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Question : 1.9 kg/s of a liquid containing 10 per cent by mass of disso...

1.9 kg/s of a liquid containing 10 per cent by mass of dissolved solids is fed at 338 K to a forward-feed double-effect evaporator. The product consists of 25 per cent by mass of solids and a mother liquor containing 25 per cent by mass of dissolved solids. The steam fed to the first effect is dry and saturated at 240 kN/m^{2} and the pressure in the second effect is 20 kN/m^{2}. The specific heat capacity of the solid may be taken as 2.5 kJ/kg K,both in solid form and in solution, and the heat of solution may be neglected. The mother liquor exhibits a boiling point rise of 6 deg K. If the two effects are identical, what area is required if the heat transfer coefficients in the first and second effects are 1.7 and 1.1 kW/m^{2} K respectively?

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The percentage by mass of dissolved and undissolved solids in the final product = (0.25 × 75) + 25 = 43.8 per cent and hence the mass balance becomes:

Solids (kg/s) Liquor
(kg/s)		 Total (kg/s)
Feed 0.19 1.71 1.9
Product 0.19 0.244 0.434
Evaporation __ (D_{1}+D_{2})=1.466 1.466

At 240 kN/m^{2}: T0 = 399 K At 20 kN/m^{2}, T_{2}=333 K

Thus: T'_{2}= (333 + 6) = 339 K

and, allowing for the boiling-point rise in the first effect:

\Delta T_{1}+\Delta T_{2}= (399 – 339) – 6 = 54 deg K

From equation 14.8: 1.7\Delta T_{1}=1.1\Delta T_{2}

and: \Delta T_{1} = 21 deg K and \Delta T_{2} = 33 deg K

Modifying these values to allow for the cold feed, it will be assumed that:

\Delta T_{1} = 23 deg K and \Delta T_{2} = 31 deg K

Assuming that the liquor exhibits a 6 deg K boiling-point rise at all concentrations, then, with T'_{1} as the temperature of boiling liquor in the first effect and T'_{2} that in the second effect:

T_{0} = 399 K at which \lambda _{0} = 2185 kJ/kg

T'_{1}= (399 – 23) = 376 K

T_{1} = (376 – 6) = 370 K at which \lambda _{1} = 2266 kJ/kg

T'_{2}= 339 K

T_{2} = 333 K at which \lambda _{2} = 2258 K

Making a heat balance over each effect:

(1) D_{0}\lambda _{0}=WC_{p}(T'_{1}-T_{f})+D_{1}\lambda _{1}

or 2185D_{0} = (1.90 × 2.5)(376 – 338) + 2266D_{1}

(2) D_{1}\lambda _{1}+(W-D_{1})C_{p}(T'_{1}-T'_{2})=D_{2}\lambda _{2}

or 2358D_{2} = (1.90 – D_{1})2.5(376 – 339) + 2266D_{1}

Solving: D_{0} = 0.833 kg/s, D_{1} = 0.724 kg/s, and D_{2} = 0.742 kg/s

The areas are then given by:

A_{1}=D_{0}\lambda _{0}/U_{1}(T_{0}-T'_{1})=0.833 × 2185/[1.7(399 – 376)] = 46.7 m^{2}

A_{2}=D_{1}\lambda _{1}/U_{2}(T_{1}-T'_{2})=0.724 × 2266/[1.1(370 – 339)] = 48.0 m^{2}

which are close enough for design purposes.

The area to be specified for each effect is approximately 47.5 m^{2} .

 

 

equation 14.8: U_{1}\Delta T_{1}=U_{2}\Delta T_{2}=U_{3}\Delta T_{3}