To find the pressure reading at the larger diameter of the tapered pipe, we can use the Bernoulli equation, which states that the total mechanical energy along a streamline is conserved for an incompressible, non-viscous fluid. The equation for two points along the streamline is given by:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \]
Since the pipe is horizontal, we can ignore the height term (\(gh\)). Thus, the equation simplifies to:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]
Where:
- \(P_1\) is the pressure at the larger diameter,
- \(P_2\) is the pressure at the smaller diameter (given as 16 kPa),
- \(v_1\) is the velocity at the larger diameter,
- \(v_2\) is the velocity at the smaller diameter (given as 20 m/s),
- \(\rho\) is the density of the fluid (1,000 kg/m³).
First, we need to find the velocity at the larger diameter (\(v_1\)). We can apply the principle of conservation of mass (continuity equation):
\[ A_1 v_1 = A_2 v_2 \]
Where:
- \(A_1\) is the cross-sectional area of the larger diameter,
- \(A_2\) is the cross-sectional area of the smaller diameter.
The area \(A\) of the pipe can be calculated using the formula for the area of a circle \(A = \frac{\pi d^2}{4}\).
- Calculate \(A_1\) (for diameter 0.65 m):
\[ A_1 = \frac{\pi (0.65)^2}{4} \approx 0.33257 , \text{m}^2 \]
- Calculate \(A_2\) (for diameter 0.45 m):
\[ A_2 = \frac{\pi (0.45)^2}{4} \approx 0.15804 , \text{m}^2 \]
Now, we can find \(v_1\):
\[ A_1 v_1 = A_2 v_2 \Rightarrow v_1 = \frac{A_2 v_2}{A_1} \]
Substituting in the known values:
\[ v_1 = \frac{0.15804 , \text{m}^2 \cdot 20 , \text{m/s}}{0.33257 , \text{m}^2} \approx 9.48 , \text{m/s} \]
Now that we have \(v_1\), we can substitute \(P_2\), \(v_2\), \(\rho\), and \(v_1\) back into the Bernoulli equation:
- \(P_2 = 16 , \text{kPa} = 16,000 , \text{Pa}\)
- \(v_2 = 20 , \text{m/s}\)
- \(v_1 \approx 9.48 , \text{m/s}\)
The Bernoulli equation becomes:
\[ P_1 + \frac{1}{2} (1,000 , \text{kg/m}^3)(9.48)^2 = 16,000 + \frac{1}{2} (1,000) (20)^2 \]
Calculating the kinetic energy terms:
\[ \frac{1}{2} (1,000)(9.48)^2 \approx 44,908.8 , \text{Pa} \] \[ \frac{1}{2} (1,000)(20)^2 = 200,000 , \text{Pa} \]
Substituting back into the equation:
\[ P_1 + 44,908.8 = 16,000 + 200,000 \]
Rearranging to find \(P_1\):
\[ P_1 = 16,000 + 200,000 - 44,908.8 \] \[ P_1 = 171,091.2 , \text{Pa} \] \[ P_1 \approx 171.1 , \text{kPa} \]
Thus, the pressure reading at the larger diameter of the pipe is approximately \(171.1 , \text{kPa}\).
1 answer