Question 14.2: Refer to Example 14-1, and find the geometric configuration ......

To do this, the NTU and fluid capacity rates must be computed. For the air,

\dot{m}  =  ρ\dot{Q}  =  \frac{14.7  (144)}{53.35  (510)}  (2000)  (60)  =  9,336  lbm/hr

and

C_{air}  =  C_{c}  =  0.24  (9336)  =  2241  Btu/(hr-F)

For the water,

\dot{q}  =  C_{w}  (t_{wi}  –  t_{wo})  =  C_{air}  (t_{ao}  –  t_{ai})

and

C_{w}  =  C_{h}  =  C_{air}  \frac{t_{ao}  –  t_{ai}}{t_{wi}  –  t_{wo}}

C_{w}  =  2241  \frac{100  –  50}{150  –  140}  =  11,205  Btu/(hr-F)

Since C_{w}  >  C_{air},  we  have  C_{air}  =  C_{min}  =  C_{c},  C_{w}  =  C_{h}  =  C_{max}, and

\frac{C_{min}}{C_{max}}  =  \frac{2241}{11,205}  =  0.20

The effectiveness ε is given by

ε  =  \frac{t_{co}  –  t_{ci}}{t_{hi}  –  t_{ci}}  =  \frac{100  –  50}{150  –  50}  =  0.50

Assuming that the flow arrangement is crossflow, the NTU is read from Fig. 14-18 at ε = 0.5 and C_{min}/C_{max} = 0.2 as 0.74. Assuming counterflow would yield very near the same value of NTU, then

NTU  =  \frac{U_{o}  A_{o}}{C_{min}}

A_{o}  =  \frac{0.74  (2241)}{10.2}  =  163  ft^{2}

The total volume of the heat exchanger is given by

V  =  \frac{A_{o}}{α}  =  \frac{163}{170}  =  0.96  ft^{3}

Since a face velocity of 900 ft/min was assumed, the face area is

A_{fr}  = \frac{Q}{V_{fr}}  = \frac{2000}{900}  =  2.22  ft^{2}

and the depth is

L  =  \frac{V}{A_{fr}}  =  \frac{0.96}{2.22}  =  0.43  ft  =  5.18  in.

The number of rows of tubes N_{r} will then be

N_{r}  =  \frac{L}{\chi_{b}}  =  \frac{5.18}{1.083}  =  4.78

Since N_{r} must be an integer and a multiple of two for the flow arrangement of Fig. 14-18, six rows must be used. This will overdesign the heat exchanger. Another possibility is to use five rows with a two-pass per row circuiting arrangement so that the water connections are on the same end of the coil. This will be considered in Example 14-4 when the lost head on the water side is computed.