Make a preliminary design for a vertical thermosyphon reboiler for the column specified in Example 11.9. Take the vapour rate required to be 36 kmol/h. From example 8.3:
Operating pressure 8.3 (neglecting pressure drop over column).
Bottoms composition: C _{3} 0.001, iC _{4} 0.001, nC _{4} 0.02, iC _{5} 0.34, nC _{5} 0.64, kmol.
Bubble point of mixture, approximately, 120°C.
The concentrations of C _{3} \text { and } iC _{4} are small enough to be neglected. Take the liquid: vapour ratio as 3 : 1.
Estimate the liquid and vapour compositions leaving the reboiler:
Vapour rate, V D 36/3600 D 0.1 kmol/s
L / V=3, so liquid rate, L=3 V =0.3 kmol/s and feed, F=L+V=0.4 kmol/s. The vapour and liquid compositions leaving the reboiler can be estimated using the same procedure as that for a flash calculation; see Section 11.3.3.
| K_{i} | A_{i}=K_{i} \times L / V | V_{i}=z_{i} /\left(1+A_{I}\right) | y_{i}=V_{i} / V | x_{i}=\left(F z_{i}-V_{i}\right) / L | |
| nC _{4} | 2.03 | 6.09 | 0.001 | 0.01 | 0.023 |
| iC _{5} | 1.06 | 3.18 | 0.033 | 0.324 | 0.343 |
| nC _{5} | 0.92 | 2.76 | 0.068 | 0.667 | 0.627 |
| Totals | 0.102 | 1.001 | 0.993 | ||
| (near enough correct) | |||||
Enthalpies of vaporisation, from Figures (b) and (c) Example 11.9, kJ/mol
| x_{i} | H_{i} | h_{i} | H_{i}-h_{i} | x_{i}\left(H_{i}-h_{i}\right) | |
| nC _{4} | 0.02 | 50 | 34 | 16 | 0.32 |
| iC _{5} | 0.35 | 58 | 41 | 17 | 5.95 |
| nC _{5} | 0.63 | 61 | 42 | 19 | 11.97 |
| Totals | 18.24 | ||||
Exchanger duty, feed to reboiler taken as at its boiling point
= vapour flow-rate × heat of vaporisation
=0.1 \times 10^{3} \times 18.24=\underline{\underline{1824}} kW
Take the maximum flux as 37,900 W / m ^{2}; see Section 12.11.5.
Heat transfer area required =1,824,000 / 37,900=48.1 m ^{2}
Use 25 mm i.d., 2.5 m long tubes, a popular size for vertical thermosyphon reboilers.
Area of one tube =25 \times 10^{-3} \pi \times 2.5=0.196 m ^{2}
Number of tubes required =48.1 / 0.196=246
Liquid density at base of exchanger =520 kg / m ^{3}
Relative molecular mass at tube entry =58 \times 0.02+72(0.34+0.64)=71.7
vapour at exit =58 \times 0.02+72(0.35+0.63)=71.7
Two-phase fluid density at tube exit:
volume of vapour =0.1 \times(22.4 . / 8.3) \times(393 / 273)=0.389 m ^{3}
volume of liquid =(0.3 \times 71.7) / 520=0.0413 m ^{3}
total volume =0.389+0.0413=0.430 m ^{3}
exit density =\frac{(0.4 \times 71.7)}{0.430} \times 71.7=66.7 kg / m ^{3}