Question 14.3: Referring to Examples 14-1 and 14-2, determine the pressure ......

The lost head on the air side of the exchanger is given by Eq. 14-38, where the ratio A/A_{c} is given by

\frac{ΔP_{0}}{P_{01}}  =  \frac{G^{2}_{c}}{2 g_{c}  ρ_{1}  P_{01}}\left[(1  +  \sigma ^{2}) \left\lgroup\frac{\rho _{1}}{\rho _{2}}  –  1\right\rgroup  +  f  \frac{\rho _{1}}{\rho _{m}} \right]                      (14-38)

\frac{A}{A_{c}}  =  \frac{αV}{σ A_{fr}}  =  \frac{170  (0.96)}{0.555  (2.22)}  =  132

The mass velocity G_{c} was previously computed as 7569 lbm/(hr-ft²), and using the perfect gas law the mean density ρ_{m} is approximately

ρ_{m}  =  \frac{P}{2  R}  \left\lgroup\frac{1}{T_{ci}}  +  \frac{1}{T_{co}} \right\rgroup

ρ_{m}  =  \frac{14.7  (144)}{2  (53.35)}  \left\lgroup\frac{1}{510}  +  \frac{1}{560} \right\rgroup  =  0.074  lbm/ft^{3}

The friction factor is read from Fig. 14-15 with FP computed from Eq. 14-44. Using Eq. 14-45,

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

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

\frac{D^{*} }{D}  =  \frac{11.6}{1  +  (1.25  –  0.525)/0.125}  =  1.71

and

FB  =  (7526)^{-0.25}  (1.71)^{-0.25} \left\lgroup\frac{1.25 – 0.525}{4 (0.125 – 0.006)} \right\rgroup^{-0.4} \left\lgroup\frac{1.25}{0.898}  –  1   \right\rgroup^{-0.4} = 0.130

then from Fig. 14-15 we have f = 0.027, and

∆P_{o}  =  \frac{(7569)^{2}}{2 (0.078) (32.2) (3600)^{2}}

×\left\{\left[1  + (0.555)^{2}\right] \left\lgroup\frac{0.078}{0.071}  –  1\right\rgroup  +  0.027 (132) \left\lgroup\frac{0.078}{0.074} \right\rgroup \right\}

∆P_{o}  =  3.42  lbf/ft^{2}

or

∆P_{o}  =  \frac{3.42}{62.4}  =  0.055  ft  wg  =  0.66  in.wg