General Information on Main Equations of State:
1. Ideal Gas Law
𝑃𝑉 = 𝑛𝑅𝑇
2. Van der Waals
𝑎
(𝑃 + 2 ) (𝑉𝑚 − 𝑏) = 𝑅𝑇
𝑉𝑚
𝑤ℎ𝑒𝑟𝑒 𝑉𝑚 𝑖𝑠 𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
− 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 𝒂 𝑎𝑛𝑑 𝒃 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠 𝑃𝑐 , 𝑇𝑐 𝑎𝑛𝑑 𝑉𝑐
27(𝑅𝑇𝑐 )2
2
𝑎 = 3𝑃𝑐 𝑉𝑐 =
64𝑃𝑐
𝑉𝑐 𝑅𝑇𝑐
𝑏= =
3
8𝑃𝑐
3. Redlich-Kwong
- Superior to Van der Waals, but performs poorly with respect to liquid phase and
therefore cannot be used for calculating vapor-liquid equilibria
𝑅𝑇
𝑎
𝑃=
−
𝑉𝑚 − 𝑏 √𝑇𝑉𝑚 (𝑉𝑚 + 𝑏)
0.42748𝑅 2 𝑇𝑐2.5
𝑎=
𝑃𝑐
0.08662𝑅𝑇𝑐
𝑏=
𝑃𝑐
- Good for calculations of gas phase properties when the ratio of the pressure to
the critical pressure is less than about one-half of the ratio of the temperature to
the critical temperature:
𝑃
𝑇
<
𝑃𝑐 2𝑇𝑐
4. Peng-Robinson
𝑅𝑇
𝑎𝛼
𝑃=
− 2
𝑉𝑚 − 𝑏 𝑉𝑚 + 2𝑏𝑉𝑚 − 𝑏 2
0.457235𝑅 2 𝑇𝑐2
𝑎=
𝑃𝑐
0.077796𝑅𝑇𝑐
𝑏=
𝑃𝑐
2
𝛼 = (1 + 𝜅(1 − 𝑇𝑟0.5 ))
𝜅 = 0.37464 + 1.54226𝜔 − 0.26992𝜔2
𝑇
𝑇𝑟 =
𝑇𝑐
Question 3
Small hydrocarbons are often liquefied for ease of transport. Consider the liquefaction of
methane by the Linde process as shown in the diagram below. Methane is fed at 4 bar and
295 K where it is mixed with uncondensed vapor at 4 bar and 290 K. The stream is then fed
to a compressor where it is compressed to 60 bar and then cooled isobarically (ΔP=0) to
295 K. This stream then exchanges heat with the vapor stream leaving the throttle (no
mass between the streams is exchanged; process is adiabatic). The vapor stream leaving
the heat exchanger is then throttled isenthalpically (ΔH=0) and adiabatically (ΔQ=0) to
4 bar, dropping the temperature of the vapor and liquid streams leaving the throttle in
equilibrium sufficient to liquefy some of the methane. The uncondensed methane vapor
exchanges heat with the incoming stream and then is mixed with the feed. Using a
volumetric flow rate of 100 L/s leaving the compressor as a basis of calculation, answer the
following questions:
a. (10 pts) Solve for the molar flow rate of the methane leaving the compressor.
Explicitly state what equation of state you are using and why.
Given: P = 6bar = 5.921atm , T = 295K, R=0.082L*atm/K*mol
Use ideal gas law
𝑷𝑽̇ = 𝒏̇ 𝑹𝑻
𝑷𝑽̇
(𝟓. 𝟗𝟐𝟏𝒂𝒕𝒎)(𝟏𝟎𝟎 𝑳⁄𝒔)
𝒏̇ =
=
= 𝟐𝟒. 𝟒𝟖 𝒎𝒐𝒍⁄𝒔
𝑹𝑻 (𝟎. 𝟎𝟖𝟐 𝑳𝒂𝒕𝒎⁄𝑲𝒎𝒐𝒍)(𝟐𝟗𝟓𝑲)
b. (10 pts) What is the temperature (in K) of the liquid methane leaving the
throttler? Hint: this liquid stream and the leaving vapor stream share the
same temperature and pressure.
Heat Exchanger is isenthalpic and adiabatic
Use the following equation to solve for outlet temp of heat exchanger
𝒅𝑯 = 𝟎 = 𝑪𝒑 𝒅𝑻
𝑪𝑷 𝒇𝒐𝒓 𝒎𝒆𝒕𝒉𝒂𝒏𝒆 𝒈𝒂𝒔 = 𝟎. 𝟎𝟏𝟗𝟖𝟕 + 𝟓. 𝟎𝟐𝟏 ∗ 𝟏𝟎−𝟓 𝑻
𝑻𝒇
𝟎 = ∫ (𝟎. 𝟎𝟏𝟗𝟖𝟕 + 𝟓. 𝟎𝟐𝟏 ∗ 𝟏𝟎−𝟓 𝑻)𝒅𝑻
𝟐𝟗𝟓
= (𝟎. 𝟎𝟏𝟗𝟖𝟕𝑻𝒇 + 𝟏. 𝟎𝟎𝟒 ∗ 𝟏𝟎−𝟒 𝑻𝟐𝒇 )
− (𝟎. 𝟎𝟏𝟗𝟖𝟕𝑻𝟐𝟗𝟓 + 𝟏. 𝟎𝟎𝟒 ∗ 𝟏𝟎−𝟒 𝑻𝟐𝟐𝟗𝟓 )
𝟎 = 𝟏. 𝟎𝟎𝟒 ∗ 𝟏𝟎−𝟒 𝑻𝟐𝒇 + 𝟎. 𝟎𝟏𝟗𝟖𝟕𝑻𝒇 − 𝟓. 𝟖𝟗𝟏
Use quadratic equation to solve for Tf
−𝟎. 𝟎𝟏𝟗𝟖𝟕 ± √𝟎. 𝟎𝟎𝟎𝟑𝟗𝟒𝟖 − 𝟒 ∗ 𝟎. 𝟎𝟎𝟎𝟏𝟎𝟎𝟒 ∗ −𝟓. 𝟖𝟗𝟏
𝑻𝒇 =
𝟐 ∗ 𝟎. 𝟎𝟎𝟎𝟏𝟎𝟎𝟒
−𝟎. 𝟎𝟏𝟗𝟖𝟕 ± 𝟎. 𝟎𝟓𝟐𝟓
𝑻𝒇 =
= 𝟏𝟔𝟑. 𝟏𝟓𝑲
𝟎. 𝟎𝟎𝟎𝟐
Throttle is adiabatic
𝑸 = 𝟎 = 𝒎𝒄𝚫𝑻
No change in temperature because both m & c are held constant
𝑻𝒐𝒖𝒕 = 𝟏𝟔𝟑. 𝟏𝟓𝑲
c. (20 pts) Use an energy balance to find the moles of liquid methane that leave
the throttler. Approximate heat capacity date is listed. You can neglect
isothermal pressure changes for the liquid, but since the methane vapor is
not ideal you will have to calculate the enthalpy change for isothermal
pressure changes using the following empirical relation:
𝑘𝐽
𝑏𝑎𝑟 ∙ 𝑚𝑜𝑙
The hat above the H means it’s the partial enthalpy of methane
̂ = 𝑎 [𝑃2 − 𝑃1 ]; 𝑎 = −0.018
𝐻
𝑸 = 𝟎 = 𝚫𝑯 = ∑ 𝒏𝒊 𝑯𝒊 − ∑ 𝒏𝒊 𝑯𝒊
𝒐𝒖𝒕
𝒊𝒏
= 𝒏𝒐𝒖𝒕,𝒗 𝑯𝒐𝒖𝒕,𝒍 + 𝒏𝒐𝒖𝒕,𝒍 𝑯𝒐𝒖𝒕,𝒍 − 𝒏𝒊𝒏,𝒗 𝑯𝒊𝒏,𝒗 − 𝒏𝒐𝒖𝒕,𝒍 𝑯𝒐𝒖𝒕,𝒍
̂
𝟎 = 𝒏𝒗 𝑯𝒗 − 𝒏𝒍 𝚫𝑯𝒍
̂ 𝒗 = 𝒂(𝑷𝟐 − 𝑷𝟏 )
𝑯
Antoine equation constants: Log10(P*)=A – B/(T+C) P* in bar; T in K
Substance
A
B
C
Methane
3.9895 443.028 -0.49
Substance
Methane liquid
Methane vapor
Cp (kJ/mol-K)
.05293
.01987+5.021e-05*T (T in K)
)
Question 5
An air stream at 25oC and 1 atm is contaminated with 4 mol% benzene vapor. To
satisfy EPA and OSHA requirements, 85% by mass of the incoming benzene must be
stripped from the air stream. To accomplish this, the on-site engineer has
constructed a compressor/condenser to pressurize the incoming stream
isothermally (𝚫𝑻 = 𝟎). Two streams emerge from the compressor/condenser in
equilibrium at 25oC: an air stream with the reduced concentration of benzene, and a
stream of pure liquid benzene. Using a basis of 100 L(STP)/s of the incoming air
stream, answer the following questions:
a. (5 pts) What is the molar flow rate of the incoming air stream in mol/s?
Basis = 100L/s for incoming air stream
Ideal gas at STP: 24L/mol
𝟏𝟎𝟎𝑳 𝟏𝒎𝒐𝒍
∗
= 𝟒. 𝟏𝟔𝟕𝒎𝒐𝒍/𝒔
𝒔
𝟐𝟒𝑳
Molar flow rate of air:
𝟒. 𝟏𝟔𝟕 ∗ 𝟎. 𝟗𝟔 = 𝟒𝒎𝒐𝒍/𝒔
Sidenote:
MW of benzene (C6H6) = 78.11g/mol
MW of Air (79% N2, 21% O2)=
(0.79*28g/mol)+(0.21*32g/mol)=28.84g/mol
b. (15 pts) What pressure (in atm) are the benzene liquid and benzene/air
stream leaving the compressor? To solve this, first compute the molar
fraction of benzene in the exit air stream.
Molar flow rate of benzene in inlet stream = 0.167mol/s
Mass flow rate of benzene in inlet stream = 0.167*78.11=13.04g/s
Mass flow rate of benzene in outlet stream = 13.04*0.15=1.96g/s
Molar flow rate of benzene in outlet stream = 1.96/78.11=0.025mol/s
For isothermal process we can use the following total balance:
𝑷𝒊𝒏 𝑽𝒊𝒏 = 𝑷𝒐𝒖𝒕 𝑽𝒐𝒖𝒕 = 𝑷𝒍 𝑽𝒍 + 𝑷𝒗 𝑽𝒗
(𝟏𝒂𝒕𝒎)(𝟏𝟎𝟎𝑳) = 𝑷𝒍 𝑽𝒍 + 𝑷𝒗 𝑽𝒗
We also know the following V & P values:
𝑽𝒗 = 𝑽𝒗,𝑪𝟔 𝑯𝟔 + 𝑽𝒗 𝒂𝒊𝒓
𝑽𝒊𝒏,𝒂𝒊𝒓 = 𝑽𝒗,𝒂𝒊𝒓 = (𝟒𝒎𝒐𝒍)(𝟐𝟒 𝑳⁄𝒎𝒐𝒍) = 𝟗𝟔𝑳
⇒ 𝑽𝒗,𝑪𝟔 𝑯𝟔 = 𝑽𝒗 + 𝟗𝟔
𝑽𝒊𝒏 = 𝟏𝟎𝟎𝑳 = 𝑽𝒊𝒏,𝑪𝟔 𝑯𝟔 + 𝑽𝒊𝒏,𝒂𝒊𝒓
𝑽𝒊𝒏,𝑪𝟔 𝑯𝟔 = 𝑽𝒗,𝑪𝟔 𝑯𝟔 + 𝑽𝒍
𝟐𝟒𝑳
𝑽𝒊𝒏,𝑪𝟔 𝑯𝟔 = 𝟎. 𝟏𝟔𝟕𝒎𝒐𝒍 (
) = 𝟒. 𝟎𝟎𝟖𝑳
𝟏𝒎𝒐𝒍
⇒ 𝟒. 𝟎𝟎𝟖 = 𝑽𝒗,𝑪𝟔 𝑯𝟔 + 𝑽𝒍
𝑽𝒗,𝑪𝟔 𝑯𝟔 = 𝟒. 𝟎𝟎𝟖 − 𝑽𝒍
𝒄𝒐𝒎𝒃𝒊𝒏𝒆 𝒓𝒆𝒅 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒂𝒏𝒅 𝒕𝒐𝒕𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒂𝒏𝒅
𝒔𝒐𝒍𝒗𝒆 𝒂 𝒔𝒆𝒕 𝒐𝒇 𝒕𝒘𝒐 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒘𝒊𝒕𝒉 𝒕𝒘𝒐 𝒖𝒏𝒌𝒐𝒏𝒘𝒏:
𝟏𝟎𝟎 = 𝑽𝒗 + 𝑽𝒍
𝟒. 𝟎𝟎𝟖 − 𝑽𝒍 = 𝑽𝒗 + 𝟗𝟔
𝑽𝒗 =
Use ideal gas law and volumes above to compute pressures:
c. (20 pts) Estimate the energy requirement (in kW) required to drive the
condenser and cool the gases if the compressor does 2.5 kW of work to the
gases. Assume negligible enthalpy changes for compressing gases. (𝚫𝑯 =
𝟎)To help your calculations, please use the inlet-outlet enthalpy table on the
next page and explicitly state your reference state(s) directly on the
table!!
Substance
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
Air(g)
Benzene(v)
Benzene(l)
r.s._________________________________________________________________________
Hout (kJ/mol)
6. (35 pts) In the Haber process, ammonia can be produced from nitrogen and hydrogen
gas. The unbalanced reaction is as follows and occurs isothermally at 400oC and 1 atm:
𝐻2(𝑔) + 𝑁2(𝑔) → 𝑁𝐻3(𝑣)
In order to allow the reaction to proceed isothermally, 320 kW of heat is transferred from
the reactor. Nitrogen and hydrogen are stoichiometrically balanced, and there is no initial
ammonia. For a basis of a fresh feed rate of 100 mol/s of nitrogen, determine the single
pass conversion of nitrogen by answering the following questions:
d. (5 pts) Write a stoichiometrically balanced reaction for the process.
𝟑𝑯𝟐(𝒈) + 𝑵𝟐(𝒈) → 𝟐𝑵𝑯𝟑(𝒗)
e. (5 pts) Solve the material balances as completely as possible. You won’t be
able to solve all of them– leave the unknown variables in terms of a single
unknown.
Reactive mechanism so do material balances in terms of atomic
components rather than molecules
Nitrogen Balance: input = output
𝟏𝟎𝟎𝒎𝒐𝒍 𝑵𝟐
𝟐𝒎𝒐𝒍 𝑵
𝒏̇ 𝟐 (𝒎𝒐𝒍 𝑵𝑯𝟑 )
𝟏 𝒎𝒐𝒍 𝑵
𝒏̇ 𝟐 (𝒎𝒐𝒍 𝑵𝟐 ) 𝟐 𝒎𝒐𝒍 𝑵
(
)(
)=(
)(
)+(
)(
)
𝒔
𝟏 𝒎𝒐𝒍 𝑵𝟐
𝒔
𝟏 𝒎𝒐𝒍 𝑵𝑯𝟑
𝒔
𝟏 𝒎𝒐𝒍 𝑵𝟐
𝟐𝟎𝟎 = 𝒏̇ 𝟐,𝑵𝑯𝟑 + 𝟐𝒏̇ 𝟐,𝑵𝟐
Hydrogen Balance: input = output
𝒏̇ 𝟏,𝑯𝟐 (𝒎𝒐𝒍𝑯𝟐 )
𝟐𝒎𝒐𝒍 𝑯
)(
)
𝒔
𝟏 𝒎𝒐𝒍 𝑯𝟐
𝒏̇ 𝟐 (𝒎𝒐𝒍𝑵𝑯𝟑 )
𝟑𝒎𝒐𝒍 𝑯
𝒏̇ 𝟐 (𝒎𝒐𝒍 𝑯𝟐 )
𝟐𝒎𝒐𝒍 𝑯
=(
)(
)+(
)(
)
𝒔
𝟏𝒎𝒐𝒍 𝑵𝑯𝟑
𝒔
𝟏 𝒎𝒐𝒍 𝑯𝟐
(
̇ 𝟐 = 𝟑𝒏̇ 𝟐,𝑵𝑯𝟑 + 𝟐𝒏̇ 𝟐,𝑯𝟐
𝟐𝒏𝟏,𝑯
𝟑𝒏̇ 𝟏,𝑯𝟐 = 𝒏̇ 𝟏,𝑵𝟐 = 𝟏𝟎𝟎
𝟔𝟔. 𝟔𝟕 = 𝟑(𝟐𝟎𝟎 − 𝟐𝒏̇ 𝟐,𝑵𝟐 ) + 𝟐𝒏̇ 𝟐,𝑯𝟐
−𝟓𝟑𝟑. 𝟑 = −𝟔𝒏̇ 𝟐,𝑵𝟐 + 𝟐𝒏̇ 𝟐,𝑯𝟐
Overall Balance: 𝒏̇ 𝟏,𝑯𝟐 + 𝒏̇ 𝟏,𝑵𝟐 = 𝒏̇ 𝟐,𝑵𝑯𝟑 + 𝒏̇ 𝟐,𝑵𝟐 + 𝒏̇ 𝟐,𝑯𝟐
𝟐𝒎𝒐𝒍 𝑵𝑯𝟑
𝟏𝟎𝟎𝒎𝒐𝒍 𝑵𝟐 (
) = 𝟐𝟎𝟎𝒎𝒐𝒍 𝑵𝑯𝟑
𝟏𝒎𝒐𝒍 𝑵𝟐
𝟑𝟑. 𝟑 + 𝟏𝟎𝟎 = 𝟐𝟎𝟎 + 𝒏̇ 𝟐,𝑵𝟐 + 𝒏̇ 𝟐,𝑯𝟐
f. (20 pts) Use an energy balance around the reactor to solve for the remaining
material balances. To help your calculations, please use the inlet-outlet
enthalpy table below and explicitly state your reference state(s) directly
on the table!!
Use reference temperature of 25C for all components
Substance
nin (mol/s)
Hin (kJ/mol)
nout (mol/s)
H2
N2
NH3
r.s._________________________________________________________________________
Hout (kJ/mol)
g. (5 pts) Calculate the single pass conversion of nitrogen.