Water is conveyed through a uniform tube of 8 cm in diameter and 3140 m in length at the rate `2 xx 10^(-3) m^(3)` per second. The pressure required to maintain the flow is ( viscosity of water =`10^(-3)` )

Options

A. 1. `6.25 xx 10^(3) Nm^(-2)`
B. 2. `0.625 N m^(-2)`
C. 3. `0.0625 Nm^(-2)`
D. 4. `0.00625 N m^(-2)`

Explanation

To solve the problem of determining the pressure required to maintain the flow of water through a uniform tube, we will use Poiseuille's equation. Here’s the step-by-step solution:

### Step 1: Understand the Given Data
- Diameter of the tube (d) = 8 cm = 0.08 m (conversion from cm to m)
- Radius of the tube (r) = d/2 = 0.08 m / 2 = 0.04 m
- Length of the tube (L) = 3140 m
- Volume flow rate (V) = 2 x 10^(-3) m³/s
- Viscosity of water (η) = 10^(-3) Pa·s

### Step 2: Use Poiseuille's Equation
Poiseuille's equation relates the pressure difference (ΔP) required to maintain a flow rate through a cylindrical pipe:
\[ V = \frac{\pi \Delta P r^4}{8 \eta L} \]
Where:
- \( V \) = flow rate (m³/s)
- \( \Delta P \) = pressure difference (Pa)
- \( r \) = radius of the tube (m)
- \( \eta \) = viscosity (Pa·s)
- \( L \) = length of the tube (m)

### Step 3: Rearranging the Equation for Pressure
To find the pressure difference (ΔP), we rearrange the equation:
\[ \Delta P = \frac{8 \eta L V}{\pi r^4} \]

### Step 4: Substitute the Values
Now we will substitute the known values into the equation:
- \( \eta = 10^{-3} \) Pa·s
- \( L = 3140 \) m
- \( V = 2 \times 10^{-3} \) m³/s
- \( r = 0.04 \) m

Substituting these values:
\[ \Delta P = \frac{8 \times (10^{-3}) \times 3140 \times (2 \times 10^{-3})}{\pi \times (0.04)^4} \]

### Step 5: Calculate the Denominator
First, calculate \( r^4 \):
\[ r^4 = (0.04)^4 = 0.000256 \text{ m}^4 \]
Now calculate \( \pi r^4 \):
\[ \pi r^4 = \pi \times 0.000256 \approx 0.000804 \text{ m}^4 \]

### Step 6: Calculate the Numerator
Now calculate the numerator:
\[ 8 \times (10^{-3}) \times 3140 \times (2 \times 10^{-3}) = 8 \times 10^{-3} \times 3140 \times 2 \times 10^{-3} = 50.24 \times 10^{-6} = 0.00005024 \text{ Pa·m}^3/s \]

### Step 7: Calculate the Pressure
Now substitute back into the equation:
\[ \Delta P = \frac{0.00005024}{0.000804} \approx 62.5 \text{ Pa} \]

### Step 8: Convert to Newton per square meter
Since 1 Pa = 1 N/m², we can write:
\[ \Delta P \approx 6.25 \times 10^3 \text{ N/m}^2 \]

### Final Answer
The pressure required to maintain the flow is:
\[ \Delta P \approx 6.25 \times 10^3 \text{ N/m}^2 \]