Required Pump Gauge Pressure for a Pipe Flow System
Problem Statement
The pipe flow in the figure is driven by a pump. Determine the gauge pressure that must be supplied by the pump to provide a water flow rate of Q = 60 m³/h. Neglect the head loss from A to B. The head loss from C to D is given by 30(VCD²)/(2g) and from D to E by 20(VCD²)/(2g). The diameters are: dAB = dCD = 5 cm and dDE = 2 cm. (VCD is the velocity in pipe CD.)
Given Data
| Flow Rate (Q) | 60 m³/h = 60/3600 ≈ 0.0167 m³/s |
| Diameter of AB and CD (dAB, dCD) | 5 cm = 0.05 m |
| Area of AB and CD (AAB, ACD) | A = (π/4) × (0.05)² ≈ 0.001963 m² |
| Diameter of DE (dDE) | 2 cm = 0.02 m |
| Area of DE (ADE) | A = (π/4) × (0.02)² ≈ 0.000314 m² |
| Pressure at Point A (PA) | Atmospheric (gauge = 0) |
| Pressure at Point B (PB) | Atmospheric (gauge = 0) |
| Head Loss from C to D | hCD = 30(VCD²)/(2g) |
| Head Loss from D to E | hDE = 20(VCD²)/(2g) |
| Specific Weight of Water (γ) | 9810 N/m³ |
| Acceleration due to Gravity (g) | 9.81 m/s² |
| Elevations |
(For the purpose of applying Bernoulli’s equation, assume: – Point 1: Z₁ = 10 m – Point E: ZE = 80 m) |
1. Determining Velocities
The discharge Q is constant:
– In pipe AB (and CD):
VAB = VCD = Q / AAB ≈ 0.0167 / 0.001963 ≈ 8.5 m/s
– In pipe DE:
VDE = Q / ADE ≈ 0.0167 / 0.000314 ≈ 53.2 m/s
2. Calculating Head Losses
The head loss from C to D is:
hCD = 30(VCD²)/(2g) = 30 × (8.5²)/(2×9.81)
8.5² ≈ 72.25 → hCD ≈ 30 × 72.25 / 19.62 ≈ 110.5 m
The head loss from D to E is:
hDE = 20(VCD²)/(2g) = 20 × (8.5²)/(2×9.81) ≈ 73.65 m
Total head loss, hL = 0 (from A to B) + 110.5 + 73.65 ≈ 184.15 m
3. Determining the Pump Head (hₚ)
Applying Bernoulli’s equation between point 1 (in the reservoir) and point E (at the discharge) with the datum taken through A:
P₁/γ + (V₁²)/(2g) + Z₁ + hₚ = PE/γ + (VE²)/(2g) + ZE + hL
Since both P₁ and PE are atmospheric (gauge = 0) and V₁ = 0:
0 + 0 + 10 + hₚ = (53.2²)/(2×9.81) + 80 + 184.15
(53.2²)/(2×9.81) ≈ 144.3 m, so:
10 + hₚ = 144.3 + 80 + 184.15
hₚ = (144.3 + 80 + 184.15) − 10 ≈ 398.45 m
4. Calculating the Gauge Pressure
To determine the gauge pressure required from the pump, apply Bernoulli’s equation between points B and C. Since the velocities and elevations at these points are equal, the pressure difference is:
PC − PB = γ hₚ
Substituting:
PC − PB = 9810 N/m³ × 398.45 m ≈ 3,908,304 Pa
Converting to kPa: 3,908,304 Pa ≈ 3908.3 kPa
Physical Interpretation
In this system, the pump must overcome not only the kinetic energy increases due to velocity changes (from the 5 cm to the 2 cm pipe sections) but also significant head losses along the pipe segments C–D and D–E. The large velocity in the small-diameter section (DE) contributes to a high dynamic head, and when combined with the frictional head losses (expressed here as a function of VCD²), the overall head required from the pump becomes very high. The gauge pressure required is then calculated from this head using the relation P = γh.
Detailed Explanation for Students
Step 1: Velocity Calculation
The flow rate is constant throughout the system. Calculate the velocity in the 5 cm pipe using Q = A V, then determine the much higher velocity in the 2 cm pipe.
Step 2: Head Loss Evaluation
Using the given head loss coefficients, compute the losses in the sections C–D and D–E. Add these to obtain the total head loss in the system.
Step 3: Pump Head Requirement
Apply Bernoulli’s equation between the reservoir (point 1) and the discharge (point E), including the pump head hₚ. Solve for hₚ, which represents the extra energy per unit weight needed from the pump.
Step 4: Gauge Pressure Determination
Finally, the pump must provide a pressure increase equivalent to γhₚ. Multiply the calculated pump head by the specific weight of water to determine the required gauge pressure.
