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Ket [755]
3 years ago
6

A distillation column with a partial reboiler and a total condenser is being used to separate a mixture of benzene, toluene, and

cumene. The feed is 40 mol% benzene, 30 mol% toluene and 30 mol% cumene. The feed is input as a saturated vapor. We desire 99% recovery of the toluene in the bottoms and 98% recovery of the benzene in the distillate. The reflux is returned as a saturated liquid, and CMO can be assumed. Equilibrium can be represented as constant relative volatilities. Choosing toluene as the reference component, ?benzene-toluene = 2.25 and ?cumene-toluene = 0.210. Use the Fenske equation to find the number of equilibrium stages required at total reflux and the recovery fraction of cumene in the bottoms. b. For the distillation problem given in part a, find the minimum reflux ratio by use of the Underwood equations. Use a basis of 100 moles of feed/h. Clearly state your assumptions. d. For L/D = 1.25(L/D)min , find the total number of equilibrium stages required for the distillation problem presented in parts a and b. Use the Gilliland correlation. Estimate the optimum feed plate location.

Engineering
1 answer:
marshall27 [118]3 years ago
7 0

Answer:

See attached pictures.

Explanation:

Hello,

In this case, considering the given information and that the feed is a saturated liquid, the solution is shown on the attached pictures with the proper information and procedure.

Best regards.

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A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

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Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l}  }

Th=20+273=293 K

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COP_{reversible}=\frac{293}{293-283}=29.3

4 0
3 years ago
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4 0
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Igoryamba

Answer:

true

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