Question 12.1: Design an exchanger to sub-cool condensate from a methanol c...

Only the thermal design will be considered.

This example illustrates Kern’s method.

Coolant is corrosive, so assign to tube-side.

 

$\text { Heat capacity methanol }=2.84 kJ / kg ^{\circ} C$

 

$\text { Heat load }=\frac{100,000}{3600} \times 2.84(95-40)=4340 kW$

 

$\text { Heat capacity water }=4.2 kJ / kg ^{\circ} C$

 

$\text { Cooling water flow }=\frac{4340}{4.2(40-25)}=68.9 kg / s$

 

$\Delta T_{\operatorname{lm}}=\frac{(95-40)-(40-25)}{\ln \frac{(95-40)}{(40-25)}}=31^{\circ} C$ (12.4)

 

Use one shell pass and two tube passes

 

$R=\frac{95-40}{40-25}=3.67$ (12.6)

 

$S=\frac{40-25}{95-25}=0.21$ (12.7)

 

From Figure 12.19

 

$F_{t}=0.85$

 

$\Delta T_{m}=0.85 \times 31=26^{\circ} C$

 

From Figure 12.1

 

$U=600 W / m ^{2}{ }^{\circ} C$

 

Provisional area

 

$A=\frac{4340 \times 10^{3}}{26 \times 600}=278 m ^{2}$ (12.1)

 

Choose 20 mm o.d., 16 mm i.d., 4.88-m-long tubes $\left(\frac{3}{4} in . \times 16 ft \right)$, cupro-nickel.

Allowing for tube-sheet thickness, take

L = 4.83 m

 

$\text { Area of one tube }=4.83 \times 20 \times 10^{-3} \pi=0.303 m ^{2}$

 

$\text { Number of tubes }=\frac{278}{0.303}=\underline{\underline{918}}$

 

As the shell-side fluid is relatively clean use 1.25 triangular pitch.

 

$\text { Bundle diameter } D_{b}=20\left(\frac{918}{0.249}\right)^{1 / 2.207}=826 mm$ (12.3b)

 

Use a split-ring floating head type.

From Figure 12.10, bundle diametrical clearance = 68 mm,

 

$\text { shell diameter, } D_{s}=826+68=894 mm$

 

(Note. nearest standard pipe sizes are 863.6 or 914.4 mm).

Shell size could be read from standard tube count tables.

Tube-side coefficient

 

$\text { Mean water temperature }=\frac{40+25}{2}=33^{\circ} C$

 

$\text { Tube cross-sectional area }=\frac{\pi}{4} \times 16^{2}=201 mm ^{2}$

 

$\text { Tubes per pass }=\frac{918}{2}=459$

 

$\text { Total flow area }=459 \times 201 \times 10^{-6}=0.092 m ^{2}$

 

$\text { Water mass velocity }=\frac{68.9}{0.092}=749 kg / s m ^{2}$

 

$\text { Density water }=995 kg / m ^{3}$

 

$\text { Water linear velocity }=\frac{749}{995}=0.75 m / s$

 

$h_{i}=\frac{4200(1.35+0.02 \times 33) 0.75^{0.8}}{16^{0.2}}=3852 W / m ^{2{ }^{\circ} C }$ (12.17)

 

The coefficient can also be calculated using equation 12.15; this is done to illustrate use of this method.

 

$\frac{h_{i} d_{i}}{k_{f}}=j_{h} \operatorname{RePr} r^{0.33}\left(\frac{\mu}{\mu_{w}}\right)^{0.14}$ (12.15)

 

$\frac{h_{i} d_{i}}{k_{f}}=j_{h} \operatorname{RePr}{ }^{0.33}\left(\frac{\mu}{\mu_{w}}\right)^{0.14}$

 

$\text { Viscosity of water }=0.8 mNs / m ^{2}$

 

$\text { Thermal conductivity }=0.59 W / m ^{\circ} C$

 

$R e=\frac{\rho u d_{i}}{\mu}=\frac{995 \times 0.75 \times 16 \times 10^{-3}}{0.8 \times 10^{-3}}=14,925$

 

$\operatorname{Pr}=\frac{C_{p} \mu}{k_{f}}=\frac{4.2 \times 10^{3} \times 0.8 \times 10^{-3}}{0.59}=5.7$

 

$\text { Neglect }\left(\frac{\mu}{\mu_{w}}\right)$

 

$\frac{L}{d_{i}}=\frac{4.83 \times 10^{3}}{16}=302$

 

$\text { From Figure } 12.23, j_{h}=3.9 \times 10^{-3}$

 

$h_{i}=\frac{0.59}{16 \times 10^{-3}} \times 3.9 \times 10^{-3} \times 14,925 \times 5.7^{0.33}=3812 W / m ^{2{ }^{\circ} C }$

 

Checks reasonably well with value calculated from equation 12.17; use lower figure.

Shell-side coefficient

 

$\text { Choose baffle spacing }=\frac{D_{s}}{5}=\frac{894}{5}=178 mm$

 

$\text { Tube pitch }=1.25 \times 20=25 mm$

 

$\text { Cross-flow area } A_{s}=\frac{(25-20)}{25} 894 \times 178 \times 10^{-6}=0.032 m ^{2}$ (12.21)

 

$\text { Mass velocity, } G_{S}=\frac{100,000}{3600} \times \frac{1}{0.032}=868 kg / s m ^{2}$

 

$\text { Equivalent diameter } d_{e}=\frac{1.1}{20}\left(25^{2}-0.917 \times 20^{2}\right)=14.4 mm$ (12.23)

 

$\text { Mean shell side temperature }=\frac{95+40}{2}=68^{\circ} C$

 

$\text { Methanol density }=750 kg / m ^{3}$

 

$\text { Viscosity }=0.34 mNs / m ^{2}$

 

$\text { Heat capacity }=2.84 kJ / kg ^{\circ} C$

 

$\text { Thermal conductivity }=0.19 W / m ^{\circ} C$

 

$R e=\frac{G_{s} d_{e}}{\mu}=\frac{868 \times 14.4 \times 10^{-3}}{0.34 \times 10^{-3}}=36,762$ (12.24)

 

$\operatorname{Pr}=\frac{C_{p} \mu}{k_{f}}=\frac{2.84 \times 10^{3} \times 0.34 \times 10^{-3}}{0.19}=5.1$

 

Choose 25 per cent baffle cut, from Figure 12.29

 

$j_{h}=3.3 \times 10^{-3}$

 

Without the viscosity correction term

 

$h_{s}=\frac{0.19}{14.4 \times 10^{-3}} \times 3.3 \times 10^{-3} \times 36,762 \times 5.1^{1 / 3}=2740 W / m ^{2}{ }^{\circ} C$

 

Estimate wall temperature

 

$\text { Mean temperature difference }=68-33=35^{\circ} C$

 

across all resistances

 

$\text { across methanol film }=\frac{U}{h_{o}} \times \Delta T=\frac{600}{2740} \times 35=8^{\circ} C$

 

$\text { Mean wall temperature }=68-8=60^{\circ} C$

 

$\mu_{w}=0.37 mNs / m ^{2}$

 

$\left(\frac{\mu}{\mu_{w}}\right)^{0.14}=0.99$

 

which shows that the correction for a low-viscosity fluid is not significant.

Overall coefficient

 

Thermal conductivity of cupro-nickel alloys D 50 $W / m ^{\circ} C$.

Take the fouling coefficients from Table 12.2; methanol (light organic) 5000$Wm ^{-2 \circ} C ^{-1}$,

brackish water (sea water), take as highest value, 3000 $Wm ^{-2 \circ} C ^{-1}$

 

 

$\frac{1}{U_{o}}=\frac{1}{2740}+\frac{1}{5000}+\frac{20 \times 10^{-3} \ln \left(\frac{20}{16}\right)}{2 \times 50}$ $+\frac{20}{16} \times \frac{1}{3000}+\frac{20}{16} \times \frac{1}{3812}$ $U_{o}=738 W / m ^{2}{ }^{\circ} C$ (12.2)

 

$\text { well above assumed value of } 600 W / m ^{2}{ }^{\circ} C \text {. }$

 

Pressure drop

Tube-side

 

$\text { From Figure } 12.24 \text {, for } \operatorname{Re}=14,925$

 

$j_{f}=4.3 \times 10^{-3}$

 

Neglecting the viscosity correction term

 

$\Delta P_{t}=2\left(8 \times 4.3 \times 10^{-3}\left(\frac{4.83 \times 10^{3}}{16}\right)+2.5\right) \frac{995 \times 0.75^{2}}{2}$ $=7211 N / m ^{2}=7.2 kPa (1.1 psi )$ (12.20)

 

low, could consider increasing the number of tube passes.

Shell side

 

$\text { Linear velocity }=\frac{G_{s}}{\rho}=\frac{868}{750}=1.16 m / s$

 

$\text { From Figure } 12.30, \text { at } \operatorname{Re}=36,762$

 

$j_{f}=4 \times 10^{-2}$

 

Neglect viscosity correction

 

$\Delta P_{s}=8 \times 4 \times 10^{-2}\left(\frac{894}{14.4}\right)\left(\frac{4.83 \times 10^{3}}{178}\right) \frac{750 \times 1.16^{2}}{2}$ (12.26)

 

$\begin{array}{l}=272,019 N / m ^{2} \\=272 kPa (39 psi ) \text { too high, }\end{array}$

 

could be reduced by increasing the baffle pitch. Doubling the pitch halves the shell-side velocity, which reduces the pressure drop by a factor of approximately $(1 / 2)^{2}$

 

$\Delta P_{s}=\frac{272}{4}=68 kPa (10 psi ), \text { acceptable }$

 

This will reduce the shell-side heat-transfer coefficient by a factor of $(1 / 2)^{0.8}\left(h_{o} \propto\right.$ $\left.R e^{0.8} \propto u_{s}^{0.8}\right)$

 

$h_{o}=2740 \times\left(\frac{1}{2}\right)^{0.8}=1573 W / m ^{2}{ }^{\circ} C$

 

This gives an overall coefficient of 615 $W / m ^{2}{ }^{\circ} C -\text { still }$ above assumed value of 600 $W / m ^{2}{ }^{\circ} C$.