Analysis of Turbine System with Reservoir and Tailrace
Problem Statement
A pipe connects a reservoir to a turbine which discharges water to the tailrace through another pipe. The head loss between the reservoir and the turbine is 8 times kinetic head in the pipe and that from the turbine to tailrace is 0.4 times the kinetic head in the pipe. The rate of flow is 1.36 m³/s and the pipe diameter in both cases is 1 m. The elevation of the reservoir water level is 54 m, the turbine inlet is at elevation 49 m, and the tailrace water level is at elevation 4 m.
Determine:
(a) The pressure at inlet and exit of turbine
(b) The power generated by the turbine
Given Data
| Diameter of pipes (d) | 1 m |
| Discharge (Q) | 1.36 m³/s |
| Reservoir water level (point 1) | 54 m (elevation) |
| Turbine inlet (point 2) | 49 m (elevation) |
| Turbine outlet (point 3) | 49 m (elevation, same as inlet) |
| Tailrace water level (point 4) | 4 m (elevation) |
| Head loss between reservoir and turbine | 8 × (kinetic head in pipe) |
| Head loss between turbine and tailrace | 0.4 × (kinetic head in pipe) |
| Acceleration due to Gravity (g) | 9.81 m/s² |
| Density of Water (ρ) | 1000 kg/m³ |
1. Determining Velocity and Kinetic Head
Since the diameter is the same for both pipes (inlet and outlet), the velocity will be the same in both sections.
Calculate the velocity:
V = Q / A = Q / (π × d² / 4)
V = 1.36 / (π × 1² / 4)
V = 1.36 / 0.7854
V = 1.73 m/s
Calculate the kinetic head:
Kinetic head = V² / (2g) = 1.73² / (2 × 9.81)
Kinetic head = 2.99 / 19.62
Kinetic head = 0.152 m
2. Calculating Head Losses
Head loss between reservoir and turbine (hL,1-2):
hL,1-2 = 8 × (V² / 2g) = 8 × 0.152
hL,1-2 = 1.22 m
Head loss between turbine and tailrace (hL,3-4):
hL,3-4 = 0.4 × (V² / 2g) = 0.4 × 0.152
hL,3-4 = 0.06 m
Total head loss:
hL,total = hL,1-2 + hL,3-4 = 1.22 + 0.06 = 1.28 m
3. Determining Pressure at Turbine Inlet (Point 2)
Applying Bernoulli’s equation between points 1 (reservoir) and 2 (turbine inlet):
P₁/ρg + V₁²/2g + Z₁ = P₂/ρg + V₂²/2g + Z₂ + hL,1-2
Where:
P₁ = 0 (atmospheric pressure at free surface)
V₁ = 0 (velocity at reservoir surface ≈ 0)
Z₁ = 54 m
V₂ = 1.73 m/s
Z₂ = 49 m
hL,1-2 = 1.22 m
Rearranging to solve for P₂/ρg:
P₂/ρg = Z₁ – Z₂ – V₂²/2g – hL,1-2
P₂/ρg = 54 – 49 – 0.152 – 1.22
P₂/ρg = 5 – 1.372
P₂/ρg = 3.628 m
P₂ = 3.628 × ρg = 3.628 × 1000 × 9.81
P₂ = 35,590 Pa = 35.59 kPa
4. Determining Pressure at Turbine Exit (Point 3)
Applying Bernoulli’s equation between points 3 (turbine exit) and 4 (tailrace):
P₃/ρg + V₃²/2g + Z₃ = P₄/ρg + V₄²/2g + Z₄ + hL,3-4
Where:
P₄ = 0 (atmospheric pressure at free surface)
V₄ = 0 (velocity at tailrace surface ≈ 0)
Z₄ = 4 m
V₃ = 1.73 m/s
Z₃ = 49 m
hL,3-4 = 0.06 m
Rearranging to solve for P₃/ρg:
P₃/ρg = Z₄ – Z₃ + hL,3-4 + V₃²/2g
P₃/ρg = 4 – 49 + 0.06 + 0.152
P₃/ρg = 4 – 49 + 0.212
P₃/ρg = -44.788 m
P₃ = -44.788 × ρg = -44.788 × 1000 × 9.81
P₃ = -439,370 Pa = -439.37 kPa (gauge pressure)
Pressure at Turbine Exit (P₃) = -439.37 kPa (gauge)
5. Determining Power Generated by Turbine
Calculate the head extracted by the turbine:
Applying Bernoulli’s equation between points 2 and 3 and solving for turbine head (hT):
P₂/ρg + V₂²/2g + Z₂ = P₃/ρg + V₃²/2g + Z₃ + hT
Since Z₂ = Z₃ (same elevation) and V₂ = V₃ (same velocity):
P₂/ρg = P₃/ρg + hT
hT = P₂/ρg – P₃/ρg
hT = 3.628 – (-44.788)
hT = 48.416 m
Calculate the power generated by the turbine:
Power = ρ × g × Q × hT
Power = 1000 × 9.81 × 1.36 × 48.416
Power = 644,680 W = 644.68 kW
Conclusion
In this turbine system analysis, we determined that:
1. Pressure Conditions: The pressure at the turbine inlet is 35.59 kPa (gauge), which is positive as expected due to the reservoir elevation. The pressure at the turbine exit is -439.37 kPa (gauge), which is negative due to the suction effect created by the elevation difference between the turbine and the tailrace.
2. Power Generation: The turbine generates 644.68 kW of power by extracting energy from the 48.416 m of head available between its inlet and outlet.
3. Energy Conversion: This system effectively converts potential energy (due to elevation differences) into mechanical energy through the turbine, with some energy losses due to friction in the pipes.
