Water with a density of 1 000 kg/m³ flows into the smaller diameter (0,15 m) end of

Question

a 10,5 m tapered pipe, with a velocity of 20 m/s under 15 kPa of pressure. The water flows out of the bigger diameter (0,3 m) end of the pipe. If the inlet side of the pipe is 3,5 m higher than the outlet side, calculate the pressure exerted at the outlet end of the tapered pipe

Answer 1

To determine the pressure at the outlet end of the tapered pipe, we will use Bernoulli's equation, which relates the pressure, velocity, and height (potential energy) of a fluid in steady flow. The equation can be expressed as:

\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]

Where:

• \( P \) = pressure
• \( \rho \) = density of the fluid
• \( v \) = velocity of the fluid
• \( g \) = acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \))
• \( h \) = height above a reference level

Given data:

• \( \rho = 1000 , \text{kg/m}^3 \)
• Inlet diameter \( d_1 = 0.15 , \text{m} \), thus cross-sectional area \( A_1 = \frac{\pi}{4} d_1^2 = \frac{\pi}{4} (0.15)^2 = 0.01767 , \text{m}^2 \)
• Inlet velocity \( v_1 = 20 , \text{m/s} \)
• Inlet pressure \( P_1 = 15 , \text{kPa} = 15000 , \text{Pa} \)
• Height difference \( h_1 - h_2 = 3.5 , \text{m} \) (so, \( h_1 = 3.5 \) m and \( h_2 = 0 \) m)
• Outlet diameter \( d_2 = 0.3 , \text{m} \), thus cross-sectional area \( A_2 = \frac{\pi}{4} d_2^2 = \frac{\pi}{4} (0.3)^2 = 0.07069 , \text{m}^2 \)

First, we need to find the outlet velocity \( v_2 \) using the continuity equation:

\[ A_1 v_1 = A_2 v_2 \]

Solving for \( v_2 \):

\[ v_2 = \frac{A_1 v_1}{A_2} \]

Substituting the known values:

\[ v_2 = \frac{0.01767 , \text{m}^2 \cdot 20 , \text{m/s}}{0.07069 , \text{m}^2} = \frac{0.3534}{0.07069} \approx 5.00 , \text{m/s} \]

Now substitute all known values into Bernoulli’s equation:

\[ 15000 , \text{Pa} + \frac{1}{2} (1000) (20^2) + (1000)(9.81)(3.5) = P_2 + \frac{1}{2} (1000) (5^2) + (1000)(9.81)(0) \]

Calculating each term on the left side:

1. \( P_1 = 15000 , \text{Pa} \)
2. Kinetic energy term: \( \frac{1}{2} \times 1000 \times 20^2 = \frac{1}{2} \times 1000 \times 400 = 200000 , \text{Pa} \)
3. Potential energy term: \( 1000 \times 9.81 \times 3.5 = 34335 , \text{Pa} \)

Adding these contributions from the inlet side:

\[ 15000 + 200000 + 34335 = 249335 , \text{Pa} \]

Now for the outlet side:

1. Kinetic energy term at outlet: \( \frac{1}{2} \times 1000 \times 5^2 = \frac{1}{2} \times 1000 \times 25 = 12500 , \text{Pa} \)
2. There is no potential energy at the outlet since \( h_2 = 0 \).

The Bernoulli equation becomes:

\[ 249335 = P_2 + 12500 \]

Solving for \( P_2 \):

\[ P_2 = 249335 - 12500 = 236835 , \text{Pa} \]

Finally, converting to kPa:

\[ P_2 \approx 236.8 , \text{kPa} \]

Thus, the pressure exerted at the outlet end of the tapered pipe is approximately 236.8 kPa.