by Jeffrey Kantor (jeff at nd.edu). The latest version of this notebook is available at https://github.com/jckantor/CBE20255.

This Jupyter notebook shows how to use Antoine's equation and Raoult's law to compute flammability limits for methanol.

We'd like to estimate range of temperatures over which this methanol fueled fire starter will successfully operate.

In [1]:

```
from IPython.display import YouTubeVideo
YouTubeVideo("8gFqzbnUO-Y")
```

Out[1]:

The flammability limits of methanol in air at 1 atmosphere pressure correspond to vapor phase mole fractions in the range

$$ 6.7 \mbox{ mol%} \leq y_{MeOH} \leq 36 \mbox{ mol%} $$

Assuming the pure methanol located in the wick of this fire starter reaches a vapor-liquid with air in the device, find the lower and upper operating temperatures for this device.

The first thing we'll do is define a simple python function to calculate the saturation pressure of methanol at a given temperature using Antoine's equation

$$\log_{10} P^{sat} = A - \frac{B}{T + C}$$

Constants for methanol can be found in the back of the course textbook for the case where pressure is given in units of mmHg and temperature in degrees centigrade.

In [2]:

```
# Pressure in mmHg, Temperature in degrees C
A = 7.89750
B = 1474.08
C = 229.13
def Psat(T):
return 10**(A - B/(T+C))
```

To test the function, we'll plot the saturation pressure of methanol for a limited range of temperatures.

In [3]:

```
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
T = np.linspace(-14.0, 65.0, 200)
plt.plot(T,Psat(T))
plt.xlabel('Temperature [deg C]')
plt.ylabel('Pressure [mmHg]')
plt.title('Saturation Pressure of Methanol')
plt.grid()
```

By Dalton's law, the partial pressure of pure methanol is equal to the saturation pressure,

$$y_{MeOH} P = P^{sat}_{MeOH}(T)$$

Solving for the mole fraction of methanol in the vapor phase

$$y_{MeOH} = \frac{P^{sat}_{MeOH}(T)}{P}$$

In [9]:

```
print("Methanol Vapor Pressure at 25 deg C =",round(Psat(25.0),2),"mmHg")
```

At atmospheric pressure of 760 mmHg, the mole fraction of methanol is given by

In [10]:

```
print("Mole Fraction Methanol at 25 deg C and 1 atmosphere =",round(Psat(25.0)/760.0,3))
```

Since this is within the flammability limits of methanol, the fire starter should work at room temperatures.

At the lower flammability limit, the partial pressure of methanol will be

$$P_{MeOH} = y_{MeOH}P$$

In [11]:

```
P_MeOH = 0.067*760
print(P_MeOH, "mm Hg")
```

Next we need to solve for the temperature at which the partial pressure of methanol is at the lower flammability limit.

$$P^{sat}_{MeOH}(T) = P_{MeOH}$$

We'll first attempt a solution by trial and error, then we'll use an equation solve to get an accurate solution. Let's start with a guess of 15 deg C.

In [12]:

```
print(Psat(15.0), "mmHg")
```

That's too high. Let's try some more values

In [13]:

```
print(Psat(10.0), "mmHg")
```

In [17]:

```
print(Psat(5.0), "mmHg")
```

In [18]:

```
print(Psat(7.5), "mmHg")
```

In [19]:

```
print(Psat(8.75), "mmHg")
```

In [20]:

```
print(Psat(9.0), "mmHg")
```

So $T = 9$ deg C is pretty close. To get a precise answer, use the `brentq`

equation solving function from `scipy.optimize`

.

In [21]:

```
from scipy.optimize import brentq
def f(T):
return Psat(T) - P_MeOH
Tlow = brentq(f,5,15)
print("Lower Flammability Temperature = ", Tlow, "deg C")
print("Lower Flammability Temperature = ", 32.0 + 9.0*Tlow/5.0, "deg F")
```

Calculate the upper limit on temperature at which this lighter can operate.