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```					                            Enthalpy
• Form of energy important in flow systems where temperature
and chemical composition play an important role

h = u + Pv

h : specific enthalpy, (kJ/kg, J/mol, Btu/lb etc.)
u : specific internal energy
P : pressure
v : specific volume

• Units for u and Pv obviously need to be consistent with h

• Total enthalpy H (kJ) = (m)(h) ; i.e. (kg) (kJ/kg) etc.
Specific enthalpy for a pure substance

• From phase rule, C=1, P=1, F=2
Thus,       h=h(T,P)

• For ideal gases: h=h(T)
Holds for many other substances as well

• Notable exception: STEAM
(That’s why we need steam tables to look up properties of
superheated steam and compressed liquid as a function of both
T and P)
Change of enthalpy with temperature
dh
 C p heat capacity at constant pressure
dT
 kJ   Btu 
 kg.C  ,  lb - mol.F  etc.
      
                     
ith
Thus, if C p is constant w temperature :
Δh  C p ΔT ; i.e. hT2  hT1  C p (T2  T1 )
ith
if C p is NOT constant w temperature :
T2
hT2  hT1   C p dT
T1
Change of enthalpy with phase

The (latent) heat of vaporization at T:

hvT  hvapourT  hliquid T
Heat of fusion at T:

h fT  hliquid T  hsolid T
Heat of sublimation at T:

hsT  hvapourT  hsolid T
Note: we have
h vs T,
for water,
used h for
including         specific
phase             enthalpy
Changes

from Fig. 23.5,
Himmelblau,
7th ed.
hv100C  hE  hD

h f 0 C  hC  hB

hs30C  hG  hA    G   .
hF  hE  C p H O vapour (130  100)
2

hD  hC  C p H O liquid (100  0)
2

hB  hA  C p H2O ice (0  (30))
hvT  hvapourT  hliquid T
h fT  hliquid T  hsolid T
hsT  hvapourT  hsolid T
•   The above formulation implies that each of the phase changes (vaporization,
fusion (melting) and sublimation) can take place at different temperatures and
the enthalpy change associated with the phase change is a function of
temperature.

•   In practice it is only the vaporization temperature and enthalpy that shows any
variation which we will be interested in.

•   The enthalpy of vaporization is at its maximum value at 0 C (the normal melting
point) and it is 0 at _____

•   For water, the value of 40.65 kJ/mol is the value at the normal boiling
temperature
Absolute (relative ?) value of h
choosing a DATUM
• We can calculate changes in specific enthalpy by dh=CpdT
• What about the absolute value? h=h(T)
• Remember h is what we need, to find H=mh

We have,
T2
hT2  hT1  C p (T2  T1 ) or hT2  hT1   C p dT
T1

Define hT1  0 ; this is called " choosing a datum"
Then,
T2
hT2  C p (T2  T1 ) or hT2   C p dT
T1

relative to T1
Enthalpy of a mixture - ideal mixtures
• Simply add the enthalpy of the components to get the total
enthalpy
mmixture  m A  mB  mC  ...
m: moles or mass
H mixture  H A  H B  H C  ...
H : enthalpy (kJ, Btu etc.)
H A  m A hA
 kJ 
kJ  kg  or
 kg         lb  mol 

Btu 
 etc.
                        lb  mol 
H mixture   mi h i
H mixture
hmixture   
mmixture
Enthalpy of a mixture - nonideal mixtures
1+12 !

• Examples: strong acid and base solutions

(1 L of 2 M HCl @ 25C, 1atm)
 (1 L of H 2O @ 25C, 1atm)
 (2 L of 1 M HCl @ 25 C, 1 atm)  heat

• The resulting solution must have less energy than the sum of
the energies of its constituents at the same temperature and
pressure.
Properties of humid air
expressed per unit mass of BDA

BDA         H2O (vapour)                   Humid air
Mass (Kg)
1                 w                         1+w

Cp, kJ/kg.C,
(assumed constant          1.00              1.88                   1.00+1.88w
over ambient
temperature range)

h, kJ/kg, at T (C)
Relative to BDA and
T
H2O (liquid) at 0 C, 1
T       hv , H 2O ,0 C  1.88T   (hv, H 2O,0 C  1.88T ) w
atm

air

hhumid  Cp BDA (T  0)  hv , H 2O ,0 C  C p H 2Ovapour T  0 w 
kJ        kJ         kJ           kJ        kg H 2O
             C                     C
kg BDA (kg BDA)(C )    kg H 2O (kg H 2O)(C )  kg BDA
Properties of humid air
expressed per unit mass of BDA

BDA          H2O (vapour)              Humid air
Mass (lb)
1                  w                    1+w

Cp, Btu/lb.F,
(assumed constant          0.25               0.45               0.25+0.45w
over ambient
temperature range)

h, Btu/lb, at T (F)
Relative to BDA at 0 F       T                                 0.25T 
and H2O (liquid) at 32                hv, H 2O,32 F  0.45T
F, 1 atm                                                       (1061  0.45T ) w

                                   
hhumid  Cp BDA (T  0)  hv , H 2O ,32 F  C p H 2Ovapour T  32 w
air

Btu       Btu         Btu          Btu       lb H 2O
             F                      F
lb BDA (lb BDA)(F )     lb H 2O (lb H 2O)(F )  lb BDA
Temperatures
and temperature differences
• Temperature:
C    -17.8   0     10    20    25
K    290.8   273   283   293   298
F      0     32    50    68    77
R     460    492   510   528   537

• Temperature difference:
C = K
F = R
F = 1.8 C
Energy balance
Conservation of energy in a flow system dominated by thermal and
composition effects
(i.e. kinetic and potential energy differences are relatively small)
Hin + Q = Hout + W
Where:
Hin, Hout : The enthalpy of the material entering and leaving the
system

Q : heat (thermal energy) added to the system by heat transfer
Q = UA(Tsurroundings - Tsystem)

W : Work done by the system on the surroundings (via some
mechanical means like a shaft)

• Each or these terms has units of energy, kJ, Btu, etc.
Energy balance
for a flow system dominated by thermal and composition effects
(i.e. kinetic and potential energy differences are relatively small)

Hin + Q = Hout + W
W : Work done by the system on the
surroundings
Hin                                                                         Hout
(with material coming in)                                                   (with material leaving)

Q : heat (thermal energy)
heat transfer
Energy balance
Hin + Q = Hout + W
This is an expression of the first law of thermodynamics
(conservation of energy).

Usual form in MAAE2400: ΔU = Q – W

What is different here?
-   flow vs open system; if the system is closed we are only dealing
with the changes in the internal energy of the material in the
system, when material enters and leaves the system the pv
term in h = u + pv represents the the work that is involved in
pushing the material into the system and pushing it out again.
Another way of looking at the situation is to consider pv as the
material’s ability to do work (hence energy) by virtue of its
pressure.

-   The Δ in ΔU indicates change from initial condition to final
condition. In a flow system we can have multiple streams and
so we are not looking at the difference in energy of the same
material. We need to look at all the energy in, and all the
energy out instead of defining a difference between two streams
Strategy for Analyzing Energy Balance problems

 mh      in
 Q   mhout             (assuming W  0)

1) See strategy for analyzing material balance problems.
-   Usually the material balances have to be solved first before the energy
balance is attempted, so that the m’s are known.
-   Sometimes they are simultaneous.
-   No problem that involves just an energy balance.
2) Choose a DATUM for all components going in and out of your
system, so that the h’s can be calculated.
3) Determine the status of Q:
-   Unknown: the energy balance has to be solved for Q
-   Known : the energy balance can be used as one more equation to solve for
an unknown (a material balance unknown, temperature, or state
-   Often we are interested in the adiabatic, Q=0, case
• SYSTEM: Wet Bulb
Large amount of air blown over the wet bulb (temperature and humidity
of the air in and out are the same)
System is really not at steady state since the water on the wick is
evaporating. However, the wet bulb temperature is established pretty
quickly and is maintained until all the water is evaporated. So we can
analyze the flows at steady state.
• MASS BALANCES:
– air: IN = OUT
– water: DEPLETION = OUT
Wet bulb
•   ENERGY BALANCE: Hin + Q = Hout
•   Datum: BDA @ Tair, H2O (liquid) @ Twb

Hair + Q = Hair + H evaporated water

            
Rate of water evaporation : k g wTwb  wT mass/time

             
H evaporated water  mh  k g wTwb  wT hv energy/time

Q, Rate of heat transfer : hc (T  Twb ) energy/time

kg : mass transfer coefficient, hc : heat transfer coefficient
Wet bulb energy balance

Q  H evaporated water
hc (T  Twb )  k g ( wTwb  wT )hv
wTwb  wT   hc 1

Twb  T    k g hv
For the conditions of the wet bulb system:
hc
 0.25 Btu / lb.F ; 1.04 kJ / kg.K
kg
•   MASS BALANCES:
•   BDA: 1 kg BDA in = 1 kg BDA out (BASIS)
•   Water: w kg in with air + m kg make-up = ws kg out with air
• ENERGY BALANCE: Hin = Hout
• Datum: BDA @ Ts, H2O (liquid) @ Ts

H IN
BDA : C pBDA (T  Ts )
water vapour : [hv  C pvapour (T  Ts )]w
make  up water : mC p liquid (Ts  Ts )
H OUT
BDA : C pBDA (Ts  Ts )
water vapour : [hv  C pvapour (Ts  Ts )]ws

C pBDA (T  Ts )  [hv  C pvapour (T  Ts )]w  hv ws
(C pBDA  C pvapour w)(T  Ts )  hv ( ws  w)
(CS )(T  Ts )  hv ( ws  w)
wTs  wT   Cs

Ts  T    hv
• Energy balance on wet bulb          • Energy balance for adiabatic
saturation

wTwb  wT   hc 1                         wTs  wT   Cs
                                      
Twb  T    k g hv                       Ts  T    hv
hc
 1.04 kJ / kg.K (water air systemin wet bulb scenario)
kg
Cs  C pBDA  C pvapour w  1.0  1.88w kJ / kgBDA.K
w  0.01  0.1
Cs  1.02  1.19 ; very close to the1.04 kJ / kg.K value
Thus there is little error in taking the wet bulb lines
to also represent adiabatic saturation lines
on the w vs T coordinates of the water- air system,
the PSYCHROMET RIC CHART.
Energy balance
for systems with chemical reaction
• So far we have considered the change of enthalpy only with
temperature and phase (solid, liguid, vapour)
• Thus, in choosing a DATUM we have only specified a
temperature and a state, e.g. H2O (liquid) at 0 C, BDA at 0 C
etc.
• What about the following system?

1 mol CH4                                1 mol CO2
2 mol O2                                 2 mol H2O
At 25 C, 1 atm                           At 25 C, 1 atm
CH 4  2O2  CO2  2 H 2O
1 mol CH4                                        1 mol CO2
2 mol O2                                         2 mol H2O (g)
Q
At 25 C, 1 atm                                   At 25 C, 1 atm

• If we set DATUM: 25 C for all gases
Then, Hin = Hout = 0, and Q = Hout – Hin = 0

• However, when we carry out this process, we observe that
Q = - 802.32 kJ, i.e. the system gives off energy to the
surroundings

• It looks like there is “generation” of energy in the system.
So we revise our energy balance to read:
Hin + Q + heat generated = Hout
• We know that energy is neither created nor destroyed but can
change from one form to another
• So what we are observing in a chemical reaction is the
transformation of energy from one form to another
• We are transforming the bond energies in the molecules to
thermal energy
• We cannot measure bond energies by measuring the
temperature or phase of a substance, so it appears as if energy
is being “created”
• Other reactions may appear to “consume” energy.
CH 4  2O2  CO2  2 H 2O              H rxn  802 .32 kJ
0

i.e. the reaction creates energy
• The opposite is an endothermic reaction, consuming energy
• We calculate H0rxn by:

 H = ( mi  h )products - ( mi  h )reactants
0
rxn
0
fi
0
fi
• Where the h0f are the standard heats of formation for each
compound, tabulated in chemistry texts.
• The superscript 0 refers to the standard state (25 C, 1 atm)
• The bar over the H indicates that the value has been calculated
for the indicated quantity of reactans/products, i.e. 1 mol of
methane
Energy balance with chemical reaction –
“The Heat of reaction” approach
We revise the energy balance to include a term to account for the heat
effects of chemical reactions:
Hin + Q + heat generated = Hout
Using the convention for exothermic and endothermic reactions, we we
can write this as:

H IN  Q  H rxn  H OUT
When Hrxn is negative we end up with a positive value for “heat
generated”, and vice versa
The Hrxn term must account for the actual amount of reaction that took
place.
It must account for reactants going to products at the specified DATUM
The DATUM does not have to be at 25 C but it is almost always more
convenient to choose a DATUM of 25 C
• The standard heats of formation are themselves the heats of
reactions that form the compounds from the constituent
elements in their stable forms at 25 C and 1 atm.
• e.g.
C  O2  CO2            h  393.51 kJ
0
f

1
H 2  O2  H 2O ( g ) h 0  241.83 kJ
f
2
1
H 2  O2  H 2O (l ) h f  285.84 kJ
0

2
C  2 H 2  CH 4 h 0  74.84 kJ
f
• The heat of formation of elemental species are 0! (by definition)
Energy balance with chemical reaction –
“The heat of formation” approach
We retain the energy balance developed for non-reacting systems:
Hin + Q = Hout
However, we recognize that the enthalpy of any material entering
or leaving the system must account for differences in enthalpy
due to the chemical bond structure of the material. We therefore
add a term to the specific enthalpy of substances:
 energy due to          
 energy in                                 
 chemical bonds    temperature or phase, 
h                
                  i.e. " sensible" enthalpy
                        
T
h  h  
0
f          C p dT
25 C
“The heat of formation” approach
DATUM: Elements in their natural states at 25 C and 1 atm

hO2 , 25C  0
hN 2 , 25C  0
hH 2 , 25C  0
hCO2 , 25C  h   0
f CO 2     393.51 kJ / mole
hH 2O ( g ), 25C  h   0
f H 2O ( g )    241.83 kJ / mole
hH 2O (l ), 25C  h   0
f H 2O ( l )    285.84 kJ / mole

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