Question 12.12: Design a vaporiser to vaporise 5000 kg/h n-butane at 5.84 ba...

Only the thermal design and general layout will be done. Select kettle type.

Physical properties of n-butane at 5.84 bar:

latent heat = 326 kJ/kg

Heat loads:

add 5 per cent for heat losses

From Figure 12.1 assume U = 1000 W / m ^{2}{ }^{\circ} C \text {. }

Mean temperature difference; both sides isothermal, steam saturation temperature at 1.7 bar = 115.2{ }^{\circ} C

Select 25 mm i.d., 30 mm o.d. plain U-tubes,

Nominal length 4.8 m (one U-tube)

Draw a tube layout diagram, take minimum bend radius

Proposed layout gives 26 U-tubes, tube outer limit diameter 420 mm.

Boiling coefficient

Use Mostinski’s equation:

heat flux, based on estimated area,

\begin{aligned}h_{n b} &=0.104(38)^{0.69}\left(59.2 \times 10^{3}\right)^{0.7}\left[1.8\left(\frac{5.84}{38}\right)^{0.17}+4\left(\frac{5.84}{38}\right)^{1.2}+10\left(\frac{5.84}{38}\right)^{10}\right] \\&=4855 W / m ^{2}{ }^{\circ} C\end{aligned} (12.63)

Take steam condensing coefficient as 8000 W / m ^{2}{ }^{\circ} C, fouling coefficient 5000 W / m ^{2}{ }^{\circ} C;

butane fouling coefficient, essentially clean, 10,000 W / m ^{2}{ }^{\circ} C.

Tube material will be plain carbon steel, k_{w}=55 W / m ^{\circ} C

\begin{aligned}\frac{1}{U_{o}} &=\frac{1}{4855}+\frac{1}{10,000}+\frac{30 \times 10^{-3} \ln \frac{30}{25}}{2 \times 55}+\frac{30}{25}\left(\frac{1}{5000}+\frac{1}{8000}\right) \\U_{o} &=1341 W / m ^{2}{ }^{\circ} C\end{aligned} (12.2)

Close enough to original estimate of 1000 W / m ^{2}{ }^{\circ} C for the design to stand.

Myers and Katz (Chem. Eng. Prog. Sym. Ser. 49(5) 107 114) give some data on the boiling of n-butane on banks of tubes. To compare the value estimate with their values an estimate of the boiling film temperature difference is required:

Myers data, extrapolated, gives a coefficient of around 3000 Btu/h ft ^{2} F \text { at } a 29^{\circ} F temperature difference = 17,100 W / m ^{2}{ }^{\circ} C, so the estimated value of 4855 is certainly on the safe side.

Check maximum allowable heat flux. Use modified Zuber equation.

\begin{aligned}q_{c} &=0.44 \times 1.5 \times \frac{326 \times 10^{3}}{\sqrt{52}}\left[9.7 \times 10^{-3} \times 9.81(550-12.6) 12.6^{2}\right]^{0.25} \\&=283,224 W / m ^{2} \\&=280 kW / m ^{2}\end{aligned} (12.74)

Applying a factor of 0.7, maximum flux should not exceed 280 x 0.7 = 196 kW / m ^{2}.

Actual flux of 59.2 kW / m ^{2} is well below maximum allowable.

Layout

Take shell diameter as twice bundle diameter

Take liquid level as 500 mm from base,

freeboard = 840 – 500 = 340 mm, satisfactory.

From sketch, width at liquid level = 0.8 m.

Surface area of liquid = 0.8 x 2.4 = 1.9 m ^{2}.

Maximum allowable velocity

\hat{u}_{v}=0.2\left[\frac{550-12.6}{12.6}\right]^{1 / 2}=\underline{\underline{\underline{1.3} m / s }} (12.75)

so actual velocity is well below maximum allowable velocity. A smaller shell diameter could be considered.