Question 11.7: Design a finned-tube condenser for the service of Example 11...

(a) Make initial specifications.
The following changes are made to the initial specifications of Example 11.6.
(i) Shell and head types
Based on the results of Example 11.6 and the fact that a smaller shell will result from the use of finned tubes, difficulty in meeting the shell-side pressure drop specification while providing adequate tube support is anticipated. Therefore, a cross-flow shell is selected and an AXU configuration is specified.
(ii) Tubing
Referring to Table B.5, 3/4-in., 16 BWG, 26 fpi tubing is selected; the corresponding part number is 265065. A tube length of 16 ft and a triangular layout with tube pitch of 15/16 in. are also specified.
(iii) Baffles
Tube support plates are used in a cross-flow exchanger rather than standard baffles. A sufficient number of plates must be provided for adequate tube support and suppression of tube vibration. Considering the potential for tube vibration problems in this application, a plate spacing of 24 in. is a reasonable initial estimate, but this figure has no effect on the thermal or hydraulic calculations.
(b) Energy balances.
From Example 11.6 we have:

q=25.74 \times 10^{6}  Btu / h

 

\dot{m}_{\text {water }}=735,729  lbm / h

(c) Mean temperature difference.
The following data are obtained from Example 11.6:

\left(\Delta T_{ln }\right)_{c f}=72.8^{\circ} F

 

R=0.443 \quad P=0.355

The LMTD correction factor chart for an X-shell in Appendix 11.A is for a single tube pass; no chart is available for an X-shell with an even number of tube passes. However, with the given values of R and P, F will not differ greatly from the value of approximately 0.98 for a single tube pass. Therefore,

\Delta T_{m} \cong 0.98 \times 72.8=71.3^{\circ} F

(d) Approximate overall heat-transfer coefficient.
Based on the results of Example 11.6, the following values are estimated for the film coefficients:

h_{i} \cong 1000  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F \quad h_{o} \cong 250  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

The value of ho for finned tubes is expected to be significantly higher than the value of 132 Btu/h .ft². ºF calculated for plain tubes, while the value of h_{i} should be similar to that for plain tubes (1085 Btu/h . ft². ºF). The following data are obtained from Table B.5 for #265065 finned tubes:

A_{T o t} / L=0.596 ft ^{2} / ft   = external surface area per unit length

A_{\text {Tot }} / A_{i}=4.35

 

D_{r}=0.652 \text { in. }=0.0543  ft = root-tube diameter

D_{i}=0.522 \text { in. }=0.0435  ft

Using Equation (4.25) and assuming \eta_{w} \cong 1.0 we obtain:

\frac{1}{U_{D}A_{Tot}}=\frac{1}{h_{i} A_{i}}+\frac{R_{D i} }{A_{i}}+\frac{\ln \left(D_{0} / D_{i}\right)}{2 \pi k_{\text {pipe }} L}+\frac{1}{h_{o} \eta_{w}A_{Tot}}+\frac{R_{D o}}{\eta_{w}A_{Tot}}     (4.25)

 

U_{D}=\left[\frac{A_{T o t}}{h_{i} A_{i}}+\frac{R_{D i} A_{T o t}}{A_{i}}+\frac{A_{T o t} \ln \left(D_{r} / D_{i}\right)}{2 \pi k_{\text {tube }} L}+\frac{1}{h_{o} \eta_{w}}+\frac{R_{D o}}{\eta_{w}}\right]^{-1}

 

=\left[\frac{4.35}{1000}+0.001 \times 4.35+\frac{0.596 \ln (0.652 / 0.522)}{2 \pi \times 30}+\frac{1}{250 \times 1.0}+\frac{0.0005}{1.0}\right]^{-1}

 

U_{D}=72  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(e) Heat-transfer area and number of tubes.

A=\frac{q}{U_{D} \Delta T_{m}}=\frac{25.74 \times 10^{6}}{72 \times 71.3}=5014  ft ^{2}

 

n_{t}=\frac{A}{\left(A_{\text {Tot }} / L\right) \times L}=\frac{5014}{0.596 \times 16}=526

 

(f) Number of tube passes.
Assuming two tube passes, the velocity is:

V=\frac{\dot{m}\left(n_{P} / n_{ f }\right)}{\rho \pi\left(D_{i}^{2}\right)}=\frac{(735,429 / 3600)(2 / 526)}{0.99 \times 62.43 \times \pi(0.0435)^{2} / 4}=8.5  ft / s

Since the velocity is in the acceptable range for the tubing material, two tube passes will be used.

(g) Shell size and tube count.
From Table C.2, for a U-tube exchanger with 2 tube passes, the closest tube count is 534 tubes in a 25-in. shell.

(h) Required overall coefficient

U_{\text {req }}=\frac{q}{n_{ r }\left(A_{\text {Tot }} / L\right) \Delta T_{m}}=\frac{25.74 \times 10^{6}}{534 \times 0.596 \times 16 \times 71.3}=71  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

(i) Calculate h_{i} assuming \phi _{i} = 1.0.

R e=\frac{4 \dot{m}\left(n_{p} / n_{t}\right)}{\pi D_{i} \mu}=\frac{4 \times 735,429(2 / 534)}{\pi \times 0.0435 \times 0.72 \times 2.419}=46,289

 

h_{i}=\left(k / D_{i}\right) \times 0.023 R e^{0.8} P_{r}^{1 / 3}\left(\mu / \mu_{w}\right)^{0.14}

 

=(0.37 / 0.0435) \times 0.023(46,289)^{0.8}(4.707)^{1 / 3} \times 1.0

 

h_{i}=1770  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

(j) Calculate h_{0}.
For finned tubes, the equivalent diameter defined by Equation (11.65) is required:

D_{e}^{-0.25}=\frac{1.30 \eta_{f} A_{\text {fins }} E^{-0.25}+A_{\text {prime }} D_{r}^{-0.25}}{\eta_{w} A_{\text {Tot }}}

The fin efficiency for low-fin tubes is usually quite high unless the material of construction has a relatively low thermal conductivity. Therefore, \eta_{f}and \eta_{w} can be set to unity as a first approximation. The remaining parameters are calculated from the tube and fin dimensions:

E=\pi\left(r_{2}^{2}-r_{1}^{2}\right) / 2 r_{2}

 

r_{1}=D_{r} / 2=0.652 / 2=0.326  in

 

r_{2}=0.75 / 2=0.375  in

 

E=\pi\left\{(0.375)^{2}-(0.326)^{2}\right\} /(2 \times 0.375)

 

E=0.14388 \text { in }=0.0120  ft

For convenience, the fin and prime surface areas are calculated per inch of tube length:

A_{\text {fins }}=2 N_{f} \pi\left(r_{2}^{2}-r_{1}^{2}\right)=2 \times 26 \pi\left\{(0.375)^{2}-(0.326)^{2}\right\}=5.611 \text { in. }^{2}

 

\tau=\text { fin thickness }=0.013 \text { in. } (Table B.5)

 

A_{\text {prime }}=2 \pi r_{1}\left(L-N_{f \tau}\right)=2 \pi \times 0.326(1.0-26 \times 0.013)=1.356 \text { in. }^{2}

 

A_{\text {fins }} / A_{T o t}=5.611 /(5.611+1.356)=0.805

 

A_{\text {prime }} / A_{\text {Tot }}=1-A_{\text {fins }} / A_{T o t}=0.195

Substituting into the above equation for D_{e} gives:

D_{e}^{-0.25}=\frac{1.30 \times 1.0 \times 0.805(0.0120)^{-0.25}+0.195(0.0543)^{-0.25}}{1.0}=3.5658  ft ^{-0.25}

 

D_{e}=(3.5658)^{-4}=0.0062  ft

From Example 11.6, a tube wall temperature of 112ºF is assumed, which gives T_{f} = 128ºF and μ_{L} = 0.161 cp. The shell-side heat-transfer coefficient is calculated using Equation (11.68) with ⌈ replaced by \Gamma*:

h_{o}=0.609\left[\frac{k_{L}^{3} \rho_{L}\left(\rho_{L}-\rho_{V}\right) g \eta_{w}\left(A_{T o t} / L\right)}{\mu_{L} D_{e} \Gamma^{*}}\right]^{1 / 3}

 

=0.609\left[\frac{(0.057)^{3} \times 35.5(35.5-0.845) \times 4.17 \times 10^{8} \times 1.0 \times 0.596}{0.161 \times 2.419 \times 0.0062 \times 170.92}\right]^{1 / 3}

 

h_{o}=314  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

(k) Calculate T_{p}, T_{wtd}  and T_{f}.
For η_{w} = 1.0, Equations (4.38) and (4.39) give:

T_{p}=\frac{h_{i} t_{\text {ave }}+h_{o}\eta _w\left(A_{\text {Tot }} / A_{i}\right) T_{\text {ave }}}{h_{i}+h_{o}\eta _w\left(A_{\text {Tot }} / A_{i}\right)}      (4.38)

 

T_{wtd}=\frac{h_{i}\eta _w t_{\text {ave }}+\left[h_i(1-\eta _w)h_{o}\eta _w\left(A_{\text {Tot }} / A_{i}\right)\right]T_{\text {ave }}}{h_{i}+h_{o}\eta _w\left(A_{\text {Tot }} / A_{i}\right)}     (4.39)

 

T_{p}=T_{\text {wtd }}=\frac{h_{i} t_{\text {ave }}+h_{o}\left(A_{\text {Tot }} / A_{i}\right) T_{\text {ave }}}{h_{i}+h_{o}\left(A_{\text {Tot }} / A_{i}\right)}=\frac{1770 \times 102.5+314 \times 4.35 \times 175.75}{1770+314 \times 4.35}

 

T_{p}=T_{\text {wtd }}=134.4  ^{\circ}F

 

T_{f}=0.75 T_{w t d}+0.25 T_{V}=0.75 \times 134.4+0.25 \times 175.75 \cong 145  ^{\circ}F

At this value of T_{f}, μ_{L} = 0.148 cp. Using this value to recalculate ho and the wall temperatures gives:

h_{0}=323  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

T_{p}=T_{\text {wtd }}=134.9  ^{\circ} F

 

T_{f}=145.1  ^{\circ} F

Since the two values of T_{f} are essentially the same, no further iteration is required.
(l) Calculate fin efficiency.
Equation (2.27) is used to calculate the fin efficiency.

\eta _f\cong \frac{\tanh(m\psi )}{m\psi }            (2.27)

 

r_{2 c}=r_{2}+\tau / 2=0.375+0.013 / 2=0.3815  in .

 

\tau=0.013 / 12=0.001083  ft

 

\psi=\left(r_{2 c}-r_{1}\right)\left[1+0.35 \ln \left(r_{2 c} / r_{1}\right)\right]=(0.3815-0.326)[1+0.35 \ln (0.3815 / 0.326)]

 

\psi=0.5855  in .=0.004879  ft

 

m=\left(2 h_{o} / k \tau\right)^{0.5}=(2 \times 323 / 30 \times 0.001083)^{0.5}=141.0  ft ^{-1}

 

m \psi=141.0 \times 0.004879=0.6879

 

\eta_{f}=\tanh (m \psi) /(m \psi)=\tanh (0.6879) / 0.6879=0.867

The weighted efficiency of the finned surface is computed using Equation (2.31):

\eta_{w}=\left(A_{\text {prime }} / A_{\text {Tot }}\right)+\eta_{f}\left(A_{\text {fins }} / A_{\text {Tot }}\right)=0.195+0.867 \times 0.805 \cong 0.893

 

(m) Recalculate h_{o} and fin efficiency.
Repeating steps (j) and (k) with \eta_{f} = 0.867 and \eta_{w} = 0.893 yields the following results:

Repeating step (1) with the new value of h_{o} gives:

Since the new values of efficiency are essentially the same as the previous set, no further iteration is required and the above results are accepted as final.

(n) Viscosity correction factor for cooling water.
At 132ºF, the viscosity of water is 0.52 cp from Figure A.1. Hence,

h_{i}=1770(0.72 / 0.52)^{0.14}=1853  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

(o) Overall heat-transfer coefficient.

U_{D}=\left[\frac{A_{T o t}}{h_{i} A_{i}}+\frac{R_{D i} A_{T o t}}{A_{i}}+\frac{\left(A_{T o t} / L\right) \ln \left(D_{r} / D_{i}\right)}{2 \pi k_{t u b e}}+\frac{1}{h_{o} \eta_{w}}+\frac{R_{D o}}{\eta_{w}}\right]^{-1}

 

=\left[\frac{4.35}{1852}+0.001 \times 4.35+\frac{0.596 \ln (0.652 / 0.522)}{2 \pi \times 30}+\frac{1}{311 \times 0.896}+\frac{0.0005}{0.896}\right]^{-1}

 

U_{D}=86.6  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

(p) Correction for sensible heat transfer.
The effect of sensible heat transfer is neglected here since the effects of interfacial shear and condensate subcooling have also been neglected, which will compensate for this factor. Therefore, since U_{D} > U_{req} , the condenser is thermally acceptable, but somewhat over-sized (over-design = 22%).

(q) Tube-side pressure drop

f=0.4137 R e^{-0.2585}=0.4137(46,289)^{-0.2585}=0.0257

 

G=\frac{\dot{m}\left(n_{p} / n_{ t }\right)}{A_{f}}=\frac{735,429(2 / 534)}{(\pi / 4)(0.0435)^{2}}=1,853,366  lbm / h \cdot ft ^{2}

 

\Delta P_{f}=\frac{f n_{p} L G^{2}}{7.50 \times 10^{12} D_{i} s \phi}=\frac{0.0257 \times 2 \times 16(1,853,366)^{2}}{7.50 \times 10^{12} \times 0.0435 \times 0.99 \times 1.0}

 

\Delta P_{f}=8.40  psi

 

\Delta P_{r}=1.334 \times 10^{-13}\left(1.6 n_{p}-1.5\right) G^{2} / s

 

=1.334 \times 10^{-13}(1.6 \times 2-1.5)(1,853,366)^{2} / 0.99

 

\Delta P_{r}=0.79  psi

Assuming 10-in. schedule 40 nozzles are used, the nozzle losses will be the same as calculated in Example 11.6, namely:

\Delta P_{n}=0.36  psi

The total tube-side pressure drop is:

\Delta P_{i}=\Delta P_{f}+\Delta P_{r}+\Delta P_{n}=8.4+0.79+0.36

 

\Delta P_{i} \cong 9.6  psi

(r) Shell-side pressure drop.
The ideal tube bank correlations in Chapter 6 can be used to calculate the pressure drop in a cross-flow shell. The cross-flow area, S_{m}, is approximated by a_{s} with the baffle spacing equal to the length of the shell. For finned tubes, an effective clearance is used that accounts for the area occupied by the fins:

C_{e f f}^{\prime}=P_{T}-D_{r e}

where

D_{r e}=D_{r}+2 n_{f} b \tau = equivalent root-tube diameter
η_{f} = number of fins per unit length
Note that B C_{e ff}^{\prime} is the flow area between two adjacent tubes in one baffle space. Substituting the values of the parameters in the present problem gives:

D_{r e}=0.652+2 \times 26 \times 0.049 \times 0.013=0.6851 \text { in } .

 

C_{e f f}^{\prime}=15 / 16-0.6851=0.2524  in.

 

a_{s}=\frac{d_{s} C_{e f f}^{\prime} B}{144 P_{T}}=\frac{25 \times 0.2524 \times(16 \times 12)}{144 \times 15 / 16}=8.974  ft ^{2}

 

G=\dot{m}_{V} / a_{5}=180,000 / 8.974=20,058   lbm / h \cdot ft ^{2}

In calculating the Reynolds number for finned tubes, the equivalent root-tube diameter is used in place of D_{0} [32]:

\operatorname{Re}_{V}=\frac{D_{r e} G}{\mu_{V}}=\frac{(0.6851 / 12) \times 20,058}{0.0085 \times 2.419}=55,694

The friction factor for plain tubes is obtained from Figure 6.2: f_{\text {ideal }} = 0.10. The friction factor for finned tubes is estimated as 1.4 times the value for plain tubes [32]. Thus

f_{\text {ideal }}^{\prime}=1.4 f_{\text {ideal }}=1.4 \times 0.10=0.14

The number of tube rows crossed is estimated using Equation (6.8) with the baffle cut taken as zero.

N_{c}=\frac{D_{S}\left(1-2 B_{c}\right)}{P_{T}^{\prime}}     (6.8)

 

N_{c}=\frac{d_{S}\left(1-2 B_{c}\right)}{P_{T} \cos \theta_{t p}}=\frac{25}{(15 / 16) \cos 30^{\circ}}=30.8

The pressure drop for all-vapor flow is calculated using Equation (6.7). The effect of the bundle bypass flow is neglected here:

\Delta P_{\text{ideal}}=\frac{2 f_{\text {ideal }}^{\prime} N_{c} G^{2}}{g_{c} \rho\phi }      (6.7)

 

\left(\Delta P_{f}\right)_{V O}=\frac{2 f_{\text {ideal }}^{\prime} N_{c} G^{2}}{g_{c} \rho_{V} \phi}=\frac{2 \times 0.14 \times 30.8(20,058)^{2}}{4.17 \times 10^{8} \times 0.845 \times 1.0}

 

\left(\Delta P_{f}\right)_{V O}=9.85  lbf / ft ^{2}=0.068  psi

The two-phase friction loss is calculated using an average two-phase multiplier of 0.33 for a total condenser:

\Delta P_{f}=\bar{\phi}_{V O}^{2}\left(\Delta P_{f}\right)_{V O}=0.33 \times 0.068=0.022  psi

As would be expected, the pressure drop in the X-shell is very small. Assuming two 10-in. schedule 40 inlet nozzles are used, the nozzle losses will be the same as calculated in Example 11.6, namely

\Delta P_{n, \text { in }}=0.266  psi

Two condensate nozzles are specified, corresponding to the two inlet nozzles. The condensate nozzles are sized for self-venting operation. Since each nozzle handles half the condensate, the volumetric flow rate is:

\dot{v}_{L}=\dot{m} / \rho=(90,000 / 3600) / 35.5=0.704  ft ^{3} / s

Using Inequality (11.5a), we obtain:

D_{n}>0.89  \dot{v}_{L}^{0.4}=0.89(0.704)^{0.4}=0.773  ft =9.28  in

From Table B.2, for schedule 40 pipe the smallest size that satisfies this condition is 10 in. Therefore, two 10-in. outlet nozzles are required.
The pressure drop across the self-venting condensate nozzles will be very small and can be neglected. Hence, the total shell-side pressure drop is:

\Delta P_{o}=\Delta P_{f}+\Delta P_{n, \text { in }}=0.022+0.266 \cong 0.29  psi

All design criteria are satisfied; however, the condenser is somewhat over-sized. The number of tubes cannot be reduced because the tube-side pressure drop is close to the maximum. Therefore, we consider using shorter tubes. The required tube length is:

L_{\text {req }}=\frac{q}{\left(A_{\text {Tot }} / L\right) n_{t} U_{D} \Delta T_{m}}=\frac{25.74 \times 10^{6}}{0.596 \times 534 \times 86.6 \times 71.3}

 

L_{\text {req }}=13.1  ft

Hence, the tube length is reduced to 14 ft, which gives an over-design of about 7%. This change will decrease the tube-side pressure drop and increase the shell-side pressure drop slightly. Although these changes will not affect the viability of the design, the new pressure drops are calculated here for completeness. For the tube side, we have

\Delta P_{f}=8.40(14 / 16)=7.35  psi

 

\Delta P_{i}=\Delta P_{f}+\Delta P_{r}+\Delta P_{n}=7.35+0.79+0.36

 

\Delta P_{i}=8.5  psi

For the shell side:

a_{s}=\frac{25 \times 0.2524(14 \times 12)}{144(15 / 16)}=7.85  ft ^{2}

 

G=180,000 / 7.85=22,930  lbm / h \cdot ft ^{2}

 

R e=\frac{(0.6851 / 12) \times 22,930}{0.0085 \times 2.419}=63,668

 

f_{\text {ideal }} \cong 0.095 (Figure 6.2)

 

f_{\text {ideal }}^{\prime}=1.4 f_{\text {ideal }}=1.4 \times 0.095=0.133

 

\left(\Delta P_{f}\right)_{V O}=\frac{2 \times 0.133 \times 30.8(22,930)^{2}}{4.17 \times 10^{8} \times 0.845 \times 1.0}

 

\left(\Delta P_{f}\right)_{V O}=12.2 lbf / ft ^{2}=0.085  psi

 

\Delta P_{f}=0.33 \times 0.085=0.028  psi

 

\Delta P_{o}=0.028+0.266 \cong 0.29  psi

The final design parameters are summarized below.
Final design summary
Tube-side fluid: cooling water
Shell-side fluid: condensing hydrocarbon
Shell: type AXU, 25-in. ID, oriented horizontally
Number of tubes: 534
Tube size: 0.75 in., 16 BWG, radial low-fin tubes, 14 ft long
Fins: 26 fpi, 0.049 in. high, 0.013 in. thick
Tube layout: 15/16-in. triangular pitch
Tube passes: 2
Baffles/support plates: 6 support plates
Sealing strips: as needed based on tube layout
Tube-side nozzles: 10-in. schedule 40 inlet and outlet
Shell-side nozzles: two 10-in. schedule 40 inlet (top), two 10-in. schedule 40 outlet (bottom)
Materials: tubes and tubesheets, 90/10 copper-nickel alloy; all other components, carbon steel