Question 10.2: 96,000 lb/h of a distillation bottoms having the following c...

(a) Make initial specifications.
(i) Fluid placement
There is no choice here; the boiling fluid must be placed in the shell and the heating medium in the tubes.
(ii) Tubing
One-inch, 14 BWG, U-tubes with a length of 16 ft are specified. A tubing diameter of  ¾ in. could also be used.
(iii) Shell and head types
A TEMA K-shell is chosen for a kettle reboiler, and a type B head is chosen since the tube-side fluid (steam) is clean. Thus, a BKU configuration is specified.
(iv) Tube layout
A square layout with a tube pitch of 1.25 in. is specified to permit mechanical cleaning of the external tube surfaces. Although this service should be quite clean, contaminants in distillation feed streams tend to concentrate in the bottoms, and kettle reboilers are very prone to fouling.
(v) Baffles and sealing strips
None are specified for a kettle reboiler. Support plates will be used to provide tube support and vibration suppression. Four plates are specified to give an unsupported tube length that is safely below the maximum of 73 in. listed in Table 5.C1.

(vi) Construction materials
Since neither stream is corrosive, plain carbon steel is specified for all components.
(b) Energy balance and steam flow rate.
The reboiler duty is obtained from an energy balance on the process stream (boiling fluid):

q=\dot{m_V}H_V+\dot{m_L}H_L-\dot{m_F}H_F

where the subscripts F, L, and V refer to the reboiler feed, liquid overflow, and vapor return streams, respectively. Substituting the appropriate enthalpies and flow rates gives:

q=48,000 \times 216.4+48,000 \times 109.9-96,000 \times 106.7

 

q \cong 5.42 \times 10^{6}  Btu / h

From Table A.8, the latent heat of condensation for steam at 20 psia is 960.1 Btu/lbm. Therefore, the steam flow rate will be:

\dot{m}_{\text {steam }}=q / \lambda_{\text {steam }}=5.42 \times 10^{6} / 960.1=5645  lbm / h

(c) Mean temperature difference.
The effective mean temperature difference is computed as if the boiling-side temperature were constant at the vapor exit temperature, which in this case is 202.4ºF. The temperature of the condensing steam is also constant at the saturation temperature, which is 228.0ºF at 20 psia from Table A.8. Therefore, the effective mean temperature difference is:

\Delta T_{m}=228.0-202.4=25.6  ^{\circ}F

(d) Approximate overall coefficient.
Referring to Table 3.5, it is seen that for light hydrocarbons boiling on the shell side with condensing steam on the tube side, 200 ≤ U_{d} ≤  300 Btu/h . ft².º F. Taking the mid-range value gives U_{D} = 250 Btu/h · ft²· ºF. for preliminary design purposes.

TABLE 3.5 Typical Values of Overall Heat-Transfer Coefficients in Tubular Heat Exchangers. U = Btu/h · ft²· ºF.

NC: non-condensable gas present; V: vacuum; A: atmospheric pressure.
Dirt (or fouling factor) units are (h) (ft²) (ºF)/Btu

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

A=\frac{q}{U_{D} \Delta T_{m}}=\frac{5.42 \times 10^{6}}{250 \times 25.6} \cong 847  ft ^{2}

 

n_{t}=\frac{A}{\pi D_{0} L}=\frac{847}{\pi(1 / 12) \times 16}=202

Note that n_{t} represents the number of straight sections of tubing in the bundle, i.e., the number of tube holes in the tubesheet. For U-tubes, this is twice the actual number of tubes. However, it corresponds to the value listed in the tube-count tables, and so will be referred to as the number of tubes.
(f) Number of tube passes.
For condensing steam, two passes are sufficient.
(g) Actual tube count and bundle diameter.
From Table C.5, the closest tube count is 212 tubes in a 23.25 in. shell. This shell size is the smaller diameter of the K-shell at the tubesheet. The bundle diameter will, of course, be somewhat smaller, but a value of 23 in. will be sufficiently accurate for design calculations.
(h) Required overall coefficient.
The required overall heat-transfer coefficient is calculated in the usual manner:

U_{\text {raq }}=\frac{q}{n_{t} \pi D_{0} L  \Delta T_{m}}=\frac{5.42 \times 10^{6}}{212 \times \pi \times(1 / 12) \times 16 \times 25.6}=238  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(i) Inside coefficient, h_{i} .
For condensing steam we take

\left[\left(D_{o} / D_{i}\right)\left(1 / h_{i}+R_{D i}\right)\right]^{-1} \cong 1500  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(j) Outside coefficient, h_{o} = h_{b} .
Palen’s [1] method, which was presented in Chapter 9, will be used in order to ensure a safe (i.e., conservative) design. It is based on the Mostinski correlation for the nucleate boiling heat-transfer coefficient, to which correction factors are applied to account for mixture effects and convection in the tube bundle.
(i) Nucleate boiling coefficient, h_{nb}
The first step is to compute the pseudo-critical and pseudo-reduced pressures for the mixture, which will be used in place of the true values in the Mostinski correlation:

P_{p c}=\sum x_{i} P_{c, i}=0.15 \times 616.3+0.25 \times 529.0+0.60 \times 551.1=555.4  psi

 

P_{p r}=P / P_{p c}=250 / 555.4=0.45

The Mostinski correlation is used as given in Equation (9.2a), along with the mixture correction factor as given by Equation (9.17a). Also, since P_{p r}  > 0.2, Equation (9.18) is used to calculate the pressure correction factor. Thus,

h_{n b}=0.00622  P_{c}^{0.69} \hat{q}^{0.7} F_{P} F_{m}

 

F_{P}=1.8  P_{r}^{0.17} = 1.8(0.45)^{0.17}=1.5715

 

F_{m}=\left(1+0.0176  \hat{q}^{0.15}  B R^{0.75}\right)^{-1}

 

B R=T_{D}-T_{B}=205.6-197.6=8.0^{\circ} F

The heat flux is computed using the required duty and the heat-transfer area for the initial configuration:

\hat{q}=\frac{q}{n_{ t } \pi D_{o} L}=\frac{5.42 \times 10^{6}}{212 \pi(1 / 12) \times 16}=6103  Btu / h \cdot ft ^{2}

 

F_{m}=\left[1+0.0176(6103)^{0.15}(8)^{0.75}\right]^{-1}=0.7636

 

h_{n b}=0.00622(555.4)^{0.69}(6103)^{0.7} \times 1.5715 \times 0.7636

 

h_{n b}=261  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(ii) Bundle boiling coefficient, h_{b}

The boiling heat-transfer coefficient for the tube bundle is given by Equation (9.19):

h_{b}=h_{n b} F_{b}+h_{n c}

Although the tube wall temperature is unknown, with an overall temperature difference of 25.6ºF, the heat transfer by natural convection should be small compared to the boiling component. Therefore, h_{nc} is roughly estimated as 44 Btu/h · ft²·ºF. The bundle convection factor is computed using Equation (9.20) with D_{b} ≅  23 in.:

F_{b}=1.0+0.1\left[\frac{0.785 D_{b}}{C_{1}\left(P_{T} / D_{o}\right)^{2} D_{o}}-1.0\right]^{0.75}

 

=1.0+0.1\left[\frac{0.785 \times 23}{1.0(1.25 / 1.0)^{2} \times 1.0}-1.0\right]^{0.75}

 

F_{b}=1.5856

The outside coefficient is then:

h_{o}=h_{b}=261 \times 1.5856+44 \cong 458  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(k) Overall coefficient.

U_{D}=\left[\left(1 / h_{i}+R_{D i}\right)\left(D_{o} / D_{i}\right)+\frac{D_{o} \ln \left(D_{o} / D_{i}\right)}{2 k_{\text {tube }}}+1 / h_{o}+R_{D o}\right]^{-1}

Based on the values in Table 10.2, a boiling-side fouling allowance of 0.0005 h · ft² · ºF/Btu is deemed appropriate for this service. For 1-in. 14 BWG tubes, D_{i} = 0.834 in. from Table B.1. Taking k_{tube } ≅ 26 Btu/h · ft² · ºF for carbon steel, we obtain:

TABLE 10.2 Recommended Fouling Factors for Reboiler Design

U_{D}=\left[(1 / 1500)+\frac{(1.0 / 12) \ln (1.0 / 0.834)}{2 \times 26}+1.0 / 458+0.0005\right]^{-1}

 

U_{D}=275  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(l) Check heat flux and iterate if necessary.
A new value of the heat flux can be obtained using the overall coefficient

\hat{q}=U_{D} \Delta T_{m}=275 \times 25.6=7040  Btu / h \cdot ft ^{2}

Since this value differs significantly from that used to calculate h_{nb} , steps (j) and (k) should be repeated until consistent values for \hat{q} and U_{D} are obtained. Due to the uncertainty in both the heat-transfer coefficient and the mean temperature difference, exact convergence is not required. The following values are obtained after several more iterations:

h_{ b } \cong 523  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

U_{D} \cong 297  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

\hat{q} \cong 7600  Btu / h \cdot ft ^{2}

The overall coefficient exceeds the required coefficient of 238 Btu/h · ft²·ºF by a significant amount (over-design = 25%), indicating that the reboiler is over-sized.
(m) Critical heat flux.
The critical heat flux for nucleate boiling on a single tube is calculated using the Mostinski correlation, Equation (9.23a)

\hat{q}_{c}=803 P_{c} P_{r}^{0.35}\left(1-P_{r}\right)^{0.9}

 

=803 \times 555.4(0.45)^{0.35}(1-0.45)^{0.9}

 

\hat{q}_{c}=196,912  Btu / h \cdot ft ^{2}

The critical heat flux for the bundle is obtained from Equation (9.24):

\hat{q}_{c, \text { bundile }}=\hat{q}_{c, \text { tube }} \phi_{b}=196,912 \phi_{b}

The bundle geometry parameter is given by:

\psi_{b}=\frac{D_{b}}{n_{ t } D_{o}}=\frac{23}{212 \times 1.0}=0.1085

Since this value is less than 0.323, the bundle correction factor is calculated as:

\phi_{b}=3.1 \psi_{b}=3.1 \times 0.1085=0.3364

Therefore, the critical heat flux for the bundle is:

\hat{q}_{c, \text { bundle }}=196,912 \times 0.3364 \cong 66,240  Btu / h \cdot ft ^{2}

Now the ratio of the actual heat flux to the critical heat flux is:

\hat{q} / \hat{q}_{c, b u n d l e}=7600 / 66,240 \cong 0.11

This ratio should not exceed 0.7 in order to provide an adequate safety margin for reliable operation of the reboiler. In the present case, this criterion is easily met.
Note: In addition, the process-side temperature difference, ΔT_{e}, must be in the nucleate boiling range. In operation, the value of ΔT_{e} may exceed the maximum value for nucleate boiling, particularly when the unit is clean. This situation can usually be rectified by adjusting the steam pressure. Maximum values of ΔT_{e} are tabulated for a number of substances in Ref. [11], and they provide guidance in specifying an appropriate design temperature for the heating medium in these and similar cases. For a given substance, the critical ΔT_{e} decreases markedly with increasing pressure. It is sometimes stated that the overall ΔT should not exceed about 90ºF to 100ºF in order to ensure nucleate boiling. However, this rule is not generally valid owing (in part) to the effect of pressure on the critical ΔT_{e}.
(n) Design modification.
The simplest way to modify the initial design in order to reduce the amount of heat-transfer area is to shorten the tubes. The required tube length is calculated as follows:

L_{\text {req }}=\frac{q}{n_{ t } \pi D_{0} U_{D} \Delta T_{m}}=\frac{5.42 \times 10^{6}}{212 \pi(1 / 12) \times 297 \times 25.6}

 

L_{\text {req }}=12.8  ft

Therefore, a tube length of 13 ft will be sufficient. (Note that a length of 12.8 ft is consistent with a heat flux of 7600 Btu/h . ft².) A second option is to reduce the number of tubes. From the tube-count table, the next smallest standard bundle (21.25 in.) contains 172 tubes. This modification will not be pursued here; it is left as an exercise for the reader to determine the suitability of this configuration.
(o) Number of nozzles.
Equation (10.1) gives the number of pairs of nozzles:

N_{n}=\frac{L}{5 D_{b}}=\frac{13}{5(23 / 12)}=1.36

Rounding upward to the next largest integer gives two pairs of inlet and outlet nozzles. They will be spaced approximately 4.4 ft apart. Alternatively, a single pair of nozzles could be used with the inlet nozzle located near the front of the shell.
(p) Shell diameter.
We first use Equation (10.2) to calculate the vapor loading

V L=2290 \rho_{V}\left(\frac{\sigma}{\rho_{L}-\rho_{V}}\right)^{0.5}=2290 \times 2.76\left(\frac{3.59}{28.4-2.76}\right)^{0.5}

 

V L=2365  lbm / h \cdot ft ^{3}

The required dome segment area is then found using Equation (10.3):

S A=\frac{\dot{m}_{V}}{L \times V L}=\frac{48,000}{13 \times 2365} \cong 1.56  ft ^{2}

Next, the effective liquid height in the reboiler is estimated by adding 4 in. to the approximate bundle diameter (23 in.) to account for foaming, giving a value of 27 in. Assuming as a first approximation that the liquid height is 60% of the shell diameter, we obtain:

D_{s}=\frac{27}{0.6}=45.0 \text { in. } \cong 3.75  ft

 

h / D=1-0.6=0.4

The sector area factor is obtained from Appendix 10.A:

h: height; D: diameter; and A: area.
Rules for using table: (1) Divide height of segment by the diameter; multiply the area in the table corresponding to the quotient, height/diameter, by the diameter squared. When segment exceeds a semicircle, its area is: area of circle minus the area of a segment whose height is the circle diameter minus the height of the given segment. (2) To find the diameter when given the chord and the segment height: the diameter = [(½ chord)²/height] + height.
Source: Ref. [19]

Multiplying this factor by the square of the diameter gives the segment area:

S A=0.29337(3.75)^{2}=4.13  ft ^{2}

Since this is much greater than the required area, a smaller diameter is needed. Assuming (after several more trials) that the effective liquid height is 73% of the shell diameter, the next trial gives:

D_{s}=\frac{27}{0.73}=36.99 \text { in. } \cong 3.08  ft

 

h / D=1-0.73=0.27

 

A=0.17109 (Appendix 10.A)

S A=0.17109(3.08)^{2}=1.62  ft ^{2}

This value is slightly larger than the required dome segment area, which is acceptable. Therefore, a shell diameter of about 37 in. will suffice.
(q) Liquid overflow reservoir.
The reservoir is sized to provide adequate holdup time for control purposes. We first calculate the volumetric flow rate of liquid over the weir:
volumetric flow rate =\frac{48,000  lbm / h }{\left(28.4  lbm / ft ^{3}\right)(60  min / h )}=28.17  ft ^{3} / min

Next, the cross-sectional area of the shell sector below the weir is calculated. The sector height is equal to the weir height, which is about 23 in. Therefore

h / D=23 / 37=0.62

1-h /D = 0.38

The sector area factor corresponding to this value is 0.27386 from Appendix 10.A. Hence,
sector area above weir = 0.27386(37 / 12)² = 2.60 ft²

sector area below weir = \pi(37 / 12)^{2} / 4-2.60=4.87  ft ^{2}

Now the shell length required is:

L_{s}=\frac{28.17  ft ^{3} / min }{4.87  ft ^{2}} \cong 5.8  ft /

Therefore, a reservoir length of 3 ft will provide a holdup time of approximately 30 s, which is adequate to control the liquid level using a standard cascaded level-to-flow control loop. With allowances for U-tube return bends and clearances, the overall length of the shell will then be about 17 ft. It is assumed that relatively large fluctuations in the bottom product flow rate are acceptable in this application.
(r) Feed and return lines.
The available liquid head between the reboiler inlet and the surface of the liquid in the column sump is 9 ft. The corresponding pressure difference is:

\Delta P_{\text {available }}=\rho_{L}\left(g / g_{c}\right) \Delta h_{L}=28.4(1.0) \times 9

 

\Delta P_{\text {ayailable }}=255.6  lbf / ft ^{2}=1.775 psi

This pressure difference must be sufficient to compensate for the friction losses in the feed line, vapor return line, and the reboiler itself; the static heads in the reboiler and return line; and the pressure loss due to acceleration of the fluid in the reboiler resulting from vapor formation. Of these losses, only the friction losses in the feed and return lines can be readily controlled, and these lines must be sized to meet the available pressure drop. We consider each of the pressure losses in turn.
(i) Static heads
The static head consists of two parts, namely, the two-phase region between the reboiler inlet and the surface of the boiling fluid, and the vapor region from the surface of the boiling fluid through the return line and back down to the  liquid surface in the column sump. We estimate the two-phase head loss using the average vapor fraction in the boiling region, x_{ave} = 0.25. The average density is calculated using the homogeneous model, which is sufficiently accurate for the present purpose:

\rho_{\text {ave }}=\left[\frac{1-x_{\text {ave }}}{\rho_{L}}+\frac{x_{\text {ave }}}{\rho_{V}}\right]^{-1}=\left[\frac{0.75}{28.4}+\frac{0.25}{2.76}\right]^{-1} \cong 8.55  lbm / ft ^{3}

The vertical distance between the reboiler inlet and the surface of the boiling fluid is approximately 23 in. The corresponding static pressure difference is:

\Delta P_{t p}=\frac{8.55 \times(23 / 12)}{144}=0.114  psi

The elevation difference between the boiling fluid surface in the reboiler and the liquid surface in the column sump is:

\Delta h=9-23 / 12=7.08  ft

The pressure difference corresponding to this head of vapor is:

\Delta P_{V}=\frac{2.76 \times 7.08}{144} \cong 0.136  psi

The total pressure difference due to static heads is the sum of the above values:

\Delta P_{\text {staric }}=0.114+0.136=0.250  psi

(ii) Friction and acceleration losses in reboiler
The friction loss is small due to the low circulation rate characteristic of kettle reboilers. The large vapor volume provided in the kettle results in a relatively low vapor velocity, and therefore the acceleration loss is also small. Hence, both these losses can be neglected. However, as a safety factor, an allowance of 0.2 psi will be made for the sum of these losses.(A range of 0.2–0.5 psi is typical for thermosyphon reboilers, so an allowance of 0.2 psi should be more than adequate for a kettle.)
(iii) Friction loss in feed lines
We begin by assuming the configuration shown below for the feed lines. The total length of the primary line between the column sump and the tee is approximately 23 ft as given in the problem statement. Each branch of the secondary line between the tee and the reboiler has a horizontal segment of length 2.2 ft and a vertical segment of length 1.0 ft.

The pipe diameter is chosen to give a liquid velocity of about 5 ft/s. Thus, for the primary line:

D_{i}=\left(\frac{4 \dot{m}}{\pi \rho V}\right)^{1 / 2}=\left[\frac{4(96,000 / 3600)}{\pi \times 28.4 \times 5}\right]^{1 / 2}

 

D_{i}=0.49  ft =5.87  in

From Table B.2, a 6-in. schedule 40 pipe with an inside diameter of 6.065 in. is appropriate.
For the secondary line, the flow rate is halved. Therefore

D_{i}=\left[\frac{4(48,000 / 3600)}{\pi \times 28.4 \times 5}\right]^{1 / 2}=0.0346  ft =4.15  in

A 4-in. schedule 40 pipe with an inside diameter of 4.026 in. is the closest match. However, with 4-in. inlet nozzles, the value of ρV² will exceed the TEMA erosion prevention limit of 500 lbm/ft . s² for bubble-point liquids. Therefore, in order to avoid the need for impingement protection, 5-in. nozzles with inside diameter of 5.047 in. and matching piping will be used.
The pressure drop is computed using the equivalent pipe lengths for flow resistance of fittings given in Appendix D. The equivalent lengths for the two pipe sizes are tabulated below. Note that only one branch of the 5-in. pipe is used because the pressure drop is the same for each parallel branch.

The Reynolds number for the 6-in. pipe is:

R e=\frac{4 \dot{m}}{\pi D_{i} \mu}=\frac{4 \times 96,000}{\pi(6.065 / 12) \times 0.074 \times 2.419}=1.351 \times 10^{6}

The friction factor is calculated using Equation (4.8):

f=0.3673 R e^{-0.2314}=0.3673\left(1.351 \times 10^{6}\right)^{-0.2314}

f = 0.014

The pressure drop is given by Equation (4.5) with the equivalent pipe length used in place of the actual length. The mass flux and specific gravity are computed first:

G=\dot{m} / A_{\text {flow }}=\frac{96,000}{(\pi / 4)(6.065 / 12)^{2}}=478,500  lbm / h \cdot ft ^{2}

 

s=\rho / \rho_{\text {tater }}=28.4 / 62.43=0.455

 

\Delta P_{f}=\frac{f L G^{2}}{7.50 \times 10^{12} D_{i} s \phi}=\frac{0.014 \times 91(478,500)^{2}}{7.50 \times 10^{12}(6.065 / 12) \times 0.455 \times 1.0}

 

\Delta P_{f} \cong 0.169  psi

The calculations for the 5-in. pipe are similar:

R e=\frac{4 \times 48,000}{\pi(5.047 / 12) \times 0.074 \times 2.419}=811,768

 

f=0.3673(811,768)^{-0.2314} \cong 0.0158

 

G=\frac{48,000}{(\pi / 4)(5.047 / 12)^{2}}=345,499  lbm / h \cdot ft ^{2}

 

\Delta P_{f}=\frac{0.0158 \times 43.7(345,499)^{2}}{7.50 \times 10^{12}(5.047 / 12) \times 0.455 \times 1.0}

 

\Delta P_{f} \cong 0.0574  psi

The total friction loss in the feed lines is therefore:

\Delta P_{\text {feed }}=0.169+0.0574 \cong 0.226  psi

(iv) Friction loss in return lines
A return line configuration similar to that of the feed line is assumed as shown below. The primary line has a total length of 20 ft as given in the problem statement. Each branch of the line connected to the reboiler has a vertical segment of length 1.0 ft and a horizontal segment of length 2.2 ft.

We begin by calculating the maximum recommended vapor velocity using Equation (5.B.1):

V_{\max }=\frac{1800}{(P M)^{0.5}}=\frac{1800}{(250 \times 55.48)^{0.5}}=15.3  ft / s

The lines will be sized for a somewhat lower velocity of about 12 ft/s. For the main line, the required diameter is:

D_{i}=\left(\frac{4 \dot{m}}{\pi \rho V}\right)^{1 / 2}=\left[\frac{4(48,000 / 3600)}{\pi \times 2.76 \times 12}\right]^{1 / 2}

 

D_{i}=0.716  ft =8.59  in

From Table B.2, the closest match is an 8-in. schedule 40 pipe with an internal diameter of 7.981 in.
For the split-flow section, we have:

D_{i}=\left[\frac{4(29,000 / 3600)}{\pi \times 2.76 \times 12}\right]^{1 / 2}=0.506 ft =6.07 in

Six-inch schedule 40 pipe (ID =6.065 in.) is appropriate for this section. Equivalent pipe lengths are summarized in the following table:

The calculations for the 8-in. line are as follows:

R e=\frac{4 \dot{m}}{\pi D_{i} \mu}=\frac{4 \times 48,000}{\pi(7.981 / 12) \times 0.0095 \times 2.419}=3.999 \times 10^{6}

 

f=0.3673 R e^{-0.2314}=0.3673\left(3.999 \times 10^{6}\right)^{-0.2314} \cong 0.0109

 

G=\dot{m} / A_{f l o w}=\frac{48,000}{(\pi / 4)(7.981 / 12)^{2}}=138,165  lbm / h \cdot ft ^{2}

 

s=\rho / \rho_{\text {water }}=2.76 / 62.43=0.0442

 

\Delta P_{f}=\frac{f L G^{2}}{7.50 \times 10^{12} D_{i} s \phi}=\frac{0.0109 \times 122(138,165)^{2}}{7.50 \times 10^{12}(7.981 / 12) \times 0.0442 \times 1.0}

 

\Delta P_{f} \cong 0.115  psi

The calculations for the 6-in. line are similar, but the flow rate is halved:

R e=\frac{4 \times 24,000}{\pi(6.065 / 12) \times 0.0095 \times 2.419}=2.631 \times 10^{6}

 

f=0.3673\left(2.631 \times 10^{6}\right)^{-0.2314}=0.012

 

\Delta P_{f}=\frac{0.012 \times 38.2(119,625)^{2}}{7.50 \times 10^{12}(6.065 / 12) \times 0.0442 \times 1.0}

 

\Delta P_{f} \cong 0.0392  psi

The total friction loss in the return lines is thus:

\Delta P_{\text {return }}=0.115+0.0392 \cong 0.154  psi

(v) Total pressure loss
The total pressure loss is the sum of the individual losses calculated above:

\Delta P_{\text {total }}=\Delta P_{\text {static }}+\Delta P_{\text {relouler }}+\Delta P_{\text {feed }}+\Delta P_{\text {retum }}

= 0.191 + 0.2 + 0.226 + 0.154

\Delta P_{\text {total }}=0.770  psi

Since this value is less than the available pressure drop of 1.775 psi, the piping configuration is acceptable. However, a smaller feed line can be used. Replacing the 6-in. line with a 5-in. line gives a fluid velocity of 6.8 ft/s and a total pressure drop of 1.0 psi, both of which are still comfortably on the safe side. In actual operation, the liquid level in the column sump will self-adjust to satisfy the pressure balance.
(s) Tube-side pressure drop.
The pressure drop for condensing steam is usually small due to the low flow rate compared with sensible heating media. For completeness, however, the pressure drop is estimated here. For a condensing vapor, the two-phase pressure drop in the straight sections of tubing can be approximated by half the pressure drop calculated at the inlet conditions (saturated steam at 20 psia, vapor fraction = 1.0). The requisite physical properties of steam are obtained from Tables A.8 and A.9:

\rho=1 / 20.087=0.0498  lbm / ft ^{3}

 

s=\rho / \rho_{\text {water }}=0.0498 / 62.43=0.000797

 

\mu=0.012 cp

\dot{m}=5645  lbm / h (from step (b))

\dot{m}_{\text {per tule }}=\dot{m}\left(n_{p} / n_{ t }\right)=5645(2 / 212)=53.25   lbm / h

 

G=\frac{\dot{m}_{\text {per tule }}}{(\pi / 4) D_{i}^{2}}=\frac{53.25}{(\pi / 4)(0.834 / 12)^{2}}=14,037  lbm / h \cdot ft ^{2}

 

R e=\frac{D_{1} G}{\mu}=\frac{(0.834 / 12) \times 14,037}{0.012 \times 2.419}=33,608

The friction factor is calculated using Equation (5.2):

f=0.4137 R e^{-0.2585}=0.4137(33,608)^{-0.2585}

f = 0.0280

The pressure drop is calculated by incorporating a factor of 1/2 on the right side of Equation (5.1):

\Delta P_{f} \cong \frac{1}{2}\left[\frac{f n_{p} L G^{2}}{7.50 \times 10^{12} D_{i} 5 \phi}\right]=\frac{1}{2}\left[\frac{0.0280 \times 2 \times 13(14,037)^{2}}{7.50 \times 10^{12}(0.834 / 12) \times 0.000797 \times 1.0}\right]

 

\Delta P_{f} \cong 0.173  psi

To this degree of approximation, the pressure drop in the return bends can be neglected. However, the pressure drop in the nozzles will be calculated to check the nozzle sizing. Based on Table 10.3, 6-in. and 3-in. schedule 40 nozzles are selected for steam and condensate, respectively. For the steam nozzle we have:

TABLE 10.3 Guidelines for Sizing Steam and Condensate Nozzles

G_{n}=\frac{\dot{m}}{(\pi / 4) D_{1}^{2}}=\frac{5645}{(\pi / 4)(6.065 / 12)^{2}}=28,137  lbm / h \cdot ft ^{2}

 

R e_{n}=\frac{D_{i} G_{n}}{\mu}=\frac{(6.065 / 12) \times(28,137)}{0.012 \times 2.419}=489,903

Since the flow is turbulent, allow 1 velocity head for the inlet nozzle loss. From Equation (4.11), we obtain:

\Delta P_{n, \text { steam }}=1.334 \times 10^{-13} G_{n}^{2} / s=\frac{1.334 \times 10^{-13}(28,137)^{2}}{0.000797}

 

\Delta P_{n \text { steam }}=0.133  psi

For the condensate at 20 psia, the physical properties are obtained from Tables A.8 and A.9.

\rho=1 / 0.016834=59.40  lbm / ft ^{3}

 

s=\rho / \rho_{\text {water }}=59.40 / 62.43=0.9515

 

\mu=0.255  cP

 

G_{n}=\frac{\dot{m}}{(\pi / 4) D_{i}^{2}}=\frac{5645}{(\pi / 4)(3.068 / 12)^{2}}=109,958  lbm / h \cdot ft ^{2}

 

R e_{n}=\frac{D_{1} G_{n}}{\mu}=\frac{(3.068 / 12) \times 109,958}{0.255 \times 2.419}=45,575

Since the flow is turbulent, allow 0.5 velocity head for the loss in the exit nozzle:

\Delta P_{\text {neandensate }}=\frac{0.5 \times 1.334 \times 10^{-13}(109,958)^{2}}{0.9515}=0.00085  psi

The total tube-side pressure drop is estimated as:

\Delta P_{i} \cong \Delta P_{f}+\Delta P_{n, \text { steam }}+\Delta P_{n, \text { condensate }}

 

\Delta P_{i}=0.173+0.133+0.00085 \cong 0.3  psi

The pressure drop is small, as it should be for condensing steam. Therefore, the tubing and nozzle configurations are acceptable. The final design parameters are summarized below.
Design summary
Shell type: BKU
Port diameter/Shell ID: 23.25 in./37 in.
Shell length: approximately 17 ft
Length beyond weir: 3 ft
Weir height: approximately 23 in.
Tube bundle: 212 tubes (106 U-tubes), 1 in. OD, 14 BWG, 13 ft long on 1.25 in. square pitch
Baffles: none
Support plates: 3 (One less plate is used due to the reduced tube length.)
Shell-side nozzles: two 5-in. schedule 40 inlet, two 6-in. schedule 40 vapor outlet, one 4-in. schedule 40 liquid outlet
Tube-side nozzles: 6-in. schedule 40 inlet, 3-in. schedule 40 outlet
Feed lines: 5-in. schedule 40 from column to inlet tee, 5-in. schedule 40 from tee to reboiler
Return lines: 6-in. schedule 40 from reboiler to outlet tee, 8-in. schedule 40 from tee to column
Materials: plain carbon steel throughout
Note: Due to the elevated pressure on the shell side of the reboiler, it is particularly important in this application to check the required wall thickness of the shell-side nozzles. By running a mechanical design program, it was found that schedule 40 pipe is inadequate for these nozzles. Schedule 80 pipe is required for the vapor exit nozzles, while schedule 120 pipe is needed for the liquid exit and feed nozzles.