INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
8th Edition
ISBN: 9781260940961
Author: SMITH
Publisher: INTER MCG
Question
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Chapter 3, Problem 3.26P
Interpretation Introduction

Interpretation:

It is to derive an equation for final temperature of gas remaining in the tank contains gas at some given temperatureinitially after tank has been bled from an initial pressure to final temperature at given initial and final pressure P1 and P2 .

Concept Introduction:

Since temperature is involved in this problem so energy balance is required. Since assumption is that gas is ideal with constant properties, so energy balance shall be

  Q=n'H'+d(nU)dt

Where n' and H' are the molar rate and enthalpy of exit streams so energy balance has positive sign.

This is a throttling process since stream bled from the tank is throttled. So, ΔH=0H=constant

Or H1=H'.....(a)

In terms of moles or material balance is

  dndt+Δn=0

Where dndt is the rate of change of moles of air inside the room and Δn is difference between exit and entrance of number of moles of air.

For steady state dndt that implies Δn=0

Or dn1=dn'.....(b)

To solve the problem, it is needed to find a relation which relates heat transfer of the tank to quantity of air supplies to it and its temperaturechange. Hence,

  n1=P1VtRT

And

  n2=P2VtRT

Where Vt is the total volume of the tank, n1 and n2 are the number of moles at different pressure and T is constant temperature.

So, the quantity of air supplies to tank is

  n'=n2n1=(P2P1)VtRT

For non adiabatic, no shaft work steady state neglecting kinetic and potential terms, the resultant energy balance shall be

  Q=n'H'+d(nU)dt....(1)

Multiplying by dt and integrating with respect to time implies

  Q=n2U2n1U1n'H'Q=n2(U2H')n1(U1H')Q=n2(H2RTH')n1(H1RTH')Q=n2(CP(TT')RT)n1(CP(TT')RT)

Hence

  Q=n'(CP(TT')RT)....(2)

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Chapter 3 Solutions

INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<

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