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23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2014

Question 3 of 8: Multi-Unit Material Balance on a Casein Plant

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, May 2014 — 04-Chem-A1 Process Balances and Chemical Thermodynamics, three-hour, open book, any non-communicating calculator. The paper is in four parts: Part A (Q1–Q2, 20%), Part B (Q3–Q4, 30%), Part C (Q5–Q6, 20%) and Part D (Q7–Q8, 30%); a candidate answers ONE question from each part. For completeness all eight questions are fully worked below.

Reference texts: Smith, Van Ness & Abbott, Introduction to Chemical Engineering Thermodynamics (8th ed.) for the energy balances, residual properties, VLE and reaction equilibrium; Felder & Rousseau, Elementary Principles of Chemical Processes (4th ed.) for the material and energy balances; steam properties from the ASME/NIST steam tables (Cengel & Boles appendix). SI throughout; ideal-gas constant \(R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}\).

Question 3: Multi-Unit Material Balance on a Casein Plant (Part B — 30%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Flows (kg/h): coagulum 49,600 (2.76 % casein, 3.68 % lactose); raw whey from screening 41,360 (0.012 % casein, 4.1 % lactose); whey (cycloned) 39,460 (0.007 % casein); fresh wash water 18,694 (solute-free); wash water 27,772 (0.026 % casein, 0.8 % lactose); waste wash water 25,572 (0.008 % casein); dried product 11.9 % moisture. The cookers heat indirectly (cycloned whey in Cooking 1, steam in Cooking 2), so no mass is added to the product line through the cookers; acidulation precipitates casein without changing the stream mass.

Find. (a) flow and composition of the dewheyed curd; (b) flow and composition of the dried casein product.

Cooking 1Cooking 2AcidulationScreeningWashingPressingDryingHydrocyclone 1Hydrocyclone 2Coagulum 49,600dewheyed curdWhey 39,460Raw whey 41,360Wash water 27,772Fresh wash 18,694Waste 25,572Dried product
Figure 3 — casein plant flowsheet. Coagulum is cooked and acidified, screened into dewheyed curd and raw whey; the curd is washed, pressed and dried. Two hydrocyclones recover casein fines from the raw whey and the spent wash water and recycle the combined fines to the screen feed. (Cooking heat is supplied indirectly, so the cycloned-whey and steam loops add no mass to the product line.)

Approach

The two hydrocyclones fix the recycle (“fines mix”) stream by difference and by species balances; a casein + lactose balance around the screen then gives the curd; a balance around the washer plus the fixed 11.9 % product moisture gives the dried product. Lactose, being fully soluble, keeps the same concentration in every stream split from a given whey/water solution.

  1. Hydrocyclone 1 (raw whey → cycloned whey + whey fines). By total balance the whey-fines underflow is \(41{,}360-39{,}460 = 1{,}900\) kg/h. A casein balance gives the casein it carries: $$m_{cas} = 41{,}360(0.00012) - 39{,}460(0.00007) = 4.963 - 2.762 = 2.201\ \mathrm{kg/h}.$$
  2. Hydrocyclone 2 (spent wash water → waste + second fines). Second-fines flow \(=27{,}772-25{,}572 = 2{,}200\) kg/h; casein $$m_{cas} = 27{,}772(0.00026) - 25{,}572(0.00008) = 7.221 - 2.046 = 5.175\ \mathrm{kg/h}.$$
  3. Combine the recycle (fines mix). The two fines streams merge and return to the screen feed: $$\dot F = 1{,}900 + 2{,}200 = 4{,}100\ \mathrm{kg/h},\quad m_{cas,F} = 2.201+5.175 = 7.376\ \mathrm{kg/h}.$$ Because lactose is fully soluble, the whey-fines carry lactose at the raw-whey concentration (4.1 %) and the second fines at the wash-water concentration (0.8 %): $$m_{lac,F} = 1{,}900(0.041) + 2{,}200(0.008) = 77.9 + 17.6 = 95.5\ \mathrm{kg/h}.$$
  4. (a) Balance around the screen. The acidulated coagulum (49,600 kg/h, unchanged in mass and composition through the indirect cookers and acidulation) plus the fines mix enter; dewheyed curd and raw whey leave: $$\dot C_{curd} = 49{,}600 + 4{,}100 - 41{,}360 = 12{,}340\ \mathrm{kg/h}.$$ Casein balance: \(49{,}600(0.0276)+7.376-41{,}360(0.00012) = 1368.96+7.376-4.963 = 1371.4\) kg/h.
    Lactose balance: \(49{,}600(0.0368)+95.5-41{,}360(0.041) = 1825.28+95.5-1695.76 = 225.0\) kg/h.
    Water by difference: \(12{,}340-1371.4-225.0 = 10{,}743.6\) kg/h.

    Dewheyed curd ≈ 12,340 kg/h: 11.11 % casein, 1.82 % lactose, 87.06 % moisture.

  5. (b) Balance around the washer. Curd + fresh wash water in; pressed cake to the dryer + spent wash water (27,772) out: $$\dot P = 12{,}340 + 18{,}694 - 27{,}772 = 3{,}262\ \mathrm{kg/h}.$$ Casein to dryer: \(1371.4 - 27{,}772(0.00026) = 1371.4-7.22 = 1364.2\) kg/h.
    Lactose to dryer: \(225.0 - 27{,}772(0.008) = 225.0-222.2 = 2.84\) kg/h.
  6. Drying to 11.9 % moisture. Drying removes only water, so all casein and lactose report to the product, which is \((100-11.9)=88.1\%\) solids: $$\dot D = \frac{1364.2 + 2.84}{0.881} = \frac{1367.0}{0.881} = 1{,}551.6\ \mathrm{kg/h}.$$

    Dried product ≈ 1,552 kg/h: 87.9 % casein, 0.18 % lactose, 11.9 % moisture.

StreamFlow (kg/h)CaseinLactoseMoisture
Fines-mix recycle4,1000.18 %2.33 %97.5 %
(a) Dewheyed curd12,34011.11 %1.82 %87.06 %
Pressed cake to dryer3,26241.8 %0.09 %58.1 %
(b) Dried product1,55287.9 %0.18 %11.9 %
Check / assumptions

The cooker heating loops are taken as indirect (as the problem states the cycloned whey and steam heat by indirect exchange), so they add no mass to the product line and the acidulated coagulum reaches the screen at its original 49,600 kg/h and composition. The fines-stream casein is obtained by casein balance on each hydrocyclone from the tabulated overflow concentrations; lactose (fully soluble) keeps the parent-solution concentration in every split. Both product compositions close to 100.0 %, confirming the stream network is internally consistent.