Total Energy Head Calculation for Different Fluids
Problem Statement
A fluid is flowing in a 20cm diameter pipe at a pressure of 28 KN/m² with a velocity of 2.4m/s. The elevation of center of pipe above a given datum is 4m. Find the total energy head above the given datum if the fluid is:
- Water
- Oil of specific gravity 0.82
- Gas with a specific weight of 6.4 N/m³
Given Data
| Pipe diameter (D) | 20 cm = 0.2 m |
| Pressure (P) | 28 kN/m² = 28 kPa |
| Velocity (V) | 2.4 m/s |
| Elevation (Z) | 4 m |
| Acceleration due to gravity (g) | 9.81 m/s² |
1. The Energy Equation
The total energy head in a fluid is the sum of three components:
E = P/γ + V²/(2g) + Z
Where:
E = Total energy head above datum (m)
P = Pressure (N/m²)
γ = Specific weight of fluid (N/m³)
V = Velocity (m/s)
g = Acceleration due to gravity (9.81 m/s²)
Z = Elevation above datum (m)
The velocity head V²/(2g) is constant for all fluids at 2.4 m/s:
V²/(2g) = (2.4 m/s)²/(2 × 9.81 m/s²) = 5.76 m²/s²/(19.62 m/s²) = 0.294 m
The elevation head Z is also constant at 4 m.
The pressure head P/γ will vary for each fluid due to different specific weights.
2. Case (a): Water
Step 2.1: Calculate the specific weight of water
γwater = ρwater × g = 1000 kg/m³ × 9.81 m/s² = 9810 N/m³
Step 2.2: Calculate the pressure head
Pressure head = P/γwater = 28000 N/m² ÷ 9810 N/m³ = 2.854 m
Step 2.3: Calculate the total energy head
Ewater = P/γwater + V²/(2g) + Z
Ewater = 2.854 m + 0.294 m + 4 m = 7.148 m
3. Case (b): Oil (Specific Gravity = 0.82)
Step 3.1: Calculate the specific weight of oil
γoil = S.G. × γwater = 0.82 × 9810 N/m³ = 8044.2 N/m³
Step 3.2: Calculate the pressure head
Pressure head = P/γoil = 28000 N/m² ÷ 8044.2 N/m³ = 3.480 m
Step 3.3: Calculate the total energy head
Eoil = P/γoil + V²/(2g) + Z
Eoil = 3.480 m + 0.294 m + 4 m = 7.774 m
4. Case (c): Gas (Specific Weight = 6.4 N/m³)
Step 4.1: Use the given specific weight of gas
γgas = 6.4 N/m³
Step 4.2: Calculate the pressure head
Pressure head = P/γgas = 28000 N/m² ÷ 6.4 N/m³ = 4375 m
Step 4.3: Calculate the total energy head
Egas = P/γgas + V²/(2g) + Z
Egas = 4375 m + 0.294 m + 4 m = 4379.294 m
5. Comparison of Results
Note: The scale for gas is broken as the value is much larger than for water and oil.
Physical Interpretation
The results demonstrate several important physical principles:
- Inverse relationship between specific weight and pressure head: As the specific weight of the fluid decreases (from water to oil to gas), the pressure head increases dramatically.
- Constant components: The velocity head and elevation head remain the same for all fluids as they are independent of fluid properties.
- Gas vs. Liquids: The total energy head for the gas (4379.3 m) is significantly higher than for liquids (7.15 m for water and 7.77 m for oil) due to the much lower specific weight of gas.
- Practical implications: The higher energy head in gas systems means that pressure energy is converted to a much greater equivalent height, which explains why gas can easily travel long distances in pipelines with relatively small pressure differences.
- Engineering considerations: When designing systems with different fluids, engineers must account for these differences in energy heads to ensure proper system performance.
Conclusion
We have calculated the total energy head for three different fluids flowing in the same pipe with identical conditions of pressure, velocity, and elevation:
- For water (γ = 9810 N/m³): E = 7.15 m
- For oil with specific gravity 0.82 (γ = 8044.2 N/m³): E = 7.77 m
- For gas (γ = 6.4 N/m³): E = 4379.3 m
The variation in total energy head is primarily due to differences in the pressure head component, which depends inversely on the specific weight of the fluid. The velocity head and elevation head remain constant for all fluids under the given conditions.
