SYSTEM ANALYSIS & PRELIMINARY DESIGN
Given values:
Population = 200 residents.
Domestic use = 150 L/person/day.
Peak hour factor = 3.0.
Simultaneous use factor = 0.6.
Fire protection requirement = 30 L/s for 30 min.
Cooling tower makeup (if applicable) = 5 L/s average.
DN = 100 mm (chosen).
Pipe roughness and friction factor for head-loss estimation in Part 1b: use f = 0.02
(assignment instruction).
Minor loss coefficient K_total = 8.
Water: ρ = 1000 kg/m³, g = 9.81 m/s².
Floor height = 3.5 m → elevation of 8th floor = 8 × 3.5 m = 28.0 m
above ground.
Required minimum pressure at fixture = 150 kPa.
Municipal supply at ground = 250 kPa (gage).
1a) Peak-hour water demand (L/s)
Average daily domestic consumption:
150 L/person/day × 200 people = 30,000 L/day
Average flow rate (over 24 h):
30000
Qavg = =0.34722 L / S
86400
Peak domestic flow using peak and simultaneous factors:
Q peak , dom =Q avg ∗( peak h our factor )∗( simultaneous use factor )
=0.34722∗3∗0.6=0.625 L/S
Total peak system demand (domestic + cooling) (L/s):
Q peak , total = 0.625+5=5.625 L / s
Fire flow (separate) and worst-case simultaneous:
Fire requirement = 30 L/s for 30 minutes.
Worst-case simultaneous (peak domestic + fire):
Q worst =5.625+30=35.625 L / s .
Summary
Peak domestic demand (after peak & simultaneous factors): 0.6250 L/s.
Peak system demand (domestic + cooling): 5.6250 L/s.
Fire flow requirement: 30.000 L/s (30 minutes).
Worst-case simultaneous flow (domestic + fire): 35.6250 L/s
,1(b) Minimum required pump discharge pressure to guarantee 150 kPa at 8th-floor
fixtures
Convert required fixture pressure to head (m):
Required pressure = 150 kPa = 150,000 Pa.
150000 150000
hreq = = =15.2905 m .
pg 1000∗9.81
Static elevation head to 8th floor (m):
Floor height = 3.5 m (floor to floor). Elevation to 8th floor = 8 × 3.5 = 28.000 m.
1(c) Velocity in DN100 at peak flow (used for head-loss estimates):
DN=0.100 m
2
π (0.100) 2
A= =0.00785 m
4
Q = 5.625 L/s = 0.005625 m³/s.
V = Q/A = 0.005625/0.0078539 = 0.7162 m/s.
1(d) Major (friction) head loss using Darcy–Weisbach with f = 0.02
2
LV
Darcy–Weisbach: hf = f
D ∗2 g
Given f = 0.02, L = 80 m, D = 0.100 m, V = 0.71619867 m/s, g = 9.81 m/s²:
V2 (0.71612)2
kinetic term: = = 0.026137 m
2 g 2 ∗ 9.81
Numerically V 2 = 0.71622 =0.512966 m 2 / s 2
2
V /2 g =0.512966/(2∗9.81)=0.512966/19.62= 0.026137304666 m .
80
hf =0.02× × 0.026137304666
0.100
=0.02×800×0.026137304666 = 0.418301 m.
Minor losses (fittings, valves):
Minor-head coefficient K tot = 8 (assignment). Minor loss head:
2
V
hm = Ktot 8×0.026137304666=0.209151 m
2g
Convert required head to pressure (gage, kPa):
P = ρgH = 1000×9.81×43.917972=430835.309 Pa=430.835 kPa(gage).
Interpretation: The pump must be able to deliver ≈ 43.92 m of head at the building
peak flow (5.625 L/s) so that 8th-floor fixtures see ≥150 kPa. The municipal supply at
ground is 250 kPa (gage); that means the pump must provide a net boost if tied to
municipal suction
, Pump option check, energy consumption and costs
Pump A: H_max = 40 m; Q_max = 15 L/s; electrical rating = 8 kW; cost = QR
11,000; efficiency η = 70% (0.70).
Pump B: H_max = 50 m; Q_max = 12 L/s; electrical rating = 10 kW; cost = QR
16,500; efficiency η = 75% (0.75).
Pump C: H_max = 60 m; Q_max = 10 L/s; electrical rating = 12 kW; cost = QR
22,000; efficiency η = 78% (0.78).
Feasibility
Required design flow = 5.625 L/s and required head ≈ 43.918 m.
Pump A: H_max = 40 m < 43.918 m ⇒ CANNOT meet required head →
infeasible.
Pump B: H_max = 50 m ≥ 43.918 m and Q_max = 12 L/s ≥ 5.625 L/s ⇒
FEASIBLE.
Pump C: H_max = 60 m ≥ 43.918 m but Q_max = 10 L/s ≥ 5.625 L/s ⇒
FEASIBLE.
Hydraulic power at design conditions
Hydraulic power (W):
P h yd =ρgQH
P h yd =1000×9.81×0.005625×43.917972=2423.4486147 W=2.423449 kW
Electrical input power (kW) for each pump:
Pelec = ❑
η
· Pump A (η = 0.70): Pelec ,A=2.423449/0.70=3.462069 kW.
· Pump B (η = 0.75): Pelec ,B=2.423449/0.75=3.231265 kW.
· Pump C (η = 0.78): Pelec ,C=2.423449/0.78=3.106985 kW.
Daily and annual energy consumption (pump runs 8 h/day on average):
Daily energy (kWh/day) = (kW)×8 h. Annual energy = daily × 365.
· Pump A: daily = 3.462069 × 8 = 27.696556 kWh/day; annual = 27.696556 × 365 =
10,109.243 kWh/yr.
· Pump B: daily = 3.231265 × 8 = 25.850120 kWh/day; annual = 25.850120 × 365 =
9,435.293 kWh/yr.
· Pump C: daily = 3.106985 × 8 = 24.855883 kWh/day; annual = 24.855883 × 365 =
9,072.397 kWh/yr.
Given values:
Population = 200 residents.
Domestic use = 150 L/person/day.
Peak hour factor = 3.0.
Simultaneous use factor = 0.6.
Fire protection requirement = 30 L/s for 30 min.
Cooling tower makeup (if applicable) = 5 L/s average.
DN = 100 mm (chosen).
Pipe roughness and friction factor for head-loss estimation in Part 1b: use f = 0.02
(assignment instruction).
Minor loss coefficient K_total = 8.
Water: ρ = 1000 kg/m³, g = 9.81 m/s².
Floor height = 3.5 m → elevation of 8th floor = 8 × 3.5 m = 28.0 m
above ground.
Required minimum pressure at fixture = 150 kPa.
Municipal supply at ground = 250 kPa (gage).
1a) Peak-hour water demand (L/s)
Average daily domestic consumption:
150 L/person/day × 200 people = 30,000 L/day
Average flow rate (over 24 h):
30000
Qavg = =0.34722 L / S
86400
Peak domestic flow using peak and simultaneous factors:
Q peak , dom =Q avg ∗( peak h our factor )∗( simultaneous use factor )
=0.34722∗3∗0.6=0.625 L/S
Total peak system demand (domestic + cooling) (L/s):
Q peak , total = 0.625+5=5.625 L / s
Fire flow (separate) and worst-case simultaneous:
Fire requirement = 30 L/s for 30 minutes.
Worst-case simultaneous (peak domestic + fire):
Q worst =5.625+30=35.625 L / s .
Summary
Peak domestic demand (after peak & simultaneous factors): 0.6250 L/s.
Peak system demand (domestic + cooling): 5.6250 L/s.
Fire flow requirement: 30.000 L/s (30 minutes).
Worst-case simultaneous flow (domestic + fire): 35.6250 L/s
,1(b) Minimum required pump discharge pressure to guarantee 150 kPa at 8th-floor
fixtures
Convert required fixture pressure to head (m):
Required pressure = 150 kPa = 150,000 Pa.
150000 150000
hreq = = =15.2905 m .
pg 1000∗9.81
Static elevation head to 8th floor (m):
Floor height = 3.5 m (floor to floor). Elevation to 8th floor = 8 × 3.5 = 28.000 m.
1(c) Velocity in DN100 at peak flow (used for head-loss estimates):
DN=0.100 m
2
π (0.100) 2
A= =0.00785 m
4
Q = 5.625 L/s = 0.005625 m³/s.
V = Q/A = 0.005625/0.0078539 = 0.7162 m/s.
1(d) Major (friction) head loss using Darcy–Weisbach with f = 0.02
2
LV
Darcy–Weisbach: hf = f
D ∗2 g
Given f = 0.02, L = 80 m, D = 0.100 m, V = 0.71619867 m/s, g = 9.81 m/s²:
V2 (0.71612)2
kinetic term: = = 0.026137 m
2 g 2 ∗ 9.81
Numerically V 2 = 0.71622 =0.512966 m 2 / s 2
2
V /2 g =0.512966/(2∗9.81)=0.512966/19.62= 0.026137304666 m .
80
hf =0.02× × 0.026137304666
0.100
=0.02×800×0.026137304666 = 0.418301 m.
Minor losses (fittings, valves):
Minor-head coefficient K tot = 8 (assignment). Minor loss head:
2
V
hm = Ktot 8×0.026137304666=0.209151 m
2g
Convert required head to pressure (gage, kPa):
P = ρgH = 1000×9.81×43.917972=430835.309 Pa=430.835 kPa(gage).
Interpretation: The pump must be able to deliver ≈ 43.92 m of head at the building
peak flow (5.625 L/s) so that 8th-floor fixtures see ≥150 kPa. The municipal supply at
ground is 250 kPa (gage); that means the pump must provide a net boost if tied to
municipal suction
, Pump option check, energy consumption and costs
Pump A: H_max = 40 m; Q_max = 15 L/s; electrical rating = 8 kW; cost = QR
11,000; efficiency η = 70% (0.70).
Pump B: H_max = 50 m; Q_max = 12 L/s; electrical rating = 10 kW; cost = QR
16,500; efficiency η = 75% (0.75).
Pump C: H_max = 60 m; Q_max = 10 L/s; electrical rating = 12 kW; cost = QR
22,000; efficiency η = 78% (0.78).
Feasibility
Required design flow = 5.625 L/s and required head ≈ 43.918 m.
Pump A: H_max = 40 m < 43.918 m ⇒ CANNOT meet required head →
infeasible.
Pump B: H_max = 50 m ≥ 43.918 m and Q_max = 12 L/s ≥ 5.625 L/s ⇒
FEASIBLE.
Pump C: H_max = 60 m ≥ 43.918 m but Q_max = 10 L/s ≥ 5.625 L/s ⇒
FEASIBLE.
Hydraulic power at design conditions
Hydraulic power (W):
P h yd =ρgQH
P h yd =1000×9.81×0.005625×43.917972=2423.4486147 W=2.423449 kW
Electrical input power (kW) for each pump:
Pelec = ❑
η
· Pump A (η = 0.70): Pelec ,A=2.423449/0.70=3.462069 kW.
· Pump B (η = 0.75): Pelec ,B=2.423449/0.75=3.231265 kW.
· Pump C (η = 0.78): Pelec ,C=2.423449/0.78=3.106985 kW.
Daily and annual energy consumption (pump runs 8 h/day on average):
Daily energy (kWh/day) = (kW)×8 h. Annual energy = daily × 365.
· Pump A: daily = 3.462069 × 8 = 27.696556 kWh/day; annual = 27.696556 × 365 =
10,109.243 kWh/yr.
· Pump B: daily = 3.231265 × 8 = 25.850120 kWh/day; annual = 25.850120 × 365 =
9,435.293 kWh/yr.
· Pump C: daily = 3.106985 × 8 = 24.855883 kWh/day; annual = 24.855883 × 365 =
9,072.397 kWh/yr.
