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Fundamentals of
Bioengineering 2
CONVERSIO N OF ENERGY (SECOND PARD
ASST. PROF. BETOL GORONLO
4.5 Calculation of Enthalpy in Nonreactive
Processes
•A change in enthalpy can occur as a result of a
temperature change, a pressure change, a phase
change, mixing, or a reaction.
•The conservation of energy equation formulated with
enthalpy terms is better suited for solving problems
involving open systems commonly found in
biomedical applications; therefore, the focus in this
section is exclusively on enthalpy.
4.5.l Enthalpy as a State Function
•A state function or property is an intensive
property that depends only on the current state
properties include temperature, pressure,
composition, specific enthalpy, and specific
volume. Heat and work are not state functions;
they are path functions, because they depend on
the path or method used to deliver the
energy.
of the existence of the system and not on the
path taken to reach it. Examples of state
4.5. l E nthalpy as a State Function
• Consider the state property of temperature (Figure
4.6). If you begin with a system of water at l5°C, heat it to
85°C, and then cool it to 60°C (path 1). the temperature of
the system is the same as if you begin with the exact
same system of water at 1 5°C and heat it directly to 60°C
(path 2) .
• In other w ords, the temperature of the system does
not dep end on the path taken to warm the water
from l5°C to 60°C; it measures 60°C at the final
condition for both scenarios.
• On the other hand, the amount of heat required for
the two paths does differ. More heat is re quired to
••
gu..4.
6 warm the water from l5°C to 85°C (path 1) than from l5°C
to 60°C (path 2) .
•Since all the heat is not recovered when cooling the
water from 85°C to 60°C (path 1), the amount of heat for
path 1 is greater than for path 2. Hence, heat is
considered a path function.
Two r
p:uM w tbt bc.1r
icg, of
w.........c ""'" "°"''S'C•
•W
'C
.
PJ.t l l - -
4.5.l Enthalpy as a State Function
• In order to apply the conservation of total energy equation, you
often need to calculate the values of internal energy or enthalpy.
Internal energy captures molecular motion and potential energy from
forces within and between molecules, none of which can be accurately
measured. Enthalpy is a function of internal energy and depends on the
same immeasurable processes. For these reasons,
the absolute values of internal energy and enthalpy can never be known.
However, specific internal energy, 0 , and specific enthalpy, R , are
both state properties. Like all state functions, {} and Fl
depend on the state of the system, specifically its temperature, phase
(gas, liquid. solid, or crystal). and pressure. The fact that U and fl
are state functions has an important consequence for the
application of the total energy conservation equation. Specifically, the
difference in internal energy or enthalpy between any two
states can be calculated.
• For example, recall the d ifferentia l eq uation :
. JE
J
F
,,c.., w nonRow = dt
[4.5-1]
•The change in the rate of enthalpy (6f/) is defined as:
a i-I = - 2:7 iif t . + 2:m,H;
' ;
[4 .5 -2 ]
• The value of 6H is the total enthalpy rate of the output minus the total
enthalpy rate of the input. For calculations involving the qifferential form of the total energy
conservation equation, the definition for t.H in equation [4.5-2) is criticalbecause it
offers a method to calculate the change in specific enthalpy as the difference between the inlets
and outlets of the system. Sometimes, specific enthalpies are given on a permole basis. In this
case, the change in the rate of specific enthalpy across a system is:
[4.5-3]
L T + - Laj + + LQ
+
' i
• For a system with a single input,a single output, and no accumulat ion, the
mass conservation equation (equation (3.3- 1OJ) reduces to:
.
1n, t1
"
1· = 0
I (4 .5-4]
• Thus,we can reduce equation [4.5-2] for a system with a single input and
single output:
- 111
I
A, + ,;,,tf,= ll i-1
ri1,(H, - H,) = 1i1,t:.H = t:.i-1
[4.5-5)
{4.5-6]
• where specific enthalpy is given on a per-mass basis. The change in the
rate of enthalpy can also be written:
n H- f/.) = nI·AH = 6.H
1
j  I [4.5-7]
• where specific enthalpy is given on a per-mole basis. This same discussion can
be applied to the algebraic equation [4.3-17]. The change in the enthalpy of a
system, tiH,is defined as:
A
H= - "
2
:
,1
n
,H
1 + "
2
:
,1
n
;
J/;
i I
[4.5-8J
• where specific enthalpy is given on a per-mass basis. For calculations
involving the algebraic form of the total energy conservation equation, the definition
for t.H in equation [4.5-8] is critical because it offers a method to calculate the
change in specific enthalpy as the difference between the inputs and outputs of
the system.
• Since Fi is a state function, the path to convert a system from one state to another
can be accomplished using the most convenient pathway or transition. For
example, consider the specific enthalpy change from state A to state B:
6fl
stare A - - + srare B [4.5-9]
• This process requires the selection of a reference state-an arbitrarily chosen
phase, temperature,and pressure that is usually assigned a specific enthalpy of
zero. The specific enthalpy at state A is FIA- Href • where Href is an arbitrarily
assigned reference value. The specific enthalpy at state B is
F
i8 - Href · Since Href
enthalpy, t.Fl, is:
is the same for both states, the change in specific
[4.5-10)
I
Since cf dropped out of the equation, its absolute value is irrelevant. To calculate enthalpy
changes, you often need to develop a hypothetical path that transitions
changes do not depend on the path, a series of hypothetical steps from one condition to
another that are convenien t for calculation can be constructed . Inother words, the
hypothetical path does not have to accurately reflect how the process actually occurs in a
physical sense beca use enthalpy is a state function.
Best practice is to have one change along each step of the system. Usually, each step of the
path in a nonreacting system has a change in one of the following: tem
perature, pressure, or phase. The change in the specific enthalpy across a system (e.g.,
state A state B) is the sum of all the steps in the hypothericaJ path:
[4.5-11]
where k is the number of steps in the hypothetical path. Specific enthalpies of phase changes,
mixing, and reactions are often calcula ted at certain tempera tures and pres sure such as room
temperature (25°C) and pressure (1 atm). Thus, when calculating the enthalpy change of a
system, it is good practice to follow a hypothetical path that uses these certain temperatures
and pressures.
To i ll u strate how to constru ct a hy pothetica l pa th to ca lcula te the entha l py change, consider the creation
of a diamond from graphite in an industrial setting. Dia mon ds a ire na tu ra lly formed u nd er pressure i nside the
Ea rth over mi ll ions of years; those found near the Earth's surfa ce are care and valua ble. Foe industrial pu
rposes, such as cutting roots, lower-qu n lity na tu ra l d ia monds ca n be replaced wich synthetic ones created in a
much shorter period of time under extren'lely high pressu re a nd tcmpcra ru re (a pproxi ma rely 1 million psi ::ind
1800°C). To calcu la te
1
1
fl,
overall path
the changeinenthalpy for the industrial
reaction, a hypothetical path needs to be
constructed that allows for the use of the
tabulated enthalpy phase change value at I
million psi and 18000C ffigure 4.7). One
path could i11clude the following
Graphite Diamond
25 c.14.7 E
J
S
i 25 c,1
4.7 psi
+ .
i easeT f
i
v
estep
s:(a
)stepI, H
h
istheb
e
a
r
i
n
go
fg
r
ap
hiteto800C
;!b
)ste
p2
16H
1
,
i r
e
p
resentsthei
n
c
r
e
a
s
ei
np
re
s
s
u
r
e
;(c
)step3
,A
H
1
,sh
o
w
sth
ep
h
a
s
ec
h
a
n
g
e&
o
m
? 1
4
.7
psi graphitetod
i
a
m
o
n
datrnoo
°
C
a
n
d1m
i
l
lio
np
si;(d)step4
,6 ,isth
ed
e
c
re
a
s
e
t . inp
re
s
s
u
r
e
;a
n
d(e
)s
r
e
p5,tiH
5, isthec
o
o
l
i
n
go
fthediam
o
c
db
a
c
ktothestart·
t llH4
: 0ecre:1
se r
I
i
n
gco
n
d
i
t
i
o
n
s.N
o
tethate
a
c
hsteph
a
sach
a
n
g
ei
no
n
l
yo
n
eo
fthef
o
l
l
o
w
i
n
g
:
: t
e
m
i
J
e
r
a
r
u
re,pressure,orp
h
a
s
e.
I
• I
M /11
Increase Ti
t
Graphite
1800'(', 14.7
1P5i t
. I
ll//2 I
Increase P 1
t
I
'
Gmphilc - - - - - - - - - Diamond
l!IOO"C,I million psi llHJ
Phase
change
ISOO"C, l million psi
Figure 4.7
Hypothetica l path for rhc
enthalpy change
assodatcd with the
production of
industrial diamond.
- - - - - . . llypolhctical path
I
M lI
...
".
"
, 'Idnurogcn
Nilr"S"• gar
2911K Overall p•Hh 77 K
'
.
,
'
11
11
,'" '
,JI(
/ ,
f;/
Nilr<>sen gas
77 K
- - - • Hypolhcllcal
path
Figure 4.8
Hypochet:ical path for the
encbalpy change associated
with cbe cooling of
nitrogen gas to its liquid
scare.
EXAMPLE 4 .6 Cooling of Liquid Nitrogen
Problem: Liquid ofrrogen is used in a number of mcdjca [ appHcations, such as
removing rumors during surgery. Suppose gaseous Ni ar room c:emperature (298 K) is cooled ro liquid
N i jusr below irs boiling poin r. N i Uquefies or 77 K. The specific enthalpy change ro cool
nitrogen from 298 K co n K is - 1435 cal/mol. The heat of vaporiuition, the specific enthalpy
change from the liquid to the vapor form, of nitrogen is 1336 caVmol. (Hear of vaporization
is discussed further in Section 4.5.4.) Whar would be the overall change in specific entha lpy
for this process?
Solutio n : Remember that enthalpy is not dependent on path, so we can break up the pro
cess into steps. W e can ser up a hyporJ1etical process involving rwo steps, since ir involves
nvo changes: tempera ture and phase (Figure 4.8). We can ser up step 1 as rhe cooUng of the
nitrogen from 298 K ro 77 K and srep 2 as cbe liquefaction (vapor phase to liqwd phase) of
A
nirrogen. The entha lpy change for the overall process, t:.f-1, is the sum o f the two steps:
ti.it = ti.it• = ti.F
t, + ti.A2
The specific entha lpy change for step 1 when rucrogen is cooling is t:.H1 = - 1435 cal/mol.
Liquefaction is the reverse of vaporizarion, so the specific enthalpy for step 2 is:
A A ca l
t:.Hi = - t:.Hv = - 13 36 -
mol
Theo
v
erallchangeinspecificenthalp
yis:
.. .. .. cal cal cal
A
H= dH
1+A
H
2= - 1
4
3
5- - 1
336- = -277 0 -
m
o
l n
1
ol m
o
l
Note that the tvo enthalpy terms are approxin1ately the
san1e order of n1agnitude. As Ve discuss later,
contributions fron1 phase changes tend to be larger than
contributions from ten1perature changes unless the
ten1perature change is significa nt (over 200 K in this
example). The negative sign for the enthalpy change mea
ns that energy n1ust be ren1oved fron1the systen1
inordertocoolan
dliquefynitrogen. 1
4.5.2 Change inTemperature
In a system, the energy transferred to raise or lov.ier the temperature of a
material is sometimes called sensible bea t. This energy transfer is equal to
the enthalpy change that occurs when a change in temperature occurs. For a
system with a temperature change, the rate at which sensible heat is added
or removed is equal to the rate of enthalpy change.
Consider an open, steady-state system with no changes in potentia l or
kinetic energy and no nonflow work. This system can be mathematica lly
described with the following reduced differential and algebraic accounting
equa tions:
(4.5-12)
H;- + Q = O
I I
(4.5-13)
Inthiscase,thesensibleheatisequ
altotheenthalpydifferencebetv
eenoutletand
inlet conditions caused by an increase or decrease in
ten1perature. Using the defini- tions of LiH and Li H
from equations [4.5-2] and [4.5-9],the sensi ble heat
is:
•
[4.5-14)
[4.5-15]
For a systen1in which the rnateria l has a temperature
change, the specific entha lpy
•
difference is quantified in Li H or tl H. Thus, in the
special case where there is no
potentia l or kinetic energy changesand no changes in
nonf!ov work, sensible heat is
I
The specific enthalpy of a substance
dependsstrongly on temperature.Figure4.9 shows a
hypothetical plot of specific entha lpy as a function
of temperature for a system under constant
pressure. In mathema tical terms,a temperature
change,AT, leads to a change in specific enthalpy,
AA. As AT goes to zero,the ratio AA/AT
approaches the slope of the curve,which is the heat
capacity:
c - lim
AA
p(J1 -
AT...O AT
1
4
.5-16]
F l g - . . r e 4 - . 9
l cni, i.• L•c- rc:l a•C:;;J> ni.:.:l,,p
bccw-c-cn spc.ci£ic c.:nrb.oJ._py
n n c .1 t c
:
"
"m p c - rut_ '-• r C".. - 1 l"'tC"
-.f c:-.p c <-.r ..-1,c 11n c , .., r-h c::
b c o c c n p a c _i_ry .a.c
c o _ o . s c n ..o.c
p r c .q'-' r e .
where CP is the heat capacity at constant pressure. Note tha t specific enthalpy increases as
temperatu re increases in a nonlinea r fashion; therefore, Cp is given as a function of temperature
and is represented by Cp(7). Heat ca pacity is typically given in u nits of ca l/( mol .°C), J/( mol
.°C), or j/(g .°C).
Equation [4.5-16J can be written i n its integra l form as:
(4.5- J 7]
where 7i is the first temperature and 12 is the second temperature at constan t pres sure. Given
equa tion [4.5-15), the integral of the heat capacity across a tempera tu re range is equ al to tbe
sensible beat required to warm or cool a material.
C11
(n = a + bT + cT2 -1- dT 3 [4.5-18)
I
- - -
Heat capacities for most substances vary with temperature. This means tha t
when calcula ting the enthalpy change due to changes in temperature, the c,, at that
particular temperature must be evaluated. Heat capacities are physical properties that are
often tabulated as polynomial functions of temperature, such as:
The coefficien ts a, b, c,an d d used to calculated the C11 for several gases and water at 1atm
are given in Table 4.1.For this table, CP is calculated using equation [4.5-18) in units of
J/(mol ·°C), and temperature, T,is in units of degrees Celsius.Heat capacities for other
materials are given in Appendices E.1-E.3, E.7, and E.8.
For solids and liquids in most biological systems, heat capacities are not a func
tion of temperature. Therefore, the heat capaci ties for solids and liquids can usually be
approxima ted with just the first term of equa tion [4.5-18):
Cp = a
Since CPis constant, equation (4.5-17] is integrated as follows:
[4.5-19]
tlH = C11
(12 - 7i) [4.5-20)
For example,tbebeatcapacity for liquid water (75.4j/(moJ·°C), or 1 cal/(g ·°C))
is not a function of temperature in tbe range of 0-100°C. Tbe first term, a, of
equa tion [4.5-18] is predominant in gases, as well as in the temperature
range of most bioJogicaJ systems, as shown in Example 4.7.
TABLE 4.1
Coeff icient Heat Capacity Values for Severa l Gases and Water at 1 atm•t
Temperature
Species State a b x 102 d x 109
c x 1<
>
5 range
"Hear capacity is in un i ts of j/mol ·°C, and temperatures m ust be in Celsius. The values of b, c,and d
are very small and have been scaled by multiple orders of magnin1de in this ta ble (ind icated in rhe
col umn headers). Example 4.7 illustrates the proper way ro use these va lues. tExcerpred from Appendix E.1.
Air gas 28.94 0.41
47 0.31
91 -1
.965 0-1
500°C
Carbon dioxide gas 36.1
1 4.233 -2.887 7.
464 0-1
500°c
Hydrogen gas 28.84 0.
00765 0.3288 -0.8698 0-1
soo0
c
Nirrogen gas 29.00 0.21
99 0.5723 -2.871 0-1
500°c
Oxygen gas 29.1
0 1
.1
58 -0'.6076 1
.311 0-1
500°c
Water vapor 33.46 0.688 0.7604 -3.593 o-1
500°c
Water liquid 75.4 0-1
00 °C
The heat ca pacity at constant volume, Cv(1), is given in an analogous eq uation to [4.5-16)
as:
[4.5-2]]
Internal energy changes that result only from changes in temperarure can becalculated as:
h
T2
6.D = C.,(1 )dT
T1
[4.5-22)
wbere T1 is tbe fust temperature and T2 is tbe second temperature at constant vol ume. For
tiquids and solids, C.,, is approxima tely equal to CP" For gases:
[4.5-23)
where R is the ideal gas constant.
Since values of heat capaci6es are tabulated on a per-mass and per-mole basis,
6.H and 6.0 may have units of energy per mass or energy per mole. To ca lculate the
absolute change of enthalpy of the system for this step, either amount of mass or moles may
be used:
6.H = m6.H
6.H = n6.H
[4.5-? 4 )
[4.5-25]
I
Therateofchangeofenthalpyofthesystemforthisstepmaybecalculatedsimilarly,
usingm
assflowrateorm
olar£lov
1rate:
[4.5-26]
[4.5-27]
. "'
6H = i16H
For some liquids and gases, such as liquid water in equilibri
um with saturated steam, charts have been prepa red that give
the specificenthalpy as a function of temperature and
pressure. Tabula ted values for sanl!ated stearn are given in
Appendices E.5 and
EXAMPLE 4.7 Warming of A ir During Inhalation
Problem : The air you breathe is immediately warmed up from che ambient temperature to
37°C before entering your lungs.Calcu late the specific enthalpy change when the
tcn1peraturc of air is ra ised from 20°1
C to 37°C. Assun1e that the air is bone-dry, which is
defined as an environment in vhich no water is vapori zed in the a ir or in vh ich the
humidity is Oo/o.
Solution: Figure 4.10 shows a path to raise the tempera ture of air from 20°C to 37°C. The
coefficients to calcu late Cp as a function of temperature are found in Table 4.1; remember
to account for the scaJing of coefficients b, c,and d. To caJcu late the change in speci fic
enthalpy for warming the air, Ve can use equation [4.5-17]:
fl.Fl= rTiCp(T) dr
JT
,
37"C
= j{ (28.94 + 0.4147 X 10 2 T + 0.3191 X lO 5 T2 - 1.965 X 10 9 T3)dT oc
20"C n10 ·
J
n10
1
J
mo
1
+ 0.045 J I - 0.00084 J
mo
1
= 491.98 + 2.01
mo
J
= 494-
mol
I
Figure 4.10
Varming of
dry air in the
lungs.
•
A
ira
t2
5
°C- - -
H
-
1
1
- A
ira
t3
7
°C
O
v
erailp
a
th
The first term in the heat capacify equation is
predominant,which implies that Cp has very little
temperatu re dependence within the specified
range.The specific enthalpy change to raise the
temperature of the air from 20°C to 37°C is 494
j/mol.
During one breath, the number of moles of gas is
constant,while the temperature of the air
increases.Therefo re, by the ideal gas law, the
pressure of air must have increased slightly,
d
i
s
c
u
s
s
e
dinS
e
c
t
i
o
n4.5
.3. 1
4.5.3 Change in Pressure
Changes i n enthalpy as a result of cha nges in pressure are nor as important in mosr
biological and medical systems as the other changes considered, but a discussion is
warranted for completeness . Reca ll equation [4.2-25] and consider the d ifference
between the outlet and inlet conditions:
A A A
tl H = tl U + A (PV) [4.5-28]
For solids and liq uids, it bas been observed experimentally that specific internal energy
( U) and speci fic volume (V) are nea rly independ ent of pressure. Therefore equation
[4.5-25] reduces to the following for solids and liquids:
[4.5-29J
A
ln biological systems, pressure changes are usualJy not signif ica nt, so tl H is unaf fected .
For ideal gases, specific entha lpy does not depend on pressure; therefore, you can assume
that tlH is zero when considering changes i n pre ssure. This approxima tion brea ks dow n
for ideal gases whose temperatures are below 0°C or whose pressures
are well above 1 atm, bur these situations rarely occur in biomedical calculations.
Consu l t more advanced textbooks to handle nonideal gases (e.g., Reid RC, Pra usnitz
JM, and Poling BE, The Properties of Gases and LiqH ids, 1987).
4.5.4 Change in Phase
Phase changes are accompanied by rela tively large changes in i nterna l
energy and enthalpy as noncovalent bonds between molecules,such as hydrogen
bonds, are broken and formed. Breaking and forming noncovalenr bonds req
uires and releases energy, respectively. Consider the conversion of water among
itsvapor, liquid, and solid phases. 1n the vapor phase, water molecules move
around most freely and have a high specific enthalpy. In the l iquid and solid
phases, water molecules are more densely packed and are held together by
attractive forces between the niolecules. ln the solid phase, they have little
rotation or freedom of motion. As shown in Table 4.2, the specific enthalpy of
liquid water is lower than that of saturated water vapor at 100°C.
TABLE 4.2
Specific Enthalpy of Water at 100°C and 1 atm
Phase Specific enthalpy U/g)
Saturated
Liquid
Saturated
vapor
419.1
2676
Entha lpy cannot be known in absol u te terms and is determined relative to a reference point or state
(see Section 4.5.1). Inthe case of water, the specific enthalpy is defined relative to its triple point, the
temperature and pressure at which the liq uid, vapor, and solid phases are in equilibrium (0.01°C, 0.00611
bar). The specific enthalpy of water at the triple point is arbitrarily defined to be zero. Remember that values for
specific enthalpy can be used only when calculating the difference between two conditions.
The specific enthalpy change associated with the transition of a substance from one phase
to another at constant pressure and tempera ture is sometimes known as the latent heat of the
phase change. As with sensi ble beat, the term heat is used to descri be an enthalpy change.
A derivation simjlar to equation [4.5-14] shows that latent heat is equal to the enthalpy
associated with the phase change under specified conditions. Transitions between liquid and
vapor, solid and liquid, and solid and vapor are su m marized in Table 4.3.
A
The latent heat of vaporization, t:..Hv, is the specific enthalpy difference between the liquid and
vapor forms of a species at a given temperature and pressure. It describes the specific
enthalpy change for the process of evaporation. Evaporation requires the input of energy
(e.g., boiling a pot of water). Since condensation is the reverse of vaporiza tion and enthalpy is
a state property, the latent bea t of condensa tion is the negative of the latent bear of vaporiza
tion (-t:..Hv).Thus, the condensation of a gas ro its liquid phase requires removal of energy.
The latent heat of melting or fusion, 6.H
M
, is the specific enthalpy difference
between the solid and liquid forms of a species at a given temperature and
pressure. Melting requires the input of energy (e.g., melting an ice cube). Since
freezing is the reverse of melting and enthalpy is a stat...e. proper ty, the la tent
bear of freezing is the negative of the latent heat of melting {AHM). The
freezing of a liquid to its solid phase requires removal of energy.
The latent heat of sublima tion, fiH"s, is the specific enthalpy difference
between tbe solid and vapor forms of a species at a given tempera ture and
pressure. Sublima
tion requires the input of energy [e.g., sublimation of a block of solid carbon
dioxide (i.e., dry ice) to gaseous carbon dioxide]. Since deposition is the revers,e
of sublima
tion, the latent beat of deposition is the negative of the latent heat of sublimation
....
(-tiHs). Enthalpy changes associated with phase changes between solids and vapors
and between different solid phases are not considered in this textbook..
While several terms such as sensible heat, latent heat, and heat of vaporization
include the word heat, usng heat in 1these phrases ca n be confusing. These
"heats" describe a change in enthalpy not necessarily a change in !heat. For
example, con sider the heat of vaporization. Energy in the form of hea t can be
added to a syste m to change the phase from liq uid to a gas. This is strictly
the entha lpy change of
TABLE4.3
PhaseChangeProcessesandTransitions
Processname Specificenthalpychange
t
Initialphase Finalphase
Vaporization or boiling
Condensation or
liquefaction Melting or
fusion
Freezing Sublimation
Deposition
ilH
"v
A
Liqui
d
Vapo
r
Solid
Liqui
d
Solid
Vapo
r
V
apor
Liquid
- ilHv
..
il
H
M
A
-tl
H
M
il
H"
s
A
L
iquid
S
olid
V
apor
S
olid
- flH
s
•Enthalpychangesaredefined intext.
vaporization;altnougn some engineers use tne
term neat of vaporization. WorK can also oe
aooeo to change pnase. In tnis case, tne term
heat of vaporization ooes not
maKe sense, wnereas entnalpy cnange of
vaporization aoesl For tnis reason, tnis
textDOOK pically uses tne term ''entnalpy
cnange of Iii I,, ratner tnan 'heat of . ,, unless neat is
explicitly transferrea aieross a te1nperature
graaient.
. . . . . . . .
EXA M PLE 4 . 8 Va p o riz a t io n o f W a t e r a t 3 7 ° C
P ro b l e m :
a l 3 7 ° C .
S o l u"tio n :
C a l cu la te t h e en tha lp y c h a n g e o n a per-gra m basis f o r t h e va pori za t i o n of w a te r
T h is p r o b l e m c a n b e s o l ve d in c w o w a ys. T h e firsr '
Va y in v o l ves usi n g r h e In ten t
hcac of va pori:r.nti on for wn tcr a t 37°C, wh ich from Appen dix F...5 is 241 4 k.Jlkg. ( N ote: A n in rerpolnrion between 36°C
and 38°C is req u ired.) The speci fic hea r of vn pori:r.n rion is:
fv = 2414 J ( 1 cal ) = S77 cal
g 4. 184 ] g
The second "va y req u i res you to consider n h y pothetical pa th. I f the hen r of vn pori za rion for wa ter is knowin only at
100°C (its boiling point). you can find it at 37°C by considering rh e followi ng hypothetical pa t h (Figure 4. 11 ):
6H = 6 H, + 6H,, + 6H1
where 6H1 is the specific entha lpy cha n ge to raise the tempera ture of wa ter fron> 37°C to 100°C, 6 f l2 is rbe specdic cn
tbalpy ch ange of vapor Lln r1on of wa ter at 100°C, a nd 6 H3 is dlc specific en thalpy change to lower the tem pera tu re of water
from I00°C ro 37°C.
For liqu id waler, th e h ent co paciry, Cp , is not n fun ction of tem pera ru n:n nd is a consta n t:
6 f t, - c ,,(72 ( 7 5 .4
7i ) _ J )(1m ot ) ( 1c a l )<JO
o
o
c
m o l · °C 18 g 4 . 1 8 4 J
3 7 °C ) - 63.Ical
g
·
r b c hcn t o f v a p o r i z a t ion of w n tc r a t 1 0 0 ° C is f o u n d to A p p e n d i x E .S :
6 / l i - ( 2 2 5 7 : ! .) ( 1
ca l ) - 539 cal
g
iS 4 . .18 4 J
I
Figure 4.11
Hypoiht!K:al
parh for w
mrhllpy ch3ngt
associat!d with
che vaporiunon
of w:ircr ar 37"C.
f..IJuill lH
<Mnlpllh
'ilpor
at37'C al3T
'{'
.: ..
l l /11
•Mli
flqllid:
I •
l.iq
uK!_
atIWC
:Cidl3pOI
I
- - - - -
Y
ap.W
.itw. atI
O
O
'C
Thecoefficientstocakulate c,as a function of cempuaturr
for water vapor are found in Table 4.1;remember to accouni
for chc scaling of cocffidcms b, c, and J. When cooling the
wacer vapor back down, the specific emhalpy change is:
J
T"C J?'C
&
H
1= C
,1
nJT= { (a+b
T+cT2+JT
1)JT
IO
O
'C jIfll'C
rt
= { (3
3.46+0.688X 10-?Tt 0.7604X 10-
5T1- 3.593X U
r' T
1
)1f J
J1
rt1C m
o
l
= _2140 J {1mol )( 1cal )
= _
.4c
al
g
28
;;j 1
8g 4.1
84J
Theo
v
e
r
a
Uchangeinspecificenthalpyisrhesumofthethrees
upsinthepath:
A• =A• +ll • +A• =6
3
.1
cal
H H
1 H
i H
1 - + 53
9 - - 2
8.4- =57
4-
g g g g
T
herefo
re,th
elatemh
e
atofv
aporizationac37
°C,calculate
ddtroughthehypothetic
alpath,
is5
74cal/g,w
h
ichisv
eryclosetothefirstsolution(577cal/g). 1

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week 6 (1).ppt

  • 1. Fundamentals of Bioengineering 2 CONVERSIO N OF ENERGY (SECOND PARD ASST. PROF. BETOL GORONLO
  • 2. 4.5 Calculation of Enthalpy in Nonreactive Processes •A change in enthalpy can occur as a result of a temperature change, a pressure change, a phase change, mixing, or a reaction. •The conservation of energy equation formulated with enthalpy terms is better suited for solving problems involving open systems commonly found in biomedical applications; therefore, the focus in this section is exclusively on enthalpy.
  • 3. 4.5.l Enthalpy as a State Function •A state function or property is an intensive property that depends only on the current state properties include temperature, pressure, composition, specific enthalpy, and specific volume. Heat and work are not state functions; they are path functions, because they depend on the path or method used to deliver the energy. of the existence of the system and not on the path taken to reach it. Examples of state
  • 4. 4.5. l E nthalpy as a State Function • Consider the state property of temperature (Figure 4.6). If you begin with a system of water at l5°C, heat it to 85°C, and then cool it to 60°C (path 1). the temperature of the system is the same as if you begin with the exact same system of water at 1 5°C and heat it directly to 60°C (path 2) . • In other w ords, the temperature of the system does not dep end on the path taken to warm the water from l5°C to 60°C; it measures 60°C at the final condition for both scenarios. • On the other hand, the amount of heat required for the two paths does differ. More heat is re quired to •• gu..4. 6 warm the water from l5°C to 85°C (path 1) than from l5°C to 60°C (path 2) . •Since all the heat is not recovered when cooling the water from 85°C to 60°C (path 1), the amount of heat for path 1 is greater than for path 2. Hence, heat is considered a path function. Two r p:uM w tbt bc.1r icg, of w.........c ""'" "°"''S'C• •W 'C . PJ.t l l - -
  • 5. 4.5.l Enthalpy as a State Function • In order to apply the conservation of total energy equation, you often need to calculate the values of internal energy or enthalpy. Internal energy captures molecular motion and potential energy from forces within and between molecules, none of which can be accurately measured. Enthalpy is a function of internal energy and depends on the same immeasurable processes. For these reasons, the absolute values of internal energy and enthalpy can never be known. However, specific internal energy, 0 , and specific enthalpy, R , are both state properties. Like all state functions, {} and Fl depend on the state of the system, specifically its temperature, phase (gas, liquid. solid, or crystal). and pressure. The fact that U and fl are state functions has an important consequence for the application of the total energy conservation equation. Specifically, the difference in internal energy or enthalpy between any two states can be calculated.
  • 6. • For example, recall the d ifferentia l eq uation : . JE J F ,,c.., w nonRow = dt [4.5-1] •The change in the rate of enthalpy (6f/) is defined as: a i-I = - 2:7 iif t . + 2:m,H; ' ; [4 .5 -2 ] • The value of 6H is the total enthalpy rate of the output minus the total enthalpy rate of the input. For calculations involving the qifferential form of the total energy conservation equation, the definition for t.H in equation [4.5-2) is criticalbecause it offers a method to calculate the change in specific enthalpy as the difference between the inlets and outlets of the system. Sometimes, specific enthalpies are given on a permole basis. In this case, the change in the rate of specific enthalpy across a system is: [4.5-3] L T + - Laj + + LQ + ' i
  • 7. • For a system with a single input,a single output, and no accumulat ion, the mass conservation equation (equation (3.3- 1OJ) reduces to: . 1n, t1 " 1· = 0 I (4 .5-4] • Thus,we can reduce equation [4.5-2] for a system with a single input and single output: - 111 I A, + ,;,,tf,= ll i-1 ri1,(H, - H,) = 1i1,t:.H = t:.i-1 [4.5-5) {4.5-6] • where specific enthalpy is given on a per-mass basis. The change in the rate of enthalpy can also be written: n H- f/.) = nI·AH = 6.H 1 j I [4.5-7]
  • 8. • where specific enthalpy is given on a per-mole basis. This same discussion can be applied to the algebraic equation [4.3-17]. The change in the enthalpy of a system, tiH,is defined as: A H= - " 2 : ,1 n ,H 1 + " 2 : ,1 n ; J/; i I [4.5-8J • where specific enthalpy is given on a per-mass basis. For calculations involving the algebraic form of the total energy conservation equation, the definition for t.H in equation [4.5-8] is critical because it offers a method to calculate the change in specific enthalpy as the difference between the inputs and outputs of the system.
  • 9. • Since Fi is a state function, the path to convert a system from one state to another can be accomplished using the most convenient pathway or transition. For example, consider the specific enthalpy change from state A to state B: 6fl stare A - - + srare B [4.5-9] • This process requires the selection of a reference state-an arbitrarily chosen phase, temperature,and pressure that is usually assigned a specific enthalpy of zero. The specific enthalpy at state A is FIA- Href • where Href is an arbitrarily assigned reference value. The specific enthalpy at state B is F i8 - Href · Since Href enthalpy, t.Fl, is: is the same for both states, the change in specific [4.5-10)
  • 10. I Since cf dropped out of the equation, its absolute value is irrelevant. To calculate enthalpy changes, you often need to develop a hypothetical path that transitions changes do not depend on the path, a series of hypothetical steps from one condition to another that are convenien t for calculation can be constructed . Inother words, the hypothetical path does not have to accurately reflect how the process actually occurs in a physical sense beca use enthalpy is a state function. Best practice is to have one change along each step of the system. Usually, each step of the path in a nonreacting system has a change in one of the following: tem perature, pressure, or phase. The change in the specific enthalpy across a system (e.g., state A state B) is the sum of all the steps in the hypothericaJ path: [4.5-11] where k is the number of steps in the hypothetical path. Specific enthalpies of phase changes, mixing, and reactions are often calcula ted at certain tempera tures and pres sure such as room temperature (25°C) and pressure (1 atm). Thus, when calculating the enthalpy change of a system, it is good practice to follow a hypothetical path that uses these certain temperatures and pressures.
  • 11. To i ll u strate how to constru ct a hy pothetica l pa th to ca lcula te the entha l py change, consider the creation of a diamond from graphite in an industrial setting. Dia mon ds a ire na tu ra lly formed u nd er pressure i nside the Ea rth over mi ll ions of years; those found near the Earth's surfa ce are care and valua ble. Foe industrial pu rposes, such as cutting roots, lower-qu n lity na tu ra l d ia monds ca n be replaced wich synthetic ones created in a much shorter period of time under extren'lely high pressu re a nd tcmpcra ru re (a pproxi ma rely 1 million psi ::ind 1800°C). To calcu la te 1 1 fl, overall path the changeinenthalpy for the industrial reaction, a hypothetical path needs to be constructed that allows for the use of the tabulated enthalpy phase change value at I million psi and 18000C ffigure 4.7). One path could i11clude the following Graphite Diamond 25 c.14.7 E J S i 25 c,1 4.7 psi + . i easeT f i v estep s:(a )stepI, H h istheb e a r i n go fg r ap hiteto800C ;!b )ste p2 16H 1 , i r e p resentsthei n c r e a s ei np re s s u r e ;(c )step3 ,A H 1 ,sh o w sth ep h a s ec h a n g e& o m ? 1 4 .7 psi graphitetod i a m o n datrnoo ° C a n d1m i l lio np si;(d)step4 ,6 ,isth ed e c re a s e t . inp re s s u r e ;a n d(e )s r e p5,tiH 5, isthec o o l i n go fthediam o c db a c ktothestart· t llH4 : 0ecre:1 se r I i n gco n d i t i o n s.N o tethate a c hsteph a sach a n g ei no n l yo n eo fthef o l l o w i n g : : t e m i J e r a r u re,pressure,orp h a s e. I • I M /11 Increase Ti t Graphite 1800'(', 14.7 1P5i t . I ll//2 I Increase P 1 t I ' Gmphilc - - - - - - - - - Diamond l!IOO"C,I million psi llHJ Phase change ISOO"C, l million psi Figure 4.7 Hypothetica l path for rhc enthalpy change assodatcd with the production of industrial diamond. - - - - - . . llypolhctical path
  • 12. I M lI ... ". " , 'Idnurogcn Nilr"S"• gar 2911K Overall p•Hh 77 K ' . , ' 11 11 ,'" ' ,JI( / , f;/ Nilr<>sen gas 77 K - - - • Hypolhcllcal path Figure 4.8 Hypochet:ical path for the encbalpy change associated with cbe cooling of nitrogen gas to its liquid scare. EXAMPLE 4 .6 Cooling of Liquid Nitrogen Problem: Liquid ofrrogen is used in a number of mcdjca [ appHcations, such as removing rumors during surgery. Suppose gaseous Ni ar room c:emperature (298 K) is cooled ro liquid N i jusr below irs boiling poin r. N i Uquefies or 77 K. The specific enthalpy change ro cool nitrogen from 298 K co n K is - 1435 cal/mol. The heat of vaporiuition, the specific enthalpy change from the liquid to the vapor form, of nitrogen is 1336 caVmol. (Hear of vaporization is discussed further in Section 4.5.4.) Whar would be the overall change in specific entha lpy for this process? Solutio n : Remember that enthalpy is not dependent on path, so we can break up the pro cess into steps. W e can ser up a hyporJ1etical process involving rwo steps, since ir involves nvo changes: tempera ture and phase (Figure 4.8). We can ser up step 1 as rhe cooUng of the nitrogen from 298 K ro 77 K and srep 2 as cbe liquefaction (vapor phase to liqwd phase) of A nirrogen. The entha lpy change for the overall process, t:.f-1, is the sum o f the two steps: ti.it = ti.it• = ti.F t, + ti.A2 The specific entha lpy change for step 1 when rucrogen is cooling is t:.H1 = - 1435 cal/mol. Liquefaction is the reverse of vaporizarion, so the specific enthalpy for step 2 is: A A ca l t:.Hi = - t:.Hv = - 13 36 - mol
  • 13. Theo v erallchangeinspecificenthalp yis: .. .. .. cal cal cal A H= dH 1+A H 2= - 1 4 3 5- - 1 336- = -277 0 - m o l n 1 ol m o l Note that the tvo enthalpy terms are approxin1ately the san1e order of n1agnitude. As Ve discuss later, contributions fron1 phase changes tend to be larger than contributions from ten1perature changes unless the ten1perature change is significa nt (over 200 K in this example). The negative sign for the enthalpy change mea ns that energy n1ust be ren1oved fron1the systen1 inordertocoolan dliquefynitrogen. 1
  • 14. 4.5.2 Change inTemperature In a system, the energy transferred to raise or lov.ier the temperature of a material is sometimes called sensible bea t. This energy transfer is equal to the enthalpy change that occurs when a change in temperature occurs. For a system with a temperature change, the rate at which sensible heat is added or removed is equal to the rate of enthalpy change. Consider an open, steady-state system with no changes in potentia l or kinetic energy and no nonflow work. This system can be mathematica lly described with the following reduced differential and algebraic accounting equa tions: (4.5-12) H;- + Q = O I I (4.5-13)
  • 15. Inthiscase,thesensibleheatisequ altotheenthalpydifferencebetv eenoutletand inlet conditions caused by an increase or decrease in ten1perature. Using the defini- tions of LiH and Li H from equations [4.5-2] and [4.5-9],the sensi ble heat is: • [4.5-14) [4.5-15] For a systen1in which the rnateria l has a temperature change, the specific entha lpy • difference is quantified in Li H or tl H. Thus, in the special case where there is no potentia l or kinetic energy changesand no changes in nonf!ov work, sensible heat is
  • 16. I The specific enthalpy of a substance dependsstrongly on temperature.Figure4.9 shows a hypothetical plot of specific entha lpy as a function of temperature for a system under constant pressure. In mathema tical terms,a temperature change,AT, leads to a change in specific enthalpy, AA. As AT goes to zero,the ratio AA/AT approaches the slope of the curve,which is the heat capacity: c - lim AA p(J1 - AT...O AT 1 4 .5-16] F l g - . . r e 4 - . 9 l cni, i.• L•c- rc:l a•C:;;J> ni.:.:l,,p bccw-c-cn spc.ci£ic c.:nrb.oJ._py n n c .1 t c : " "m p c - rut_ '-• r C".. - 1 l"'tC" -.f c:-.p c <-.r ..-1,c 11n c , .., r-h c:: b c o c c n p a c _i_ry .a.c c o _ o . s c n ..o.c p r c .q'-' r e . where CP is the heat capacity at constant pressure. Note tha t specific enthalpy increases as temperatu re increases in a nonlinea r fashion; therefore, Cp is given as a function of temperature and is represented by Cp(7). Heat ca pacity is typically given in u nits of ca l/( mol .°C), J/( mol .°C), or j/(g .°C). Equation [4.5-16J can be written i n its integra l form as: (4.5- J 7] where 7i is the first temperature and 12 is the second temperature at constan t pres sure. Given equa tion [4.5-15), the integral of the heat capacity across a tempera tu re range is equ al to tbe sensible beat required to warm or cool a material.
  • 17. C11 (n = a + bT + cT2 -1- dT 3 [4.5-18) I - - - Heat capacities for most substances vary with temperature. This means tha t when calcula ting the enthalpy change due to changes in temperature, the c,, at that particular temperature must be evaluated. Heat capacities are physical properties that are often tabulated as polynomial functions of temperature, such as: The coefficien ts a, b, c,an d d used to calculated the C11 for several gases and water at 1atm are given in Table 4.1.For this table, CP is calculated using equation [4.5-18) in units of J/(mol ·°C), and temperature, T,is in units of degrees Celsius.Heat capacities for other materials are given in Appendices E.1-E.3, E.7, and E.8. For solids and liquids in most biological systems, heat capacities are not a func tion of temperature. Therefore, the heat capaci ties for solids and liquids can usually be approxima ted with just the first term of equa tion [4.5-18): Cp = a Since CPis constant, equation (4.5-17] is integrated as follows: [4.5-19] tlH = C11 (12 - 7i) [4.5-20)
  • 18. For example,tbebeatcapacity for liquid water (75.4j/(moJ·°C), or 1 cal/(g ·°C)) is not a function of temperature in tbe range of 0-100°C. Tbe first term, a, of equa tion [4.5-18] is predominant in gases, as well as in the temperature range of most bioJogicaJ systems, as shown in Example 4.7. TABLE 4.1 Coeff icient Heat Capacity Values for Severa l Gases and Water at 1 atm•t Temperature Species State a b x 102 d x 109 c x 1< > 5 range "Hear capacity is in un i ts of j/mol ·°C, and temperatures m ust be in Celsius. The values of b, c,and d are very small and have been scaled by multiple orders of magnin1de in this ta ble (ind icated in rhe col umn headers). Example 4.7 illustrates the proper way ro use these va lues. tExcerpred from Appendix E.1. Air gas 28.94 0.41 47 0.31 91 -1 .965 0-1 500°C Carbon dioxide gas 36.1 1 4.233 -2.887 7. 464 0-1 500°c Hydrogen gas 28.84 0. 00765 0.3288 -0.8698 0-1 soo0 c Nirrogen gas 29.00 0.21 99 0.5723 -2.871 0-1 500°c Oxygen gas 29.1 0 1 .1 58 -0'.6076 1 .311 0-1 500°c Water vapor 33.46 0.688 0.7604 -3.593 o-1 500°c Water liquid 75.4 0-1 00 °C
  • 19. The heat ca pacity at constant volume, Cv(1), is given in an analogous eq uation to [4.5-16) as: [4.5-2]] Internal energy changes that result only from changes in temperarure can becalculated as: h T2 6.D = C.,(1 )dT T1 [4.5-22) wbere T1 is tbe fust temperature and T2 is tbe second temperature at constant vol ume. For tiquids and solids, C.,, is approxima tely equal to CP" For gases: [4.5-23) where R is the ideal gas constant. Since values of heat capaci6es are tabulated on a per-mass and per-mole basis, 6.H and 6.0 may have units of energy per mass or energy per mole. To ca lculate the absolute change of enthalpy of the system for this step, either amount of mass or moles may be used: 6.H = m6.H 6.H = n6.H [4.5-? 4 ) [4.5-25]
  • 20. I Therateofchangeofenthalpyofthesystemforthisstepmaybecalculatedsimilarly, usingm assflowrateorm olar£lov 1rate: [4.5-26] [4.5-27] . "' 6H = i16H For some liquids and gases, such as liquid water in equilibri um with saturated steam, charts have been prepa red that give the specificenthalpy as a function of temperature and pressure. Tabula ted values for sanl!ated stearn are given in Appendices E.5 and
  • 21. EXAMPLE 4.7 Warming of A ir During Inhalation Problem : The air you breathe is immediately warmed up from che ambient temperature to 37°C before entering your lungs.Calcu late the specific enthalpy change when the tcn1peraturc of air is ra ised from 20°1 C to 37°C. Assun1e that the air is bone-dry, which is defined as an environment in vhich no water is vapori zed in the a ir or in vh ich the humidity is Oo/o. Solution: Figure 4.10 shows a path to raise the tempera ture of air from 20°C to 37°C. The coefficients to calcu late Cp as a function of temperature are found in Table 4.1; remember to account for the scaJing of coefficients b, c,and d. To caJcu late the change in speci fic enthalpy for warming the air, Ve can use equation [4.5-17]: fl.Fl= rTiCp(T) dr JT , 37"C = j{ (28.94 + 0.4147 X 10 2 T + 0.3191 X lO 5 T2 - 1.965 X 10 9 T3)dT oc 20"C n10 · J n10 1 J mo 1 + 0.045 J I - 0.00084 J mo 1 = 491.98 + 2.01 mo J = 494- mol
  • 22. I Figure 4.10 Varming of dry air in the lungs. • A ira t2 5 °C- - - H - 1 1 - A ira t3 7 °C O v erailp a th The first term in the heat capacify equation is predominant,which implies that Cp has very little temperatu re dependence within the specified range.The specific enthalpy change to raise the temperature of the air from 20°C to 37°C is 494 j/mol. During one breath, the number of moles of gas is constant,while the temperature of the air increases.Therefo re, by the ideal gas law, the pressure of air must have increased slightly, d i s c u s s e dinS e c t i o n4.5 .3. 1
  • 23. 4.5.3 Change in Pressure Changes i n enthalpy as a result of cha nges in pressure are nor as important in mosr biological and medical systems as the other changes considered, but a discussion is warranted for completeness . Reca ll equation [4.2-25] and consider the d ifference between the outlet and inlet conditions: A A A tl H = tl U + A (PV) [4.5-28] For solids and liq uids, it bas been observed experimentally that specific internal energy ( U) and speci fic volume (V) are nea rly independ ent of pressure. Therefore equation [4.5-25] reduces to the following for solids and liquids: [4.5-29J A ln biological systems, pressure changes are usualJy not signif ica nt, so tl H is unaf fected . For ideal gases, specific entha lpy does not depend on pressure; therefore, you can assume that tlH is zero when considering changes i n pre ssure. This approxima tion brea ks dow n for ideal gases whose temperatures are below 0°C or whose pressures are well above 1 atm, bur these situations rarely occur in biomedical calculations. Consu l t more advanced textbooks to handle nonideal gases (e.g., Reid RC, Pra usnitz JM, and Poling BE, The Properties of Gases and LiqH ids, 1987).
  • 24. 4.5.4 Change in Phase Phase changes are accompanied by rela tively large changes in i nterna l energy and enthalpy as noncovalent bonds between molecules,such as hydrogen bonds, are broken and formed. Breaking and forming noncovalenr bonds req uires and releases energy, respectively. Consider the conversion of water among itsvapor, liquid, and solid phases. 1n the vapor phase, water molecules move around most freely and have a high specific enthalpy. In the l iquid and solid phases, water molecules are more densely packed and are held together by attractive forces between the niolecules. ln the solid phase, they have little rotation or freedom of motion. As shown in Table 4.2, the specific enthalpy of liquid water is lower than that of saturated water vapor at 100°C. TABLE 4.2 Specific Enthalpy of Water at 100°C and 1 atm Phase Specific enthalpy U/g) Saturated Liquid Saturated vapor 419.1 2676
  • 25. Entha lpy cannot be known in absol u te terms and is determined relative to a reference point or state (see Section 4.5.1). Inthe case of water, the specific enthalpy is defined relative to its triple point, the temperature and pressure at which the liq uid, vapor, and solid phases are in equilibrium (0.01°C, 0.00611 bar). The specific enthalpy of water at the triple point is arbitrarily defined to be zero. Remember that values for specific enthalpy can be used only when calculating the difference between two conditions. The specific enthalpy change associated with the transition of a substance from one phase to another at constant pressure and tempera ture is sometimes known as the latent heat of the phase change. As with sensi ble beat, the term heat is used to descri be an enthalpy change. A derivation simjlar to equation [4.5-14] shows that latent heat is equal to the enthalpy associated with the phase change under specified conditions. Transitions between liquid and vapor, solid and liquid, and solid and vapor are su m marized in Table 4.3. A The latent heat of vaporization, t:..Hv, is the specific enthalpy difference between the liquid and vapor forms of a species at a given temperature and pressure. It describes the specific enthalpy change for the process of evaporation. Evaporation requires the input of energy (e.g., boiling a pot of water). Since condensation is the reverse of vaporiza tion and enthalpy is a state property, the latent bea t of condensa tion is the negative of the latent bear of vaporiza tion (-t:..Hv).Thus, the condensation of a gas ro its liquid phase requires removal of energy.
  • 26. The latent heat of melting or fusion, 6.H M , is the specific enthalpy difference between the solid and liquid forms of a species at a given temperature and pressure. Melting requires the input of energy (e.g., melting an ice cube). Since freezing is the reverse of melting and enthalpy is a stat...e. proper ty, the la tent bear of freezing is the negative of the latent heat of melting {AHM). The freezing of a liquid to its solid phase requires removal of energy. The latent heat of sublima tion, fiH"s, is the specific enthalpy difference between tbe solid and vapor forms of a species at a given tempera ture and pressure. Sublima tion requires the input of energy [e.g., sublimation of a block of solid carbon dioxide (i.e., dry ice) to gaseous carbon dioxide]. Since deposition is the revers,e of sublima tion, the latent beat of deposition is the negative of the latent heat of sublimation .... (-tiHs). Enthalpy changes associated with phase changes between solids and vapors and between different solid phases are not considered in this textbook.. While several terms such as sensible heat, latent heat, and heat of vaporization include the word heat, usng heat in 1these phrases ca n be confusing. These "heats" describe a change in enthalpy not necessarily a change in !heat. For example, con sider the heat of vaporization. Energy in the form of hea t can be added to a syste m to change the phase from liq uid to a gas. This is strictly the entha lpy change of
  • 27. TABLE4.3 PhaseChangeProcessesandTransitions Processname Specificenthalpychange t Initialphase Finalphase Vaporization or boiling Condensation or liquefaction Melting or fusion Freezing Sublimation Deposition ilH "v A Liqui d Vapo r Solid Liqui d Solid Vapo r V apor Liquid - ilHv .. il H M A -tl H M il H" s A L iquid S olid V apor S olid - flH s •Enthalpychangesaredefined intext.
  • 28. vaporization;altnougn some engineers use tne term neat of vaporization. WorK can also oe aooeo to change pnase. In tnis case, tne term heat of vaporization ooes not maKe sense, wnereas entnalpy cnange of vaporization aoesl For tnis reason, tnis textDOOK pically uses tne term ''entnalpy cnange of Iii I,, ratner tnan 'heat of . ,, unless neat is explicitly transferrea aieross a te1nperature graaient. . . . . . . . .
  • 29. EXA M PLE 4 . 8 Va p o riz a t io n o f W a t e r a t 3 7 ° C P ro b l e m : a l 3 7 ° C . S o l u"tio n : C a l cu la te t h e en tha lp y c h a n g e o n a per-gra m basis f o r t h e va pori za t i o n of w a te r T h is p r o b l e m c a n b e s o l ve d in c w o w a ys. T h e firsr ' Va y in v o l ves usi n g r h e In ten t hcac of va pori:r.nti on for wn tcr a t 37°C, wh ich from Appen dix F...5 is 241 4 k.Jlkg. ( N ote: A n in rerpolnrion between 36°C and 38°C is req u ired.) The speci fic hea r of vn pori:r.n rion is: fv = 2414 J ( 1 cal ) = S77 cal g 4. 184 ] g The second "va y req u i res you to consider n h y pothetical pa th. I f the hen r of vn pori za rion for wa ter is knowin only at 100°C (its boiling point). you can find it at 37°C by considering rh e followi ng hypothetical pa t h (Figure 4. 11 ): 6H = 6 H, + 6H,, + 6H1 where 6H1 is the specific entha lpy cha n ge to raise the tempera ture of wa ter fron> 37°C to 100°C, 6 f l2 is rbe specdic cn tbalpy ch ange of vapor Lln r1on of wa ter at 100°C, a nd 6 H3 is dlc specific en thalpy change to lower the tem pera tu re of water from I00°C ro 37°C. For liqu id waler, th e h ent co paciry, Cp , is not n fun ction of tem pera ru n:n nd is a consta n t: 6 f t, - c ,,(72 ( 7 5 .4 7i ) _ J )(1m ot ) ( 1c a l )<JO o o c m o l · °C 18 g 4 . 1 8 4 J 3 7 °C ) - 63.Ical g · r b c hcn t o f v a p o r i z a t ion of w n tc r a t 1 0 0 ° C is f o u n d to A p p e n d i x E .S : 6 / l i - ( 2 2 5 7 : ! .) ( 1 ca l ) - 539 cal g iS 4 . .18 4 J
  • 30. I Figure 4.11 Hypoiht!K:al parh for w mrhllpy ch3ngt associat!d with che vaporiunon of w:ircr ar 37"C. f..IJuill lH <Mnlpllh 'ilpor at37'C al3T '{' .: .. l l /11 •Mli flqllid: I • l.iq uK!_ atIWC :Cidl3pOI I - - - - - Y ap.W .itw. atI O O 'C Thecoefficientstocakulate c,as a function of cempuaturr for water vapor are found in Table 4.1;remember to accouni for chc scaling of cocffidcms b, c, and J. When cooling the wacer vapor back down, the specific emhalpy change is: J T"C J?'C & H 1= C ,1 nJT= { (a+b T+cT2+JT 1)JT IO O 'C jIfll'C rt = { (3 3.46+0.688X 10-?Tt 0.7604X 10- 5T1- 3.593X U r' T 1 )1f J J1 rt1C m o l = _2140 J {1mol )( 1cal ) = _ .4c al g 28 ;;j 1 8g 4.1 84J Theo v e r a Uchangeinspecificenthalpyisrhesumofthethrees upsinthepath: A• =A• +ll • +A• =6 3 .1 cal H H 1 H i H 1 - + 53 9 - - 2 8.4- =57 4- g g g g T herefo re,th elatemh e atofv aporizationac37 °C,calculate ddtroughthehypothetic alpath, is5 74cal/g,w h ichisv eryclosetothefirstsolution(577cal/g). 1