∆H298 411153−285830−425609−92307−()−[]J⋅:=
∆H298 1.791−105
×J=
NaOH(s) + HCl(g) —> NaCl(s) + H2O(l) (1)
NaOH(inf H2O) —> NaOH(s) + inf H2O (2)
HCl(9 H2O) —> HCl(g) + 9 H2O(l) (3)
NaCl(s) + inf H2O —> NaCl(inf H2O) (4)
—————————————————————————————-
NaOH(inf H2O) + HCl(9 H2O) —> NaCl(inf H2O)
(b) First step: m140 lb⋅:= x11:= H114 BTU
lbm
⋅:=
m275 lb⋅:= x20.25:= H27−BTU
lbm
⋅:=
12.59 BASIS: 1 mol NaOH neutralized.
For following reaction; data from Table C.4:
NaOH(s) + HCl(g) —> NaCl(s) + H2O(l)
469
∆H 45.259−kJ
mol
=∆H∆H
x1
molwt:=
However, for 1 mol of NaOH, it becomes:
molwt 40.00 gm
mol
⋅:=T 295.65 K⋅:=
Correct NaOH enthalpy to 77 degF with heat capacity at 72.5 degF
(295.65 K); Table C.2:
x119.789 %=x1
1 40.00⋅
1 40.00⋅9 18.015⋅+
:=
Weight % of 10 mol-% NaOH soln:
First, find heat of solution of 1 mole of NaOH in 9 moles of H2O
at 25 degC (77 degF).
12.60
470
HHE
x1x2
⋅
→
⎯
⎯
:=x21x
1
−
()
→
⎯
⎯
⎯
:=HE
73.27
144.21
208.64
262.83
302.84
323.31
320.98
279.58
237.25
178.87
100.71
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
−kJ
kg
:=x1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.85
0.9
0.95
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
Note: The derivation of the equations in part a) can be found in Section B
of this manual.
12.61
NaCl + inf H2O —> NaCl(inf H2O) (4)
NaOH(9 H2O) —> NaOH(s) + 9 H2O (3)
HCl(inf H2O) —> HCl(g) + inf H2O (2)
NaOH(s) + HCl(g) —> NaCl(s) + H2O(l) (1)
Now, on the BASIS of 1 mol of HCl neutralized:
471
In order to take the necessary derivatives of H
, we will fit the data to a
third order polynomial of the form HHE
x1x2
⋅
=ab
x.1
+cx
12
⋅+ dx
13
⋅+=
⎛
⎜
⎝
⎞
⎠ .
Use the Mathcad regress function to find the parameters a, b, c and d.
w
a
⎢
⎥
⎛
⎜
⎞
⎟
regress x1
H
,3,
⎡
⎢
⎤
⎥
:=
w
a
⎛
⎜
⎞
⎟
3
735.28−
⎛
⎜
⎞
⎟
=
Using the equations given in the problem statement and taking the
derivatives of the polynomial analytically:
472
⎜
⎜
⎜
1500
0
H/x1x2
HEbar1
HEbar2
x1
(kJ/kg)
12.62 Note: This problem uses data from problem 12.61
0.2
0.4
0.8
0.9
0.95
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
144.21
262.83
279.58
178.87
100.71
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
473
At time θ, let:
x1 = mass fraftion of H2SO4 in tank
m = total mass of 90% H2SO4 added up to time θ
H = enthalpy of H2SO4 solution in tank at 25 C
Hbar2x1
()
x12Hx1
()
1x
1
−
()
b2c⋅x1
⋅+ 3d⋅x12
⋅+
()
kJ
kg
⋅
⎡
⎢
⎣
⎤
⎥
⎦
⋅−
⎡
⎢
⎣
⎤
⎥
⎦
⋅:=
Hbar1x1
()
1x
1
−
()
2Hx1
()
x1b2c⋅x1
⋅+ 3d⋅x12
⋅+
()
kJ
kg
⋅
⎡
⎢
⎣
⎤
⎥
⎦
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
⋅:=
Hx
1
()
Hx1
()
x1
⋅1x
1
−
()
⋅:=
Hx1
()
abx
1
⋅+ cx
12
⋅+ dx
13
⋅+
()
kJ
kg
⋅:=
By the equations given in problem 12.61
w
⎜
⎜
⎜
a
⎜
⎢
⎥
c
d
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
3
735.28−
195.199
914.579−
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
=
w
a
c
d
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
regress x1
H
kJ
kg
⎛
⎝
⎞
⎠
,3,
⎡
⎢
⎣
⎤
⎥
⎦
:=
Fit a third order polynomial of the form HE
x1x2
⋅ab
x.1
+cx
12
⋅+ dx
13
⋅+=
⎛
⎜
⎝
⎞
⎠ .
Use the Mathcad regress function to find the parameters a, b, c and d.
474
rx
1
()
q
0.9 Hbar1x1
()
0.1 Hbar2x1
()
+H3
−
:=
When 90% acid is added to the tank it undergoes an enthalpy change equal
to: 0.9Hbar1+0.1Hbar2-H3, where Hbar1 and Hbar2 are the partial
enthalpies of H2SO4 and H2O in the solution of mass fraction x1 existing in
the tank at the instant of addition. This enthalpy change equals the heat
required per kg of 90% acid to keep the temperature at 25 C. Thus,
qQt x1
()
:=
Since the heat transfer rate q is constant:
Qx
1
()
4000kg m x1
()
+
()
Hx
1
()
⋅mx
1
()
H3
⋅−
⎡
⎣
⎤
⎦
:=
Define quantities as a function of x1
Applying these equations to the overall process, for which:
475
mdot32mdot1
:=mdot2mdot1
:=
Guess:
Use mass balances to find feed rate of cold water and product rate.a)
H370−BTU
lb
:=T3140degF:=x30.5:=
H27BTU
lb
:=T240degF:=x20.0:=
Enthalpies from Fig. 12.17
H192−BTU
lb
:=T1120degF:=x10.8:=mdot120000 lb
hr
:=
12.64
800
900
rx
1
()
kg
hr
hr
x10 0.01,0.5..:=
Plot the rate as a function of time
476
Ldx1
dt
⋅x1y1
−
()
Vdot⋅=
Rearranging this equation gives:
dLx
1
⋅
()
dt y1
−Vdot⋅=
An unsteady state species balance on water yields:
dL
dt Vdot−=
An unsteady state mole balance yields:
Let L = total moles of liquid at any point in time and Vdot = rate at
which liquid boils and leaves the system as vapor.
12.65
For an adiabatic process, Qdot is zero. Solve the energy balance to find H3
c)
Apply an energy balance on the mixerb)
477
The water can be removed but almost 16% of the organic liquid will
be removed with the water.
K115.459=K1γinf1
Psat1
P
⋅:=
x1f
50
106
:=x10
600
106
:=L01mol:=
For this problem the following values apply:
dx1
K11−
()
x1
dL
L
=
Substituting gives:
K1γinf1
Psat1
P
⋅=
where:
y1γinf1
Psat1
P
⋅x1
⋅
⎛
⎜
⎝
⎞
⎠
=K1x1
⋅=
At low concentrations y1 and x1 can be related by:
478
For NRTL equation
b12 184.70 cal
mol
:= b21 222.64 cal
mol
:= α 0.3048:=
τ12
b12
RT⋅
:= τ12 0.288= τ21
b21
RT⋅
:= τ21 0.347=
12.69 1 – Acetone 2- Methanol T 50 273.15+()K:=
For Wilson equation
a12 161.88−cal
mol
:= a21 583.11 cal
mol
:= V174.05 cm3
mol
:= V240.73 cm3
mol
:=
Λ12
V2
V1
exp a12
−
RT⋅
⎛
⎜
⎝
⎞
⎠
⋅:= Λ12 0.708= Λ21
V1
V2
exp a21
−
RT⋅
⎛
⎜
⎝
⎞
⎠
⋅:= Λ21 0.733=
479
Use the values of γ1 and γ2 at x1=0.253 and Eqs. (12.10a) and (12.10b) to
find A12 and A21.
Guess: A12 0.5:= A21 0.5:=
12.71 Psat1183.4kPa:= Psat296.7kPa:=
x10.253:= y10.456:= P 139.1kPa:=
Check whether or not the system is ideal using Raoult’s Law (RL)
PRL x1Psat1
⋅1x
1
−
()
Psat2
⋅+:= PRL 118.635 kPa=
Since PRL<P, γ1 and γ2 are not equal to 1. Therefore, we need a model for
GE/RT. A two parameter model will work.
Find γ1 and γ2 at x1=0.253 from the given data.
480
12.72 P 108.6kPa:= x10.389:=
T 35 273.15+()K:= Psat1120.2kPa:= Psat273.9kPa:=
Check whether or not the system is ideal using Raoult’s Law (RL)
PRL x1Psat1
⋅1x
1
−
()
Psat2
⋅+:= PRL 91.911 kPa=
Since PRL < P, γ1 and γ2 are not equal to 1. Therefore, we need a model for
GE/RT. A one parameter model will work.
b) γ1inf exp A12
()
:= γ1inf 1.904= γ2inf exp A21
()
:= γ2inf 1.614=
481
γ2x1
()
exp A x12
⋅
()
:=γ1x1
()
exp A 1 x1
−
()
2
⋅
⎡
⎣
⎤
⎦
:=
A 0.677=A Find A():=
Px
1exp A 1 x1
−
()
2
⋅
⎡
⎣
⎤
⎦
⋅Psat1
⋅1x
1
−
()
exp A x1
⋅
()
2
⎡
⎣
⎤
⎦
⋅Psat2
⋅+=Given
A1:=
Guess:
Use the data to find the value of A
482