1.25 kg/s of a solution is concentrated from 10 to 50 per cent solids in a triple-effect evaporator using steam at 393 K, and a vacuum such that the boiling point in the last effect is 325 K. If the feed is initially at 297 K and backward feed is used, what is the steam consumption, the temperature distribution in the system and the heat transfer area in each effect, each effect being identical?
For the purpose of calculation, it may be assumed that the specific heat capacity is 4.18 kJ/kg K, that there is no boiling point rise, and that the latent heat of vaporisation is constant at 2330 kJ/kg over the temperature range in the system. The overall heat transfer coefficients may be taken as 2.5, 2.0 and 1.6 kW/m^{2} K in the first, second and third effects, respectively.
Question : 1.25 kg/s of a solution is concentrated from 10 to 50 per ce...
Making a mass balance:
| Solid (kg/s) | Liquor (kg/s) | Total (kg/s) | |
| Feed | 0.125 | 1.125 | 1.25 |
| Product | 0.125 | 0.125 | 0.25 |
| Evaporation | __ | 1 | 1 |
Thus: D_{1}+D_{2}+D_{3}=1.0 kg/s (i)
\sum \Delta T=(T_{0}-T_{3})= (393 – 325) = 68 deg K (ii)
From equation 14.8: 2.5\Delta T_{1}=2.0\Delta T_{2}=1.6\Delta T_{3} (iii)
and from equations (ii) and (iii):
\Delta T_{1} = 18 deg K, \Delta T_{2} = 22 deg K, and \Delta T_{3} = 28 deg K
Modifying the figures to take account of the effect of the feed temperature, it will be assumed that:
\Delta T_{1} = 19 deg K, \Delta T_{2} = 24 deg K, and \Delta T_{3} = 25 deg K
The temperatures in each effect and the corresponding latent heats are then:
T_{0} = 393 K, \lambda _{0} = 2202 kJ/kg
T_{1} = 374 K, \lambda _{1} = 2254 kJ/kg
T_{2} = 350 K, \lambda _{2} = 2315 kJ/kg
T_{3} = 325 K, \lambda _{3} = 2376 kJ/kg
Making a heat balance for each effect:
(1) D_{0}\lambda _{0}=(W-D_{3}-D_{2})C_{p}(T_{1}-T_{2})+D_{1}\lambda _{1}
or 2202D_{0} = (1.25 – D_{2}-D_{3})4.18(374 – 350) + 2254D_{1} (iv)
(2) D_{1}\lambda _{1}=(W-D_{3})C_{p}(T_{2}-T_{3})+D_{2}\lambda _{2}
or 2254D_{1} = (1.25 – D_{3})4.18(350 – 325) + 2315D_{2} (v)
(3) D_{2}\lambda _{2}=WC_{p}(T_{3}-T_{f})+D_{3}\lambda _{3}
or 2315D_{2} = (1.25 × 4.18)(325 – 297) + 2376D_{3} (vi)
Solving equations (i), (iv), (v), and (vi) simultaneously:
D_{0} = 0.432 kg/s, D_{1} = 0.393 kg/s, D_{2} = 0.339 kg/s, and D_{3} = 0.268 kg/s.
The areas of transfer surface are:
A_{1}=D_{0}\lambda _{0}/U_{1}\Delta T_{1}= (0.432 × 2202)/(2.5 × 19) = 20.0 m^{2}
A_{2}=D_{1}\lambda _{1}/U_{2}\Delta T_{2}= (0.393 × 2254)/(2.0 × 24) = 18.5 m^{2}
A_{1}=D_{2}\lambda _{2}/U_{3}\Delta T_{3}= (0.268 × 2315)/(1.6 × 25) = 15.5 m^{2}
which are probably sufficiently close for design purposes; the mean area being 18.0 m^{2}
The steam consumption is therefore, D_{0} = 0.432 kg/s
The temperatures in each effect are: (1) 374 K, (2) 350 K, and (3) 325 K.
equation 14.8: U_{1}\Delta T_{1}=U_{2}\Delta T_{2}=U_{3}\Delta T_{3}