1. In the production of alternative and natural sweeteners from Stevia, the cleaned leaves at a rate of 875 kg/h are first passed through a mechanical sorter. 23.0% of the fresh feed are small leaves and can go straight to the drier. They are dried to 8.00% moisture, from the original 46.7% moisture. After drying, they are ground to a fine powder. The remaining large leaves are first fed to a grinder where the leaves are ground to a pulp. They are then fed to an extractor where hexane is also added to extract steviol glycosides, the “sweet component” of the stevia leaves. Fresh hexane at 27.2 kg/h is added to a recycled condensed hexane stream and is then fed to the extractor at 2.75 kg hexane per kg of ground stevia leaves. The exit stream of the extractor then passes through a filter where the filter cake (predominantly containing pulp) is separated from the liquid part of the stream (filtrate). The filtrate is then fed to a decanter where the hexane stream is separated from the aqueous stream. The aqueous stream is virtually pure water. The hexane stream is fed to an evaporator where all of the hexane is evaporated, leaving pure steviol glycosides. The evaporated hexane is then condensed and recycled back to the extractor. Given below are useful information for the composition of some streams and materials:
   1. Fresh Stevia leaves contain 46.7% moisture, 3.80% steviol glycosides, and the remainder is pulp.
   2. The filter cake is 92.4% pulp, the remainder is a mixture of hexane and steviol glycoside.
   3. Two final products are produced in the process: dried ground Stevia leaves and pure steviol glycoside.

Calculate the following:

1. Rate of production of the two final products in tons/year (assume that 1 year of operation is equal to 300 days of continuous operation)
2. Rate of hexane recycle (kg/h) and recycle ratio (recycled hexane/fresh hexane)

In the production of alternative and natural sweeteners from Stevia, the cleaned leaves at a rate of 875 kg/h are first passed through a mechanical sorter. 23.0% of the fresh feed are small leaves and can go straight to the drier. They are dried to 8.00% moisture, from the original 46.7% moisture. After drying, they are ground to a fine powder. The remaining large leaves are first fed to a grinder where the leaves are ground to a pulp. They are then fed to an extractor where hexane is also added to extract steviol glycosides, the “sweet component” of the stevia leaves. Fresh hexane at 27.2 kg/h is added to a recycled condensed hexane stream and is then fed to the extractor at 2.75 kg hexane per kg of ground stevia leaves. The exit stream of the extractor then passes through a filter where the filter cake (predominantly containing pulp) is separated from the liquid part of the stream (filtrate). The filtrate is then fed to a decanter where the hexane stream is separated from the aqueous stream. The aqueous stream is virtually pure water. The hexane stream is fed to an evaporator where all of the hexane is evaporated, leaving pure steviol glycosides. The evaporated hexane is then condensed and recycled back to the extractor. Some information for the composition of streams and materials are: Fresh Stevia leaves contain 46.7% moisture, 3.80% steviol glycosides, and the remainder is pulp; The filter cake is 92.4% pulp, the remainder is a mixture of hexane and steviol glycoside.; Two final products are produced in the process: dried ground Stevia leaves and pure steviol glycoside.
Find the following:
a. Complete diagram

b. Rate of production of the two final products in tons/year (assume that 1 yr. operation = 300 days of continuous operation)

c. Rate of hexane recycle (kg/h) and recycle ratio (recycled hexane/fresh hexane)

What is the significance of Bernoulli equation and Reynolds number in Fluid Mechanics?

1. Using pictures and/or diagrams describe and explain in detail the difference between conventional surface mining and solution mining. Ensure to include the advantages and disadvantages of both of these mining methods. (2.5 pages minimum.) (10 points)

A steam turbine expands steam from 70 bar, 300 °C, to 6.5 bar. The flow rate of steam is 2.5 kg/s and the turbine generates 500 kW of power.

a) Determine the temperature at the outlet.
b) Determine the efficiency of the turbine.
c) Determine the entropy generation.

1. Define BOD. Describe the measurement of BOD. Give an estimate of possible BOD values for the following three water types: pristine river, sewage and stormwater.

2. Classify microorganisms according to their source of energy and carbon. In your answer, define chemotroph, autotroph, heterotroph, phototroph, decomposer and producer.

3. In which class(es) are you likely to find algae, bacteria and fungi?

2. “Famine is a natural occurrence.” Explain what is wrong with this statement by drawing on examples of famines in two different world regions. 500 words

a) A natural draft cooling tower is used to cool the condenser of a 100 MW power plant. The water enters the tower at 40 °C at a rate of 80 kg/s. The range value of cooling tower is 15 °C. The water is cooled by air stream and the air entering the tower is at 1.01325 bar, 23°C and 60% relative humidity. The air leaves the tower as saturated at 32°C. According to the given data, using pyschrometric chart and steam table, determine the volumetric flow rate of air flowing into the cooling tower.

b) Describe briefly the definition and function of cooling tower fill material.

Ethanol-water pressure swing distillation is to be designed with the use of FUG method. first column is at 1 bar and second is at 7 bar.

Column 1 results: minimium number of theoretical stages = 7, Rmin = 0.99 ,

Column 2 results: minimuim number of theoretical stages = 40 Rmin =8.4. How come the second distillation column is so costly due large amount of stages needed . This may be due to the relative volatility being close to 1 in the second column.

Is Pressure swing distillation a wise choice for the separation of water and ethanol

a. Explain why magnetic materials that are to be used in information storage should have either square or rectangular-shaped hysteresis loops.

b. Why cannot pure iron metal be used in transformer cores?

15,000 kg / h of colloidal solution of composition 5% by weight enter a simple evaporator and must be concentrated up to 20% by weight. Heating is effected by means of live steam which condenses at 110 ° C and the condensate leaves the condensation chamber without cooling. An absolute pressure of 250 mm Hg is maintained in the evaporation chamber. The solution has no appreciable increase in boiling point, and its specific heat can be considered equal to unity for all con. centrations. The feed enters at 20 ° C. The integral coefficient of heat transmission for the operating conditions is 1,800 Kcal / m2.h. ° C. The steam from the solution is condensed in a direct contact condenser with water that enters the condenser at 20 ° C and exits at 50 ° C. Calculate: a) The heating surface b) the hourly consumption of live steam c) hourly water consumption for condesacion

Hydrogen chloride is made by the action of sulfuric acid on sodium chloride. The hydrogen chloride, being readily soluble in water, will form a solution of hydrochloric acid. Calculate the weight in grams of hydrogen chloride formed by the action of excess sulfuric acid on 1 kg of salt which is 99.5% pure NaCl.

a. 620.5 b. 320.1 c. 380.2 d.120.5

A sample of pure oxalic acid (H₂C₂O₄.2H₂O) weighs 0.200 g and requires 30.12 mL of KOH solution for complete neutralization. What is the normality of the KOH solution?

a. 0.2106 N

b. 0.0567 N

c. 0.2108 N

d. 0.1053 N

What are the differences between the mechanism of free radical and coordination polymerization?

141m3 / min of hydrogen measured at normal conditions and nitrogen in 300% excess enter an isobaric reactor. Inlet gases at 460 ° C are reacted in the presence of a catalyst to synthesize ammonia. The process has an efficiency of 26%. The exhaust gases are cooled to -40 ° C to condense the ammonia. Assume that the Cp of the gas phase species can be used in the temperature range, and that Cp NH3 (l) = 0.071938 kJ / mol K. Calculate the overall heat flow of the system in kW.