Material Balances

Abstract

Material balance calculations are employed in tracing the inflow and outflow of material in a process and thus establish quantities of components or the whole process stream. The procedures are useful in formulating products to specified compositions from available raw materials, evaluating final compositions after blending, evaluating processing yields, and evaluating separation efficiencies in mechanical separation systems.

Problems

• 2.1. A frankfurter formulation is to be made from the following ingredients:

$$ {\displaystyle \begin{array}{l}\mathrm{Lean}\ \mathrm{beef}:14\%\mathrm{fat},67\%\mathrm{water},19\%\mathrm{protein}\\ {}\mathrm{Pork}\ \mathrm{fat}:89\%\mathrm{fat},8\%\mathrm{water},3\%\mathrm{protein}\\ {}\mathrm{Soy}\ \mathrm{protein}\ \mathrm{isolate}:90\%\mathrm{protein},8\%\mathrm{water}\end{array}} $$

Water needs to be added (usually in the form of ice) to achieve the desired moisture content. The protein isolate added is 3% of the total weight of the mixture. How much lean beef, pork fat, water, and soy isolate must be used to obtain 100 kg of a formulation having the following composition: protein, 15%; moisture, 65%; and fat, 20%.

• 2.2. If 100 kg of raw sugar containing 95% sucrose, 3% water, and 2% soluble uncrystallizable inert solids is dissolved in 30 kg of hot water and cooled to 20°C, calculate:

  • (a) Kilograms of sucrose that remains in solution.

  • (b) Crystalline sucrose.

  • (c) The purity of the sucrose (in % sucrose) obtained after centrifugation and dehydration to 0% moisture. The solid phase contained 20% water after separation from the liquid phase in the centrifuge.

A saturated solution of sucrose at 20 °C contains 67% sucrose (w/w).

• 2.3. Tomato juice flowing through a pipe at the rate of 100 kg/min is salted by adding saturated salt solution (26% salt) into the pipeline at a constant rate. At what rate would the saturated salt solution be added to have 2% salt in the product?

• 2.4. If fresh apple juice contains 10% solids, what would be the solids content of a concentrate that would yield single-strength juice after diluting one part of the concentrate with three parts of water? Assume densities are constant and are equal to the density of water.

• 2.5. In a dehydration process, the product, which is at 80% moisture, initially has lost half its weight during the process. What is the final moisture content?

• 2.6. Calculate the quantity of dry air that must be introduced into an air drier that dries 100 kg/h of food from 80% moisture to 5% moisture. Air enters with a moisture content of 0.002 kg water per kg of dry air and leaves with a moisture content of 0.2 kg H₂O per kg of dry air.

• 2.7. How much water is required to raise the moisture content of 100 kg of a material from 30% to 75%?

• 2.8. In the section “Multistage Process,” Example 2.18, solve the problem if the meat to solvent ratio is 1:1. The solubility of fat in the water-solvent mixture is such that the maximum fat content in the solution is 10%.

• 2.9. How many kilograms of peaches would be required to produce 100 kg of peach preserves? The standard formula of 45 parts fruit to 55 parts sugar is used, the soluble solids content of the finished product is 65%, and the peaches have 12% initial soluble solids content. Calculate the weight of 100 grade pectin required and the amount of water removed by evaporation.

• 2.10. The peaches in Problem 9 come in a frozen form to which sugar has been added in the ratio of three parts fruit to one part sugar. How much peach preserves can be produced from 100 kg of this frozen raw material?

• 2.11. Yeast has a proximate analysis of 47% C, 6.5% H, 31% O, 7.5% N, and 8% ash on a dry weight basis. Based on a factor of 6.25 for converting protein nitrogen to protein, the protein content of yeast on a dry basis is 46.9%. In a typical yeast culture process, the growth medium is aerated to convert substrate primarily to cell mass. The dry cell mass yield is 50% of a sugar substrate. Nitrogen is supplied as ammonium phosphate.

The cowpea is a high-protein, low-fat legume that is a valuable protein source in the diet of many Third World nations. The proximate analysis of the legume is 30% protein, 50% starch, 6% oligosaccharides, 6% fat, 2% fiber, 5% water, and 1% ash. It is desired to produce a protein concentrate by fermenting the legume with yeast. Inorganic ammonium phosphate is added to provide the nitrogen source. The starch in cowpea is first hydrolyzed with amylase and yeast is grown on the hydrolyzate.

• (a) Calculate the amount of added inorganic nitrogen as ammonium phosphate to provide the stoichiometric amount of nitrogen to convert all the starch present to yeast mass. Assume none of the cowpea protein is utilized by the yeast.

• (b) If the starch is 80% converted to cell mass, calculate the proximate analysis of the fermented cowpea on a dry basis.

• 2.12. This whey is spray dried to a final moisture content of 3%, and the dry whey is used in an experimental batch of summer sausage.

  In summer sausage, the chopped meat is inoculated with a bacterial culture that converts sugars to lactic acid as the meat is allowed to ferment prior to cooking in a smokehouse. The level of acid produced is controlled by the amount of sugar in the formulation. The lactic acid level in the summer sausage is 0.5 g/100 g dry solids. Four moles of lactic acid (CH₃ CHOHCOOH) is produced from 1 mole of lactose (C₁²H²²O₁₁). The following formula is used for the summer sausage:

  $$ {\displaystyle \begin{array}{l}3.18\ \mathrm{kg}\ \mathrm{lean}\ \mathrm{beef}\ \\ {}\kern1em \left(16\%\mathrm{fat},16\%\mathrm{protein},67.1\%\mathrm{water},0.9\%\mathrm{ash}\right)\\ {}1.36\ \mathrm{kg}\ \mathrm{pork}\ \\ {}\kern1em \left(25\%\mathrm{fat},12\%\mathrm{protein},62.4\%\mathrm{water},0.6\%\mathrm{ash}\right)\\ {}0.91\ \mathrm{kg}\ \mathrm{ice}\\ {}0.18\ \mathrm{kg}\ \mathrm{soy}\ \mathrm{protein}\ \mathrm{isolate}\\ {}\kern1em \left(5\%\mathrm{water},1\%\mathrm{ash},94\%\mathrm{protein}\right)\end{array}} $$

  Calculate the amount of dried whey protein that can be added into this formulation in order that, when the lactose is 80% converted to lactic acid, the desired acidity will be obtained.

• 2.13. Osmotic dehydration of blueberries was accomplished by contacting the berries with a corn syrup solution containing 60% soluble solids for 6 hours and draining the syrup from the solids. The solid fraction left on the screen after draining the syrup is 90% of the original weight of the berries. The berries originally contained 12% soluble solids, 86.5% water, and 1.5% insoluble solids. The sugar in the syrup penetrated the berries such that the berries themselves remaining on the screen, when washed free of the adhering solution, showed a soluble solids gain of 1.5% based on the original dry solids content. Calculate:

  • (a) The moisture content of the berries and adhering solution remaining on the screen after draining the syrup.

  • (b) The soluble solids content of the berries after drying to a final moisture content of 10%.

  • (c) The percentage of soluble solids in the syrup drained from the mixture. Assume none of the insoluble solids are lost in the syrup.

• 2.14. The process for producing dried mashed potato flakes involves mixing wet mashed potatoes with dried flakes in a 95:5 weight ratio, and the mixture is passed through a granulator before drying on a drum dryer. The cooked potatoes after mashing contained 82% water and the dried flakes contained 3% water. Calculate:

  • (a) The amount of water that must be removed by the dryer for every 100 kg of dried flakes produced.

  • (b) The moisture content of the granulated paste fed to the dryer.

  • (c) The amount of raw potatoes needed to produce 100 kg of dried flakes; 8.5% of the raw potato weight is lost on peeling.

  • (d) Potatoes should be purchased on a dry matter basis. If the base moisture content is 82% and potatoes at this moisture content cost $200/ton, what would be the purchase price for potatoes containing 85% moisture.

• 2.15. Diafiltration is a process used to reduce the lactose content of whey recovered using an ultrafiltration membrane. The whey is passed through the membrane first and concentrated to twice the initial solids content, rediluted, and passed through the membrane a second time. Two membrane modules in series, each with a membrane surface area of 0.5 m², are to be used for concentrating and removing lactose from acid whey that contains 7.01% total solids, 5.32% lactose, and 1.69% protein. The first module accomplishes the initial concentration, and the retentate is diluted with water and reconcentrated in the second module to a solids content of 14.02%. Under the conditions used in the process, the membrane has an average water permeation rate of 254 kg/(h ≅ m²). The rejection factor for lactose by the membrane based on the arithmetic mean of the feed and retentate lactose concentration and the mean permeate lactose concentration is 0.2. Protein rejection factor is 1. The rejection factor is defined as: F_r = (C_rC_p)/C_r where C_r is the concentration on the retentate side and C_p is the concentration on the permeate side of the membrane. Use Visual BASIC to determine:

  • (a) The amount of 14.02% solids delactosed whey concentrate produced from the second module per hour.

  • (b) The lactose content of the delactosed whey concentrate.

• 2.16. An orange juice blend containing 42% soluble solids is to be produced by blending stored orange juice concentrate with the current crop of freshly squeezed juice. The following are the constraints:

The soluble solids to acid ratio must equal 18; and the currently produced juice may be concentrated first before blending, if necessary. The currently produced juice contains 14.5% soluble solids, 15.3% total solids, and 0.72% acid. The stored concentrate contains 60% soluble solids, 62% total solids, and 4.3% acid. Calculate:

• (a) The amount of water that must be removed or added to adjust the concentration of the soluble solids in order to meet the specified constraints.

• (b) The amounts of currently processed juice and stored concentrate needed to produce 100 kg of the blend containing 42% soluble solids.

• 2.17. The process for extraction of sorghum juice from sweet sorghum for production of sorghum molasses, which is still practiced in some areas in the rural south of the United States, involves passing the cane through a three-roll mill to squeeze the juice out. Under the best conditions, the squeezed cane (bagasse) still contains 50% water.

  • (a) If the cane originally contained 13.4% sugar, 65.6% water, and 32% fiber, calculate the amount of juice squeezed from the cane per 100 kg of raw cane, the concentration of sugar in the juice, and the percentage of sugar originally in the cane that is left unrecovered in the bagasse.

  • (b) If the cane is not immediately processed after cutting, moisture and sugar loss occurs. Loss of sugar has been estimated to be as much as 1.5% within a 24-hour holding period, and total weight loss for the cane during this period is 5.5%. Assume sugar is lost by conversion to CO²; therefore, the weight loss is attributable to water and sugar loss. Calculate the juice yield based on the freshly harvested cane weight of 100 kg, the sugar content in the juice, and the amount of sugar remaining in the bagasse.

• 2.18. In a continuous fermentation process for ethanol from a sugar substrate, the sugar is converted to ethanol and part of it is converted to yeast cell mass. Consider a 1000 L continuous fermentor operating at a steady state. Cell-free substrate containing 12% glucose enters the fermentor. The yeast has a generation time of 1.5 hours, and the concentration of yeast cells within the fermentor is 1 × 10⁷ cells/mL. Under these conditions, a dilution rate (F/V, where F is the rate of feeding of cell-free substrate and V is the volume of the fermentor) that causes the cell mass to stabilize at a steady state results in a residual sugar content in the overflow of 1.2%. The stoichiometric ratio for sugar to dry cell mass is 1:0.5 on a weight basis, and that for sugar to ethanol is based on 2 moles of ethanol produced per mole of glucose. A dry cell mass of 4.5 of the g/L fermentor is equivalent in g to ethanol/(La cell count ≅ h) of 1.6 × 10⁷ cells/mL. Calculate the ethanol productivity of the fermentor in g ethanol/(L ≅ h).

• 2.19. A protein solution is to be demineralized by dialysis. The solution is placed inside a dialysis tube immersed in continuously flowing water. For all practical purposes, the concentration of salt in the water is zero, and the dialysis rate is proportional to the concentration of salt in the solution inside the tube. The contents of the tube may be considered to be well mixed. A solution that initially contained 500 μg/mL of salt and 15 mg protein/mL contained 400 μg/mL salt and 13 mg protein/mL at the end of 2 hours. Assume no permeation of protein through the membrane, and density of the solution is constant at 0.998 g/mL. The rate of permeation of water into the membrane and the rate of permeation of salt out of the membrane are both directly proportional to the concentration of salt in the solution inside the membrane. Calculate the time of dialysis needed to drop the salt concentration inside the membrane to 10 μg/mL. What will be the protein concentration inside the membrane at this time?

• 2.20. A fruit juice-based natural sweetener for beverages is to be formulated. The sweetener is to have a soluble solids to acid ratio of 80. Pear concentrate having a soluble solids to acid ratio of 52 and osmotically concentrated/deacidified apple juice with a soluble solids to acid ratio of 90 are available. Both concentrates contain 70% soluble solids.

  • (a) Calculate the ratio of deacidified apple and pear concentrates that must be mixed to obtain the desired sol. solids/acid ratio in the product.

  • (b) If the mixture is to be diluted to 45% soluble solids, calculate the amounts of deacidified apple and pear juice concentrates needed to make 100 kg of the desired product.

• 2.21. A spray drier used to dry egg whites produces 1000 kg/h of dried product containing 3.5% moisture from a raw material that contains 86% moisture.

  • (a) If air used for drying enters with a humidity of 0.0005 kg water per kg dry air and leaves the drier with a humidity of 0.04 kg water per kg dry air, calculate the amount of drying air needed to carry out the process.

  • (b) It is proposed to install a reverse osmosis system to remove some moisture from the egg whites prior to dehydration to increase the drying capacity. If the reverse osmosis system changes the moisture content of the egg whites prior to drying to 80% moisture, and the same inlet and outlet air humidities are used on the drier, calculate the new production rate for the dried egg whites.

• 2.22. An ultrafiltration system for concentrating milk has a membrane area of 0.5 m², and the permeation rate for water and low molecular weight solutes through the membrane is 3000 g/(m² ≅ min). The solids content of the permeate is 0.5%. Milk flow through the membrane system must be maintained at 10 kg/min to prevent fouling. If milk containing initially 91% water enters the system, and a concentrate containing 81% solids is desired, calculate:

  • (a) The amount of concentrate produced by the unit/h.

  • (b) The fraction of total product leaving the membrane that must be recycled to achieve the desired solids concentration in the product.

  • (c) The solids concentration of feed entering the unit after the fresh and recycled milk is mixed.

• 2.23. A recent development in the drying of blueberries involves an osmotic dewatering prior to final dehydration. In a typical process, grape juice concentrate with 45% soluble solids (1.2% insoluble solids; 53.8% water) in a ratio 2 kg juice/kg berries is used to osmoblanch the berries, followed by draining the berries and final drying in a tunnel drier to a moisture content of 12%.

During osmoblanching (the juice concentrate is heated to 80 °C, the berries added, temperature allowed to go back to 80 °C, 5-minute hold, and the juice is drained), the solids content of blueberries increases. When analyzed after rinsing the adhering solution, the total solids content of the berries was 15%. The original berries contained 10% total solids and 9.6% soluble solids. Assume leaching of blueberry solids into the grape juice is negligible.

The drained berries carry about 12% of their weight of the solution that was drained.

Calculate:

• (a) The proximate composition of the juice drained from the blueberries.

• (b) The yield of dried blueberries from 100 kg raw berries.

• (c) The drained juice is concentrated and recycled. Calculate the amount of 45% solids concentrate that must be added to the recycled concentrate to make enough for the next batch of 100 kg blueberries.

• 2.24. A dietetic jelly is to be produced. In order that a similar fruit flavor may be obtained in the dietetic jelly as in the traditional jelly, the amount of jelly that can be made from a given amount of fruit juice should be the same as in the standard pectin jelly.

If 100 kg of fruit juice is available with a soluble solids content of 14%, calculate the amount of standard pectin jelly with a soluble solids content of 65% that can be produced.

Only the sugar in jelly contributes the caloric content.

• (a) What will be the caloric equivalent of 20 g (1 tsp) of standard pectin jelly?

• (b) The dietetic jelly should have a caloric content 20% that of the standard pectin jelly. Fructose may be added to provide sweetness. The low methoxy pectin can be used in the same amount for the same quantity of fruit as in the standard jelly. Calculate the soluble solids content in the dietetic jelly to give the caloric reduction.

• (c) In part (b), calculate the amount of additional fructose that must be used.