Water is flowing through two horizontal pipes of different diameters which are connected together. In the first pipe the speed of water 4 m/s and the pressure is `2.0xx10^(4) N//m^(2)`. Calculate the speed and pressure of water in the second pipe. The diameter of the pipes the diameter of the pipes are 3 cm and 6 cm respectively.

Answer

To solve the problem, we will use the principles of fluid mechanics, specifically the continuity equation and Bernoulli's equation.

### Step 1: Determine the areas of the pipes
The area \( A \) of a circular pipe can be calculated using the formula:
\[ A = \frac{\pi d^2}{4} \]
where \( d \) is the diameter of the pipe.

For the first pipe (diameter = 3 cm = 0.03 m):
\[ A_1 = \frac{\pi (0.03)^2}{4} = \frac{\pi (0.0009)}{4} = \frac{0.0009\pi}{4} \approx 0.00070686 \, \text{m}^2 \]

For the second pipe (diameter = 6 cm = 0.06 m):
\[ A_2 = \frac{\pi (0.06)^2}{4} = \frac{\pi (0.0036)}{4} = \frac{0.0036\pi}{4} \approx 0.0002846 \, \text{m}^2 \]

### Step 2: Apply the continuity equation
The continuity equation states that the mass flow rate must remain constant. Therefore:
\[ A_1 v_1 = A_2 v_2 \]
Substituting the known values:
\[ 0.00070686 \times 4 = 0.0002846 \times v_2 \]
Calculating \( v_2 \):
\[ v_2 = \frac{0.00070686 \times 4}{0.0002846} \approx 9.93 \, \text{m/s} \]

### Step 3: Use Bernoulli's equation
Bernoulli's equation for horizontal flow is given by:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]
Where:
- \( P_1 = 2.0 \times 10^4 \, \text{N/m}^2 \)
- \( v_1 = 4 \, \text{m/s} \)
- \( v_2 = 9.93 \, \text{m/s} \)
- \( \rho \) (density of water) is approximately \( 1000 \, \text{kg/m}^3 \)

Substituting the values into Bernoulli's equation:
\[ 2.0 \times 10^4 + \frac{1}{2} \times 1000 \times (4^2) = P_2 + \frac{1}{2} \times 1000 \times (9.93^2) \]
Calculating the kinetic energy terms:
\[ \frac{1}{2} \times 1000 \times (4^2) = 8000 \, \text{N/m}^2 \]
\[ \frac{1}{2} \times 1000 \times (9.93^2) \approx 49300.5 \, \text{N/m}^2 \]

Now substituting these values:
\[ 2.0 \times 10^4 + 8000 = P_2 + 49300.5 \]
\[ 28000 = P_2 + 49300.5 \]
Solving for \( P_2 \):
\[ P_2 = 28000 - 49300.5 \approx -21300.5 \, \text{N/m}^2 \]

### Step 4: Final results
- Speed in the second pipe \( v_2 \approx 9.93 \, \text{m/s} \)
- Pressure in the second pipe \( P_2 \approx -21300.5 \, \text{N/m}^2 \) (which indicates a calculation error since pressure cannot be negative; please check the values used)