Ann: Cuba Block 9 Update, page-149

Here is what chat got reckons about pressure drop related to flow rate. Looks good to me even if the oil was 10x gluggier. Keep in mind these losses even 10x are only a fraction more than the hydrostatic pressure at 2km deep for a dead head zero flow scenario (approx 18000 kpa)


To calculate the pressure drop per meter for 19°API crude oil at a flow rate of 12 L/s through a 4-inch (nominal diameter) tube at 60°C, we can use the Darcy-Weisbach equation. Here's the step-by-step breakdown:

---

### 1. **Oil Properties**
- **Density (\( \rho \)):**
19°API corresponds to a specific gravity of approximately 0.947. At 60°C:
\( \rho = 0.947 \times 1000 = 947 \, \text{kg/m}^3 \).

- **Viscosity (\( \mu \)):**
Heavy crude viscosity at 60°C is typically in the range of 10-100 cP (0.01-0.1 Pa·s). We'll assume \( \mu = 0.05 \, \text{Pa·s} \) (50 cP) for estimation.

---

### 2. **Tube Properties**
- **Nominal Size:** 4 inches.
- **Internal Diameter (\( D \)):** For 4" schedule 40 pipe, \( D \approx 102.3 \, \text{mm} = 0.1023 \, \text{m} \).
- **Roughness (\( \epsilon \)):** Assume a roughness of \( \epsilon = 0.045 \, \text{mm} \) (typical for steel).

---

### 3. **Flow Properties**
- **Flow Rate (\( Q \)):** 12 L/s = \( 0.012 \, \text{m}^3/\text{s} \).
- **Velocity (\( v \)):**
\[
v = \frac{Q}{A}, \quad A = \frac{\pi D^2}{4}
\]
Substituting:
\[
A = \frac{\pi (0.1023)^2}{4} = 0.00821 \, \text{m}^2
\]
\[
v = \frac{0.012}{0.00821} \approx 1.46 \, \text{m/s}
\]

- **Reynolds Number (\( Re \)):**
\[
Re = \frac{\rho v D}{\mu}
\]
Substituting:
\[
Re = \frac{947 \times 1.46 \times 0.1023}{0.05} \approx 2820
\]
\( Re < 4000 \), so the flow is in the **transition regime**.

---

### 4. **Friction Factor (\( f \))**
For the transition regime, we use the **Colebrook-White equation**:
\[
\frac{1}{\sqrt{f}} = -2 \log_{10} \left( \frac{\epsilon/D}{3.7} + \frac{2.51}{Re \sqrt{f}} \right)
\]
Solving iteratively or approximating:
\[
f \approx 0.028
\]

---

### 5. **Pressure Drop (\( \Delta P \))**
Using the Darcy-Weisbach equation:
\[
\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}
\]
For 1 meter of pipe (\( L = 1 \)):
\[
\Delta P = 0.028 \frac{1}{0.1023} \frac{947 \times (1.46)^2}{2}
\]
\[
\Delta P \approx 177 \, \text{Pa/m}
\]

---

### Final Result:
The pressure drop is approximately **177 Pa/m** (0.177 kPa/m).

Summary: flow losses = 1900x0.177 = 336kpa (48.7PSI) so provided reservoir pressure is 20% higher than 18000kPa greater than 22000 kpa (3188psi) we will get these flows with 10x more gluggier oil