Water is conveyed thourth a horizontal tube 8 cm in diameter and 4 kilometer in length at the rate of 20 litre/s Assumming only viscous resistance , callcalute the pressure required to maintain the flow . Coefficient of viscosity of water is 0.001 pa s

Answer

To calculate the pressure required to maintain the flow of water through the horizontal tube, we can use the Hagen-Poiseuille equation, which describes the flow of a viscous fluid through a pipe. The equation is given by:

\[ \Delta P = \frac{8 \eta L Q}{\pi r^4} \]

Where:
- \(\Delta P\) = pressure drop (Pa)
- \(\eta\) = dynamic viscosity (Pa·s)
- \(L\) = length of the pipe (m)
- \(Q\) = volumetric flow rate (m³/s)
- \(r\) = radius of the pipe (m)

### Step 1: Convert the given quantities to SI units
- Diameter of the tube = 8 cm = 0.08 m
- Radius of the tube, \(r = \frac{0.08}{2} = 0.04\) m
- Length of the tube = 4 km = 4000 m = 4000 m
- Volumetric flow rate, \(Q = 20\) liters/s = \(20 \times 10^{-3}\) m³/s = 0.02 m³/s
- Coefficient of viscosity, \(\eta = 0.001\) Pa·s

### Step 2: Substitute the values into the Hagen-Poiseuille equation
Now we can substitute the values into the equation:

\[ \Delta P = \frac{8 \times 0.001 \, \text{Pa·s} \times 4000 \, \text{m} \times 0.02 \, \text{m³/s}}{\pi \times (0.04 \, \text{m})^4} \]

### Step 3: Calculate the denominator
Calculate \( \pi \times (0.04)^4 \):

\[ (0.04)^4 = 0.00000256 \, \text{m}^4 \]
\[ \pi \times (0.04)^4 \approx 3.14159 \times 0.00000256 \approx 0.000008042 \, \text{m}^4 \]

### Step 4: Calculate the numerator
Calculate \( 8 \times 0.001 \times 4000 \times 0.02 \):

\[ 8 \times 0.001 \times 4000 \times 0.02 = 0.64 \, \text{Pa·m³/s} \]

### Step 5: Calculate the pressure drop
Now substitute the values into the pressure drop equation:

\[ \Delta P = \frac{0.64}{0.000008042} \approx 79607.78 \, \text{Pa} \]

### Step 6: Final result
Thus, the pressure required to maintain the flow is approximately:

\[ \Delta P \approx 79608 \, \text{Pa} \text{ or } 79.6 \, \text{kPa} \]