Food Engineering
Fundamental of Engineering Calculation 1. Determine the following unit conversions to SI units: a. Density value of 60 lbft3 to kgm3, Ans: 961.1 kg/m3 b. Energy value of 1.7 ×103Btu to kJ, Ans: 1.8 x 103 kJ c. Enthalpy value of 2475 BtulbtokJkg, Ans: 5756 kJ/kg d. Pressure value of 14.69 psig to kPa, Ans: 201.88 kPa e. Viscosity value of 20 cp to Pa∙s, Ans: 2 x 10-2 Pa∙s
Table of Conversion Factor 1 lb | 0.454 kg | 1 ft | 0.305 m | 1 Btu | 1.055 kJ | 1 psia | 1 psig, pound-force per square inch gauge - 14.69 | 1 psia, pounds per square inch absolute | 6.895 kPa | 1 in | 2.54 x 10-2 m | 1 Pa | 1 N/m2 | 1 cp, centipois | 10-3 Pa∙s |
2. The sugar solution is prepared by dissolving 10 kg of sucrose in 90 kg of water. The density of the solution is 1040 kg/m3. Molecular Weight of Sucrose: 342.30 g/mol, Water: 18 g/mol. Determine the following, f. Concentration, weight per unit weight, Ans: 0.1 kg solute/kg solvent g. Concentration, weight per unit volume, Ans: 104 kg solute/m3 solvent h. Molarity, Ans: 0.3 mole solute/L solution i. Mole Fraction, Ans: 0.0058 j. Molality, Ans: 0.325 mole solute/kg solvent k. Recalculate (a) – (e) if the, i. Sucrose solution contains 20 kg of sucrose in 80 kg of water, and density of the solution is 1083 kg/m3 ii. Sucrose solution contains 30 kg of sucrose in 70 kg of water, and density of the solution is 1129 kg/m3
Table 1: Answers for Question (f) Unit | Scenario I | Scenario II | kg solute/kg solvent | 0.2 | 0.3 | kg solute/m3 solvent | 216.6 | 338.7 | mole solute/L solution | 0.63 | 0.99 | - | 0.0130 | 0.0221 | mole solute/kg solvent | 0.731 | 1.253 |
3. Convert a moisture content of 85 % wet basis to moisture content dry basis.
Ans: MCdb = 5.67 or 567 %
4. A food is initially at a moisture content of 90 % dry basis. Calculate the moisture
Table of Conversion Factor 1 lb | 0.454 kg | 1 ft | 0.305 m | 1 Btu | 1.055 kJ | 1 psia | 1 psig, pound-force per square inch gauge - 14.69 | 1 psia, pounds per square inch absolute | 6.895 kPa | 1 in | 2.54 x 10-2 m | 1 Pa | 1 N/m2 | 1 cp, centipois | 10-3 Pa∙s |
2. The sugar solution is prepared by dissolving 10 kg of sucrose in 90 kg of water. The density of the solution is 1040 kg/m3. Molecular Weight of Sucrose: 342.30 g/mol, Water: 18 g/mol. Determine the following, f. Concentration, weight per unit weight, Ans: 0.1 kg solute/kg solvent g. Concentration, weight per unit volume, Ans: 104 kg solute/m3 solvent h. Molarity, Ans: 0.3 mole solute/L solution i. Mole Fraction, Ans: 0.0058 j. Molality, Ans: 0.325 mole solute/kg solvent k. Recalculate (a) – (e) if the, i. Sucrose solution contains 20 kg of sucrose in 80 kg of water, and density of the solution is 1083 kg/m3 ii. Sucrose solution contains 30 kg of sucrose in 70 kg of water, and density of the solution is 1129 kg/m3
Table 1: Answers for Question (f) Unit | Scenario I | Scenario II | kg solute/kg solvent | 0.2 | 0.3 | kg solute/m3 solvent | 216.6 | 338.7 | mole solute/L solution | 0.63 | 0.99 | - | 0.0130 | 0.0221 | mole solute/kg solvent | 0.731 | 1.253 |
3. Convert a moisture content of 85 % wet basis to moisture content dry basis.
Ans: MCdb = 5.67 or 567 %
4. A food is initially at a moisture content of 90 % dry basis. Calculate the moisture