Transcript ConcepTests in Chemical Engineering Thermodynamics Unit 2: Generalized Analysis of Fluid Properties Note: Slides marked with JLF were adapted from the ConcepTests of John L.

ConcepTests in Chemical
Engineering Thermodynamics
Unit 2: Generalized Analysis of
Fluid Properties
Note: Slides marked with JLF were adapted from the ConcepTests
of John L. Falconer, U. Colorado. Cf. Chem. Eng. Ed. 2004,2007
Day 22 MRs
22.1. Transform the expressions below in terms of Cp, Cv ,
T, P, and V. Your answer may include absolute values of S
if it not associated with a derivative.
(S/V)T
(a) (P/V)S
(b) (T/V)U
(c) (U/T)V
(d) (P/T)V
Day 22 MRs
22.2. Transform the expressions below in terms of
Cp, Cv , T, P, and V. Your answer may include
absolute values of S if it not associated with a
derivative.
(S/P)V
(a) Cv(T/P)V/T
(b) (T/V)U
(c) (U/T)V
(d) (P/T)V
Day 22 MRs and EOSs
22.3. Transform the expressions below in terms of
Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(V/S)P
(a) Cv(T/P)V/T
(b) T(V/T)P/Cp
(c) (T/P)S
(d) -(V/T)P
Day 22 MRs and EOSs
22.4. Transform the expressions below in terms of
Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(G/S)P
(a) V(P/S)V/T
(b) TS(V/T)P/Cv
(c) -TS/Cp
(d) -S(T/S)P
Day 22 MRs and EOSs
22.5. Use the vdW EOS to describe the following
derivative.
-T(Z/T)V
FYI vdw EOS is: Z = [1/(1-br)] – [ar/RT]
(a) [1/(1-br)]
(b) [1/(1-br)2]
(c) –[ar/RT]
(d) [ar/RT2]
QikQiz2.1
Q2.1.1. Transform the expression below in terms
of Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(A/P)V
(a) -S(T/P)V
(b) Cp(T/P)V
(c) TS/P
(d) -VS(T/V)P /Cp
QikQiz2.1
Q2.1.2. Transform the expression below in terms
of Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(H/P)T
(a) V
(b) V-T(V/T)P
(c) -T(S/P)T+V
(d) -T(P/T)V - P
QQ2.1
Q2.1.3. The following strange equation of state has been
proposed: P = (RT/V1.5) - a/T1.3
where a is a constant. Derive an expression for
(P/T)V
(a) RT2/(2*V1.5) + a/(0.3*T0.3)
(b) (R/V1.5) – 1.3a/T2.3
(c) -1.5(R/V2.5) – 1.3a/T2.3
(d) (R/V1.5) + 1.3a/T2.3
Day 24 MRs and EOSs
24.1. Transform the expression below in terms of
Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(H/S)V
(a) T(1+ V(P/T)V/Cv )
(b) VS(T/V)P/Cv
(c) TS/Cp
(d) -VS(T/V)P /Cp
Day 24 MRs and EOSs
24.2. Transform the expression below in terms of
Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(H/P)V
(a) Cv(T/P+ T(V/P)T/V )
(b) VS(T/V)P/Cv
(c) TS/P
(d) Cp(T/ P )V + [V-T(V/T)P]
Day 24 MRs and EOSs
24.3. Use the PR(1976) EOS to describe the
following derivative. -T(Z/T)V
FYI: PR EOS is on P204 (Eq. 6.16-6.19)
(a)

1 da  
br
 a

 bRT 2 bRT dT  1  2br  (br )2 


(b)

1 da  
br
 a
 bRT  bR dT  1  2br  (br )2 


 ac Tr
1 
a

(c)
a
bRT
(d)
ac Tr
bRT


br

2
1

2
b
r

(
b
r
)

 


br
1  2br  (br )2 


Day 24 MRs and EOSs
24.4. FOR the SRK(1972) EOS:
a  ac Tr
1 
-T(Z/T)V
bRT 
a
br
Evaluate  T  Z  d (br )
 T V br
0
=
(a)
a
bRT
 ac Tr
1 
a

  (br ) 2 


 1  br 
  br 


1

b
r

 
(b)
a
bRT
 ac Tr
1 
a

  0.5(br )2 
 ln 

1

b
r
 

(c)
a
bRT
 ac Tr
1 
a


 ln 1  br 

(d)
ac Tr
ln 1  br 
bRT
Day 25 MRs and EOSs
25.1. Transform the expression below in terms of
Cv, T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(H/T)P
(a) Cv+ T(P/T)V (V/T)P
(b) Cv+ [T(P/T)V –P ](V/T)P
(c) Cp
(d) (U/ T)P + P(V/T)P]
Day 25 MRs and EOSs
25.2. Transform the expression below in terms of
Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(A/V)T
(a) (U/V)T - T (S/V)T
(b) [(P/T)V – P] + (P/T)V
(c) -P
(d) –T (P/T)V
Day 26 Dep Funs
br
a  br 

26.1. FOR the SRK EOS: Z  1 
1  br bRT 1  br 
br
d ( br )
Evaluate
0 ( Z  1) br
(a)
a
ln(1  br ) 
ln(1  br )
bRT
(b)
 0.5(br )2 
a
 ln(1  br ) 
ln 

bRT  1  br 
(c)
 ln(1  br ) 
(d)
 ln(1  br )
a
 2
ln(1  br )
b
b RT
a
ln(1  br )
bRT
Day 26 DepFuns
26.2. FOR the PR EOS:
Evaluate
(Hint:p602)
br
 ( Z  1)
0
a
ln(1  br )
bRT
(a)
 ln(1  br ) 
(b)
 ln(1  br ) 
(c)
(d)
 ln(1  br ) 
d ( br )
br

br
a 
br
Z  1

1  br bRT 1  2br  (br )2 
a
 1  br 
arctan 

bRT 8
 2 
 1  2.414br 
a
ln 
bRT  1  0.414br 
 2  8  br 
a
 ln(1  br ) 
ln 

bRT 8  2  8  br 
Day 26 Dep Funs
4cbr
9.5qYbr
26.3. FOR the ESD EOS:
Z  1

1  1.9br 1  1.8Ybr
where Y = exp(e/kBT)-1.06
c and q are constants
br
Evaluate
d ( br )
0 ( Z  1) br
(a)
4c ln(1  1.9br ) 9.5qY

ln(1  1.8Ybr )
1.9
1.8
(b)
4c ln(1  1.9br )  9.5q ln(1  1.8Ybr )
(c)
4c ln(1  1.9br ) 9.5qY

ln(1  1.8Ybr )
1.9
1.8
(d)
4c ln(1  1.9br ) 9.5q

ln(1  1.8Ybr )
1.9
1.8
Day 27 QikQiz2.2
Q2.2.1. Transform the expression below in terms
of Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(G/S)P
(a) - V(T/V)P
(b) PS(T/P)V/Cp
(c) –ST/Cp
(d) (H/S)P –T –S(T/S)P
Day 27 QikQiz2.2
Q2.2.2. Transform the expression below in terms
of Cp, Cv , T, P, and V. Your answer may include
absolute values of S if not associated with a
derivative.
(P/S)G
(a) -T(V/T)P
(b) [-(V/T)P – CpV/(ST)]-1
(c) –(ST/Cp)(V/T)P + V
(d) -(T/V)P
Day 27 QQ2.2
Q2.2.3 FOR the ESD EOS: Z HS  1 
br
Evaluate
 (Z
0
(a)
(b)
(c)
(d)
HS
d ( br )
 1)
br
 ln(1  1.9br )
4ln(1  1.9br )
4ln(1  1.9br )
4 ln(1  1.9br )
1.9
4br
1  1.9br
Day 27 QQ2.2
Q2.2.4 For the SAFT EOS:
A  Aig
A1(br ) A2(br )
 3ln(1  br ) 

RT
T
T2
Derive an expression for (U-Uig)/RT
(a)
3
A1 A2

 2
(1  br ) T T
(b)
3
A1 A2


(1  br ) T T 2
(c)
A1
A2
2 2
T
T
(d)
A1 A2

T T2
Day 28 EOSs
28.1. Why do we write our Equation of State
models as Z(T,V) or A(T,V) when what we want is
V(T,P)?
A. because dA = PdV – SdT is more
“fundamental.”
B. because pressure is a sum of forces, but
density is not a sum of pressures.
C. to make life difficult for poor students.
D. because V(T,P) is not a function.
Day 28 EOSs
4cbr
9.5qYbr
28.2. FOR the ESD EOS:
Z  1

1  1.9br 1  1.8Ybr
where Y = exp(e/kBT)-1.06
c,q are constants
Evaluate
( A  Aig )T ,V
RT
(a)
4c ln(1  1.9br ) 9.5qY

ln(1  1.8Ybr )
1.9
1.8
(b)
4c ln(1  1.9br )  9.5q ln(1  1.8Ybr )
(c)
4c ln(1  1.9br ) 9.5qY

ln(1  1.8Ybr )
1.9
1.8
(d)
4c ln(1  1.9br ) 9.5q

ln(1  1.8Ybr )
1.9
1.8
28.3.True or false
Day 28 EOSs
____The compressibility factor Z is always less than or
equal to unity.
____The critical properties Tc and Pc are constants for a
given compound.
____A steady-state flow process is one for which the
velocities of all streams may be assumed negligible.
____The temperature of a gas undergoing a continuous
throttling process may either increase or decrease across
the throttling device, depending on conditions.
(a) FTFT
(b) TTTF
(c) TFTF
(d) FFFT
Day 29 HW
29.1. At 2.25$/gal, and 0.692 g/cm3, the price of
gasoline in $/kg is closest to:
(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.8
Day 29 HW
29.2. At 4$/gal, the price of propane in $/kg is
closest to:
(a) 1
(b) 2
(c) 3
(d) 4
Day 29 HW
29.3. Referring to problem 6.21, the resulting
equation of state at the given conditions has the
value of Z = ___
(a) 0.5
(b) 1.5
(c) 2.5
(d) 3.5
Day 33 DepFuns
33.1 FOR the Scott EOS:
Evaluate
Z
1  2 br
1  2 br
(G  G ig )
RT
(a)
2ln(1  2br )  Z  1  ln( Z )
(b)
 1  2br 
ln 
 Z  1  ln( Z )

 1  2br 
(c)
4ln(1  2br )  Z  1  ln( Z )
(d)
2ln(1  2br )  Z  1  ln( Z )
Day 33 DepFuns
Z  1
33.2 FOR the EOS:
Evaluate
(a)
( A  Aig )T ,V
RT
2ln(1  br )
a / 2
(b)
1  br 
(c)
a  br (2  br ) 

2 
2  1  br  
(d)
abr
1  br
2
abr
1  br 
3
Day 33
33.3 Which of the following would indicate a small acentric
factor?
(a) high molecular weight
(b) a noble gas
(c) strong hydrogen bonding
(d) a spherical molecule with strong hydrogen bonding
Day 33
33.4. “Boiling” is the process of transforming a liquid into a
vapor. “Sublimation” is the process of transforming a solid
into a vapor. For carbon dioxide, the heat of sublimation
(HV-HS) is roughly 24750 J/mole at the triple point
temperature and pressure of -56.6C and 5.27 bars.
Estimate the sublimation temperature at 0.5 bar.
(a) 240
(b) 225
(c) 210
(d) 195
QikQiz2.3
Q2.3.1 Vapor ethylene oxide is compressed from 25C and 1
bar to 125C and 20 bar. The change in entropy (J/mol-K) is:
(a) 8
(b) 10
(c) -12
(d) -16
QikQiz2.3
Q2.3.2. Determine the work (kW) required to continuously
compress reversibly and adiabatically 0.5kg/min of
ethylene oxide from 25C and 1 bar to 20 bar. The
temperature (K) exiting the compressor is:
(a) 425
(b) 450
(c) 470
(d) 500
QikQiz2.3
Q2.3.3. Determine the work (kW) required to continuously
compress reversibly and adiabatically 0.5kg/min of
ethylene oxide (MW=40) from 25C and 1 bar to 20 bar.
(a) 1.8
(b) 2.0
(c) 200
(d) 9000
QikQiz2.3
Q2.3.4. Ethylene oxide (MW=40) enters a throttle as
saturated liquid at 2MPa and exits at 1bar. Determine the
quality (%) at the exit.
(a) 45
(b) 35
(c) 25
(d) 15
Day 33 HW Ch 7&8
33.1 FOR the SRK EOS:
Evaluate
( H  H ig )T ,P
RT
(a)
 a
a   Tr
Z 1 
 c
bRT
 bRT

1

2
 1  br 
(b)
 a
a  Tr
Z 1 
 c
bRT
 bRT

1

 1  br 
(c)
 a ac  Tr
Z 1 

bRT
bRT


 ln 1  br 

(d)
 a ac  Tr
Z 1 

bRT
bRT


 ln 1  br   ln( Z )

QikQiz2.4
Q2.4.1. Derive the simplest form of the Gibbs energy
departure function for the following equation of state:
Z = 1 + 4br/(1-2br) - ar/RT1.7
(a) –ln(1-br) - ar/RT1.7 + Z – 1 - lnZ
(b) -2ln(1-2br) - ar/RT1.7 + Z – 1 - lnZ
(c) -2ln(1-2br) + 1.7ar/RT2.7 + Z – 1 - lnZ
(d) -4ln(1-2br) - a/RT1.7 + Z – 1 - lnZ
QikQiz2.4
Q2.4.2. Estimate the vapor pressure (bars) of n-butane at
T=40C.
(a) 1
(b) 2
(c) 3
(d) 4
QikQiz2.4
Q2.4.3. Estimate the saturation temperature (K) of nbutane at P=20bars.
(a) 300
(b) 325
(c) 350
(d) 375
QikQiz2.4
Q2.4.4. “Boiling” is the process of transforming a liquid into
a vapor. “Sublimation” is the process of transforming a
solid into a vapor. For carbon dioxide, the heat of
sublimation (HV-HS) is roughly 24750 J/mole at the triple
point temperature and pressure of -56.6C and 5.27 bars.
Estimate the sublimation temperature at 0.5 bar.
(a) 240
(b) 225
(c) 210
(d) 195
QikQiz2.5
Qq2.5.1. FOR the Scott EOS: Z 
Evaluate
1  2 br
1  2 br
(G  G ig )
RT
(a)
2ln(1  2br )  Z  1  ln( Z )
(b)
 1  2br 
ln 
 Z  1  ln( Z )

 1  2br 
(c)
4ln(1  2br )  Z  1  ln( Z )
(d)
2ln(1  2br )  Z  1  ln( Z )
QikQiz2.5
Qq2.5.2. FOR the EOS:
Evaluate
(a)
( A  Aig )T ,V
RT
2ln(1  br )
a / 2
(b)
1  br 
(c)
a  br (2  br ) 

2 
2  1  br  
(d)
abr
1  br
2
Z  1
abr
1  br 
3
QikQiz2.5
Q2.5.3. A power cycle is to run on bromine
operating at 0.1MPa in the condenser and 6MPa in
the boiler. Estimate the Carnot efficiency.
(a) 0.3
(b) 0.4
(c) 0.5
(d) 0.6
QikQiz2.5
Q2.5.4. A Rankine cycle is to operate on bromine
operating at 0.1MPa in the condenser and 6MPa in
the boiler. Estimate the turbine work (kJ/mol).
(a) 18
(b) 12
(c) 6
(d) 3
QikQiz2.5
Q2.5.5. A Rankine cycle is to operate on bromine
operating at 0.1MPa in the condenser and 6MPa in
the boiler. Estimate the Rankine efficiency.
(a) 0.3
(b) 0.4
(c) 0.5
(d) 0.6
DayNew Preview
End of File