An oil of relative density `0.9` and viscosity `0.12 kg//ms` flows through a `2.5 cm` diameter pipe with a pressure drop of `38.4 kN//m^(2)` in a length of `30 m`. Determine
Determine the discharge

Options

A. `2.16xx10^(-4)m^(3)//s`
B. `2.9xx10^(-3)m^(3)//s`
C. `1xx10^(-4)m^(3)//s`
D. `2xx10^(-4)m^(3)//s`

Explanation

To determine the discharge of oil flowing through a pipe, we can use the Poiseuille equation, which relates the discharge (Q) to the pressure drop (ΔP), the viscosity (η), the diameter of the pipe (D), and the length of the pipe (L). The equation is given by:

\[ Q = \frac{\pi D^4 \Delta P}{128 \eta L} \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Relative density of oil (not needed for discharge calculation)
- Viscosity (η) = 0.12 kg/(m·s)
- Diameter of the pipe (D) = 2.5 cm = 0.025 m (convert to meters)
- Pressure drop (ΔP) = 38.4 kN/m² = 38,400 Pa (convert to Pascals)
- Length of the pipe (L) = 30 m

2. **Convert Diameter to Meters:**
\[ D = 2.5 \, \text{cm} = 0.025 \, \text{m} \]

3. **Calculate D^4:**
\[ D^4 = (0.025)^4 = 3.90625 \times 10^{-9} \, \text{m}^4 \]

4. **Substitute Values into the Poiseuille Equation:**
\[ Q = \frac{\pi \times (3.90625 \times 10^{-9}) \times (38400)}{128 \times 0.12 \times 30} \]

5. **Calculate the Denominator:**
\[ \text{Denominator} = 128 \times 0.12 \times 30 = 460.8 \]

6. **Calculate the Numerator:**
\[ \text{Numerator} = \pi \times 3.90625 \times 10^{-9} \times 38400 \approx 4.733 \times 10^{-4} \]

7. **Calculate the Discharge (Q):**
\[ Q = \frac{4.733 \times 10^{-4}}{460.8} \approx 1.026 \times 10^{-6} \, \text{m}^3/\text{s} \]

8. **Final Result:**
\[ Q \approx 1.026 \times 10^{-6} \, \text{m}^3/\text{s} \approx 1.026 \, \text{L/s} \]

### Final Answer:
The discharge of the oil through the pipe is approximately \(1.026 \, \text{L/s}\).