Question 14.4: Referring to Examples 14-1, 14-2, and 14-3, compute the pres......

Recall that a velocity of 4 ft/sec was assumed to compute the heat-transfer coefficient $h_{i}$ . It has also been determined that at least five rows of tubes are required and the water connections must be on the same end of the exchanger. Therefore, consider the arrangement shown in Fig. 14-20. If we use two passes per row of tubes, the water enters and leaves the same end of the coil. For the coil shown there are five separate water circuits.

The flow cross-sectional area for the water may be determined from the fluid capacity rate for the water and the continuity equation:

$\dot{m}_{w} = \bar{V} A ρ = \frac{C_{h}}{c_{p}}$

and

$A = \frac{C_{h}}{\bar{V} ρc_{p}} \left\lgroup\frac{11,205}{3600} \right\rgroup \frac{1}{(4) (61.5) (1.0)}$

$A = 0 01265  ft^{2}$

For N tubes,

$A  =  N  \frac{\pi}{4} D^{2}_{i}$

and

$N  =  \frac{4  A}{\pi D^{2}_{i}}  =  \frac{4  (0.01265)  (144)}{\pi  (0.483)^{2}}  =  9.94$

Since N must be an integer, 10 tubes are required and the water velocity is reduced somewhat. This reduction in velocity will not significantly reduce the heat-transfer coefficient $h_{i}$ . To adapt to the flow arrangement of Fig. 14-20, a coil that is 20 tubes high must be used. Then the height H becomes

$H  =  20 \chi_{a}  =  20(1.25)  =  25  in.$

The frontal area $A_{fr}$ was previously found to be 2.22 ft². Then the width W is

$W  =  \frac{A_{fr}}{H}  =  \frac{2.22}{25/144}  =  12.8  in.$

This arrangement will meet all of the design requirements; however, the shape of the coil (height 25 in. and width 12.8 in.) may be unacceptable. If so, another alternative must be sought, such as using six rows of tubes or placing the headers on opposite ends.

The lost head $l_{fw}$ will be computed using Eq. 10-6. Lost head in the return bends will be allowed for by assuming a loss coefficient of 2 for each bend. The flow length $L_{w}$ is

$L_{w}  =  f  \frac{L}{D}  \frac{\bar{V}}{2g}$                  (10-6)

$L_{w}  =  2  (5)  (12.8/25)  =  10.7  ft$

and the Moody friction factor is 0.023 from Fig. 10-1 at a Reynolds number of 34,275, which takes into account the lower water velocity. There are nine return bends in each circuit. Then

$l_{fw}  =  0.023  \frac{10.7}{(0.483/12)}  \frac{(4)^{2}}{(64.4)}  +  2  (9)  \frac{(4)^{2}}{(64.4)}$

$l_{fw}  =  6  ft  of  head$