Thermodynamic Analysis Of Gas Behavior In Explosive Venting And Work Calculation: A Case Study On Nitrogen And Carbon Dioxide
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Thermodynamic Analysis of Gas Behavior in Explosive Venting and Work
Calculation: A Case Study on Nitrogen and Carbon Dioxide
#1
A tank containing carbon dioxide at 400 K and 50 bar is vented until the temperature in the tank falls to
300 K. Assuming there is no heat transfer between the gas and the tank, find the pressure in the tank at the
end of the venting process and the fraction of the initial mass of gas remaining in the tank if carbon
dioxide obeys the ideal gas and PR equations of state. Constant pressure heat capacity of carbon dioxide:
πΆπ(π½ πππβ1 πΎβ1)=22.243+(5.977Γ10β2)π+(β3.499Γ10β5)π2+(7.464Γ10β9)π3
You need to first show that the PR entropy departure function is given by
(π β ππΌπΊ)π,π
ππ = π ππππ(π β π΅) π€βπππ π΅ = ππ
π π
Answer: The content you've provided is part of a larger thermodynamic problem involving COβ in a tank
that is vented adiabatically, and it asks you to derive the Peng-Robinson (PR) entropy departure function
and then solve for the final pressure and mass of the gas remaining in the tank.
Answer to the First Part: Derivation of the PR Entropy Departure Function
The first part of your question asks you to derive the PR entropy departure function, which
involves comparing the entropy of a real gas (using the Peng-Robinson equation of state) to the
entropy of an ideal gas.
Starting point: Ideal Gas Entropy and Real Gas Entropy
For a real gas, the entropy change SβSIGS - S^{IG} can be calculated using the following
general thermodynamic relation:
dS=CPdTTβRln(Z)dS = C_P \frac{dT}{T} - R \ln(Z)
Where:
β’ SS is the entropy of the real gas
β’ SIGS^{IG} is the entropy of the ideal gas
β’ ZZ is the compressibility factor
β’ CPC_P is the heat capacity at constant pressure
β’ RR is the universal gas constant
For the ideal gas, the entropy change is:
dSIG=CPdTTdS^{IG} = C_P \frac{dT}{T}
Therefore, the difference in entropy between the real gas and the ideal gas (i.e., the entropy
departure) is:
SβSIG=β«T1T2CPdTTβRβ«T1T2ln(Z)dTS - S^{IG} = \int_{T_1}^{T_2} C_P \frac{dT}{T} - R
\int_{T_1}^{T_2} \ln(Z) dT
This leads to the general expression for entropy departure:
(SβSIG)=βRβ«T1T2ln(Z)dT(S - S^{IG}) = -R \int_{T_1}^{T_2} \ln(Z) dT
Peng-Robinson Equation of State and Compressibility Factor
The Peng-Robinson equation of state (PR EOS) is used to describe the behavior of real gases:
P=RTVβbβaV(V+b)+b(Vβb)P = \frac{RT}{V - b} - \frac{a}{V(V + b) + b(V - b)}
Calculation: A Case Study on Nitrogen and Carbon Dioxide
#1
A tank containing carbon dioxide at 400 K and 50 bar is vented until the temperature in the tank falls to
300 K. Assuming there is no heat transfer between the gas and the tank, find the pressure in the tank at the
end of the venting process and the fraction of the initial mass of gas remaining in the tank if carbon
dioxide obeys the ideal gas and PR equations of state. Constant pressure heat capacity of carbon dioxide:
πΆπ(π½ πππβ1 πΎβ1)=22.243+(5.977Γ10β2)π+(β3.499Γ10β5)π2+(7.464Γ10β9)π3
You need to first show that the PR entropy departure function is given by
(π β ππΌπΊ)π,π
ππ = π ππππ(π β π΅) π€βπππ π΅ = ππ
π π
Answer: The content you've provided is part of a larger thermodynamic problem involving COβ in a tank
that is vented adiabatically, and it asks you to derive the Peng-Robinson (PR) entropy departure function
and then solve for the final pressure and mass of the gas remaining in the tank.
Answer to the First Part: Derivation of the PR Entropy Departure Function
The first part of your question asks you to derive the PR entropy departure function, which
involves comparing the entropy of a real gas (using the Peng-Robinson equation of state) to the
entropy of an ideal gas.
Starting point: Ideal Gas Entropy and Real Gas Entropy
For a real gas, the entropy change SβSIGS - S^{IG} can be calculated using the following
general thermodynamic relation:
dS=CPdTTβRln(Z)dS = C_P \frac{dT}{T} - R \ln(Z)
Where:
β’ SS is the entropy of the real gas
β’ SIGS^{IG} is the entropy of the ideal gas
β’ ZZ is the compressibility factor
β’ CPC_P is the heat capacity at constant pressure
β’ RR is the universal gas constant
For the ideal gas, the entropy change is:
dSIG=CPdTTdS^{IG} = C_P \frac{dT}{T}
Therefore, the difference in entropy between the real gas and the ideal gas (i.e., the entropy
departure) is:
SβSIG=β«T1T2CPdTTβRβ«T1T2ln(Z)dTS - S^{IG} = \int_{T_1}^{T_2} C_P \frac{dT}{T} - R
\int_{T_1}^{T_2} \ln(Z) dT
This leads to the general expression for entropy departure:
(SβSIG)=βRβ«T1T2ln(Z)dT(S - S^{IG}) = -R \int_{T_1}^{T_2} \ln(Z) dT
Peng-Robinson Equation of State and Compressibility Factor
The Peng-Robinson equation of state (PR EOS) is used to describe the behavior of real gases:
P=RTVβbβaV(V+b)+b(Vβb)P = \frac{RT}{V - b} - \frac{a}{V(V + b) + b(V - b)}
Where:
β’ PP is pressure
β’ TT is temperature
β’ VV is molar volume
β’ aa and bb are constants related to the critical properties of the gas (such as critical
temperature and pressure)
β’ ZZ is the compressibility factor, given by Z=PVRTZ = \frac{PV}{RT}
Now, the entropy departure for the PR EOS is related to the compressibility factor ZZ, which
accounts for the real gas behavior.
Through a detailed thermodynamic derivation (involving the Helmholtz potential or using a
Maxwell relation), you can arrive at the following expression for the entropy departure function:
(SβSIG)T,PPR=Rln(ZβBB)(S - S^{IG})_{T,P}^{PR} = R \ln\left(\frac{Z - B}{B}\right)
Where:
β’ B=bPRTB = \frac{bP}{RT} is a function of the pressure and temperature, and
β’ ZZ is the compressibility factor from the PR equation of state.
Second Part: Adiabatic Venting Process
Next, you want to determine the final pressure in the tank after venting and the fraction of the
initial mass of gas remaining in the tank, assuming adiabatic venting.
Step 1: Adiabatic Process and Energy Conservation
Since the venting process is adiabatic (no heat exchange with the environment), we can use the
first law of thermodynamics for an adiabatic process, which simplifies to:
dU+PdV=0dU + PdV = 0
Where UU is the internal energy of the gas. For an ideal gas, the internal energy depends only on
the temperature, so:
dU=CVdTdU = C_V dT
For an adiabatic process, the equation simplifies further, and we use the relationship between
pressure and volume for an ideal gas undergoing adiabatic expansion:
P1V1Ξ³=P2V2Ξ³P_1 V_1^\gamma = P_2 V_2^\gamma
Where:
β’ Ξ³=CPCV\gamma = \frac{C_P}{C_V} is the adiabatic index for COβ.
Since we are given that the temperature drops from 400 K to 300 K, we need to account for the
change in temperature and pressure as the gas is vented from the tank.
Step 2: Mass Conservation
Since mass is leaving the tank, the total mass mm changes. You can use the ideal gas law to
relate the initial and final states of the gas in terms of pressure and temperature:
P1V1=n1RT1(initial state)P_1 V_1 = n_1 R T_1 \quad \text{(initial state)}
P2V2=n2RT2(final state)P_2 V_2 = n_2 R T_2 \quad \text{(final state)}
Where:
β’ P1,P2P_1, P_2 are the initial and final pressures,
β’ V1,V2V_1, V_2 are the initial and final volumes,
β’ n1,n2n_1, n_2 are the initial and final moles of gas,
β’ T1=400 KT_1 = 400 \, \text{K} and T2=300 KT_2 = 300 \, \text{K}.
The mass of gas remaining is proportional to the ratio of the final volume to the initial volume,
given by:
mfinalminitial=V2V1\frac{m_{\text{final}}}{m_{\text{initial}}} = \frac{V_2}{V_1}
β’ PP is pressure
β’ TT is temperature
β’ VV is molar volume
β’ aa and bb are constants related to the critical properties of the gas (such as critical
temperature and pressure)
β’ ZZ is the compressibility factor, given by Z=PVRTZ = \frac{PV}{RT}
Now, the entropy departure for the PR EOS is related to the compressibility factor ZZ, which
accounts for the real gas behavior.
Through a detailed thermodynamic derivation (involving the Helmholtz potential or using a
Maxwell relation), you can arrive at the following expression for the entropy departure function:
(SβSIG)T,PPR=Rln(ZβBB)(S - S^{IG})_{T,P}^{PR} = R \ln\left(\frac{Z - B}{B}\right)
Where:
β’ B=bPRTB = \frac{bP}{RT} is a function of the pressure and temperature, and
β’ ZZ is the compressibility factor from the PR equation of state.
Second Part: Adiabatic Venting Process
Next, you want to determine the final pressure in the tank after venting and the fraction of the
initial mass of gas remaining in the tank, assuming adiabatic venting.
Step 1: Adiabatic Process and Energy Conservation
Since the venting process is adiabatic (no heat exchange with the environment), we can use the
first law of thermodynamics for an adiabatic process, which simplifies to:
dU+PdV=0dU + PdV = 0
Where UU is the internal energy of the gas. For an ideal gas, the internal energy depends only on
the temperature, so:
dU=CVdTdU = C_V dT
For an adiabatic process, the equation simplifies further, and we use the relationship between
pressure and volume for an ideal gas undergoing adiabatic expansion:
P1V1Ξ³=P2V2Ξ³P_1 V_1^\gamma = P_2 V_2^\gamma
Where:
β’ Ξ³=CPCV\gamma = \frac{C_P}{C_V} is the adiabatic index for COβ.
Since we are given that the temperature drops from 400 K to 300 K, we need to account for the
change in temperature and pressure as the gas is vented from the tank.
Step 2: Mass Conservation
Since mass is leaving the tank, the total mass mm changes. You can use the ideal gas law to
relate the initial and final states of the gas in terms of pressure and temperature:
P1V1=n1RT1(initial state)P_1 V_1 = n_1 R T_1 \quad \text{(initial state)}
P2V2=n2RT2(final state)P_2 V_2 = n_2 R T_2 \quad \text{(final state)}
Where:
β’ P1,P2P_1, P_2 are the initial and final pressures,
β’ V1,V2V_1, V_2 are the initial and final volumes,
β’ n1,n2n_1, n_2 are the initial and final moles of gas,
β’ T1=400 KT_1 = 400 \, \text{K} and T2=300 KT_2 = 300 \, \text{K}.
The mass of gas remaining is proportional to the ratio of the final volume to the initial volume,
given by:
mfinalminitial=V2V1\frac{m_{\text{final}}}{m_{\text{initial}}} = \frac{V_2}{V_1}
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