Jeffrey C. Kantor ([email protected])

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

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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))
```

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 [4]:

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

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

In [5]:

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

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

$$P_{MeOH} = y_{MeOH}P$$In [6]:

```
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 [7]:

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

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

In [8]:

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

In [9]:

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

In [10]:

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

In [11]:

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

In [12]:

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

`brentq`

equation solving function from `scipy.optimize`

.

In [14]:

```
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.

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```