP  =  Re^{-0.4}_{D}  \left\lgroup \frac{A}{A_{t}}\right\rgroup ^{-0.15}                   (14-39)

JP  =  (4870)^{–0.4} (16.28)^{–0.15}  =  0.022

The Reynolds number based on the fin spacing is

Re_{s}  =  Re_{D}  \left\lgroup\frac{s}{D} \right\rgroup   =  4870  \left\lgroup\frac{0.0833}{0.525} \right\rgroup   =  773

Then

J(s)  =  0.84  +  (4  ×  10^{–5})(773)^{1.25}  =  1.003

and

J_{i}(s )  =  [0.95  +  (4  ×  10^{–5})  (773)^{1.25}]  \left[\frac{0.0833}{0.0833  –  0.006} \right] ^{2}  =  1.293

Then

JP J(s) = 0.0220(1.003) = 0.022

and

JP  J_{i}(s)  =  0.0220(1.293)  =  0.028

Using Fig. 14-14, j = 0.0071 and j_{i } = 0.0088.

From Eq. 14-29,

j  =  \frac{\bar{h}}{ G  c_{p}}  Pr^{2/3}                    (14-29)

h  =  \frac{j G_{c}  c_{p}}{Pr^{2/3}}  =  \frac{0.0071  (4900)  0.24}{(0.7)^{2/3}}  =  10.5  Btu/(hr – ft² – F)

and from Eq. 13-19

j_{m}  =  \frac{h_d}{\rho_a \overline{V}}Sc^{2/3}                (13-19)

h_{d}  =  \frac{j_{i} G_{c}}{Sc^{2/3}}  =  \frac{0.0088  (4900)}{(0.6)^{2/3}}  =  60.6  Ibma/(ft² – F)

The friction factor will be determined from Fig. 14-15. Using Eqs. 14-44 and 14-45,

FP  =  Re_{D}^{-0.25}  \left\lgroup\frac{D}{D^{*}} \right\rgroup^{0.25}  \left[\frac{\chi_{a}  –  D}{4 (s  –  y)} \right]^{-0.4}  \left[\frac{\chi_{a}}{D^{*}}  –  1\right]^{-0.5}                          (14-44)

D^{*}  = \frac{ \left(D  \frac{A}{A_{t}}\right)}{1  +  \frac{\chi_{a}  –  D}{s}}                                   (14-45)

D^{*}  =  \frac{0.525  (16.28)}{1  +  (1.25  –  0.525)/0.0833}  =  0.881

and

FP  =  (4870)^{-0.25}  \left\lgroup\frac{0.525}{0.881} \right\rgroup ^{0.25}  \times  \left[\frac{1.25  –  0.525}{4(0.0833  –  0.006)}\right] ^{-0.4} \left[\frac{1.25}{0.881}  –  1\right]  =  0.116

Using Eq. 14-53,

F (s)  =  [1  +  (Re_{s})^{-0.4}]  \left\lgroup\frac{s}{s  –  y} \right\rgroup ^{1.5}                  (14-53)

F (s)  =  (1  +  773^{-0.4})  \left\lgroup\frac{0.0833}{0.0833  –  0.006} \right\rgroup ^{1.5}=1.20

Then

FP F(s) = 0.116(1.20) = 0.14

and from Fig. 14-15, f = 0.031.

Note that the presence of moisture on the surface increases h, h_{d}, and f.