Question 13.9: Water is to be cooled from 328 to 293 K by means of a counte......

Assuming the latent heat of water at 273 K = 2495 kJ/kg
specific heat of air = 1.003 kJ/kg K
and specific heat of water vapour = 2.006 kJ/kg K

the enthalpy of the inlet air stream,
H_{G1} = 1.003(293 – 273) + \mathscr{H}[2495 + 2.006(293 – 273)]
From Figure 13.4:
at θ = 293 K and 20 per cent RH, \mathscr{H} = 0.003 kg/kg, and hence
H_{G1} = (1.003 x 20) + 0.003[2495 + (2.006 x 20)]
= 27.67 kJ/kg
In the inlet air, water vapour = 0.003 kg/kg dry air
or: \frac{(0.003/18)}{(1/29)} = 0.005 kmol/kmol dry air

Thus flow of dry air = (1 – 0.005)0.68 = 0.677 m³/m²s
Density of air at 293 K = \left({\frac{29}{22.4}}\right)\left({\frac{273}{293}}\right) = 1.206 kg/m³

and mass flow of dry air = (1.206 x 0.677) = 0.817 kg/m²s
Slope of operating line: (L^{\prime}C_{L}/G^{\prime})={\frac{(0.26\times4.18)}{0.817}} = 1.33

The coordinates of the bottom of the operating line are:
\theta _{L1} = 293 K, H_{G1} = 27.67 kJ/kg
Hence on an enthalpy-temperature diagram, the operating line of slope 1.33 is drawn through the point (293, 27.67) = (\theta _{L1}, H_{G1}).

The top point of the operating line is given by \theta _{L2} = 328 K, and H_{G2} is found to be 76.5 kJ/kg (Figure 13.19). From Figures 13.4 and 13.5 the curve representing the enthalpy of saturated air as a function of temperature is obtained and drawn in. Alternatively, this plot may be calculated from:

H_{F}=C_{a}(\theta_{f}-273)+\mathscr{H}_{0}[C_{w}(\theta_{f}-273)+\lambda]  kJ/kg

The curve represents the relation between enthalpy and temperature at the interface, that is H_{f} as a function of θ_{f}.
It now remains to evaluate the integral ∫ dH_{G}/(H_{f} — H_{G}) between the limits, H_{G1} = 27.7 kJ/kg and H_{G2} = 76.5 kJ/kg. Various values of H_{G} between these limits are selected and the value of θ obtained from the operating line. At this value of \theta _{1}, now \theta _{f}, the corresponding value of H_{f} is obtained from the curve for saturated air. The working is as follows:

A plot of 1/(H_{f} — H_{G}) and H_{G} is now made as shown in Figure 13.20 from which the area under the curve = 0.65. This value may be checked using the approximate solution of CAREY and WILLIAMSON^{(14)}.
At the bottom of the column:
H_{G1} = 27.7 kJ/kg, H_{f1} = 57.7 kJ/kg ∴ ΔH_{1} = 30 kJ/kg

At the top of the column:
H_{G2} = 76.5 kJ/kg, H_{f2} = 355 kJ/kg  ∴ ΔH_{2} = 279 kJ/kg
At the mean water temperature of 0.5(328 + 293) = 310.5 K:
H_{Gm} = 52 kJ/kg, H_{f} = 145 kJ/kg ∴ ΔH_{m} = 93 kJ/kg

and from Figure 13.16: f = 0.79

Thus: \frac{(H_{G2}-H_{G1})}{f\,\Delta H_{m}}=\frac{(76.5-27.7)}{(0.79\times93)}=0.66

which agrees well with the value (0.65) obtained by graphical integration.

Thus, in equation 13.53:

z=\int_{1}^{2}\,\mathrm{d}z=\frac{G^{\prime}}{h_{D}a\rho}\int_{1}^{2}\frac{\mathrm{d}H_{G}}{(H_{f}-H_{G})}                      (13.53)

height of packing, z = \int_{H_{G1}}^{H_{G2}}\frac{\mathrm{d}H_{G}}{\left(H_{f}-H_{G}\right)}\frac{G^{\prime}}{h_{D }a\rho}

= \frac{(0.65\times0.817)}{(0.2\times1.206)}

= \underline{\underline{2.20\ m}}

Assuming that the resistance to mass transfer lies entirely within the gas phase, the lines connecting \theta _{L} and \theta _{f} are parallel with the enthalpy axis.

In Figure 13.18 a plot of H_{G} and θ_{G} is obtained using the construction given in Section 13.6.4 and shown in Figure 13.15. From this curve, the value of θ_{G2} corresponding toH_{G2} = 76.5 kJ/kg is 300 K. From Figure 13.5, under these conditions, the exit air has a humidity of 0.019 kg/kg which from Figure 13.4 corresponds to a relative humidity of 83 per cent.