Name:
SID:
Discussion Session:
Chemical Engineering Thermodynamics 141 — Fall 2008
Thursday, November 6, 2008
Midterm II - 70 minutes
100 Points Total
Closed Book and Notes
(20 points)
1. Evaluate whether the statements are true or false. You may enter your answers on the
exam sheet.
a) Most compressors and turbines have mechanical efficiencies of less than 0.5.
FALSE. Per SV&A p269: “Values of η [for turbines] usually range from 0.7 to 0.8.” Per
SV&A p274: “Compressor efficiencies are usually in the range of 0.7 to 0.8.”
b) The regenerative cycle has a higher efficiency than the standard Rankine cycles because
the latent heat that would be lost in the condenser is partially used to preheat the water
entering the boiler.
TRUE. This is a correct description of the regenerative cycle.
c) The thermal efficiency of a vapor compression cycle depends on the choice of the
working fluid.
TRUE. The vapor compression cycle contains an irreversible adiabatic expansion; what
fraction of vapor is liquefied in this step depends on the thermodynamic properties of the
working fluid.
d) In a refrigeration cycle, the lower the evaporator temperature, the higher the COP,
retaining the rest of the cycle unchanged.
FALSE. In fact, this is exactly backward.
e) For pure species, partial molar properties equal molar properties.
TRUE. This works even for entropy, because the extra -R ln xi term in the entropy of a
mixture drops out when xi = 1.
f) At constant temperature and pressure, partial molar properties can vary independently,
based on Gibbs-Duhem relationship.
FALSE. At constant T and P, the Gibbs-Duhem relationship is
∑ x dM
i
i
= 0 for any
partial molar property M i . What this equation means is that partial molar properties
cannot vary independently, because if M i changes for one species, all the other M j ≠ i must
necessarily change in a way that keeps the sum still equal to zero.
g) As pressure approaches zero, fugacity coefficient approaches partial pressure.
FALSE. Trick question. Fugacity approaches partial pressure; fugacity coefficient
approaches 1.
h) Fugacity of pure condensed phase at temperature T and low pressure may be
approximated by saturation pressure at temperature T.
TRUE. Credit was given for “FALSE” to those students who commented that at extremely
low temperatures, the Poynting correction can no longer be neglected, even at low
pressures. Remember that all gases become ideal gases in the low pressure limit, so it is not
necessary to make an added assumption about ideal gas behavior here: that assumption is
already built in.
i) Raoult’s law describes well an ideal gas in coexistence with an ideal solution at low
pressure.
TRUE. This is exactly what Raoult’s law does.
j) Activity coefficients describe deviations from ideal solution behavior.
TRUE. This is exactly what activity coefficients do.
(30 points)
2. Freon 134a is to be used in a refrigerator that operates with an evaporator temperature of 10°C and a condenser temperature of 30°C. Saturated liquid refrigerant from the condenser
flows through an expansion valve into the evaporator, from which it emerges as saturated vapor.
The thermodynamic diagram for Freon 134a is given on the next page.
4
Condenser
1
3
Evaporator
2
a) Draw the actual process on the enclosed P-H diagram. For a refrigeration capacity of
5000 J/s, what is the circulation rate of the refrigerant?
Reading from the P-H diagram, we find:
H1 = H4 = 240 kJ/kg
H2 = 390 kJ/kg
m& =
Q& c
5kJ / s
=
= 0.033 kg/s
H 2 − H 1 390kJ / kg − 240 kJ / kg
b) Suppose the cycle in part (a) is modified by the inclusion of a countercurrent heat exchanger
between the condenser and throttle valve in which heat is transferred to vapor returning
from the evaporator. The liquid from the condenser enters the exchanger at 30°C and the
vapor from the evaporator enters the exchanger at -10°C and leaves at 20°C. Draw the
process on the enclosed P-H diagram. What is the circulation rate of the refrigerant?
4’
4
Condenser
1
3
Evaporator
2
2’
Using an energy balance on the heat exchanger gives:
H4 + H2 = H4’ + H2’
Solving for H4’ gives H4’ = H4 + H2 – H2’
The heat exchanger operates at constant pressure. Thus, H2’ = 420 kJ/kg.
H4’ = H1 = 240 kJ/kg + 390 kJ/kg – 420 kJ/kg = 210 kJ/kg
m& =
Q& c
5kJ / s
=
=0.028 kg/s
'
H 2 − H1 390kJ / kg − 210kJ / kg
Blue = part a
Purple = part b
(30 points) 3.
of state:
P=
a) Show that the fugacity of a vapor that obeys the volumetric equation
RT
V −b
bP 
is given by: f V = P exp 
.
 RT 
The general equation for fugacity in terms of volumetric properties is
 1 P
RT  
f = P exp 
V
−

∫  P  dP 
RT

0

Our equation of state can be rewritten
RT
V=
+b
P
Inserting this equation into the general fugacity equation gives
 1 P  RT
RT  
f V = P exp 
+b−

 dP 
∫
RT
P
P

 

0
P b

f V = P exp  ∫
dP 
 0 RT

bP


f V = P exp 

 RT 
b) Derive an equation for the fugacity coefficient of the liquid at saturation, ln φ Lsat , as a
function of saturation pressure.
At saturation, we have f Lsat = f Vsat. We already know the fugacity of the vapor phase:
 bP 
f V = P exp 

 RT 
Evaluating this at the saturation pressure, we have
 bP sat 

f Lsat = f Vsat = P sat exp 
 RT 
Rewriting to solve for the fugacity coefficient, we find
ln φ SAT =
bP sat
RT
c) When the substance used in part (a) is isothermally compressed, an incompressible liquid is
produced. Develop an equation for the fugacity of the liquid.
The equation for the fugacity of a liquid can be written
 1 P

f L = P φ exp 
VdP 
∫
 RT Psat

And since the liquid is incompressible, V is not a function of P. Thus, we have
 1
sat sat
sat 
f L = P φ exp 
V0 (P − P )
 RT

sat
Finally, we already know f from part (b). Plugging this in, we have
 bP sat 
1

 exp 
f L = P sat exp 
V0 P − P sat 
 RT

 RT 
sat
sat
(
)
Or
 V P − V0 P sat + bP sat
f L = P sat exp  0
RT




d) Now let’s pass supercritical CO2 over our compressed liquid. Develop an expression for
the solubility of our liquid in SCF CO2? Assume a negligible amount of CO2 dissolves into
compressed liquid.
The gaseous mixture obeys the following equation of state:
P=
RT
V − bmix
with mixing rule bmix = ∑ yi bi .
i
Solubility is a measure of how much of our liquid, call it species (1), will dissolve into the
supercritical CO2 phase at equilibrium. We assume the amount of CO2 that dissolves into
the liquid phase is negligible. Thus, the fugacity of our pure compressed liquid must be equal
to the fugacity of the same substance in the CO2 supercritical phase:
f liquid = f 1,supercrit
Assuming the Lewis-Randall rule applies in the supercritical phase, we have
f 1v = φ1v y1 P
RT
We have assumed that the vapor phase obeys the equation P =
V − bmix
To get φ1v , we start by applying the definition of partial molar fugacity:
 ∂n ln φ 
dP
ln φ = 
where ln φ = ∫ (Z − 1)

P
 ∂n1  T , P
0
We know P(V-bmix) = RT, from which we can show
P(V − bmix )
=1
RT
PV bmixP
−
=1
RT
RT
P
v
1
b P
PV
− 1 = Z − 1 = mix
RT
RT
P
dP P  bmix P  dP P bmix
(
)
ln φ = ∫ Z − 1
= 

=
dP
P ∫0  RT  P ∫0 RT
0
(Note: having done parts (a)-(c) already, you could jump right to this step as a starting
point.)
b P
ln φ = mix
RT
nb P (n b + n2 b2 )P
n ln φ = mix = 1 1
RT
RT
 ∂n ln φ 
b1 P
ln φ1v = 
 =
 ∂n1  T , P RT
(
)
b P
f 1v = P1 exp ln φ1v = y1 P exp  1 
 RT 
Now that we have the fugacity for substance 1 in the supercritical phase, we equate it with
the fugacity of substance 1 in the liquid phase, which we already know from part (c):
f 1L = y1 f 1v
P
sat
 V0 P − V0 P sat + b1 P sat 
bP
 = y1 P exp  1 
exp 
RT
 RT 


(
)
 (V0 − b1 ) P − P sat 
P sat

y1 =
exp 
P
RT


This is essentially the equation from lecture involving the “enhancement factor,” only with
expressions for the fugacity coefficients substituted in.
(20 points)
4. It has been discovered that a 20 wt% solution of acetic acid in water will effectively kill the
majority of common species of weeds, while actually encouraging the growth of desirable
plants. Vinegar has the added benefit of also discouraging ants from eating your flowers. The
molar volume of a water-acetic acid mixture at 25°C is well described by the equation V =
4.4857 x² + 34.961 x + 18.02, where x is the mole fraction of acetic acid in the solution.
(a) Find the partial molar volumes of water and acetic acid in a 20 wt% acetic acid solution.
The molecular weights of water and acetic acid are 18.015 g/mol and 60.053 g/mol,
respectively.
We begin by converting weight percent into mole fraction:
w AcOH / MW AcOH
x AcOH =
wAcOH / MW AcOH + (1 − w AcOH ) / MW H 2O
0.20 / 60.053
0.20 / 60.053 + 0.80 / 18.015
x AcOH = 0.070
x AcOH =
To find partial molar volumes, we use the formulas
 ∂V 

V AcOH = V + (1 − x AcOH )
 ∂x AcOH  T , P
 ∂V 

VH 2O = V − ( x AcOH ) 
 ∂x AcOH  T , P
Observing that V = 4.4857 x² + 34.961 x + 19.02, we find
 ∂V 

 = 8.9714 x + 34.961
 ∂x

 AcOH T , P
Plugging in that x = 0.070 gives
 ∂V 

 = (8.9714)( 0.070) + 34.961 = 35.589 mol/mL
 ∂x

 AcOH T , P
At x = 0.070, V = (4.4857)(0.070)² + (34.961)(0.070) + 18.02 = 20.489 mol/mL
V AcOH = 20.489 + (1–0.070)(35.589) = 53.587 mL/mol
VH 2O = 20.489 – (0.070)(35.589) = 17.998 mL/mol
(b) After weeding your garden, you decide to make a salad, topped (of course!) with a
vinegar-and-oil dressing. What volume of 20 wt% acetic acid should be combined with
pure water to make 100 mL of a 5 wt% acetic acid solution (eg, vinegar)?
To complete this problem, we must recognize that for non-ideal solutions, volume is not
additive. If you simply add 25 mL of 20 wt% acetic acid to 75 mL of water, you will not
get a 100 mL solution of 5 wt% acetic acid.
Luckily, mass (or equivalently, number of moles) is additive (after all, conservation of mass
is a fundamental precept of (non-relativistic non-nuclear) physics).
We wish to make 100 mL of 5 wt% acetic acid solution. Again, we begin by converting to
mole fraction:
0.05 / 60.053
0.05 / 60.053 + 0.95 / 18.015
x AcOH = 0.0155
x AcOH =
To determine how many moles we actually have, we need the molar volume.
V = 4.4857*(0.0155)² + 34.961*0.0155 + 18.02 = 18.563 mL/mol
The total number of moles is thus
100 mL / 18.563 mL/mol = 5.3871 mol
NAcOH = 0.0155*5.387 = 0.0835 moles
NH2O = (1–0.0155)*5.387 = 5.3036 moles
All 0.0835 moles of acetic acid must come from the 20 wt% (7 mol%) solution. Thus, we
need
0.0835molAcOH
= 1.1929 mol of the 20 wt% (7 mol%) solution. The molar
0.070 molAcOH
molSolution
volume of that solution was 20.489 mL/mol, so we need
1.1929 mol * 20.489 mL/mol = 24.44 mL of 20 wt% acetic acid.