Question 9.28: Using Kern's method, design a shell and tube heat exchanger ......

Since it is corrosive, the water will be passed through the tubes.
At a mean temperature of 0.5(370 + 315) = 343 K, from Table 3, Appendix A1, the thermal capacity of butyl alcohol = 2.90 kJ/kg K and hence:
Heat load = (30 x 2.90)(370 – 315) = 4785 kW
If the heat capacity of water is 4.18 kJ/kg K, then:
Flow of cooling water = 4785/(4.18(315 – 300)) = 76.3 kg/s
The logarithmic mean temperature difference,
\theta _{m} = [(370 – 315) – (315 – 300)]/ln[(370 – 315)/(315 – 300)] = 30.7 deg K
With one shell-side pass and two tube-side passes, then from equation 9.213:

X={\frac{\theta_{2}-\theta_{1}}{T_{1}-\theta_{1}}}\quad{\mathrm{and}}\quad Y={\frac{T_{1}-T_{2}}{\theta_{2}-\theta_{1}}}            (9.213)

X = (370 – 315)/(315 – 300) = 3.67 and Y = (315 – 300)/(370 – 300) = 0.21
and from Figure 9.75:
F = 0.85 and F\theta _{m} = (0.85 x 30.7) = 26.1 deg K
From Table 9.17, an estimated value of the overall coefficient is U = 500 W/m²K and hence, the provisional area, from equation 9.212, is:

Q = UAF\theta _{m}          (9.212)
A = (4785 x 10³)/(26.1 x 500) = 367 m²

It is convenient to use 20 mm OD, 16 mm ID tubes, 4.88 m long which, allowing for the tube-sheets, would provide an effective tube length of 4.83 m. Thus:
Surface area of one tube = π(20/1000) = 0.303 m²
and: Number of tubes required = (367/0.303) = 1210
With a clean shell-side fluid, 1.25 triangular pitch may be used and, from equation 9.211:

Number of tubes, N_{t}=a(d_{b}/d_{o})^{b}               (9.211)

from which: d_{b} = 937 mm
Using a split-ring floating head unit, then, from Figure 9.71. the diametrical clearance between the shell and the tubes = 68 mm and:
Shell diameter, d_{s} = (937 + 68) = 1005 mm
which approximates to the nearest standard pipe size of 1016 mm.

Tube-side coefficient

The water-side coefficient may now be calculated using equation 9.218, although here, use will be made of the j_{h} factor.

Nu=(h d/k)= j_{h}R e P r^{0.33}(\mu/\mu_{s})^{-0.14}                  (9.218)

Cross-sectional area of one tube = (π/40) x 16² = 201 mm²
Number of tubes/pass = (1210/2) = 605

Thus: Tube-side flow area = (605 x 201 x 10^{-6}) = 0.122 m²
Mass velocity of the water = (76.3/0.122) = 625 kg/m²s
Thus, for a mean water density of 995 kg/m³:
Water velocity, u = (625/995) = 0.63 m/s
At a mean water temperature of 0.5(315 + 300) = 308 K, viscosity, μ = 0.8 mN s/m² and thermal conductivity, k = 0.59 W/m K.
Thus:
Re = duρ/μ = (16 x 10^{-3} x 0.63 x 995)/(0.8 x 10^{-3}) = 12540
Pr = C_{p}μ/k = (4.18 x 10³ x 0.8 x 10^{-3})/0.59 = 5.67
l/d_{i} = 4.83/(16 x 10^{-3}) = 302
Thus, from Figure 9.77, j_{h} = 3.7 x 10^{-3}, and, in equation 9.218, neglecting the viscosity term:

Nu=(h d/k)= j_{h}R e P r^{0.33}(\mu/\mu_{s})^{-0.14}                  (9.218)

and: h_{i} = 3030 W/m²K

Shell-side coefficient
The baffle spacing will be taken as 20 per cent of the shell diameter or (1005 x 20/100) = 201 mm
The tube pitch = (1.25 x 20) = 25 mm and, from equation 9.226:

A_{s}=d_{s}l_{B}C^{\prime}/Y              (9.226)

Cross-flow area, A_{s} = [(25 – 20)/25](1005 x 201 x 10^{-6}) = 0.040 m²
Thus: Mass velocity in the shell, G_{s} = (30/0.040) = 750 kg/m²s
From equation 9.228:

For square pitch : d_{e}=4(Y^{2}-\pi d_{o}^{2})4/\pi d_{0}=\;1.27(Y^{2}-0.785d_{o}^{2})/d_{o}(9.277)

and for triangular pitch : d_{e}=4[(0.87Y\times Y/2)-(0.5\pi d_{o}^{2}/4]/(\pi d_{o}/2)

=1.10(Y^{2}-0.917d_{o}^{2})/d_{o}          (9.228)

Equivalent diameter, d_{e} = 1.10[25² – (0.917 x 20²)]/20 = 14.2 mm
At a mean shell-side temperature of 0.5(370 + 315) = 343 K, from Appendix A1:
density of butyl alcohol, ρ = 780 kg/m³, viscosity, μ = 0.75 mN s/m², heat capacity, C_{p} = 3.1 kJ/kg K and thermal conductivity, k = 0. 16 W/m K.
Thus, from equation 9.229:

R e_{s}=G_{s}^{\prime}d_{e}/\mu=u_{s}d_{e}\rho/\mu          (9.229)

Re = G_{s}d_{e}/μ = (750 x 14.2 x 10^{-3})/(0.75 x 10^{-3}) = 14200
Pr = C_{p}μ/k = (3.1 x 10³ x 0.75 x 10^{-3})/0.16 = 14.5
Thus, with a 25 per cent segmental cut, from Figure 9.81: j_{h} = 5.0 x 10^{-3}
Neglecting the viscosity correction term in equation 9.230:

N u=(h_{s}d_{e}/k_{f})=j_{h}R e P r^{0.33}(\mu/\mu_{s})^{0.14}            (9.230)

and: h_{s} = 1933 W/m²K
The mean butanol temperature = 343 K, the mean water temperature = 308 K and hence the mean wall temperature may be taken as 0.5(343 + 308) = 326 K at which \mu _{s} = 1.1 mN s/m²
Thus: (\mu/\mu_{s})^{0.14}=(0.75/1.1)^{0.14}=0.95

showing that the correction for a low viscosity fluid is negligible.

Overall coefficient
The thermal conductivity of cupro-nickel alloys = 50 W/m K and, from Table 9.16, scale resistances will be taken as 0.00020 m²K/W for the water and 0.00018 m²K/W for the organic.

Based on the outside area, the overall coefficient is given by:

= (1/1933) + 0.00020 + [0.5(20 – 16) x 10^{-3}/50] + (0.00015 x 20)/16 + 20/(3030 x 16)
= 0.00052 + 0.00020 + 0.00004 + 0.000225 + 0.00041 = 0.00140 m²K/W

and: U = 717 W/m²K
which is well in excess of the assumed value of 500 W/m²K.

Pressure drop

On the tube-side, Re = 12450 and from Figure 9.78, j_{f} = 4.5 x 10^{-3}
Neglecting the viscosity correction term, equation 9.225 becomes:

-\Delta P_{total}=N_{P}[4j_{f}(l/d_{i})(\mu/\mu_{s})^{m}+1.25](\rho u^{2})                   (9.225)

ΔP_{t} = 2(4 x 4.5 x 10^{-3}(4830/16) + 1.25)(995 x 0.63²) = 5279 N/m² or 5.28 kN/m²
which is low, permitting a possible increase in the number of tube passes. On the shell-side, the linear velocity, (G_{s}/ρ) = (750/780) = 0.96 m/s
From Figure 9.82, when Re = 14200, j _{f} = 4.6 x 10^{-2}
Neglecting the viscosity correction term, in equation 9.231:

-\Delta P_{s}=4j_{f}(d_{s}/d_{e})(l/l_{B})(\rho u_{s}^{2})(\mu/\mu_{s})^{-0.14}                  (9.231)

– ΔP_{s} = (4 x 4.6 x 10^{-2})(1005/14.2)(4830/201)(780 x 0.96²)
= 224950 N/m² or 225 kN/m²
This value is very high and thought should be given to increasing the baffle spacing. If this is doubled, this will reduce the pressure drop by approximately (1/2)² = 1/4 and:

– ΔP_{s} = (225/4) = 56.2 kN/m² which is acceptable.

Since h_{o}\propto R e^{0.8}\propto u^{0.8}

which gives an overall coefficient of \underline{\underline{561\ W/m^{2}K}} which is still in excess of the assumed value of 500 W/m²K.