Question
Water flows at speed of 5.9 m/s through a horizontal pipe of diameter 3.2 cm . Suppose a thick pipe is connected to a thinner pipe. In the first pipe the gauge pressure P1 of the water is 1.6 atm. The diameter of the smaller section is 2.2 cm. Please find the gauge pressure of the water flowing through the smaller pipe. Atmospheric pressure is 1.013 × 105 Pa. The density of water is 1000 kg/m^3. The viscosity of water is negligible. Answer in units of atm
Answers
GPT-4o mini
To find the gauge pressure in the narrower pipe, we can apply Bernoulli's equation, which states:
\[
P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2
\]
Since the pipes are horizontal, the heights \(h_1\) and \(h_2\) cancel out, simplifying the equation 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 in the wider pipe.
- \(P_2\) is the pressure in the narrower pipe.
- \(\rho\) is the density of water (1000 kg/m³).
- \(v_1\) is the velocity of water in the wider pipe (5.9 m/s).
- \(v_2\) is the velocity of water in the narrower pipe.
First, let's calculate the velocity in the narrower pipe \(v_2\). Using the principle of continuity:
\[
A_1 v_1 = A_2 v_2
\]
Where \(A\) is the cross-sectional area, calculated as:
\[
A = \frac{\pi d^2}{4}
\]
Calculating the areas for both pipes:
For the wider pipe (diameter = 3.2 cm = 0.032 m):
\[
A_1 = \frac{\pi (0.032)^2}{4} = \frac{\pi \times 0.001024}{4} = 0.0008069 \, (\text{m}^2)
\]
For the narrower pipe (diameter = 2.2 cm = 0.022 m):
\[
A_2 = \frac{\pi (0.022)^2}{4} = \frac{\pi \times 0.000484}{4} = 0.0003817 \, (\text{m}^2)
\]
Now we can find \(v_2\):
\[
0.0008069 \times 5.9 = 0.0003817 \times v_2
\]
Calculating \(v_2\):
\[
v_2 = \frac{0.0008069 \times 5.9}{0.0003817} = \frac{0.00475471}{0.0003817} \approx 12.45 \, \text{m/s}
\]
Now we can plug this into Bernoulli's equation. First, we need to convert \(P_1\) from atm to Pascals:
\[
P_1 = 1.6 \, \text{atm} \times 1.013 \times 10^5 \, \text{Pa/atm} = 162080 \, \text{Pa}
\]
Using Bernoulli's equation, we can rearrange it to find \(P_2\):
\[
P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2
\]
Calculating \(\frac{1}{2} \rho v_1^2\) and \(\frac{1}{2} \rho v_2^2\):
\[
\frac{1}{2} \rho v_1^2 = \frac{1}{2} \times 1000 \times (5.9)^2 = 0.5 \times 1000 \times 34.81 = 17405 \, \text{Pa}
\]
\[
\frac{1}{2} \rho v_2^2 = \frac{1}{2} \times 1000 \times (12.45)^2 = 0.5 \times 1000 \times 155.23 = 77615 \, \text{Pa}
\]
Now substitute back:
\[
P_2 = 162080 + 17405 - 77615
\]
Calculating \(P_2\):
\[
P_2 = 162080 + 17405 - 77615 = 162080 + 17405 - 77615 = 86070 \, \text{Pa}
\]
Finally, we convert \(P_2\) back to gauge pressure in atm:
To find gauge pressure, we don't consider atmospheric pressure, so we can find \(P_2\) in atm directly:
\[
P_2 = \frac{86070}{101300} \approx 0.85 \, \text{atm}
\]
Thus, the gauge pressure of the water flowing through the smaller pipe is approximately:
\[
\boxed{0.85 \, \text{atm}}
\]
\[
P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2
\]
Since the pipes are horizontal, the heights \(h_1\) and \(h_2\) cancel out, simplifying the equation 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 in the wider pipe.
- \(P_2\) is the pressure in the narrower pipe.
- \(\rho\) is the density of water (1000 kg/m³).
- \(v_1\) is the velocity of water in the wider pipe (5.9 m/s).
- \(v_2\) is the velocity of water in the narrower pipe.
First, let's calculate the velocity in the narrower pipe \(v_2\). Using the principle of continuity:
\[
A_1 v_1 = A_2 v_2
\]
Where \(A\) is the cross-sectional area, calculated as:
\[
A = \frac{\pi d^2}{4}
\]
Calculating the areas for both pipes:
For the wider pipe (diameter = 3.2 cm = 0.032 m):
\[
A_1 = \frac{\pi (0.032)^2}{4} = \frac{\pi \times 0.001024}{4} = 0.0008069 \, (\text{m}^2)
\]
For the narrower pipe (diameter = 2.2 cm = 0.022 m):
\[
A_2 = \frac{\pi (0.022)^2}{4} = \frac{\pi \times 0.000484}{4} = 0.0003817 \, (\text{m}^2)
\]
Now we can find \(v_2\):
\[
0.0008069 \times 5.9 = 0.0003817 \times v_2
\]
Calculating \(v_2\):
\[
v_2 = \frac{0.0008069 \times 5.9}{0.0003817} = \frac{0.00475471}{0.0003817} \approx 12.45 \, \text{m/s}
\]
Now we can plug this into Bernoulli's equation. First, we need to convert \(P_1\) from atm to Pascals:
\[
P_1 = 1.6 \, \text{atm} \times 1.013 \times 10^5 \, \text{Pa/atm} = 162080 \, \text{Pa}
\]
Using Bernoulli's equation, we can rearrange it to find \(P_2\):
\[
P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2
\]
Calculating \(\frac{1}{2} \rho v_1^2\) and \(\frac{1}{2} \rho v_2^2\):
\[
\frac{1}{2} \rho v_1^2 = \frac{1}{2} \times 1000 \times (5.9)^2 = 0.5 \times 1000 \times 34.81 = 17405 \, \text{Pa}
\]
\[
\frac{1}{2} \rho v_2^2 = \frac{1}{2} \times 1000 \times (12.45)^2 = 0.5 \times 1000 \times 155.23 = 77615 \, \text{Pa}
\]
Now substitute back:
\[
P_2 = 162080 + 17405 - 77615
\]
Calculating \(P_2\):
\[
P_2 = 162080 + 17405 - 77615 = 162080 + 17405 - 77615 = 86070 \, \text{Pa}
\]
Finally, we convert \(P_2\) back to gauge pressure in atm:
To find gauge pressure, we don't consider atmospheric pressure, so we can find \(P_2\) in atm directly:
\[
P_2 = \frac{86070}{101300} \approx 0.85 \, \text{atm}
\]
Thus, the gauge pressure of the water flowing through the smaller pipe is approximately:
\[
\boxed{0.85 \, \text{atm}}
\]