Determining Pump Gauge Pressure for Given Flow Rate
Problem Statement
The pipe flow in the figure is driven by the pump. What gauge pressure is needed to be supplied by the pump to provide water flow rate of Q = 60m³/h? Neglect head loss from A to B. Head loss from C to D = 6.5V²CD; Head loss from D to E = 0.13V²DE; dAB (diameter of pipe AB) = dCD = 5cm; dDE = 2cm. Where VCD = velocity in pipe CD and VDE = velocity in pipe DE.
Given Data
| Flow Rate (Q) | 60 m³/h |
| Diameter of AB and CD (dAB, dCD) | 5 cm = 0.05 m |
| Diameter of DE (dDE) | 2 cm = 0.02 m |
| Head Loss from C to D (hLCD) | 6.5V²CD |
| Head Loss from D to E (hLDE) | 0.13V²DE |
| Head Loss from A to B (hLAB) | 0 (Neglected) |
1. Flow Rate Conversion and Cross-sectional Areas
Converting Flow Rate:
Q = 60 m³/h = 60/3600 m³/s = 0.0167 m³/s
Cross-sectional Area of Pipes AB and CD:
AAB = ACD = (π/4) × d² = (π/4) × (0.05)² = 0.001963 m²
Cross-sectional Area of Pipe DE:
ADE = (π/4) × d² = (π/4) × (0.02)² = 0.000314 m²
2. Calculating Flow Velocities
Velocity in Pipes AB and CD:
VAB = VCD = Q/AAB = 0.0167/0.001963 = 8.5 m/s
Velocity in Pipe DE:
VDE = Q/ADE = 0.0167/0.000314 = 53.2 m/s
3. Head Loss Calculations
Head Loss from A to B:
hLAB = 0 (given – neglected)
Head Loss from C to D:
hLCD = 6.5V²CD = 6.5 × (8.5)² = 6.5 × 72.25 = 469.625 m
Head Loss from D to E:
hLDE = 0.13V²DE = 0.13 × (53.2)² = 0.13 × 2830.24 = 367.93 m
Total Head Loss:
hL = hLAB + hLCD + hLDE = 0 + 469.625 + 367.93 = 837.555 m
4. Applying Bernoulli’s Equation Between Points 1 and E
Taking datum through point A and applying Bernoulli’s equation between points 1 and E:
P₁/γ + V₁²/(2g) + Z₁ + hp = PE/γ + V²E/(2g) + ZE + hL
Where:
P₁ = 0 (atmospheric pressure at point 1)
V₁ = 0 (stagnant water surface)
Z₁ = 0 (datum level)
PE = 0 (atmospheric pressure at exit point E)
VE = VDE = 53.2 m/s
ZE = 0 (assuming same level as datum)
hL = 837.555 m (total head loss)
hp = head supplied by the pump
Substituting these values:
0 + 0 + 0 + hp = 0 + (53.2)²/(2×9.81) + 0 + 837.555
hp = (53.2)²/(2×9.81) + 837.555
hp = 2830.24/19.62 + 837.555
hp = 144.25 + 837.555
hp = 981.805 m
5. Converting Pump Head to Gauge Pressure
The gauge pressure supplied by the pump can be calculated as:
Pgauge = ρ × g × hp
Pgauge = 1000 kg/m³ × 9.81 m/s² × 981.805 m
Pgauge = 9,631,507 Pa ≈ 9,631.5 kPa
Conclusion
By calculating the velocities in different pipe sections and determining the head losses, we applied Bernoulli’s equation to find the required pump head. This head was then converted to gauge pressure, resulting in a required pump gauge pressure of 9,631.5 kPa to maintain the specified flow rate of 60 m³/h through the given pipe system.
