Let's solve this problem step-by-step.
1. Convert the diameter from centimeters to meters:
The diameter of the tube is given as [tex]\(8 \, \text{cm}\)[/tex]. To convert it to meters, we divide by 100:
[tex]\[
\text{Diameter} = \frac{8 \, \text{cm}}{100} = 0.08 \, \text{m}
\][/tex]
2. Calculate the radius of the tube:
The radius is half of the diameter:
[tex]\[
\text{Radius} = \frac{0.08 \, \text{m}}{2} = 0.04 \, \text{m}
\][/tex]
3. Use the Hagen-Poiseuille equation to find the pressure drop [tex]\(\Delta P\)[/tex]:
The Hagen-Poiseuille equation for pressure drop in a tube due to viscous flow is:
[tex]\[
\Delta P = \frac{8 \eta L Q}{\pi r^4}
\][/tex]
where:
- [tex]\(\eta\)[/tex] is the dynamic viscosity,
- [tex]\(L\)[/tex] is the length of the tube,
- [tex]\(Q\)[/tex] is the volumetric flow rate, and
- [tex]\(r\)[/tex] is the radius of the tube.
Given:
- [tex]\(\eta \approx 0.001 \, \text{Pa} \cdot \text{s}\)[/tex] (assumed typical value for water at room temperature)
- [tex]\(L = 3140 \, \text{m}\)[/tex]
- [tex]\(Q = 2 \times 10^{-3} \, \text{m}^3/\text{s}\)[/tex]
- [tex]\(r = 0.04 \, \text{m}\)[/tex]
4. Plug in the values and calculate the pressure drop [tex]\(\Delta P\)[/tex]:
[tex]\[
\Delta P = \frac{8 \times 0.001 \, \text{Pa} \cdot \text{s} \times 3140 \, \text{m} \times 2 \times 10^{-3} \, \text{m}^3/\text{s}}{\pi \times (0.04 \, \text{m})^4}
\][/tex]
5. Calculate the pressure drop in Pascals:
[tex]\[
\Delta P \approx 6246.83 \, \text{Pa}
\][/tex]
6. Convert the pressure drop to the desired units:
The problem asks for the pressure drop in terms of [tex]\(10^5 \, \text{Nm}^{-2}\)[/tex]. Therefore:
[tex]\[
\text{Pressure drop} = \frac{6246.83 \, \text{Pa}}{10^5} = 0.0624683 \times 10^5 \, \text{Nm}^{-2}
\][/tex]
Thus, the pressure required to maintain the flow is approximately [tex]\(0.0624683 \times 10^5 \, \text{Nm}^{-2}\)[/tex].
