Question 9.23: A mixture of 0.57 kg/s of steam and 0.20 kg/s of carbon diox......

Hence the partial pressure of water = (308 x 0.032/0.0365) = 270 kN/m² and from Table 11A in the Appendix, the dew point = 404 K.
Mean molecular weight of the mixture = (0.57 + 0.20)/0.0365 = 21.1 kg/kmol.

\therefore steam at the outlet = \frac{0.0045\times 11.7}{296.3} = 0.000178 kmol

and: steam condensed = (0.032 – 0.000178) = 0.03182 kmol.

The heat load is now estimated at each interval between the temperatures 404, 401, 397, 380, 339 and 322 K.
For the interval 404 to 401 K
From Table 11A in the Appendix, the partial pressure of steam at 401 K = 252.2 kN/m² and hence the partial pressure of CO_{2} = (308 – 252.2) = 55.8 kN/m². Steam remaining = (0.0045 x 252.2/55.8) = 0.0203 kmol.
\therefore Steam condensed = (0.032 – 0.0203) = 0.0117 kmol
Heat of condensation = (0.0117 x 18)(2180 + 1.93(404 – 401)) = 466 kW
Heat from uncondensed steam = (0.0203 x 18 x 1.93(404 – 401)) = 1.9 kW
Heat from carbon dioxide = (0.020 x 0.92(404 – 401)) = 0.5 kW
and the total for the interval = 468.4 kW
Repeating the calculation for the other intervals of temperature gives the following results:

and the flow of water = 1407.3/(4.187(319 – 300)) = \underline{\underline{17.7\ kg/s}}.

With this flow of water and a flow area per pass of 0.0120 m², the mass velocity of water is 1425 kg/m²s, equivalent to a velocity of 1.44 m/s at which h_{i} = 6.36 kW/m² K. Basing this on the outside area, h_{io} = 5.25 kW/m² K.
Shell-side coefficient for entering gas mixture:
The mean specific heat, C_{p}={\frac{(0.20\times0.92)+(0.57\times1.93)}{0.77}} = 1.704 kJ/kg K.

Similarly, the mean thermal conductivity k = 0.025 kW/m K and the mean viscosity μ = 0.015 mN s/m²

The area for flow through the shell = 0.0411 m² and the mass velocity on the shell side

= \frac{0.20+0.57}{0.0411} = 18.7 kg/m²s

Taking the equivalent diameter as 0.024 m, Re = 29,800
and: h_{g} = 0.107 kW/m² K or 107 W/m² K.

Now:  \left(\frac{\mu}{\rho D}\right)^{0.67}=0.62,\quad\left(\frac{C_{p}\mu}{k}\right)^{0.67}=1.01

and: k_{G}=\frac{h_{g}(C_{p}\mu/k)^{0.67}}{C_{p}P_{s F}(\mu/\rho D)^{0.67}}=\frac{107\times1.01}{1704P_{s F}\times0.62}

= \frac{0.102}{P_{sF}}

Temperature of the gas T = 404 K, partial pressure of steam P_{g} = 270 kN/m², partial pressure of the inert P_{s} = 38 kN/m², water temperature T_{w} = 319 K and ΔT = (404 – 319) = 85 K. An estimate is now made for the temperature of the condensate film of T_{c} = 391 K. In this case P_{s} = 185.4 kN/m² and P^{\prime}_{s} = (308 – 185.4) = 122.6 kN/m².
Thus: P_{s F}=\frac{122.6-38}{\ln(122.6/38)}=72.2\mathrm{~kN/m^{2}}.

In equation 9.181:

h_{g}(T_{s}-T_{c})+k_{G}\lambda(P_{g}-P_{s})=h_{o}(T_{c}-T_{c m})= U\Delta T                (9.181)

259 = 378 i.e. there is no balance.
Try T_{c} = 378 K, P_{s} = 118.5 kN/m², P_{g} = (308 – 118.5) = 189.5 kN/m²

and:  P_{s F}={\frac{189.5-38}{\ln(189.5/38)}}=94.2{\mathrm{~kN/m}}^{2}

\therefore 0.107(404-378)+\left(\frac{0.102}{94.2}\right)2172(270-118.5)=5.25(378-319)

310 = 308 which agrees well

\therefore UΔT = 309 kW/m² and U = \frac{309}{(404-319)}

= 3.64 kW/m² K.

Repeating this procedure at the various temperature points selected, the heat-exchanger area may then be obtained as the area under a plot of Σq vs. 1/UΔT, or as A = Σq/UΔT according to the following tabulation:

If no condensation takes place, the logarithmic mean temperature difference is 46.6 K. In practice the value is (1407.3/18.98) = 74.2 K.
Assuming no scale resistance, the overall coefficient is {\frac{1407.3}{34.8\times74.2}} = 0.545 kW/m² K.

The available surface area on the outside of the tubes = 0.060 m²/m
or (246 x 3.65 x 0.060) = 53.9 m²
The actual coefficient is therefore \frac{1407.3}{53.9\times74.2} = 0.352 kW/m² K

and the dirt factor is {\frac{(0.545-0.352)}{(0.545\times0.352)}} = \underline{\underline{1.01\ m^{2}K/kW}}.

As shown in Figure 9.50, the clean coefficient varies from \underline{\underline{3.64\ kW/m^{2}\ K}} at the inlet to \underline{\underline{0.092\ kW/m^{2}\ K}} at the outlet.