To find the total head or energy per unit weight of water at a cross-section that is 5 meters above the datum line, we use Bernoulli's equation. The total head [tex]\(H\)[/tex] is given by the sum of the pressure head, velocity head, and elevation head. The formula can be written as:
[tex]\[ H = \frac{P}{\rho g} + \frac{v^2}{2g} + z \][/tex]
where:
- [tex]\(P\)[/tex] is the pressure
- [tex]\(\rho\)[/tex] is the density of water (1000 kg/m³)
- [tex]\(g\)[/tex] is the acceleration due to gravity (9.81 m/s²)
- [tex]\(v\)[/tex] is the mean velocity of water
- [tex]\(z\)[/tex] is the height above datum
Given data:
- Pressure [tex]\(P = 29.43 \, M/cm²\)[/tex]
- Mean velocity [tex]\(v = 2.0 \, ml/s\)[/tex]
- Height above datum [tex]\(z = 5 \, m\)[/tex]
First, we need to convert the units to consistent SI units:
1. Pressure Conversion:
- 1 Mega Pascal (M) = [tex]\(10^6\)[/tex] Pascals (Pa)
- Therefore, [tex]\(29.43 \, M/cm² = 29.43 \times 10^6 \, Pa\)[/tex]
So, the pressure in Pascals:
[tex]\[ P = 29.43 \times 10^6 \, Pa \][/tex]
2. Velocity Conversion:
- 1 milliliter (ml) = [tex]\(10^{-6}\)[/tex] liters = [tex]\(10^{-6}\)[/tex] cubic meters (m³)
- Therefore, [tex]\(2.0 \, ml/s = 2.0 \times 10^{-6} \, m/s\)[/tex]
So, the mean velocity in meters per second:
[tex]\[ v = 2.0 \times 10^{-6} \, m/s \][/tex]
Now, let's calculate each term in the Bernoulli's equation:
1. Pressure Head Term ([tex]\(\frac{P}{\rho g}\)[/tex]):
[tex]\[ \frac{P}{\rho g} = \frac{29.43 \times 10^6 \, Pa}{1000 \, kg/m³ \times 9.81 \, m/s²} \][/tex]
[tex]\[ \frac{P}{\rho g} = \frac{29.43 \times 10^6}{1000 \times 9.81} \][/tex]
[tex]\[ \frac{P}{\rho g} = \frac{29.43 \times 10^6}{9810} \][/tex]
[tex]\[ \frac{P}{\rho g} = 3000.0 \, m \][/tex]
2. Velocity Head Term ([tex]\(\frac{v^2}{2g}\)[/tex]):
[tex]\[ \frac{v^2}{2g} = \frac{(2.0 \times 10^{-6} \, m/s)^2}{2 \times 9.81 \, m/s²} \][/tex]
[tex]\[ \frac{v^2}{2g} = \frac{4 \times 10^{-12}}{19.62} \][/tex]
[tex]\[ \frac{v^2}{2g} = 2.038735983690112 \times 10^{-13} \, m \][/tex]
3. Elevation Head Term ([tex]\(z\)[/tex]):
[tex]\[ z = 5 \, m \][/tex]
Finally, summing these components gives the total head:
[tex]\[ H = 3000.0 \, m + 2.038735983690112 \times 10^{-13} \, m + 5 \, m \][/tex]
Considering the negligible value of the velocity head term, we can state the total head as:
[tex]\[ H \approx 3005.0 \, m \][/tex]
So the detailed components are:
- Pressure Head: 3000.0 meters
- Velocity Head: [tex]\(2.038735983690112 \times 10^{-13}\)[/tex] meters
- Elevation Head: 5 meters
The total head or energy per unit weight of water at the given cross-section is approximately 3005.0 meters.
