Mixture Composition Flash Distillation
Answer:
ρV ≈ ((0.0139 + 0.986) * 65) / (8.314 * 318) ≈ 0.018 kg/m³
The composition of the vapor phase is 47.08% n-butane and 52.92% 1-propanol. The amount of the vapor phase is 23.54 kmol/hr, and the amount of the liquid phase is 26.46 kmol/hr.
The answer to the problem is that the composition of the vapor phase is 47.08% n-butane and 52.92% 1-propanol. The amount of the vapor phase is 23.54 kmol/hr, and the amount of the liquid phase is 26.46 kmol/hr.
In other words, the vapor phase will contain 47.08% n-butane and 52.92% 1-propanol. The liquid phase will contain the remaining 52.92% n-butane and 47.08% 1-propanol. The total amount of vapor phase will be 23.54 kmol/hr, and the total amount of liquid phase will be 26.46 kmol/hr.
Step-by-step explanation:
The composition and amount of the resulting equilibrium phases can be calculated using the following steps:
1. Calculate the vapor pressure of each component using modified Raoult's Law.
2. Calculate the mole fraction of each component in the vapor phase using the vapor pressure and total pressure.
3. Calculate the mole fraction of each component in the liquid phase using the total mole balance and vapor mole fraction.
4. Calculate the amount of each phase using the total mole balance.
Assumptions:
- The mixture is ideal.
- The mixture is in equilibrium.
- The vapor phase is an ideal gas.
- The liquid phase is an ideal solution.
Using modified Raoult's Law, we can calculate the vapor pressure of each component as follows:
P1 = x1 * P1^sat
P2 = x2 * P2^sat
where:
- P1 and P2 are the vapor pressures of n-butane and 1-propanol, respectively.
- x1 and x2 are the mole fractions of n-butane and 1-propanol in the liquid phase, respectively.
- P1^sat and P2^sat are the saturation pressures of n-butane and 1-propanol at 45°C, respectively.
From Antoine equation:
P = exp(A - B/(T+C))
where:
- A, B, C are Antoine constants for each component
- T is temperature in Kelvin
For n-butane:
A = 13.7815
B = 2726.81
C = -33.63
For 1-propanol:
A = 7.8786
B = 1474.08
C = -53.68
At 45°C (318 K):
P1^sat = exp(13.7815 - 2726.81/(318 - 33.63)) = 0.986 atm
P2^sat = exp(7.8786 - 1474.08/(318 - 53.68)) = 0.157 atm
Using these values, we can calculate the mole fraction of each component in the vapor phase:
y1 = P1 / P_total = (x1 * P1^sat) / P_total
y2 = P2 / P_total = (x2 * P2^sat) / P_total
where:
- y1 and y2 are the mole fractions of n-butane and 1-propanol in the vapor phase, respectively.
- P_total is the total pressure in the flash distillation unit (65 kPa).
Using a total mole balance, we can calculate the mole fraction of each component in the liquid phase:
x1 + x2 = y1 + y2
Solving for x1:
x1 = (y1 * Mw1) / ((y1 * Mw1) + (y2 * Mw2))
where:
- Mw1 and Mw2 are the molecular weights of n-butane and 1-propanol, respectively.
Mw1 = 58 g/mol
Mw2 = 60 g/mol
Substituting values:
x1 = (0.0139 * 58) / ((0.0139 * 58) + (0.986 * 60)) ≈ **0.004**
x2 ≈ **0.996**
Finally, we can calculate the amount of each phase using a total mole balance:
F = L + V
where:
- F is the total amount of feed (50 kmol/hr).
- L is the amount of liquid phase.
- V is the amount of vapor phase.
Using a mass balance:
L * ρL + V * ρV = F * ρF
where:
- ρL, ρV, and ρF are densities of liquid phase, vapor phase, and feed mixture respectively.
Assuming ideal gas behavior for vapor phase:
ρV ≈ PV / (R * T)
where:
- R is universal gas constant (8.314 J/mol-K)
- T is temperature in Kelvin
PV ≈ y_total * P_total
where:
- y_total is total mole fraction in vapor phase.
Substituting values:
ρV ≈ ((0.0139 + 0.986) * 65) / (8.314 * 318) ≈ **0.018 kg/m³**
Sure, here is the math part of the problem:
* The vapor pressures of n-butane and 1-propanol at 45 °C and 65 kPa can be calculated using modified Raoult's Law:
```
P_v = x * P_sat
```
where:
* P_v is the vapor pressure of the component
* x is the mole fraction of the component
* P_sat is the saturated vapor pressure of the component
The saturated vapor pressures of n-butane and 1-propanol at 45 °C can be found in a table of vapor pressures.
The mole fractions of the components in the vapor phase can then be calculated using the following equations:
```
y_1 = P_v1 / (P_v1 + P_v2)
y_2 = P_v2 / (P_v1 + P_v2)
```
The amount of the vapor phase can be calculated using the following equation:
```
V = y_1 * 50 / (y_1 + y_2)
```
The amount of the liquid phase can then be calculated by subtracting the amount of the vapor phase from the total amount of the mixture:
```
L = 50 - V
```
For example, the vapor pressure of n-butane at 45 °C and 65 kPa is 44.8 kPa.
The mole fraction of n-butane in the vapor phase is then:
```
y_1 = 44.8 / (44.8 + 36.4) = 0.4708
```
The mole fraction of 1-propanol in the vapor phase is then:
```
y_2 = 36.4 / (44.8 + 36.4) = 0.5292
```
The amount of the vapor phase is then:
```
V = 0.4708 * 50 / (0.4708 + 0.5292) = 23.54 kmol/hr
```
The amount of the liquid phase is then:
```
L = 50 - 23.54 = 26.46 kmol/hr
```
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