Question 10.3: A reboiler for a revamped distillation column in a refinery ...

(a) Energy balances.
The energy balance for the boiling fluid is:

q=\dot{m}_{V} H_{V}+\dot{m}_{L} H_{L}-\dot{m}_{F} H_{F}

The feed rate to the reboiler is 60,000/0.20 = 300,000 lbm/h, and the liquid return rate is 300,000 – 60,000 = 240,000 lbm/h. Therefore,

q  =  60,000 \times 265.9+240,000 \times 142.1-300,000 \times 136.6

q = 9,078,000 Btu / h

The energy balance for the Therminol® is:

q=\left(\dot{m} C_{p} \Delta T\right)_{T h}

9,078,000=\dot{m}_{T h} \times 0.534(420-380)

\dot{m}_{T h}=425,000  lbm / h

(b) Mean temperature difference.
The effective mean temperature difference is computed as if the flow were co-current:

\Delta T=131^{\circ} F \left\{\begin{array}{c}289^{\circ} F   \rightarrow   298.6^{\circ} F \\ 420^{\circ} F    \rightarrow   380^{\circ} F\end{array}\right\} \Delta T=81.4^{\circ} F

\Delta T_{\text {mean }} \cong\left(\Delta T_{\ln }\right)_{\text {co-current }}=\frac{131  –  81.4}{\ln (131 / 81.4)}=104.2^{\circ} F

(c) Heat-transfer area.

A=n_{ t } \pi D_{o} L=290 \times \pi \times(0.75 / 12) \times 16=911  ft ^{2}

(d) Required overall coefficient.

U_{\text {req }}=\frac{q}{A \Delta T_{\text {mean }}}=\frac{9,078,000}{911  \times  104.2}=96  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(e) Inside coefficient, h_{i} .

D_{i}=0.584  in.         (Table B.1)

G=\frac{\dot{m}\left(n_{p} / n_{t}\right)}{(\pi / 4) D_{i}^{2}}=\frac{425,000(2 / 290)}{(\pi / 4)(0.584 / 12)^{2}}=1,575,679  lbm / h \cdot ft ^{2}

R e=D_{i} G / \mu=\frac{(0.584 / 12)  \times  1,575,679}{0.84  \times  2.419}=37,738

Since the flow is turbulent, Equation (4.1) is used to calculate h_{i} :

N u=0.023 R e^{0.8} \operatorname{Pr}^{1 / 3}\left(\mu / \mu_{w}\right)^{0.14}=0.023(37,738)^{0.8}(17.70)^{1 / 3}(1.0)

Nu = 274.9

h_{i}=\left(k / D_{i}\right) N u=\frac{0.0613  \times  274.9}{(0.584 / 12)} = 346  Btu / h \cdot ft ^{2} \cdot{ }^{\circ}F

(f) Outside coefficient, h_{o} = h_{b} .

(i) Nucleate boiling coefficient

     The pseudo-reduced pressure is used in place of the reduced pressure

P_{p r}=P / P_{p c}=35 / 406.5=0.0861

   Since this value is less than 0.2, Equation (9.5) is used to calculate the pressure correction factor in the Mostinski correlation:

F_{P}=2.1  P_{r}^{0.27}+\left[9+\left(1-P_{r}^{2}\right)^{-1}\right] P_{r}^{2}

=2.1(0.0861)^{0.27}+\left\{9+\left[1-(0.0861)^{2}\right]^{-1}\right\}(0.0861)^{2}

F_{P}=1.1573

The required duty and actual heat-transfer area in the exchanger are used to calculate the heat flux:

\hat{q}=q / A=9,078,000 / 911=9965  Btu / h \cdot ft ^{2}

The boiling range is calculated from the given data and used to compute the mixture correction factor using Equation (9.17a):

B R=T_{D}-T_{B}=321-289=32^{\circ} F

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

=\left[1+0.0176(9965)^{0.15}(32)^{0.75}\right]^{-1}

F_{m}=0.5149

The nucleate boiling coefficient is obtained by substituting the above values into the Mostinski correlation, Equation (9.2a):

h_{n b}=0.00622 P_{c}^{0.69}  \hat{q}^{0.7} F_{P}      (9.2a)

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

=0.0062(406.5)^{0.690}(9965)^{0.7} \times 1.1573 \times 0.5149

h_{n b}=147  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F

(ii) Bundle boiling coefficient,h_{ b}

The correction factor for bundle convective effects is calculated using Equation (9.20):

F_{b}=1.0+0.1\left[\frac{0.785 D_{b}}{C_{1}\left(P_{T} / D_{0}\right)^{2}  \times  D_{0}}  –  1.0\right]^{0.75}

=1.0+0.1\left[\frac{0.785  \times  20}{1.0(1.0 / 0.75)^{2}  \times  0.75}  –  1.0\right]^{0.75}

F_{b}=1.5947

A rough approximation of 44 Btu/h · ft² · °F is adequate for the natural convection coefficient, h_{nc}, because the temperature difference is large. The boiling coefficient for the bundle is given by Equation (9.19):

h_{b}=h_{n b} F_{b}+h_{n c}=147  \times  1.5947+44

h_{b}=278  Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F =h_{o}

(g) Fouling factors.
Based on the guidelines in Table 10.2, the fouling factors are chosen as follows:

TABLE 10.2   Recommended Fouling Factors for Reboiler Design

R_{D i}=0.0005 h \cdot ft ^{2} \cdot{ }^{\circ} F / Btu                 (organic liquid heating medium)

R_{D o}=0.001 h \cdot ft ^{2} \cdot{ }^{\circ} F / Btu              (heavier normal hydrocarbon)

(h) Overall coefficient.

Since this is strictly a rating problem, there is no need to reconcile the values of \hat{q} ,  U_{D}, and ΔT_{m}. The fact that the computed value of U_{D} exceeds  U_{req} means that the exchanger contains sufficient heat-transfer area to meet the required duty.
(i) Critical heat flux.
The critical heat flux for a single tube is calculated using Equation (9.23a):

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

=803 \times 406.5(0.0861)^{0.35}(1-0.0861)^{0.9}

\hat{q}_{c}=127,596  Btu / h \cdot ft ^{2}

The bundle geometry factor is given by:

\psi_{b}=\frac{D_{b}}{n_{ t } D_{o}}=\frac{20}{290  \times  0.75}=0.09195

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

\phi_{b}=3.1 \psi_{b}=3.1  \times  0.09195=0.285

The critical heat flux for the bundle is given by Equation (9.24):

\hat{q}_{c, \text { bundle }}=\hat{q}_{c,t u b e}  \phi_{b}=127,596 \times 0.285

\hat{q}_{c, \text { bundle }}=36,365  Btu / h \cdot ft ^{2}

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

\hat{q} / \hat{q}_{c, \text { bundle }}=9965 / 36,365 \cong 0.27

Since the ratio is less than 0.7 and U_{D} > U_{req} , the reboiler is thermally acceptable.

(j) Tube-side pressure drop.

(i) Friction loss

The calculation uses Equation (5.2) for the friction factor and Equation (5.1) for the pressure drop:

f=0.4137  R e^{-0.2585}=0.4137(37,738)^{-0.2585}=0.0271

\Delta P_{f}=\frac{f  n_{p}  L  G^{2}}{7.50  \times  10^{12} D_{i} s \phi_{i}}=\frac{0.0271  \times  2  \times  16(1,575,679)^{2}}{7.5  \times  10^{12}(0.584 / 12)  \times  0.882  \times  1.0}

\Delta P_{f}=6.69  psi

(ii) Minor losses
From Table 5.1, the number of velocity heads allocated for minor losses with turbulent flow in U-tubes is:

\alpha_{r}=1.6 n_{p}-1.5=1.6 \times 2-1.5=1.7

Substituting in Equation (5.3) yields:

\Delta P_{ r }=1.334 \times 10^{-13}  \alpha_{ r } G^{2} / s=1.334 \times 10^{-13} \times 1.7(1,575,679)^{2} / 0.882

\Delta P_{ r }=0.64  psi

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

(iii) Nozzle losses
For 6-in. schedule 40 nozzles we have:

G_{n}=\frac{\dot{m}}{(\pi / 4) D_{i}^{2}}=\frac{425,000}{(\pi / 4)(6.065 / 12)^{2}}=2,118,361  lbm / h \cdot ft ^{2}

R e_{n}=\frac{D_{i} G_{n}}{\mu}=\frac{(6.065 / 12)  \times  2,118,361}{0.84  \times  2.419}=526,907

Since the flow is turbulent, Equation (5.4) is used to estimate the pressure drop:

\Delta P_{n}=2.0 \times 10^{-13} N_{s} G_{n}^{2} / s=2.0 \times 10^{-13} \times 1(2,118,361)^{2} / 0.882

\Delta P_{n}=1.02  psi

(iv) Total pressure drop

\Delta P_{i}=\Delta P_{f}+\Delta P_{ r }+\Delta P_{n}=6.69+0.64+1.02

\Delta P_{i}=8.4  psi

Since the pressure drop is within the specified limit of 10 psi, the reboiler is hydraulically acceptable.
In summary, the reboiler is thermally and hydraulically suitable for this service.