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 power required to maintain the flow

Options

A. `2.2W`
B. `3.84W`
C. `5.6W`
D. `9.3W`

Explanation

To determine the power required to maintain the flow of oil through a pipe, we can follow these steps:

### Step 1: Identify the given values
- Relative density of oil, \( \rho_r = 0.9 \)
- Viscosity of oil, \( \eta = 0.12 \, \text{kg/m·s} \)
- Diameter of the pipe, \( D = 2.5 \, \text{cm} = 0.025 \, \text{m} \)
- Pressure drop, \( \Delta P = 38.4 \, \text{kN/m}^2 = 38400 \, \text{Pa} \)
- Length of the pipe, \( L = 30 \, \text{m} \)

### Step 2: Calculate the density of the oil
The density of the oil can be calculated using the relative density:
\[ \rho = \rho_r \times \rho_{\text{water}} = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3 \]

### Step 3: Calculate the cross-sectional area of the pipe
The cross-sectional area \( A \) of the pipe can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi D^2}{4} = \frac{\pi (0.025)^2}{4} \approx 4.91 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Calculate the volume flow rate \( Q \)
Using the Poiseuille equation for laminar flow, the volume flow rate \( Q \) can be expressed as:
\[ Q = \frac{\Delta P \cdot \pi D^4}{128 \eta L} \]
Substituting the values:
\[ Q = \frac{38400 \cdot \pi (0.025)^4}{128 \cdot 0.12 \cdot 30} \]
Calculating this gives:
\[ Q \approx 1.0 \times 10^{-4} \, \text{m}^3/\text{s} \]

### Step 5: Calculate the power required to maintain the flow
The power \( P \) required to maintain the flow can be calculated using the formula:
\[ P = \Delta P \cdot Q \]
Substituting the values:
\[ P = 38400 \, \text{Pa} \cdot 1.0 \times 10^{-4} \, \text{m}^3/\text{s} = 3.84 \, \text{W} \]

### Final Answer
The power required to maintain the flow is approximately \( 3.84 \, \text{W} \).

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