Question 12.7: Design a condenser for the following duty: 45,000 kg/h of mi...

Only the thermal design will be done. The physical properties of the mixture will be taken as the mean of those for n-propane (MW = 44) and n-butane (MW = 58), at the average temperature.

 

\text { Heat transferred from vapour }=\frac{45,000}{3600}(596.5-247.0)=4368.8 kW

 

\text { Cooling water flow }=\frac{4368.8}{(40-30) 4.18}=104.5 kg / s

 

\text { Assumed overall coefficient (Table 12.1) }=900 W / m ^{2}{ }^{\circ} C

 

 

 

Mean temperature difference: the condensation range is small and the change in saturation temperature will be linear, so the corrected logarithmic mean temperature difference can be used.

 

R=\frac{(60-45)}{(40-30)}=1.5 (12.6)

 

S=\frac{(40-30)}{(60-30)}=0.33 (12.7)

 

Try a horizontal exchanger, condensation in the shell, four tube passes. For one shell pass, four tube passes, from Figure 12.19, F_{t} = 0.92

 

\Delta T_{\operatorname{lm}}=\frac{(60-40)-(45-30)}{\ln \frac{(60-40)}{(45-30)}}=17.4^{\circ} C

 

\Delta T_{m}=0.92 \times 17.4=16^{\circ} C

 

\text { Trial area }=\frac{4368.8 \times 10^{3}}{900 \times 16}=303 m ^{2}

 

\text { Surface area of one tube }=20 \times 10^{-3} \pi \times 4.88=0.305 m ^{2} (ignore tube sheet thickness)

 

\text { Number of tubes }=\frac{303}{0.305}=992

 

\text { Use square pitch, } P_{t}=1.25 \times 20 mm =25 mm \text {. }

 

Tube bundle diameter

 

D_{b}=20\left(\frac{992}{0.158}\right)^{1 / 2.263}=954 mm (12.3b)

 

\text { Number of tubes in centre row } N_{r}=D_{b} / P_{t}=954 / 25=38

 

Shell-side coefficient

Estimate tube wall temperature, T_{w}; assume condensing coefficient of 1500 W / m ^{2}{ }^{\circ} C,

Mean temperature

 

\text { Shell-side }=\frac{60+45}{2}=52.5^{\circ} C

 

\text { Tube-side }=\frac{40+30}{2}=35^{\circ} C

 

\begin{aligned}\left(52.5-T_{w}\right) 1500 &=(52.5-35) 900 \\T_{w} &=42.0{ }^{\circ} C\end{aligned}

 

\text { Mean temperature condensate }=\frac{52.5+42.0}{2}=47^{\circ} C

 

\text { Physical properties at } 47^{\circ} C

 

\mu_{L}=0.16 mNs / m ^{2}

 

\rho_{L}=551 kg / m ^{3}

 

k_{L}=0.13 W / m ^{\circ} C

 

vapour density at mean vapour temperature

 

\rho_{v}=\frac{52}{22.4} \times \frac{273}{(273+52.5)} \times \frac{10}{1}=19.5 kg / m ^{3}

 

\Gamma_{h}=\frac{W_{c}}{L N_{t}}=\frac{45,000}{3600} \times \frac{1}{4.88 \times 992}=2.6 \times 10^{-3} kg / s m

 

N_{r}=\frac{2}{3} \times 38=25

 

\begin{aligned}h_{c} &=0.95 \times 0.13\left[\frac{551(551-19.5) 9.81}{0.16 \times 10^{-3} \times 2.6 \times 10^{-3}}\right]^{1 / 3} \times 25^{-1 / 6} \\&=1375 W / m ^{2{ }^{\circ} C }\end{aligned} (12.50)

 

\text { Close enough to assumed value of } 1500 W / m ^{2}{ }^{\circ} C \text {, so no correction to } T_{w} \text { needed. }

 

Tube-side coefficient

 

\text { Tube cross-sectional area }=\frac{\pi}{4}\left(16.8 \times 10^{-3}\right)^{2} \times \frac{992}{4}=0.055 m ^{2}

 

\text { Density of water, at } 35^{\circ} C =993 kg / m ^{3}

 

\text { Tube velocity }=\frac{104.5}{993} \times \frac{1}{0.055}=1.91 m / s

 

\begin{aligned}h_{i} &=\frac{4200(1.35+0.02 \times 35) 1.91^{0.8}}{16.8^{0.2}} \\&=8218 W / m ^{2 \circ} C\end{aligned} (12.17)

 

Fouling factors: as neither fluid is heavily fouling, use 6000 W / m ^{2}{ }^{\circ} C for each side.

 

k_{w}=50 W / m ^{\circ} C

 

Overall coefficient

 

\begin{aligned}\frac{1}{U} &=\frac{1}{1375}+\frac{1}{6000}+\frac{20 \times 10^{-3} \ln \left(\frac{20}{16.8}\right)}{2 \times 50}+\frac{20}{16.8} \times \frac{1}{6000}+\frac{20}{16.8} \times \frac{1}{8218} \\U &=786 W / m ^{2}{ }^{\circ} C\end{aligned} (12.2)

 

\text { Significantly lower than the assumed value of } 900 W / m ^{2}{ }^{\circ} C \text {. }

 

\text { Repeat calculation using new trial value of } 750 W / m ^{2}{ }^{\circ} C \text {. }

 

\text { Area }=\frac{4368 \times 10^{3}}{750 \times 16}=364 m ^{2}

 

\text { Number of tubes }=\frac{364}{0305}=1194

 

D_{b}=20\left(\frac{1194}{0.158}\right)^{1 / 2.263}=1035 mm (12.36)

 

\text { Number of tubes in centre row }=\frac{1035}{25}=41

 

\Gamma_{h}=\frac{45,000}{3600} \times \frac{1}{4.88 \times 1194}=2.15 \times 10^{-3} kg / m s

 

N_{r}=\frac{2}{3} \times 41=27

 

\begin{aligned}h_{c} &=0.95 \times 0.13\left[\frac{551(551-19.5) 9.81}{0.16 \times 10^{-3} \times 2.15 \times 10^{-3}}\right]^{1 / 3} \times 27^{-1 / 6} \\&=1447 W / m ^{2}{ }^{\circ} C\end{aligned} (12.50)

 

\text { New tube velocity }=1.91 \times \frac{992}{1194}=1.59 m / s

 

h_{i}=4200(1.35+0.02 \times 35) \frac{1.59^{0.8}}{16.8^{0.2}}=7097 W / m ^{2 \circ} C (12.17)

 

\begin{aligned}\frac{1}{U}=& \frac{1}{1447}+\frac{1}{6000}+\frac{20 \times 10^{-3} \ln \left(\frac{20}{16.8}\right)}{2 \times 50} \\&+\frac{20}{16.8} \times \frac{1}{6000}+\frac{20}{16.8} \times \frac{1}{7097} \\U=& 773 W / m ^{2{ }^{\circ} C }\end{aligned} (12.2)

 

Close enough to estimate, firm up design.

Shell-side pressure drop

Use pull-through floating head, no need for close clearance.

Select baffle spacing = shell diameter, 45 per cent cut.

From Figure 12.10, clearance = 95 mm.

Shell i.d. = 1035 + 95 = 1130 mm

Use Kern’s method to make an approximate estimate.

 

\begin{aligned}\text { Cross-flow area } A_{s} &=\frac{(25-20)}{25} 1130 \times 1130 \times 10^{-6} \\&=0.255 m ^{2}\end{aligned} (12.21)

 

Mass flow-rate, based on inlet conditions

 

G_{s}=\frac{45,000}{3600} \times \frac{1}{0.255}=49.02 kg / s m ^{2}

 

\text { Equivalent diameter, } \begin{aligned}d_{e} &=\frac{1.27}{20}\left(25^{2}-0.785 \times 20^{2}\right) \\&=19.8 mm\end{aligned} (12.22)

 

\text { Vapour viscosity }=0.008 mNs / m ^{2}

 

R e=\frac{49.02 \times 19.8 \times 10^{-3}}{0.008 \times 10^{-3}}=121,325

 

\text { From Figure } 12.30, j_{f}=2.2 \times 10^{-2}

 

u_{s}=\frac{G_{s}}{\rho_{v}}=\frac{49.02}{19.5}=2.51 m / s

 

Take pressure drop as 50 per cent of that calculated using the inlet flow; neglect viscosity correction.

 

\begin{aligned}\Delta P_{s} &=\frac{1}{2}\left[8 \times 2.2 \times 10^{-2}\left(\frac{1130}{19.8}\right)\left(\frac{4.88}{1.130}\right) \frac{19.5(2.51)^{2}}{2}\right] \\&=1322 N / m ^{2} \\&=1.3 kPa\end{aligned} (12.26)

 

Negligible; more sophisticated method of calculation not justified.

Tube-side pressure drop

 

\text { Viscosity of water }=0.6 mN s / m ^{2}

 

R e=\frac{u_{t} \rho d_{i}}{\mu}=\frac{1.59 \times 993 \times 16.8 \times 10^{-3}}{0.6 \times 10^{-3}}=\underline{\underline{44,208}}

 

\text { From Figure } 12.24, i_{f}=3.5 \times 10^{-3} \text {. }

 

Neglect viscosity correction.

 

\begin{aligned}\Delta P_{t} &=4\left[8 \times 3.5 \times 10^{-3}\left(\frac{4.88}{16.8 \times 10^{-3}}\right)+2.5\right] \frac{993 \times 1.59^{2}}{2} \\&=53,388 N / m ^{2} \\&=53 kPa (7.7 psi ),\end{aligned} (12.20)

 

acceptable.