Elementary Principles of Chemical Processes, Binder Ready Version
Elementary Principles of Chemical Processes, Binder Ready Version
4th Edition
ISBN: 9781118431221
Author: Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher: WILEY
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Chapter 3, Problem 3.40P
Interpretation Introduction

(a)

Interpretation:

The flowrate of the gas stream leaving the condenser should be calculated

Concept introduction:

It is given that a gas stream contains 18 mole% hexane (rest is nitrogen), this stream is passed through a condenser where some of the hexane is liquefied. The stream properties are given as,

The hexane mole fraction of gas stream leaving the condenser is 0.05 while the liquid condensate is recovered at 1.5 L/min.

Material balance at steady state is,

Material into the system = material out of the system

Interpretation Introduction

(b)

Interpretation:

The percentage of hexane recovered as the liquid should be calculated.

Concept introduction:

It is given that a gas stream contains 18 mole% hexane (rest is nitrogen), this stream is passed through a condenser where some of the hexane is liquefied. The stream properties are given as,

The hexane mole fraction of gas stream leaving the condenser is 0.05 while the liquid condensate is recovered at 1.5 L/min.

Material balance at steady state is,

Material into the system = material out of the system

Interpretation Introduction

(c)

Interpretation:

A process improvement to increase the recovery of hexane should be suggested.

Concept introduction:

It is given that a gas stream contains 18 mole% hexane (rest is nitrogen), this stream is passed through a condenser where some of the hexane is liquefied. The stream properties are given as,

The hexane mole fraction of gas stream leaving the condenser is 0.05 while the liquid condensate is recovered at 1.5 L/min.

Material balance at steady state is,

Material into the system = material out of the system

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(11.35. For a binary gas mixture described by Eqs. (3.37) and (11.58), prove that: 4812 Pу132 ✓ GE = 812 Py1 y2. ✓ SE dT HE-12 T L = = (812 - 7 1/8/123) d² 812 Pylyz C=-T Pylyz dT dT² See also Eq. (11.84), and note that 812 = 2B12 B11 - B22. perimental values of HE for binary liquid mixtures of
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please, provide me the solution with details.

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Elementary Principles of Chemical Processes, Binder Ready Version

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