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 shear stress at the pipe wall

Options

A. `8xx10^(-6)N//m^(2)`
B. `3.9xx10^(-6)N//m^(2)`
C. `2.3xx10^(-6)N//m^(2)`
D. `10.6xx10^(-6)N//m^(2)`

Explanation

To determine the shear stress at the pipe wall, we can follow these steps:

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

### Step 2: Calculate the radius of the pipe
The radius \( R \) of the pipe is half of the diameter:
\[ R = \frac{D}{2} = \frac{0.025}{2} = 0.0125 \, \text{m} \]

### Step 3: Calculate the area of the pipe cross-section
The area \( A \) of the pipe cross-section is given by:
\[ A = \pi R^2 = \pi (0.0125)^2 \]
Calculating this gives:
\[ A \approx 4.908 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Calculate the force due to the pressure drop
The force \( F \) due to the pressure drop is given by:
\[ F = \Delta P \times A \]
Substituting the values:
\[ F = 38400 \, \text{Pa} \times 4.908 \times 10^{-4} \, \text{m}^2 \approx 18.86 \, \text{N} \]

### Step 5: Calculate the tangential area
The tangential area \( A_t \) at which the force is applied is given by:
\[ A_t = 2 \pi R L \]
Substituting the values:
\[ A_t = 2 \pi (0.0125) (30) \approx 2.356 \, \text{m}^2 \]

### Step 6: Calculate the shear stress
The shear stress \( \tau \) at the pipe wall is given by:
\[ \tau = \frac{F}{A_t} \]
Substituting the values:
\[ \tau = \frac{18.86 \, \text{N}}{2.356 \, \text{m}^2} \approx 8.01 \, \text{N/m}^2 \]

### Final Answer
The shear stress at the pipe wall is approximately \( 8.01 \, \text{N/m}^2 \).

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