Question

Water with a density of 1 000 kg/m^3 flows through a tapered pipe, which has a large diameter of 0,65 m and a small diameter of 0,45 m. The pressure gauge at the smaller diameter of the pipe reads 16 kPa, and the flow velocity is 20 m/s. Find the pressure reading at the larger diameter of the pipe.

Answer 1

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 \(A = \frac{\pi d^2}{4}\).

1. Calculate \(A_1\) (for diameter 0.65 m):

\[
A_1 = \frac{\pi (0.65)^2}{4} \approx 0.33257 \, \text{m}^2
\]

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 \(171.1 \, \text{kPa}\).