Question 5.1: A kerosene stream with a flow rate of 45,000 lb/h is to be c...

(a) Make initial specifications.
(i) Fluid placement
Kerosene is not corrosive, but crude oil may be, depending on salt and sulfur contents and temperature. At the low temperature of the oil stream in this application, however, corrosion should not be a problem provided the oil has been desalted (if necessary). Nevertheless, the crude oil should be placed in the tubes due to its relatively high fouling tendency. Also, the kerosene should be placed in the shell due to its large ΔT of 140ºF according to the guidelines given in Table 3.4.

TABLE 3.4 Criteria for Fluid Placement, in Order of Priority

(ii) Shell and head types
The recommended fouling factor for kerosene is 0.001–0.003 h · ft² · ºF/Btu (Table 3.3), indicating a significant fouling potential. Therefore, a floating-head exchanger is selected to permit mechanical cleaning of the exterior tube surfaces. Also, the floating tubesheet will allow for differential thermal expansion due to the large temperature difference between the two streams. Hence, a type AES exchanger is specified.

TABLE 3.3 Typical Values of Fouling Factors (h . ft² . ºF/Btu)dcont’d

^aAssumes water velocity greater than 3 ft/s. Lower values of ranges correspond to water temperature below about 120ºF and hot stream temperature below about 250ºF.
^bAssumes desalting at approximately 250ºF and a minimum oil velocity of 2 ft/s.

(iii) Tubing
Following the design guidelines for a fouling oil service, 1 in., 14 BWG tubes are selected with a length of 20 ft.
(iv) Tube layout
Since cleaning of the tube exterior surfaces will be required, square pitch is specified to provide cleaning lanes through the tube bundle. Following the design guidelines, for 1-in. tubes a tube pitch of 1.25 in. is specified.
(v) Baffles
Segmental baffles with a 20% cut are required by the Simplified Delaware method, but this is a reasonable starting point in any case. In consideration of Figure 5.3, a baffle spacing of 0.3 shell diameters is chosen, i.e., B/d_{s} = 0.3.

(vi) Sealing strips
One pair of sealing strips per ten tube rows is specified in accordance with the requirements of the Simplified Delaware method and the design guidelines.
(vii) Construction materials
Since neither fluid is corrosive, plain carbon steel is specified for tubes, shell, and other components.

(b) Energy balances

q=\left(\dot{m} C_{P} \Delta T\right)_{\text {ker }}=45,000 \times 0.59 \times 140=3,717,000  Btu / h

 

3,717,000=\left(\dot{m} C_{p} \Delta T\right)_{\text {oil }}=150,000 \times 0.49 \times \Delta T_{\text {oil }}

 

\Delta T_{\text {oil }}=50.6^{\circ} F

outlet oil temperature = 150.6º F

(c) LMTD.

\left(\Delta T_{\ln }\right)_{c f}=\frac{239.4-150}{\ln (239.4 / 150)}=191.2^{\circ} F

(d) LMTD correction factor.

R=\frac{T_{a}-T_{b}}{t_{b}-t_{a}}=\frac{390-250}{150.6-100}=2.77

 

P=\frac{t_{b}-t_{a}}{T_{a}-t_{a}}=-\frac{150.6-100}{390-100}=0.174

From Figure 3.14 or Equation (3.15), for a 1-2 exchanger F ≅ 0.97. Therefore, one shell pass is required.

F=\frac{\sqrt{R^2+1}\ln\left(\frac{1-S}{1-RS} \right) }{(R-1)\ln\left[\frac{2-S\left(R+1-\sqrt{R^2+1}\right)}{2-S\left(R+1-\sqrt{R^2+1}\right)} \right] }             (3.15)

(e) Estimate U_{D} .
In order to obtain an initial estimate for the size of the exchanger, an approximate value for the overall heat-transfer coefficient is used. From Table 3.5, for a kerosene/oil exchanger, it is found that 20 ≤ U_{D} ≤ 35 Btu/h . ft² . ºF. A value near the middle of the range is selected: U_{D} = 25 Btu/h . ft² . ºF

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

(f) Calculate heat-transfer area and number of tubes.

A=\frac{q}{U_{D} F\left(\Delta T_{ ln }\right)_{c f}}=\frac{3,717,000}{25 \times 0.97 \times 191.2}

 

A=801.7  ft ^{2}

 

n_{t}=\frac{A}{\pi D_{0} L}=\frac{801.7}{\pi \times(1 / 12) \times 20}

 

n_{ t }=153

(g) Number of tube passes.
The number of tube passes is chosen to give fully developed turbulent flow in the tubes and a reasonable fluid velocity

R e=\frac{4 \dot{m}\left(n_{p} / n_{ f }\right)}{\pi D_{i} \mu}

 

D_{i}=0.834 \text { in. }=0.0695  ft  (Table B.1)

R e=\frac{4 \times 150,000\left(n_{p} / 153\right)}{\pi \times 0.0695 \times 8.7}=2064.5  n_{p}

We want Re 104 and an even number of passes. Therefore, take n_{p} = 6. Checking the fluid velocity,

V=\frac{\dot{m}\left(n_{p} / n_{t}\right)}{\rho \pi D_{i}^{2} / 4}=\frac{(150,000 / 3600)(6 / 153)}{0.85 \times 62.43 \pi(0.0695)^{2} / 4}=8.1  ft / s

The velocity is at the high end of the recommended range, but still acceptable. Therefore, six tube passes will be used.
(h) Determine shell size and actual tube count.
From the tube-count table for l in. tubes on 1 ¼ -in. square pitch (Table C.5), with six tube passes and a type S head, the listing closest to 153 is 156 tubes in a 21 ¼ -in. shell. Hence, the number of tubes is adjusted to n_{t} = 156 and the shell ID is taken as d_{s} = 21.25 in.
This completes the initial design of the heat exchanger. The initial design must now be rated to determine whether it is adequate for the service. Since the temperature dependence of the fluid properties is not available, they will be assumed constant; the viscosity correction factors will be set to unity and the tube wall temperature will not be calculated.

(i) Calculate the required overall coefficient.

U_{\text {req }}=\frac{q}{n_{ t } \pi D_{o} L F\left(\Delta T_{\ln }\right)_{c f}}=\frac{3,717,000}{156 \times \pi \times(1.0 / 12) \times 20 \times 0.97 \times 191.2}

 

U_{\text {req }}=24.5   Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(j) Calculate h_{i}.

R e=\frac{4 \dot{m}\left(n_{p} / n_{t}\right)}{\pi D_{i} \mu}=\frac{4 \times 150,000(6 / 156)}{\pi \times 0.0695 \times 8.7}=12,149

 

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

 

=(0.077 / 0.0695) \times 0.023(12,149)^{0.8}(55.36)^{1 / 3}(1.0)

 

h_{i}=180  Btu / h  \cdot ft ^{2} \cdot{ }^{\circ} F

(k) Calculate h_{0}.

B=0.3  d_{s}=0.3 \times 21.25=6.375 in

 

a_{s}=\frac{d_{s} C^{\prime} B}{144 P_{T}}=\frac{21.25 \times 0.25 \times 6.375}{144 \times 1.25}=0.188  ft ^{2}

 

G=\dot{m} / a_{5}=45,000 / 0.188=239,362  lbm / h \cdot ft ^{2}

 

D_{e}=0.99 / 12=0.0825  ft  (from Figure 3.17)

R e=D_{e} G / \mu=0.0825 \times 239,362 / 0.97=20,358

Equation (3.21) is used to calculate the Colburn factor, j_{H}.

j_{H}=0.5\left(1+B / d_{s}\right)\left(0.08 R e^{0.6821}+0.7 R e^{0.1772}\right)

 

=0.5(1+0.3)\left[0.08(20,358)^{0.6821}+0.7(20,358)^{0.1772}\right]

 

j_H=47.8

 

h_{o}=j_{H}\left(k / D_{e}\right) Pr^{1 / 3}\left(\mu / \mu_{w}\right)^{0.14}

 

=47.8(0.079 / 0.0825)(7.24)^{1 / 3}(1.0)

 

h_{o}=88.5  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(1) Calculate the clean overall coefficient.

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

 

=\left[\frac{1.0}{180 \times 0.834}+\frac{(1.0 / 12) \ln (1.0 / 0.834)}{2 \times 26}+\frac{1}{88.5}\right]^{-1}

 

U_{C}=54.8  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

Since U_{C} > U_{req}, continue.
(m) Fouling factors
The fouling factor for the crude oil is specified as 0.003 h . ft² . ºF/Btu, and from Table 3.3, a value of 0.002 h . ft² . ºF/Btu is taken for kerosene. Hence, the total fouling allowance is:

R_{D}=\frac{R_{D i} D_{o}}{D_{i}}+R_{D o}=\frac{0.003 \times 1.0}{0.834}+0.002=0.0056  h \cdot ft ^{2} \cdot{ }^{\circ} F / Btu

 

U_{D}=\left(1 / U_{C}+R_{D}\right)^{-1}=(1 / 54.8+0.0056)^{-1}=41.9  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(n) Calculate the design overall coefficient.

Since U_{D} is much greater than U_{req}, the exchanger is thermally workable, but over-sized.

(o) Over-surface and over-design
It is convenient to perform the calculations using overall coefficients rather than surface areas. The appropriate relationships are as follows

over-surface = U_{C} / U_{r e q}-1=54.8 / 24.5-1=124 \%

over-design = U_{D} / U_{\text {req }}-1=41.9 / 24.5-1=71 \%

Clearly, the exchanger is much larger than necessary.
(p) Tube-side pressure drop
The friction factor is computed using Equation (5.2).

f=0.4137 R e^{-0.2585}=0.4137(12,149)^{-0.2585}=0.03638

 

G=\frac{\dot{m}\left(n_{p} / n_{t}\right)}{\pi D_{i}^{2} / 4}=\frac{150,000(6 / 156)}{\left[\pi(0.0695)^{2} / 4\right]}=1,520,752  lbm / h \cdot ft ^{2}

The friction loss is given by Equation (5.1):

\Delta P_{f}=\frac{f  n_{p} L G^{2}}{7.50 \times 10^{12} D_{i} s \phi}=\frac{0.03638 \times 6 \times 20(1,520,752)^{2}}{7.50 \times 10^{12} \times 0.0695 \times 0.85 \times 1.0}=22.8  psi

The tube entrance, exit, and return losses are estimated using Equation (5.3) with α_r equal to (2n_p -1.5) from Table 5.1.

TABLE 5.1 Number of Velocity Heads Allocated for Minor Losses on Tube Side

\Delta P_{r}=1.334 \times 10^{-13}\left(2 n_{p}-1.5\right) G^{2} / s =1.334 \times 10^{-13}(10.5)(1,520,752)^{2} / 0.85

 

\Delta P_{ r }=3.81  psi

The sum of the two pressure drops is much greater than the allowed pressure drop. Therefore, the nozzle losses will not be calculated.
(q) Shell-side pressure drop
The friction factor is calculated using Equations (5.7) to (5.9).

f_{1}=\left(0.0076+0.000166 d_{5}\right) R e^{-0.125}=(0.0076+0.000166 \times 21.25)(20,358)^{-0.125}

 

f_{1}=0.00322  ft ^{2} / in ^{2}

 

f_{2}=\left(0.0016+5.8 \times 10^{-5} d_{5}\right) R e^{-0.157}=\left(0.0016+5.8 \times 10^{-5} \times 21.25\right)(20,358)^{-0.157}

 

f_{2}=0.000597  ft ^{2} / in ^{2}

 

f=144\left\{f_{1}-1.25\left(1-B / d_{s}\right)\left(f_{1}-f_{2}\right)\right\}

 

=144\{0.00322-1.25(1-0.3)(0.00322-0.000597)\}

f = 0.1332

The number of baffle spaces, n_{b}+1, is estimated by neglecting the thickness of the tubesheets. The baffle spacing is commonly interpreted as the center-to-center distance between baffles, which is technically the baffle pitch. In effect, the baffle thickness is accounted for in the baffle spacing. (This interpretation is inconsistent with the equation for the flow area, as, where the face-to-face baffle spacing should be used rather than the center-to-center spacing. However, the difference is usually of little practical consequence and is neglected herein.) The result is:

n_{b}+1 \cong L / B=(20 \times 12) / 6.375=37.65 \Rightarrow 38

The friction loss is given by Equation (5.6):

\Delta P_{f}=\frac{f G^{2} d_{s}\left(n_{b}+1\right)}{7.50 \times 10^{12} d_{e} s \phi}=\frac{0.1332(239,362)^{2} \times 21.25 \times 38}{7.50 \times 10^{12} \times 0.99 \times 0.785 \times 1.0}

 

\Delta P_{f}=1.06  psi

The nozzle losses will not be calculated because the initial design requires significant modification. This completes the rating of the initial design.
In summary, there are two problems with the initial configuration of the heat exchanger:
(1) The tube-side pressure drop is too large.
(2) The exchanger is over-sized.

In addition, the pressure drop on the shell-side is quite low, suggesting a poor trade-off between pressure drop and heat transfer.
To remedy these problems, both the number of tubes and the number of tube passes can be reduced. We first calculate the number of tubes required, assuming the overall heat-transfer coefficient remains constant.

A_{\text {req }}=\frac{q}{U_{D} F\left(\Delta T_{ ln }\right)_{c f}}=\frac{3,717,000}{41.9 \times 0.97 \times 191.2}=478  ft ^{2}

 

\left(n_{ t }\right)_{\text {req }}=\frac{A_{\text {req }}}{\pi D_{o} L}=\frac{478}{\pi(1.0 / 12) \times 20}=91.3 \Rightarrow 92

Taking four tube passes, the tube-count table shows that the closest count is 104 tubes in a 17.25-in. shell. The effect of these changes on the tube-side flow and pressure drop are estimated as follows.

R e \rightarrow 12,149(4 / 6)(156 / 104)=12,149

From Equation (5.19),

\Delta P_{f} \sim \frac{Ln_{p}^{2.74}}{n_{t}^{1.74}D_i^{4.74}}  (turbulent flow)          (5.19)

\Delta P_{f} \sim n_{p}^{2.74} n_{t}^{-1.74}

 

\Delta P_{f} \rightarrow 22.8(4 / 6)^{2.74}(104 / 156)^{-1.74}=15.2  psi

Since these changes leave Re (and hence G) unchanged, the minor losses are easily calculated using Equation (5.3).

\Delta P_{ r }=1.334 \times 10^{-13}(6.5)(1,520,752)^{2} / 0.85=2.36  psi

Thus, in order to meet the pressure-drop constraint, the tube length will have to be reduced significantly, to about 15 ft when allowance for nozzle losses is included. The resulting under-surfacing will have to be compensated by a corresponding increase in h_{0}, which is problematic. (Since Re_{i} remains unchanged, so does h_i.) It is left as an exercise for the reader to check the viability of this configuration.
In order to further reduce the tube-side pressure drop, we next consider an exchanger with more tubes. Referring again to the tube-count table, the next largest unit is a 19.25-in. shell containing a maximum of 130 tubes (for four passes). The tube-side
Reynolds number for this  configuration is:

R e \rightarrow 12,149(4 / 6)(156 / 130)=9719

Reducing the number of tubes to 124 (31 per pass) gives Re_{i} = 10,189. The friction loss then becomes:

\Delta P_{f} \rightarrow 22.8(4 / 6)^{2.74}(124 / 156)^{-1.74}=11.2  psi

The minor losses will also be lower, and shorter tubes can be used in this unit, further assuring that the pressure-drop constraint will be met. (The final tube length will be determined after the overall heat-transfer coefficient has been recalculated.)
Thus, for the second trial a 19.25-in. shell containing 124 tubes arranged for four passes is specified. Due to the low shell-side pressure drop, the baffle spacing is also reduced to the minimum of 0.2 d_s. (Using a baffle spacing at or near the minimum can cause excessive leakage and bypass flows on the shell-side, resulting in reduced performance of the unit [11]. The low shell-side pressure drop in the present application should mitigate this problem.)

Second Trial
Based on the foregoing analysis, we anticipate that the exchanger will have sufficient heat-transfer area to satisfy the duty. Therefore, we will calculate the overall coefficient, U_{D}., and use it to determine the tube length that is needed. The pressure drops will then be checked.
(a) Calculate h_{i}.

R e=\frac{4 \dot{m}\left(n_{p} / n_{t}\right)}{\pi D_{i} \mu}=\frac{4 \times 150,000(4 / 124)}{\pi \times 0.0695 \times 8.7}=10,189

 

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

 

=(0.077 / 0.0695) \times 0.023(10,189)^{0.8}(55.36)^{1 / 3}(1.0)

 

h_{i}=156  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(b) Calculate h_{0}.

B=0.2 d_{s}=0.2 \times 19.25=3.85 \text { in. }

 

a_{s}=\frac{d_{5} C^{\prime} B}{144 P_{T}}=\frac{19.25 \times 0.25 \times 3.85}{144 \times 1.25}=0.103  ft ^{2}

 

G=\dot{m} / a_{5}=45,000 / 0.103=436,893  lbm / h \cdot ft ^{2}

 

R e=D_{e} G / \mu=0.0825 \times 436,893 / 0.97=37,158

 

j_{H}=0.5\left(1+B / d_{s}\right)\left(0.08 R e^{0.6821}+0.7 R e^{0.1772}\right)

 

=0.5(1+0.2)\left[0.08(37,158)^{0.6821}+0.7(37,158)^{0.1772}\right]

 

j_{H}=65.6

 

h_{o}=j_{H}\left(k / D_{e}\right) Pr^{1 / 3}\left(\mu / \mu_{w}\right)^{0.14}

 

=65.6(0.079 / 0.0825)(7.24)^{1 / 3}(1.0)

 

h_{0}=122  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(c) Calculate U_{D}.

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

 

=\left[\frac{1.0}{156 \times 0.834}+\frac{(1.0 / 12) \ln (1.0 / 0.839)}{2 \times 26}+\frac{1}{122}+0.0056\right]^{-1}

 

U_{D}=46  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(d) Calculate tube length.

q=U_{D} n_{ t } \pi D_{0} L F\left(\Delta T_{\ln }\right)_{c f}

 

L=\frac{q}{U_{D} n_{ t } \pi D_{o} F\left(\Delta T_{ ln }\right)_{q}}=\frac{3,717,000}{46 \times 124 \pi(1.0 / 12) \times 0.97 \times 191.2}=13.4  ft

Therefore, take L = 14 ft.
(e) Tube-side pressure drop.

f=0.4137 R e^{-0.2585}=0.4137(10,189)^{-0.2585}=0.03807

 

G=\frac{\dot{m}\left(n_{p} / n_{ t }\right)}{\left(\pi D_{i}^{2} / 4\right)}=\frac{150,000(4 / 124)}{\left[\pi(0.0695)^{2} / 4\right]}=1,275,469 1bm / h \cdot ft ^{2}

 

\Delta P_{f}=\frac{f n_{p} L G^{2}}{7.50 \times 10^{12} D_{i} s \phi}=\frac{0.03807 \times 4 \times 14(1,275,469)^{2}}{7.50 \times 10^{12} \times 0.0695 \times 0.85 \times 1.0}

 

\Delta P_{ r }=1.334 \times 10^{-13}\left(2 n_{p}-1.5\right) G^{2} / s=1.334 \times 10^{-13}(6.5)(1,275,469)^{2} / 0.85

 

\Delta P_{ r }=1.66  psi

Table 5.3 indicates that 4-in. nozzles are appropriate for a 19.25-in. shell. Assuming schedule 40 pipe is used for the nozzles,

TABLE 5.3 Guidelines for Sizing Nozzles

R e_{n}=\frac{4 \dot{m}}{\pi D_{n} \mu}=\frac{4 \times 150,000}{\pi(4.026 / 12) \times 8.7}=65,432

 

G_{n}=\dot{m} /\left(\pi D_{n}^{2} / 4\right)=150,000 /\left[\pi(4.026 / 12)^{2} / 4\right]=1,696,744  lbm / h \cdot ft ^{2}

Since the flow in the nozzles is turbulent, Equation (5.4) is applicable.

\Delta P_{n}=2.0 \times 10^{-13} N_{s} G_{n}^{2} / s=2.0 \times 10^{-13} \times 1 \times(1,696,744)^{2} / 0.85

 

\Delta P_{n}=0.68  psi

The total tube-side pressure drop is:

\Delta P_{i}=\Delta P_{f}+\Delta P_{ r }+\Delta P_{n}=7.83+1.66+0.68=10.17 \cong 10.2  psi

(f) Shell-side pressure drop.
Since B/d_s = 0.2, the friction factor is given by

f=144 f_{2}=144\left(0.0016+5.8 \times 10^{-5} d_{s}\right) R e^{-0.157}

 

=144\left(0.0016+5.8 \times 10^{-5} \times 19.25\right)(37,158)^{-0.157}

f = 0.07497

n_{B}+1 \cong L / B=(14 \times 12) / 3.85=43.6 \Rightarrow 43 \text { (Rounded downward to keep } B \geq B_{\min } \text { ) }

 

\Delta P_{f}=\frac{f G^{2} d_{s}\left(n_{b}+1\right)}{7.50 \times 10^{12} d_{e} s \phi}=\frac{0.07497(436,893)^{2} \times 19.25 \times 43}{7.50 \times 10^{12} \times 0.99 \times 0.785 \times 1.0}

 

\Delta P_{f}=2.03  psi

Since the flow rate of kerosene is much less than that of the crude oil, 3-in. nozzles should be adequate for the shell. Assuming schedule 40 pipe is used,

R e_{n}=\frac{4 \dot{m}}{\pi D_{n} \mu}=\frac{4 \times 45,000}{\pi(3.068 / 12) \times 0.97}=231,034 \text { (turbulent) }

 

G_{n}=\frac{\dot{m}}{\left.\left(\pi D_{n}^{2}\right) / 4\right)}=\frac{45,000}{\left[\pi(3.068 / 12)^{2} / 4\right]}=876,545  lbm / h \cdot ft ^{2}

 

\Delta P_{n}=2.0 \times 10^{-13} N_{5} G_{n}^{2} / s=2.0 \times 10^{-13} \times 1 \times(876,545)^{2} / 0.785

 

\Delta P_{n}=0.20  psi

Note: \rho V_{n}^{2} = 1210 (lbm/ft³) (ft/s)², so impingement protection for the tube bundle will not be required to prevent erosion. The total shell-side pressure drop is

\Delta P_{o}=\Delta P_{f}+\Delta P_{n}=2.03+0.20=2.23 \cong 2.2  psi

(g) Over-surface and over-design.

U_{C}=\left[1 / U_{D}-R_{D}\right]^{-1}=[1 / 46-0.0056]^{-1}=62  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

A=n_{ t } \pi D_{o} L=124 \pi \times(1.0 / 12) \times 14=454  ft ^{2}

 

U_{\text {req }}=\frac{q}{A F\left(\Delta T_{ ln }\right)_{c f}}=\frac{3,717,000}{454 \times 0.97 \times 191.2}=44  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

 

\text { over-surface }=U_{C} / U_{\text {req }}-1=62 / 44-1=41 \%

 

\text { over-design }=U_{D} / U_{\text {req }}-1=46 / 44-1=4.5 \%

All design criteria are satisfied. The shell-side pressure drop is still quite low, but the shell-side heat-transfer coefficient (122 Btu/h . ft² . ºF) does not differ greatly from the tube-side coefficient (156 Btu/h . ft². ºF). The reader can also verify that the tube-side fluid velocity is 6.7 ft/s, which is within the recommended range. Therefore, the design is acceptable.
Final design summary
Tube-side fluid: crude oil.
Shell-side fluid: kerosene.
Shell: Type AES, 19.25-in. ID
Tube bundle: 124 tubes, 1-in. OD, 14 BWG, 14-ft long, on 1.25-in. square pitch, arranged for four passes.
Heat-transfer area: 454 ft²
Baffles: 20% cut segmental type with spacing approximately 3.85 in.
Sealing strips: one pair per ten tube rows.
Nozzles: 4-in. schedule 40 on tube side; 3-in. schedule 40 on shell side.
Materials: plain carbon steel throughout