ahol olten used as a uel additive, but to Tunction in most engines, the ethanol

Question

ahol olten used as a uel additive, but to Tunction in most engines, the ethanol must verypure a Map Ontain little water. It is difficult to distill ethanol and water above 90 mole ethanol because at this concentration the vapor and the liquid phases of this mixture have the sa me composition. To get around this effect, benzene is often added to the mixture. The distillation column then separates the mixture into a very nearly pure ethanol stream, and a mixture of benzene and water. The benzene may be separated from the water, and then recycled to use again in the ethanol-water distillation. A company wishes to distill a 0.860 mole fraction ethanol (0.140 mole fraction water) solution. To aid in the distillation process, a 0.950 mole fraction benzene (0.050 mole fraction water) mixture containing fresh benzene and a recovered benzene solution is added. The plant wishes to produce 77.0 moles/hr of a 0.999 mole fraction ethanol (0.001 mole fraction benzene) mixture from a distillation column. The other product from the distillation is a water and benzene mixture. This mixture is sent to a separator which produces a 7.90 mol/hr stream of water, and a 0.9200 mole fraction benzene (0.0800 mole fraction water) mixture, some of which is recycled back to the fresh benzene stream. Label the process flow diagram below. Label any quantites as "unknown" if you have to perform a calculation to obtain the numeric value. DO NOT LEAVE ANY ANSWER SPACE BLANK. "et" is short for ethanol, "b" s short for benzene, and "w" is short for water mol/hr mol b /mol mol w /mol mol bhr mol mol/hr hr rmoo mol b mol mol /mol mol/hr /mo b/mo mol mol mol w /mol mol b w/mol /mol Imo eparato stillatio mol/hr olumn mol w /mol mol w/hr mol et/mol mnl now/mn

Explanation / Answer

Ethanol to be produced = 77 moles/hr. Mole fraction ethanol =0.99. Molar flow rate of ethanol= 77*0.99=76.23 moles/hr.

This corresponds to 0.86 mole fraction in the feed. Moles of feed to be sent to the distillation column = 76.23/0.86=88.64 moles/hr

Water in the feed = 88.64*(1-0.86)= 12.4 moles/hr

Water is removed from separator from bottom as pure water and top containing 0.08 mole fraction water.

Hence 12.4= 7.3+x*0.08, x = molar flow rate of water and benzene mixture from top of separator

Hence x= 63.75 moles/hr. Benzene in the product = 63.75*0.92=58.65 moles/hr

This is the only withdrawal of benzene from the entire system. Hence Fresh benzene = 58.65 moles/hr

Let R= Recycle flow rate

Writing Benzene balance across the column

R*0.92+ 58.65= (R+58.65)*0.95

R*(0.95-0.92)= 58.65*(1-0.95), R*0.03= 2.9325, R= 97.75 moles/hr

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