Analysis of Pump System with 90 LPS Water Flow
Problem Statement
A pump P is pumping 90 lps of water from a tank. The system consists of a tank (point A), a pump P, and two sections of pipe with diameters 16 cm (at point B) and 10 cm (at point C).
(a) What will be the pressure at points B and C when the pump delivers 14.5 KW of power to the flow? Assume the losses in the system to be negligible.
(b) What will be the pressure at C when the loss in the inlet to the pump is negligible and between the pump and the point C, a loss equal to 2 times the velocity head at B takes place.
Given Data
| Discharge (Q) | 90 lps = 0.09 m³/s |
| Diameter at Point B (dB) | 16 cm = 0.16 m |
| Diameter at Point C (dC) | 10 cm = 0.10 m |
| Power Delivered by Pump | 14.5 kW = 14,500 W |
| Pressure at Point A (PA) | 0 (atmospheric pressure) |
| Velocity at Point A (VA) | 0 (tank level) |
| Acceleration due to Gravity (g) | 9.81 m/s² |
| Density of Water (ρ) | 1000 kg/m³ |
1. Determining Pump Head from Power
First, let’s calculate the head supplied by the pump using the power equation:
Power (P) = ρ × g × Q × hp
where hp is the head supplied by the pump.
hp = P / (ρ × g × Q)
hp = 14,500 / (1000 × 9.81 × 0.09)
hp = 14,500 / 882.9
hp ≈ 16.42 m
2. Calculating Velocities at Points B and C
Using the continuity equation Q = A × V:
Velocity at Point B:
VB = Q / AB = Q / (π × dB² / 4)
VB = 0.09 / (π × 0.16² / 4)
VB = 0.09 / 0.0201
VB = 4.47 m/s
Velocity at Point C:
VC = Q / AC = Q / (π × dC² / 4)
VC = 0.09 / (π × 0.10² / 4)
VC = 0.09 / 0.00785
VC = 11.46 m/s
3. Determining Pressure at Point B
Applying Bernoulli’s equation between points A and B, taking B as datum:
PA/ρg + VA²/2g + ZA + hp = PB/ρg + VB²/2g + ZB
Given: PA = 0, VA = 0, ZA = 4 m, ZB = 0 (datum)
Rearranging to solve for PB:
PB/ρg = 4 + 16.42 – VB²/2g
PB/ρg = 20.42 – 4.47²/(2 × 9.81)
PB/ρg = 20.42 – 1.02
PB/ρg = 19.40
PB = 19.40 × ρg = 19.40 × 1000 × 9.81
PB = 19,440 Pa = 19.44 kPa
4. Determining Pressure at Point C – Without Losses
Applying Bernoulli’s equation between points A and C, taking B as datum:
PA/ρg + VA²/2g + ZA + hp = PC/ρg + VC²/2g + ZC
Given: PA = 0, VA = 0, ZA = 4 m, ZC = -5 m (below datum)
Rearranging to solve for PC:
PC/ρg = 4 + 16.42 – VC²/2g – (-5)
PC/ρg = 25.42 – 11.46²/(2 × 9.81)
PC/ρg = 25.42 – 6.69
PC/ρg = 8.73
PC = 8.73 × ρg = 8.73 × 1000 × 9.81
PC = 85,640 Pa = 85.64 kPa
5. Determining Pressure at Point C – With Losses
For part (b), we need to account for a head loss between pump and point C equal to 2 times the velocity head at B.
Loss between B and C:
hL = 2 × (VB²/2g)
hL = 2 × (4.47²/(2 × 9.81))
hL = 2 × 1.02
hL = 2.04 m
Applying Bernoulli’s equation between points A and C with head loss:
PA/ρg + VA²/2g + ZA + hp – hL = PC/ρg + VC²/2g + ZC
PC/ρg = 4 + 16.42 – 2.04 – VC²/2g – (-5)
PC/ρg = 23.38 – 6.69
PC/ρg = 16.69
PC = 16.69 × ρg = 16.69 × 1000 × 9.81
PC = 163,730 Pa = 163.73 kPa
Conclusion
In this pump system analysis, we determined the pressures at key points in the system under two different conditions:
Part (a): With negligible losses, the pressure at point B is 19.44 kPa and at point C is 85.64 kPa when the pump delivers 14.5 kW of power.
Part (b): When accounting for head losses equal to 2 times the velocity head at point B, the pressure at point C increases to 163.73 kPa.
This demonstrates how head losses in a piping system can significantly affect the pressure distribution, which is critical for proper pump and system design.
