balanceevaporaccristalacondaire

January 11, 2019 | Author: 3lks0y | Category: Crystallization, Filtration, Air Conditioning, Water, Solubility
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Balance de materia y energía QUIM-MS-2008 Caso Balance de materia en un equipo de acondicionamiento de aire

Problem statement Fresh air air containing containing 4.00 4.00 mole% mole% water water vapour vapour is to be cooled cooled and dehumidified to a water content of 1.70 mole% water. A stream of fresh air is combined with a recycle stream of previously dehumidified air and passed through the cooler. The blended stream entering the unit contains 2.30 mole% water. In the air conditioner some of the water in the feed is condensed and removed as liquid. A fraction of the dehumidified air leaving the cooler is recycled and the remainder is delivered to a room. Taking 25.0 mol/min of dehumidified air delivered to the room, calculate the fresh feed rate, rate of water condensation, and the rate of recycled dehumidified air.

Flow sheet for air conditioner Recycled cold air . n5 mol/min  yw,5 mol H2O/mol

air .n Cold mol/min 4

air .n Wet 1 mol/min 0.04 mol H2O/mol

 yw,4 mol H2O/mol

Air conditioner feed .nAirmol/min

Cold air . n = 25.0 mol/min

0.023 mol H2O/mol

0.017 mol H2O/mol

2

6

Liquid water . n3 mol/min  yW,3 mol H2O/mol

Defining systems Recycled cold air .n mol/min 5  yw,5 mol H2O/mol

air .n Wet 1 mol/min 0.04 mol H2O/mol

Overall process

air .n Cold mol/min 4

Aircon unit Mixing point

 yw,4 mol H2O/mol

Splitter Air conditioner

feed .nAirmol/min

Cold air . n = 25.0 mol/min

0.023 mol H2O/mol

0.017 mol H2O/mol

2

6

Liquid water . n3 mol/min  yW,3 mol H2O/mol

Evaluation of material balances - I Overall system: . n1= 25.6 mol/min n. 3= 0.6 mol/min . n6=25.0 mol/min

yw,1=0.040  yw,3 =1.000 yw,6 =0.017

 yDA,1=0.960  yDA,3=0.960  yDA,1=0.983

All other systems have one/two degrees of freedom. The equations governing all the other systems, can be expressed in terms of one/ two unknown variables. The most convenient way of solving the material balance in this case is to formulate the water mole balance and the overall balance for each of the other systems:

Mixing point: Aircon unit: Splitter:

n1 + n5 = n2 0.04 ⋅ n1 + yw,5 ⋅ n5 = 0.023 ⋅ n2 n2 = n3 + n4 0.023 ⋅ n2 = n3 + yw, 4 ⋅ n4 n4 = n5 + n6

This is now a system with 5 equations/8 variables, which can be solved using the results obtained on the overall system

Evaluation of material balances - II Mixing point: n1 + n5 = n2 0.04 ⋅ n1 + yw,5 ⋅ n5

=

0.023 ⋅ n2

The mole fraction of water in stream 5 equals the mol fraction in stream 6 (physical law from point around splitter). Furthermore, from the material balance around the whole system we know the molar flow rate of stream 1: 25.6 + n5 = n2 0.04 ⋅ 25.6 + 0.017 ⋅ n5 = 0.023 ⋅ n2 mol n = 72.5 ⋅ mol 5 Solving these equations simultaneously: n2 = 98.1 ⋅ min min mol Splitter: n4 = n5 + n6 = 97.5 ⋅ min

Aircon unit: Material balance not needed anymore!

Evaluation of material balances - III cold air .nRecycled = 72.5 mol/min 5

0.017 mol H2O/mol

Cold air 4 = 97.5 mol/min 0.017 mol H2O/mol

. n

Wet air .n = 25.6 mol/min 1

0.04 mol H2O/mol

Air conditioner Air feed 2 = 98.1 mol/min 0.023 mol H2O/mol

Cold air . n = 25.0 mol/min

.n

6

0.017 mol H2O/mol

water .n Liquid = 0.6 mol/min 3

1 mol H2O/mol

Answering the questions Fresh feed rate: Rate of water condensation: Rate of recycled dehumidified air:

25.6 mol/min 0.6 mol/min 72.5 mol/min

Balance de materia y energía QUIM-MS-2008 Caso Cristalización del cromato de potasio, K2CrO4

Principle of crystallisation/precipitation Crystallisation/precipitation: Recovery of materials in solution as a solid. Governing principle: Solubility product, KSP For an ionic compound, such as K2CrO4, solubility is given by: K2CrO4(s) 2 K+ + CrO42and the solubility product is defined as: KSP

=

2

[K ] [CrO42 ] +





Often in literature solubility is given in g/100 g H2O. This is the amount of pure salt dissolved in pure water. The solubility of K2CrO4 is 62.9 g/100 g water at 20oC. Thus, 62.9 ⋅ g 1 mol 100 ⋅ g mol [K ] = 2 ⋅ 100 ⋅ g ⋅ 194.19 ⋅ g ⋅ 0.1 ⋅ litre = 6.5 ⋅ litre 3 mol mol KSP = 136 ⋅ [CrO42 ] = 3.2 ⋅ litre litre 3 +



Problem statement A solution containing 33.3 wt.-% of K2CrO4 (4500 kg/h) is to be fed to an evaporative crystallisation process. Before entering the evaporator, the fresh feed is mixed with a recycle stream containing 36.4 wt% K2CrO4. The stream leaving the evaporator, which contains 49.4 wt% K2CrO4, is fed to the crystalliser, in which the solution is cooled and crystals are formed. The filter cake removed from the crystalliser consists of K2CrO4 crystals and a solution containing 36.4 wt% K 2CrO4. The crystals account for 95% of the total mass of the filter cake. The solution passing the filter, which also contains 36.4 wt% K2CrO4, is recycled back. Calculate the rate of evaporation, the rate of production of crystalline K2CrO4, the feed rates that the evaporator and the crystallizer need to handle, and the recycle ratio. What are the benefits of recycling?

Flow sheet for evaporative crystalliser feed .m Fresh = 4500 kg/h

.mFeed kg/h

0.333 kg K2CrO4/kg 0.667 kg H2O/kg

x2 kg K2CrO4/kg 1-x2 kg H2O/kg

1

2

Water vapour . m3 kg/h

Crystalliser and filter

Evaporator Crystalliser .m kg/hfeed 4 0.494 kg K2CrO4/kg 0.506 kg H2O/kg

Filtrate (recycle) . m7 kg/h 0.364 kg K2CrO4/kg 0.636 kg H2O/kg

Filter cake . Mass flow rate of crystals: m5 kg/h . Mass flow rate of solution: m6 kg/h 0.364 kg K2CrO4/kg 0.636 kg H2O/kg

Process is completely labelled (all component flow rates can be expressed in terms of given quantities)

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