conceptests in chemical engineering thermodynamics unit 2: generalized analysis of fluid properties...
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ConcepTests in Chemical Engineering ThermodynamicsUnit 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
(a) (P/V)S (b) (T/V)U (c) (U/T)V (d) (P/T)V
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) Cv(T/P)V/T (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/T)P/Cp(c) (T/P)S
(d) -(V/T)P
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) V(P/S)V/T (b) TS(V/T)P/Cv(c) -TS/Cp (d) -S(T/S)P
Day 22 MRs and EOSs
22.4. 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) [1/(1-b)] (b) [1/(1-b)2] (c) –[a/RT] (d) [a/RT2]
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-b)] – [a/RT]
(a) -S(T/P)V
(b) Cp(T/P)V
(c) TS/P (d) -VS(T/V)P /Cp
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) V(b) V-T(V/T)P
(c) -T(S/P)T+V (d) -T(P/T)V - P
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) 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
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) 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
24.1. 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) 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
24.2. 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) (b) (c) (d)
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)
2 2
1
1 2 ( )
a da b
bRT bRT dT b b
2
1
1 2 ( )
a da b
bRT bR dT b b
21
1 2 ( )c ra Ta b
bRT a b b
21 2 ( )c ra T b
bRT b b
(a) (b) (c) (d)
Day 24 MRs and EOSs
24.4. FOR the SRK(1972) EOS:
-T(Z/T)V = Evaluate
2( )1
1c ra Ta b
bRT a b
1 ln 1c ra Tab
bRT a
ln 1c ra Tb
bRT
11
c ra Ta b
bRT a b
0
( )b
V
Z d bT
T b
20.5( )1 ln
1c ra Ta b
bRT a b
(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
25.1. 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) (U/V)T - T (S/V)T (b) [(P/T)V – P] + (P/T)V (c) -P (d) –T (P/T)V
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) (b) (c) (d)
Day 26 Dep Funs
26.1. FOR the SRK EOS:
Evaluate
ln(1 ) ln(1 )a
b bbRT
ln(1 ) ln(1 )a
b bbRT
2
ln(1 )ln(1 )
b ab
b b RT
11 1
b a bZ
b bRT b
0
( )( 1)
bd b
Zb
20.5( )ln(1 ) ln
1
a bb
bRT b
(a) (b) (c) (d)
Day 26 DepFuns
26.2. FOR the PR EOS:
Evaluate(Hint:p602)
ln(1 ) ln(1 )a
b bbRT
1 2.414ln(1 ) ln
1 0.414
a bb
bRT b
2 8ln(1 ) ln
8 2 8
a bb
bRT b
211 1 2 ( )
b a bZ
b bRT b b
0
( )( 1)
bd b
Zb
1ln(1 ) arctan
8 2
a bb
bRT
(a) (b) (c) (d)
Day 26 Dep Funs26.3. FOR the ESD EOS:where Y = exp(/kBT)-1.06c and q are constants
Evaluate
4 ln(1 1.9 ) 9.5ln(1 1.8 )
1.9 1.8
c b qYYb
4 ln(1 1.9 ) 9.5ln(1 1.8 )
1.9 1.8
c b qYb
4 ln(1 1.9 ) 9.5ln(1 1.8 )
1.9 1.8
c b qYYb
4 9.511 1.9 1 1.8
cb qYbZ
b Yb
4 ln(1 1.9 ) 9.5 ln(1 1.8 )c b q Yb
0
( )( 1)
bd b
Zb
(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
Q2.2.1. 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) -T(V/T)P (b) [-(V/T)P – CpV/(ST)]-1 (c) –(ST/Cp)(V/T)P + V (d) -(T/V)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) (b) (c) (d)
Day 27 QQ2.2
Q2.2.3 FOR the ESD EOS:
Evaluate
ln(1 1.9 )b
4ln(1 1.9 )b
4ln(1 1.9 )
1.9
b
411 1.9
HS bZ
b
0
( )( 1)
bHS d bZ
b
4ln(1 1.9 )b
(a) (b) (c) (d)
Day 27 QQ2.2
Q2.2.4 For the SAFT EOS:
Derive an expression for (U-Uig)/RT
2
3 1 2
(1 )
A A
b T T
2
1( ) 2( )3ln(1 )
igA A A b A bb
RT T T
2
3 1 2
(1 )
A A
b T T
2
1 22
A A
T T
2
1 2A A
T T
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
(a) (b) (c) (d)
Day 28 EOSs28.2. FOR the ESD EOS:where Y = exp(e/kBT)-1.06c,q are constants
Evaluate
4 ln(1 1.9 ) 9.5ln(1 1.8 )
1.9 1.8
c b qYYb
4 ln(1 1.9 ) 9.5ln(1 1.8 )
1.9 1.8
c b qYb
4 ln(1 1.9 ) 9.5ln(1 1.8 )
1.9 1.8
c b qYYb
4 9.511 1.9 1 1.8
cb qYbZ
b Yb
,( )T VA Aig
RT
4 ln(1 1.9 ) 9.5 ln(1 1.8 )c b q Yb
(a) FTFT(b) TTTF (c) TFTF(d) FFFT
Day 28 EOSs28.3.True or false____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) 0.2 (b) 0.4 (c) 0.6(d) 0.8
Day 29 HW
29.1. At 2.25$/gal, and 0.692 g/cm3, the price of
gasoline in $/kg is closest to:
(a) 0.5 (b) 1.5(c) 2.5(d) 3.5
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) (b) (c) (d)
Day 33 DepFuns
33.1 FOR the Scott EOS:
Evaluate
2ln(1 2 ) 1 ln( )b Z Z
2ln(1 2 ) 1 ln( )b Z Z
4ln(1 2 ) 1 ln( )b Z Z
1 2
1 2
bZ
b
( )igG G
RT
1 2ln 1 ln( )1 2
bZ Z
b
(a) (b) (c) (d)
Day 33 DepFuns
33.2 FOR the EOS:
Evaluate
2ln(1 )b
1
ab
b
2(2 )
2 1
a b b
b
31
1
abZ
b
,( )ig T VA A
RT
2/ 2
1
a
b
(a) high molecular weight(b) a noble gas(c) strong hydrogen bonding(d) a spherical molecule with strong hydrogen bonding
Day 33
33.3 Which of the following would indicate a small acentric factor?
(a) 240(b) 225(c) 210(d) 195
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 -56.6C and 5.27 bars. Estimate the sublimation temperature at 0.5 bar.
(a) 8 (b) 10(c) -12(d) -16
QikQiz2.3
Q2.3.1 Vapor ethylene oxide is compressed from 25C and 1 bar to 125C and 20 bar. The change in entropy (J/mol-K) is:
(a) 425 (b) 450(c) 470(d) 500
QikQiz2.3
Q2.3.2. Determine the work (kW) required to continuously compress reversibly and adiabatically 0.5kg/min of ethylene oxide from 25C and 1 bar to 20 bar. The temperature (K) exiting the compressor is:
(a) 1.8 (b) 2.0(c) 200(d) 9000
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 25C and 1 bar to 20 bar.
(a) 45 (b) 35(c) 25(d) 15
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) (b) (c) (d)
Day 33 HW Ch 7&8
33.1 FOR the SRK EOS:
Evaluate
1 ln 1 ln( )c ra TaZ b Z
bRT bRT
1 ln 1c ra TaZ b
bRT bRT
,( )ig T PH H
RT
1
11
c ra TaZ
bRT bRT b
21
11
c ra TaZ
bRT bRT b
(a) –ln(1-b) - a/RT1.7 + Z – 1 - lnZ(b) -2ln(1-2b) - a/RT1.7 + Z – 1 - lnZ(c) -2ln(1-2b) + 1.7a/RT2.7 + Z – 1 - lnZ(d) -4ln(1-2b) - a/RT1.7 + Z – 1 - lnZ
QikQiz2.4
Q2.4.1. Derive the simplest form of the Gibbs energy departure function for the following equation of state: Z = 1 + 4b/(1-2b) - a/RT1.7
(a) 300(b) 325(c) 350(d) 375
QikQiz2.4
Q2.4.3. Estimate the saturation temperature (K) of n-butane at P=20bars.
(a) 240(b) 225(c) 210(d) 195
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 -56.6C and 5.27 bars. Estimate the sublimation temperature at 0.5 bar.
(a) (b) (c) (d)
Qq2.5.1. FOR the Scott EOS:
Evaluate
2ln(1 2 ) 1 ln( )b Z Z
2ln(1 2 ) 1 ln( )b Z Z
4ln(1 2 ) 1 ln( )b Z Z
1 2
1 2
bZ
b
( )igG G
RT
1 2ln 1 ln( )1 2
bZ Z
b
QikQiz2.5
(a) (b) (c) (d)
Qq2.5.2. FOR the EOS:
Evaluate
2ln(1 )b
1
ab
b
2(2 )
2 1
a b b
b
31
1
abZ
b
,( )ig T VA A
RT
2/ 2
1
a
b
QikQiz2.5
(a) 0.3(b) 0.4(c) 0.5(d) 0.6
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) 18(b) 12(c) 6(d) 3
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) 0.3(b) 0.4(c) 0.5(d) 0.6
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.