For the Amazon sale, I invested in a much, much larger bottle for bench and field refills - previously I had a dinky 0.5L bottle and wasn't very impressed.
I had set some triggers to tell me when the bottle (and carrying case) I had eyeballed, went on sale, and pulled the trigger on the new bottle before really doing the math. How many refills would I get out of the new gear? I could guess by taking what I was getting from the 0.5L bottle and scaling up, but that'd be a real shot in the dark number.
I have mathematician friends, to ask, but I didn't want to feel like a dolt in front of them... so? ChatGPT to the rescue. I know there are others here that are curious about this very thing - so here's how ChatGPT solved for refills of 150BAR each and end pressure for both bottles once they reach equilibrium pressure (150-180BAR):
To determine how many times you can refill the 300cc bottle from the 6.8L vessel and the pressure at which both vessels will equalize, we need to apply the principles of gas laws, specifically the ideal gas law and the concept of conservation of mass for gases.
Given data:
- Volume of small vessel (V1) = 300cc = 0.3L
- Volume of large vessel (V2) = 6.8L
- Initial pressure in both vessels (P0) = 300 BAR
- Refill pressure (P_refill) = 150 BAR
We will assume that the temperature remains constant and the gas behaves ideally. This means we can use the ideal gas law \( PV = nRT \), where \( n \) is the number of moles of gas, \( R \) is the gas constant, and \( T \) is the temperature.
### Step 1: Calculate the total moles of gas initially
Initially, both vessels are at 300 BAR.
The total initial volume is:
\[ V_{total} = V1 + V2 = 0.3 \text{ L} + 6.8 \text{ L} = 7.1 \text{ L} \]
Since pressure and volume are directly proportional to the number of moles at constant temperature, the total initial moles \( n_{total} \) are:
\[ n_{total} = \frac{P0 \cdot V_{total}}{RT} \]
However, we don't need to calculate \( n_{total} \) explicitly since we are considering the pressure changes relative to the volume.
### Step 2: Calculate the moles required for a single refill
Each time you refill the small vessel from 150 BAR to 300 BAR:
\[ P1_{initial} = 150 \text{ BAR} \]
\[ P1_{final} = 300 \text{ BAR} \]
The pressure difference is:
\[ \Delta P = P1_{final} - P1_{initial} = 300 \text{ BAR} - 150 \text{ BAR} = 150 \text{ BAR} \]
The volume of the small vessel is 0.3L. So, the moles needed for each refill (using \( PV = nRT \)) are proportional to:
\[ \Delta n = \frac{\Delta P \cdot V1}{RT} = \frac{150 \cdot 0.3}{RT} = \frac{45}{RT} \]
### Step 3: Calculate how many refills can be done
Let's denote the pressure in the large vessel after \( n \) refills as \( P2_n \).
After each refill, the pressure in the large vessel drops slightly. We can approximate the number of refills \( n \) by considering the total pressure drop required in the large vessel to provide the necessary amount of gas.
Since the initial pressure in the large vessel is 300 BAR, and assuming \( n \) refills, the total volume to fill for each refill is:
\[ n \cdot \Delta n = n \cdot \frac{45}{RT} \]
Each time we refill, we are effectively using the gas from the large vessel and reducing its pressure. The volume of the large vessel is 6.8L.
The total pressure drop required from the large vessel is equal to the initial pressure of 300 BAR minus the final equalized pressure \( P_{eq} \).
Let's denote the final equalized pressure as \( P_{eq} \). The pressure drop for each refill will be distributed over the large volume. Therefore, the total pressure drop is approximated by:
\[ \Delta P_{total} = n \cdot \frac{45}{6.8} \]
Since the pressure in both vessels will equalize, we can set up the following equation to find \( P_{eq} \):
\[ P_{eq} \cdot (V1 + V2) = P0 \cdot V2 - n \cdot \frac{45}{6.8} \cdot V2 \]
Given the initial and final conditions, the final pressure in the combined system must be such that:
\[ P_{eq} = \frac{300 \cdot 6.8 - n \cdot 45}{6.8 + 0.3} \]
By solving this equation iteratively or through more complex algebraic manipulation, we can find \( n \) and \( P_{eq} \).
### Step 4: Simplify and Calculate
In practical terms, for each refill, you use some gas from the large vessel, and you can solve iteratively:
After the first refill:
\[ P_{2,1} = \frac{300 \cdot (6.8 - 0.3)}{6.8} \]
Continue iteratively to find the maximum \( n \) where \( P_{2,n} \geq 150 \).
Without going into deep algebraic detail, by approximations and iterative calculations:
You can refill approximately 10-12 times before the pressure drops significantly in the large vessel.
To find exact values, we would set up a spreadsheet or a program to solve this iteratively or numerically, but an approximate answer for your question is:
- **Number of refills**: Approximately 10-12 times.
- **Equalization pressure**: Approximately between 150-180 BAR depending on the exact calculations.
For precise answers, computational tools or more iterative manual calculations would be necessary.
This concludes my TEDTalk for today!
Cheers,
Your friend Mark
I had set some triggers to tell me when the bottle (and carrying case) I had eyeballed, went on sale, and pulled the trigger on the new bottle before really doing the math. How many refills would I get out of the new gear? I could guess by taking what I was getting from the 0.5L bottle and scaling up, but that'd be a real shot in the dark number.
I have mathematician friends, to ask, but I didn't want to feel like a dolt in front of them... so? ChatGPT to the rescue. I know there are others here that are curious about this very thing - so here's how ChatGPT solved for refills of 150BAR each and end pressure for both bottles once they reach equilibrium pressure (150-180BAR):
To determine how many times you can refill the 300cc bottle from the 6.8L vessel and the pressure at which both vessels will equalize, we need to apply the principles of gas laws, specifically the ideal gas law and the concept of conservation of mass for gases.
Given data:
- Volume of small vessel (V1) = 300cc = 0.3L
- Volume of large vessel (V2) = 6.8L
- Initial pressure in both vessels (P0) = 300 BAR
- Refill pressure (P_refill) = 150 BAR
We will assume that the temperature remains constant and the gas behaves ideally. This means we can use the ideal gas law \( PV = nRT \), where \( n \) is the number of moles of gas, \( R \) is the gas constant, and \( T \) is the temperature.
### Step 1: Calculate the total moles of gas initially
Initially, both vessels are at 300 BAR.
The total initial volume is:
\[ V_{total} = V1 + V2 = 0.3 \text{ L} + 6.8 \text{ L} = 7.1 \text{ L} \]
Since pressure and volume are directly proportional to the number of moles at constant temperature, the total initial moles \( n_{total} \) are:
\[ n_{total} = \frac{P0 \cdot V_{total}}{RT} \]
However, we don't need to calculate \( n_{total} \) explicitly since we are considering the pressure changes relative to the volume.
### Step 2: Calculate the moles required for a single refill
Each time you refill the small vessel from 150 BAR to 300 BAR:
\[ P1_{initial} = 150 \text{ BAR} \]
\[ P1_{final} = 300 \text{ BAR} \]
The pressure difference is:
\[ \Delta P = P1_{final} - P1_{initial} = 300 \text{ BAR} - 150 \text{ BAR} = 150 \text{ BAR} \]
The volume of the small vessel is 0.3L. So, the moles needed for each refill (using \( PV = nRT \)) are proportional to:
\[ \Delta n = \frac{\Delta P \cdot V1}{RT} = \frac{150 \cdot 0.3}{RT} = \frac{45}{RT} \]
### Step 3: Calculate how many refills can be done
Let's denote the pressure in the large vessel after \( n \) refills as \( P2_n \).
After each refill, the pressure in the large vessel drops slightly. We can approximate the number of refills \( n \) by considering the total pressure drop required in the large vessel to provide the necessary amount of gas.
Since the initial pressure in the large vessel is 300 BAR, and assuming \( n \) refills, the total volume to fill for each refill is:
\[ n \cdot \Delta n = n \cdot \frac{45}{RT} \]
Each time we refill, we are effectively using the gas from the large vessel and reducing its pressure. The volume of the large vessel is 6.8L.
The total pressure drop required from the large vessel is equal to the initial pressure of 300 BAR minus the final equalized pressure \( P_{eq} \).
Let's denote the final equalized pressure as \( P_{eq} \). The pressure drop for each refill will be distributed over the large volume. Therefore, the total pressure drop is approximated by:
\[ \Delta P_{total} = n \cdot \frac{45}{6.8} \]
Since the pressure in both vessels will equalize, we can set up the following equation to find \( P_{eq} \):
\[ P_{eq} \cdot (V1 + V2) = P0 \cdot V2 - n \cdot \frac{45}{6.8} \cdot V2 \]
Given the initial and final conditions, the final pressure in the combined system must be such that:
\[ P_{eq} = \frac{300 \cdot 6.8 - n \cdot 45}{6.8 + 0.3} \]
By solving this equation iteratively or through more complex algebraic manipulation, we can find \( n \) and \( P_{eq} \).
### Step 4: Simplify and Calculate
In practical terms, for each refill, you use some gas from the large vessel, and you can solve iteratively:
After the first refill:
\[ P_{2,1} = \frac{300 \cdot (6.8 - 0.3)}{6.8} \]
Continue iteratively to find the maximum \( n \) where \( P_{2,n} \geq 150 \).
Without going into deep algebraic detail, by approximations and iterative calculations:
You can refill approximately 10-12 times before the pressure drops significantly in the large vessel.
To find exact values, we would set up a spreadsheet or a program to solve this iteratively or numerically, but an approximate answer for your question is:
- **Number of refills**: Approximately 10-12 times.
- **Equalization pressure**: Approximately between 150-180 BAR depending on the exact calculations.
For precise answers, computational tools or more iterative manual calculations would be necessary.
This concludes my TEDTalk for today!
Cheers,
Your friend Mark
