Energy Supplied by the Pump in a Reservoir-Pipe-Nozzle System
Problem Statement
Water is pumped from a reservoir through a 150 mm diameter pipe and is delivered at a height of 15 m from the pump centerline via a 100 mm nozzle connected to a 150 mm discharge line. The pump inlet pressure is 210 kN/m² absolute with an inlet velocity of 5 m/s, and the jet is discharged into the atmosphere. Determine the energy supplied by the pump, assuming an atmospheric pressure of 101.3 kN/m² and no friction losses.
Given Data
| Diameter at Section 1 (d₁) | 150 mm = 0.15 m |
| Area at Section 1 (A₁) | A₁ = (π/4) × (0.15)² ≈ 0.01767 m² |
| Diameter at Section 2 (d₂) | 100 mm = 0.1 m |
| Area at Section 2 (A₂) | A₂ = (π/4) × (0.1)² ≈ 0.007854 m² |
| Inlet Velocity (V₁) | 5 m/s |
| Pump Inlet Pressure (P₁) | 210 kN/m² = 210000 N/m² |
| Atmospheric Pressure (P₂) | 101.3 kN/m² = 101300 N/m² |
| Datum Head at Pump Inlet (Z₁) | 0 m |
| Elevation of Nozzle (Z₂) | 15 m |
| Acceleration due to Gravity (g) | 9.81 m/s² |
| Specific Weight of Water (γ) | 9810 N/m³ |
1. Calculating Flow Rate and Velocities
First, determine the discharge (Q) using the inlet pipe:
Q = A₁ × V₁ = 0.01767 m² × 5 m/s = 0.08835 m³/s
Next, find the velocity at the nozzle (section 2) using:
V₂ = Q / A₂ = 0.08835 m³/s / 0.007854 m² ≈ 11.25 m/s
2. Applying Bernoulli’s Equation
Apply Bernoulli’s equation between the pump inlet (section 1) and the nozzle exit (section 2):
P₁/γ + V₁²/(2g) + Z₁ + hₚ = P₂/γ + V₂²/(2g) + Z₂
Substitute the known values:
210000/9810 + 5²/(2×9.81) + 0 + hₚ = 101300/9810 + 11.25²/(2×9.81) + 15
Evaluating the terms:
Left Side: 210000/9810 ≈ 21.41 m and 5²/(2×9.81) ≈ 1.27 m, giving a total of 22.68 m + hₚ.
Right Side: 101300/9810 ≈ 10.33 m, 11.25²/(2×9.81) ≈ 6.45 m, plus the elevation 15 m gives 31.78 m.
Solving for the pump head:
hₚ = 31.78 m − 22.68 m = 9.09 m
3. Calculating Energy Supplied by the Pump
The energy (or power) supplied by the pump is calculated by:
Energy = γ × Q × hₚ
Substituting the values:
Energy = 9810 N/m³ × 0.08835 m³/s × 9.09 m ≈ 7878 W
Physical Interpretation
This problem demonstrates how a pump must supply extra energy to overcome both the pressure difference between the reservoir and the atmosphere as well as the elevation change. Notice that:
Flow Continuity:
The discharge remains constant. Thus, when the cross-sectional area decreases at the nozzle, the fluid velocity increases.
Energy Balance:
Bernoulli’s equation balances pressure energy, kinetic energy, and potential energy. The pump head (hₚ) represents the extra energy needed to achieve the required flow and elevation.
Pump Energy:
Multiplying the specific weight, flow rate, and pump head gives the energy delivered per unit time by the pump.
Detailed Explanation for Students
Step 1: Flow Rate and Velocity Calculation
Determine the flow rate using the inlet pipe’s area and velocity. Then, by applying the principle of conservation of mass, calculate the velocity in the nozzle where the area is smaller.
Step 2: Applying Bernoulli’s Equation
Write the Bernoulli’s equation between two points – the pump inlet and the nozzle exit. Include the additional head (hₚ) provided by the pump to compensate for energy changes due to pressure, kinetic energy, and elevation differences.
Step 3: Energy Calculation
With the pump head determined, compute the energy supplied by multiplying the specific weight of water, the flow rate, and the pump head. This provides a measure of the power required from the pump.
This systematic approach integrates fundamental fluid mechanics principles with practical pump operation, illustrating how energy is imparted to a fluid to achieve desired pressure and flow characteristics.
