In a countercurrent packed column, n-butanol flows down at a rate of 0.25 kg/m² s and is cooled from 330 to 295 K. Air at 290 K, initially free of n-butanol vapour, is passed up the column at the rate of 0.7 m³/m² s.
Calculate the required height of tower and the condition of the exit air.
Data:
Mass transfer coefficient per unit volume, h_{D}a = 0.1 s^{-1}
Psychrometric ratio, \frac{h_{G}}{h_{D}\rho _{A}s}= 2.34
Heat transfer coefficients, h_{L} = 3h_{G}
Latent heat of vaporisation of n-butanol, λ = 590 kJ/kg
Specific heat of liquid n-butanol, C_{L} = 2.5 kJ/kg K
Humid heat of gas, s = 1.05 kJ/kg K
Temperature (K) | Vapour pressure of butanol (kN/m²) |
295 | 0.59 |
300 | 0.86 |
305 | 1.27 |
310 | 1.75 |
315 | 2.48 |
320 | 3.32 |
325 | 4.49 |
330 | 5.99 |
335 | 7.89 |
340 | 10.36 |
345 | 14.97 |
350 | 17.50 |
The first stage is to calculate the enthalpy of the saturated gas by way of the saturated humidity, \mathscr{H}_{0} given by:
\mathscr{H} _{0}={\frac{P_{w0}}{P-P_{w0}}}{\frac{M_{w}}{M_{A}}}={\frac{P_{w0}}{(101.3-P_{w0})}}\left({\frac{74}{29}}\right)The enthalpy is then:
H_{f}=\frac{1}{(1+\mathscr{H}_{0})} x 1.001(\theta _{f} – 273) + \mathscr{H}_{0}[2.5(\theta _{f} – 273) + 590] kJ/kg
where 1.001 kJ/kg K is the specific heat of dry air.
Thus: H_{f}=\frac{1.001\theta_{f}-273.27}{(1+\mathscr{H}_{0})}+\mathscr{H}_{0}(2.5\theta_{f}-92.5) kJ/kg moist air
The results of this calculation are presented in the following table and H_{f} is plotted against \theta _{f} in Figure 13.21.
The modified enthalpy at saturation H^{\prime }_{f} is given by:
where from equation 13.70: λ’ = λ/b = (590/2.34) or 252 kJ/kg
b={\frac{\lambda}{\lambda^{\prime}}} (13.70)
∴ H_{f}^{\prime}\,=\,{\frac{(1.001\theta_{f}-273.27)}{(1+\mathscr{H}_{0})}}\,+\mathscr{H}_{0}(2.5\theta_{f}\,-\,430.5) kJ/kg moist air
These results are also given in the following Table and plotted as H^{\prime }_{f} against \theta _{f} in Figure 13.21.
The bottom of the operating line (point a) has coordinates, \theta _{L1} = 295 K and H_{G1}, where:
H_{G1} = 1.05(290 – 273) = 17.9 kJ/kg.
At a mean temperature of, say, 310 K, the density of air is:
and: G’ = (0.70 x 1.140) = 0.798 kg/m² s
Thus, the slope of the operating line becomes:
{\frac{L^{\prime}C_{L}}{G^{\prime}}}\,=\,{\frac{(0.25\times2.5)}{0.798}}\,=\,0.783 kJ/kg K
and this is drawn in as AB in Figure 13.22 and at θ_{L2} = 330 K, H_{G2} = 46 kJ/kg.
From equation 13.77: H^{\prime}_{G} = [H_{G} + (b – 1)s(\theta _{G} – \theta _{0})]/b
H_{G}^{\prime}={\frac{1}{b}}[H_{G}+(b-1)s(\theta_{G}-\theta_{0})] (13.77)
∴ H_{G1}^{'}={\frac{17.9+{(2.34-1)}1.05{(290-273)}}{2.34}}=17.87\ \mathrm{kJ/kg}
Point a coincides with the bottom of the column.
A line is drawn through a of slope – \frac{h_{L}}{h_{D}\rho b}=-\left(\frac{3h_{G}}{h_{D}\rho b}\right)\cdot\left(\frac{h_{D}\rho s}{h_{G}}\right)
= -3s = -3.15 kJ/kg
This line meets curve RS at c (\theta _{f1}, H_{f1}^{\prime }) to give the interface conditions at the bottom of the column. The corresponding air enthalpy is given by point C whose co-ordinates are:
\theta _{f1} = 293 K H_{f1} = 29.0 kJ/kg
The difference between the ordinates of c and a gives the driving force in terms of the modified enthalpy at the bottom of the column, or:
A similar construction is made at other points along the operating line with the results shown in the following table.
from which:
\int_{H_{G1}}^{H_{G2}}\frac{\mathrm{d}H_{G}}{H_{f}^{\prime}-H_{G}^{\prime}}=2.379Substituting in equation 13.78:
\int_{H_{G1}}^{H_{G2}}\frac{\mathrm{d}H_{G}}{(H_{f}^{\prime}-H_{G}^{\prime})}=\frac{b h_{D}a\rho}{G^{\prime}}z (13.78)
\frac{b h_{D}\rho a z}{G^{\prime}}=2.379and: z={\frac{(2.379\times0.798)}{(2.34\times0.1)}} = \underline{\underline{8.1\ m}}
It remains to evaluate the change in gas conditions.
Point e, (\theta _{G1} = 290 K, H_{G1} = 17.9 kJ/kg) represents the condition of the inlet gas. ec is now drawn in, and from equation 13.75, this represents dH_{G}/d\theta _{G}. As for the air-water system, this construction is continued until the gas enthalpy reaches H_{G2}. The final point is given by D at which \underline{\underline{\theta_{G2}=308\ K}}.
\frac{(H_{G}^{\prime}-H_{f}^{\prime})}{(\theta_{G}-\theta_{f})}=\frac{\mathrm{d}H_{G}}{\mathrm{d}\theta_{G}} (cf. equation 13.51) (13.75)
\frac{H_{G}-H_{f}}{\theta_{G}-\theta_{f}}=\frac{\mathrm{d}H_{G}}{\mathrm{d}\theta_{G}} (13.51)
It is fortuitous that, in this problem, H^{\prime }_{G} = H_{G}. This is not always the case and reference should be made to Section 13.7 for elaboration of this point.
\theta _{f} (k) |
P_{w0} (kN/m²) |
\mathscr{H} _{0} (kg/kg) |
(1.001\theta _{f}-273.27)/(1+\mathscr{H} _{0}) (kJ/kg) |
\mathscr{H} _{0}(2.5\theta _{f}-92.5) (kJ/kg) |
H_{f} (kJ/kg) |
\mathscr{H} _{0}(2.5\theta _{f}-430.5) (kJ/kg) |
H_{f}^{\prime } (kJ/kg) |
295 | 0.59 | 0.0149 | 21.70 | 9.61 | 31.31 | 4.57 | 26.28 |
300 | 0.86 | 0.0218 | 24.45 | 14.33 | 40.78 | 6.97 | 33.42 |
305 | 1.27 | 0.0324 | 31.03 | 21.71 | 52.74 | 10.76 | 41.79 |
310 | 1.75 | 0.0448 | 35.45 | 30.58 | 66.03 | 15.43 | 50.88 |
315 | 2.48 | 0.0640 | 39.52 | 44.48 | 84.00 | 22.85 | 62.37 |
320 | 3.32 | 0.0864 | 43.31 | 61.13 | 104.44 | 31.92 | 75.23 |
325 | 4.49 | 0.1183 | 46.55 | 85.18 | 131.73 | 45.19 | 91.74 |
330 | 5.99 | 0.1603 | 49.18 | 117.42 | 166.60 | 63.23 | 112.41 |
335 | 7.89 | 0.2154 | 51.07 | 160.47 | 211.54 | 87.67 | 138.73 |
340 | 10.36 | 0.2905 | 51.97 | 220.05 | 272.02 | 121.87 | 173.83 |
345 | 14.97 | 0.4422 | 49.98 | 340.49 | 390.47 | 191.03 | 241.01 |
350 | 17.50 | 0.5325 | 50.30 | 416.68 | 466.98 | 236.70 | 287.00 |
\theta _{f} (K) |
H_{G} (kJ/kg) |
H^{\prime }_{G} (kJ/kg) |
H^{\prime }_{f} (kJ/kg) |
(H^{\prime }_{f} – H^{\prime }_{G}) (kJ/kg) |
1/(H^{\prime }_{f} – H^{\prime }_{G}) (kg/kJ) |
Mean value in interval | Interval | Value of integral over interval |
295 | 17.9 | 17.9 | 23.9 | 6.0 | 0.167 | |||
300 | 22.0 | 22.0 | 29.0 | 7.0 | 0.143 | 0.155 | 4.1 | 0.636 |
305 | 26.0 | 26.0 | 35.3 | 9.3 | 0.108 | 0.126 | 4.0 | 0.504 |
310 | 30.0 | 30.0 | 42.1 | 12.1 | 0.083 | 0.096 | 4.0 | 0.384 |
315 | 34.0 | 34.0 | 50.0 | 16.0 | 0.063 | 0.073 | 4.0 | 0.292 |
320 | 38.1 | 38.1 | 57.9 | 19.8 | 0.051 | 0.057 | 4.1 | 0.234 |
325 | 42.0 | 42.0 | 66.7 | 24.7 | 0.041 | 0.046 | 3.9 | 0.179 |
330 | 46.0 | 46.0 | 75.8 | 29.8 | 0.034 | 0.0375 | 4.0 | 0.150 |
Value of integral = 2.379 |