Question 14.1: Design a water-to-air heating coil of the continuous plate–f......

Figure 14-19 is a schematic of a typical water-to-air heating coil that has multiple rows of tubes. Although the water may be routed through the tubes in many different ways, the circuiting is usually such that counterflow will be approached as shown. Counterflow can usually be assumed when three or more rows are used. Because the water

inlet and outlet connections must be on the same end of the coil in this case, a multiple of two rows is used; otherwise, two passes per row will be required.

Compute the overall heat-transfer coefficient U, based on the air-side area. Equation 14-11 applies where \eta _{si} is equal to one and the wall thermal resistance is negligible:

\frac{1}{U_{0}} = \frac{1}{h_{0} \eta _{s0}}  +  \frac{Δx}{k(A_{m}/A_{0})}  +  \frac{1}{h_{i}  \eta _{s0} (A_{i}/A_{0})}                     (14-11)

\frac{1}{U_{0}}  =  \frac{1}{h_{0}  \eta_{s0}}  +  \frac{1}{h_{i} (A_{i}/A_{0})}

The subscript o refers to the air side and i to the water side. Equation 14-22 will be used to find the coefficient h_{i} assuming a water velocity of 4 ft/sec. Experience has shown that velocities greater than 5 ft/sec (1.5 m/s) result in very high lost head. Since at this point the tube diameter must be established, a surface geometry must be selected.

One standard plate–fin–tube surface uses \frac{1}{2} in. tubes in a triangular layout as shown in Fig. 14-12 with \chi_{a}  of  1.25  in.  and  \chi_{b} of 1.083 in. Assume the fin pitch is 8 fins/in. and the fin thickness is 0.006 in. As a result of fabrication of the coil, the final tube outside diameter is 0.525 in. with a wall thickness of 0.015 in. Other geometric data will be given as required, and the j-factor and friction factor will be obtained from Figs. 14-14 and 14-15. The Reynolds number based on the tube inside diameter is then

\frac{\bar{h} D}{k} = 0.023(Re_{D})^{0.8} (Pr)^{n}               (14-22)

Re_{D} = \frac{ρ \bar{V} D}{µ} = \frac{61.5  (4)  (0.4831/12)}{1.04/3600}  =  34,275

where ρ and µ are evaluated at 145 F. The Prandtl number is

Pr = \frac{µ c_{p}}{k} = \frac{(1.04)  (1.0)}{0.38}  =  2.74

Then using Eq. 14-22,

\bar{h}_{i} = 0.023 \frac{k}{D} (Re_{D})^{0.8} (Pr)^{0.3}

\bar{h}_{i} = 0.023 \frac{0.38}{(0.483/12)} (34,275)^{0.8} (2.74)^{0.3}

\bar{h}_{i} = 1250  Btu/(hr-ft^{2} -F)

where the exponent on the Prandtl number is for t_{wall}  <  t_{bulk} and L/D has been assumed to be larger than 60.

To compute the air-side heat-transfer coefficient it is necessary to know the air velocity or air mass velocity inside the core. Because the coil face velocity cannot exceed 1000 ft/min, a face velocity of 900 ft/min will be assumed. Then

\dot{m}_{a}  =  G_{fr}  A_{fr}  =  G_{c}  A_{c}

and

G_{c} = G_{fr} \frac{A_{fr}}{A_{c}} = \frac{G_{fr}}{σ}

where the subscript fr refers to the face of the coil and c refers to the minimum flow area inside the coil. The ratio of minimum flow area to frontal area for this case is about 0.555 from Figure 14-12:

G_{fr}  =  ρ_{fr}  \bar{V}_{fr}  =  \frac{14.7  (144)  (900)  60}{53.35  (510)}  =  4200  lbm/(hr-ft^{2})

and

G_{c}  =  \frac{4200}{0.555}  =  7569  lbm/(hr-ft^{2})

The j-factor correlation of Fig. 14-14 is based on the parameter JP, which is defined by Eq. 14-39. The Reynolds number is then

JP  =  Re^{0.8}_{D}  \left\lgroup \frac{A}{A_{t}}\right\rgroup ^{-0.15}                   (14-39)

Re_{D}  =  \frac{G_{c} D}{µ}  =  \frac{7569  (0.525/12)}{0.044}  =  7526

and the parameter A/A_{t} defined by Eq. 14-41 is

\frac{A}{A_{t}}  = \frac{π}{4}  \frac{\chi_{a}}{D_{h}}  \frac{\chi_{b}}{D}  σ                              (14-41)

A/A_{t}  = \frac{4  (1.083)  1.25  (0.555)}{π  (0.01312)  12  (0.525)}  =  11.6

where the hydraulic diameter is another known dimension of the coil (Fig. 14-12). The parameter JP is

JP  =  (7526)^{–0.4}(11.6)^{–0.15}  =  0.0195

The j-factor is now read from Fig. 14-14 as 0.0066. Then

St  Pr^{2/3} = \left\lgroup\frac{\bar{h_{0}}}{G_{c} c_{p}}\right\rgroup  \left\lgroup\frac{μ  c_{p}}{k}\right\rgroup^{2/3}  =  0.0066

or

\bar{h_{0}}  =  0.0066  (7569 ) (0.24)  (0.71)^{-2/3}  =  15.1  Btu/(hr-ft^{2} -F)

The next step is to compute the fin efficiency and the surface effectiveness. Equations 14-16, 14-17, 14-18, and 14-20 will be used. The equivalent fin radius R_{e} is first computed from Eq. 14-20. The dimensions L and M are found as follows by referring to Fig. 14-12:

m  =  (\frac{2hL}{kLy})^{1/2}  =  (\frac{2h}{ky})^{1/2}            (14-16)

η  =  \frac{\tanh  (mrφ)}{mrφ}                (14-17)

φ  =  (\frac{R}{r}  –  1) [1 + 0.355  \ln (R/r)]                (14-18)

\frac{R_{e}}{r}  =  1.27  ψ  (β  –  0.3)^{1/2}                  (14-20)

Dim_{1}  = \frac{\chi_{a}}{2}  =  \frac{1.25}{2}  =  0.625  in.

Dim_{2}  = \frac{\left[(\chi_{a}/2)^{2}  +  \chi^{2}_{a}\right]^{1/2}}{2}

= \frac{\left[(0.625)^{2}  +  (1.083)^{2}\right]^{1/2}}{2} =  0.625  in.

However, Dim_{1}  is  equal  to  Dim_{2} in this case:

L = M = 0.625 in.

Then

\psi  = \frac{M}{r}  =  \frac{0.625}{0.525/2}  =  2.38

\beta  = \frac{L}{M}  = \frac{0.625}{0.625}  =  1.0

and

\frac{R_{e}}{r}  =  1.27(2.38)(1.0  –  0.3)^{1/2}  =  2.53

From Eq. 14-18

φ  =  (2.53  –  1)(1  +  0.35\ln 2.53)  =  2.03

and using Eq. 14-16,

m  =  \left[\frac{2(15.1)}{100  (0.006/12)}\right]^{1/2}   =  24.6  ft^{-1}

where the thermal conductivity k of the fin material has been assumed equal to 100 (Btu-ft)/(ft²-hr-F), which is typical of aluminum fins. Then from Eq. 14-17,

\eta  =  \frac{\tanh  [(24.6)  (0.525/24)  (2.03)]}{(0.525/24)  (2.03)  (24.6)} = 0.73

The surface effectiveness η_{so} is then computed using Eq. 14-8 where A_{f} /A is 0.919:

η_{s}  =  \frac{A_{b}  +  η  A_{f}}{A}  =  1  –  \frac{A_{f}}{A} (1  –  η)                      (14-8)

η_{so}  =  1  –  0.919(1  –  0.73)  =  0.75

The ratio of the water-side to air-side heat-transfer areas must finally be determined. The ratio α of the total air-side heat-transfer area to the total volume (A_{o}/V)  is  given  as  170  ft^{–1}. The ratio of the water-side heat-transfer area to the total volume (A_{i}/V) is closely approximated by

\frac{A_{i}}{V}  =  \frac{D_{i}π}{\chi _{a}  \chi _{b}}

\frac{A_{i}}{A_{0}}  =  {\left\lgroup\frac{A_{i}}{V} \right\rgroup}/{\left\lgroup\frac{A_{0}}{V} \right\rgroup}  =  \frac{π  D_{i}}{\chi _{a}  \chi _{b}  α}

\frac{A_{i}}{A_{0}}  =  \frac{π  (0.483/12)}{(1.25/12)  (1.083/12)  (170)}  =  0.079                            (14-49)

The overall coefficient U is then given by

\frac{1}{U_{0}}  =  \frac{1}{15.1(0.75)}  +  \frac{1}{1248(0.097)}  =  0.098

and

U_{0}  =  10.2  Btu/(hr-ft^{2} -F)