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Solutions, Solutions Manual For Introduction to Chemical Engineering Thermodynamics, 9th Edition, Smith, Ness, Abbott, Swihart

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Chemical Engineering Thermodynamics

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Chemical Engineering Thermodynamics

Introduction to Chemical Engineering Thermodynamics 9th Edition Solutions / Introduction to Chemical Engineering Thermodynamics Ninth Edition Solutions Manual. J.M. Smith, Hendrick Van Ness, Michael Abbott, Mark Swihart, 9781260721478 Solutions, Solutions Manual Smith 2e For Engineering Thermodyna...

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smith 9e solutions manual
jm smith
hendrick van ness
introd
introduction to chemical engineering thermodynamic
introduction to chemical engineering 9e solutions
smith 2e for engineering thermodynamics

Institution Chemical Engineering Thermodynamics
Course Chemical Engineering Thermodynamics

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Introduction to Chemical Engineering Thermodynamics 9e
Chapter No. 01: Introduction

Solution 1.1

Problem Statement

Electric current is the fundamental SI electrical dimension, with the ampere (A) as its unit.
Determine units for the following quantities as combinations of fundamental SI units.

(a) Electric power

(b) Electric charge

(c) Electric potential difference

(d) Electric resistance

(e) Electric capacitance

Solution

(a) Power is power, whether it is electrical, mechanical, or otherwise. Thus, electric power has
the usual units of power:

energy J N m kg m 2
power = = = =
time s s s3

(b) Electric current is by definition the time rate of transfer of electrical charge. Thus

charge
current =
time

charge = current*time = A s
or
(you probably recall that the Coulomb is the usual derived unit of charge, defined as 1 A s)

,
,(c) Because power is given by the product of current and electric potential,

energy
power = = current*electric potential
time

energy kg m 2
electrical potential = =
or current*time A s3
(you probably recall that this is defined as the volt)

(d) Because (by Ohm’s Law) current is electric potential divided by resistance,

electrical potential kg m 2
resistance = = 2 3
or current A s

(this is defined as the ohm)

(e) Because electric potential is electric charge divided by electric capacitance,

charge
electrical potential =
electrical capacitance

charge As A2 s4
electrical capacitance = = 2
=
electrical potential kg m kg m 2
or A s3

Solution 1.2

Problem Statement

Liquid/vapor saturation pressure Psat is often represented as a function of temperature by the
Antoine equation, which can be written in the form:

b
log10 Psat / (tor r ) = a -
t / °C + c

, Here, parameters a, b, and c are substance-specific constants. Suppose this equation is to be
rewritten in the equivalent form:

B
ln Psat / kPa = A -
T/ K+ C

Show how the parameters in the two equations are related.

Solution

We must convert both the units and the logarithm (between base 10 and natural logarithm). We
know that t in degrees Celsius is equal to T in Kelvins minus 273.15. Also 1 kPa is equal to
7.50 torr (we might have to look up this conversion factor). So, we have
æ ö
çça- b ÷
çç t / °C + c ÷
÷ æ ö
÷
= 7.5• P sat / kPa = 7.5exp çç A -
÷ B
P sat / torr = 10è ø
÷
÷
çè T / K + Cø

Next, we might recognize that 10 can be written as exp(ln(10)) or exp(2.303). That is how we
convert from base 10 log to natural log in general. So,
æ æ ö
ö÷ æ ö
exp ççç2.303çça - ÷÷= 7.5exp çç A - ÷
b B
÷÷
÷ ÷
÷
çè çè T / K - 273.15 + c ø÷
ø ç
è T / K + C ø

Here, I have also substituted T − 273.15 for t. Taking the natural log of both sides gives
æ ö
÷
2.303çça -
b B
÷
÷ = ln (7.5) + A -
èç T / K - 273.15 + cø T / K+ C

For the two functions to be equal for all values of T, each part of the functions must be the
same, so we must have

A = 2.303a −ln(7.5) or A = ln(10)*a − ln(7.5)

B = 2.303b or B = ln(10)*b

C = c − 273.15

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Solutions, Solutions Manual For Introduction to Chemical Engineering Thermodynamics, 9th Edition, Smith, Ness, Abbott, Swihart

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