Calculate the rate of flow of glycerine of density `1.25 xx 10^(3) kg//m^(3)` through the conical section of a pipe if the radii of its ends are 0.1m and 0.04 m and the pressure drop across its lengths is `10N//m^(2)`.

Answer

To solve the problem of calculating the rate of flow of glycerine through a conical section of a pipe, we will follow these steps:

### Step 1: Identify the given data
- Density of glycerine, \( \rho = 1.25 \times 10^3 \, \text{kg/m}^3 \)
- Radius at end 1, \( r_1 = 0.1 \, \text{m} \)
- Radius at end 2, \( r_2 = 0.04 \, \text{m} \)
- Pressure drop, \( \Delta P = 10 \, \text{N/m}^2 \)

### Step 2: Calculate the cross-sectional areas
The cross-sectional area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
- For end 1:
\[ A_1 = \pi (0.1)^2 = \pi \times 0.01 = 0.0314 \, \text{m}^2 \]
- For end 2:
\[ A_2 = \pi (0.04)^2 = \pi \times 0.0016 = 0.00503 \, \text{m}^2 \]

### Step 3: Use the continuity equation
According to the continuity equation:
\[ A_1 v_1 = A_2 v_2 \]
We can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = \frac{A_1}{A_2} v_1 \]

### Step 4: Apply Bernoulli's equation
For incompressible flow, Bernoulli’s equation can be simplified to:
\[ \Delta P = \frac{1}{2} \rho (v_2^2 - v_1^2) \]
Substituting \( v_2 \):
\[ \Delta P = \frac{1}{2} \rho \left( \left( \frac{A_1}{A_2} v_1 \right)^2 - v_1^2 \right) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \Delta P = \frac{1}{2} \rho \left( \frac{A_1^2}{A_2^2} v_1^2 - v_1^2 \right) \]
\[ \Delta P = \frac{1}{2} \rho v_1^2 \left( \frac{A_1^2}{A_2^2} - 1 \right) \]

### Step 6: Solve for \( v_1 \)
Rearranging for \( v_1 \):
\[ v_1^2 = \frac{2 \Delta P}{\rho \left( \frac{A_1^2}{A_2^2} - 1 \right)} \]
\[ v_1 = \sqrt{\frac{2 \Delta P}{\rho \left( \frac{A_1^2}{A_2^2} - 1 \right)}} \]

### Step 7: Calculate \( \frac{A_1^2}{A_2^2} \)
Calculating:
\[ \frac{A_1^2}{A_2^2} = \left( \frac{0.0314}{0.00503} \right)^2 \approx 39.4 \]

### Step 8: Substitute values into \( v_1 \)
Substituting values:
\[ v_1 = \sqrt{\frac{2 \times 10}{1.25 \times 10^3 \times (39.4 - 1)}} \]
\[ v_1 = \sqrt{\frac{20}{1.25 \times 10^3 \times 38.4}} \approx \sqrt{\frac{20}{48000}} \approx \sqrt{0.00041667} \approx 0.0204 \, \text{m/s} \]

### Step 9: Calculate the volumetric flow rate \( Q \)
Using the formula:
\[ Q = A_1 v_1 \]
\[ Q = 0.0314 \times 0.0204 \approx 0.00064056 \, \text{m}^3/\text{s} \approx 6.4 \times 10^{-4} \, \text{m}^3/\text{s} \]

### Final Result
The rate of flow of glycerine through the conical section of the pipe is approximately:
\[ Q \approx 6.44 \times 10^{-4} \, \text{m}^3/\text{s} \]