Questions: A gas mixture 80% methane and 20% ethane flowing at 4 MMSCFD from a gas sweetening unit is fed to a two stage compression process. The methane enters the first stage of compression at 500 kPa(g) at 70°F and leaves at 1500 kPa(g) at 190°F. The gas is then cooled to 90°F and fed to the second stage of compression. The gas leaves the second stage of compression at 4500 kPa(g) at 200°F. What is the volume flow rate of the gas, from the second stage? Express your answer in ft^3 / day.

A gas mixture 80% methane and 20% ethane flowing at 4 MMSCFD from a gas sweetening unit is fed to a two stage compression process. The methane enters the first stage of compression at 500 kPa(g) at 70°F and leaves at 1500 kPa(g) at 190°F. The gas is then cooled to 90°F and fed to the second stage of compression. The gas leaves the second stage of compression at 4500 kPa(g) at 200°F. What is the volume flow rate of the gas, from the second stage? Express your answer in ft^3 / day.
Transcript text: 4. A gas mixture $80 \%$ methane and $20 \%$ ethane flowing at 4 MMSCFD from a gas sweetening unit is fed to a two stage compression process. The methane enters the first stage of compression at $500 \mathrm{kPa}_{(\mathrm{g})}$ at $70^{\circ} \mathrm{F}$ and leaves at $1500 \mathrm{kPa}(\mathrm{g})$ at $190^{\circ} \mathrm{F}$. The gas is then cooled to $90^{\circ} \mathrm{F}$ and fed to the second stage of compression. The gas leaves the second stage of compression at $4500 \mathrm{kPa}_{(\mathrm{g})}$ at $200^{\circ} \mathrm{F}$. What is the volume flow rate of the gas, from the second stage? Express your answer in $\mathrm{ft}^{3} /$ day.
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Solution

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Solution Steps

Step 1: Determine the initial conditions and convert units

The initial conditions of the gas mixture are:

  • Flow rate: 4 MMSCFD (Million Standard Cubic Feet per Day)
  • Pressure: 500 kPa (gauge)
  • Temperature: 70°F

First, convert the temperature to Rankine: \[ T_1 = 70 + 459.67 = 529.67 \text{°R} \]

Step 2: Calculate the conditions after the first stage of compression

The gas leaves the first stage of compression at:

  • Pressure: 1500 kPa (gauge)
  • Temperature: 190°F

Convert the temperature to Rankine: \[ T_2 = 190 + 459.67 = 649.67 \text{°R} \]

Step 3: Calculate the conditions after cooling

The gas is cooled to:

  • Temperature: 90°F

Convert the temperature to Rankine: \[ T_3 = 90 + 459.67 = 549.67 \text{°R} \]

Step 4: Calculate the conditions after the second stage of compression

The gas leaves the second stage of compression at:

  • Pressure: 4500 kPa (gauge)
  • Temperature: 200°F

Convert the temperature to Rankine: \[ T_4 = 200 + 459.67 = 659.67 \text{°R} \]

Step 5: Apply the ideal gas law to find the volume flow rate

Using the ideal gas law, we can relate the initial and final conditions of the gas. The volume flow rate after the second stage of compression can be found using the following relationship:

\[ \frac{P_1 V_1}{T_1} = \frac{P_4 V_4}{T_4} \]

Where:

  • \( P_1 = 500 + 101.325 = 601.325 \text{ kPa (absolute)} \)
  • \( P_4 = 4500 + 101.325 = 4601.325 \text{ kPa (absolute)} \)
  • \( V_1 = 4 \text{ MMSCFD} \)
  • \( T_1 = 529.67 \text{°R} \)
  • \( T_4 = 659.67 \text{°R} \)

Rearranging to solve for \( V_4 \):

\[ V_4 = V_1 \cdot \frac{P_1}{P_4} \cdot \frac{T_4}{T_1} \]

Substitute the values:

\[ V_4 = 4 \times 10^6 \text{ SCFD} \cdot \frac{601.325}{4601.325} \cdot \frac{659.67}{529.67} \]

Step 6: Perform the calculation

\[ V_4 = 4 \times 10^6 \cdot \frac{601.325}{4601.325} \cdot \frac{659.67}{529.67} \]

\[ V_4 = 4 \times 10^6 \cdot 0.1307 \cdot 1.2455 \]

\[ V_4 = 4 \times 10^6 \cdot 0.1627 \]

\[ V_4 = 650,800 \text{ SCFD} \]

Final Answer

\(\boxed{V_4 = 650,800 \text{ SCFD}}\)

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