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) 5000Wm ^{-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.