Hydrostatic force
- Pressure at depth
- p = ρgh
- Force on a vertical wall
- F = ρg·hc·A
- Centre of pressure
- yp = yc + Ic/(ycA)
- Rectangle
- Ic = bH³/12
Fluid Mechanics: cram pack
Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
1Formulas and methods
Continuity & Bernoulli
- Continuity
- A1v1 = A2v2
- Bernoulli
- p1 + ½ρv1² + ρgz1 = p2 + ½ρv2² + ρgz2
- Flow rate
- Q = A·v
- Circle area
- A = πd²/4
Pipe friction & head loss
- Reynolds number
- Re = V·D/ν = ρ·V·D/μ
- Flow regime
- laminar below Re ≈ 2000, turbulent above Re ≈ 4000
- Mean velocity
- V = Q/A, A = π·D²/4
- Darcy–Weisbach
- hf = f·(L/D)·V²/(2g)
- Minor losses
- hm = ΣK·V²/(2g)
- Typical K values
- entrance 0.5, exit 1.0, 90° bend ≈ 0.3
- Pressure drop
- Δp = ρ·g·hf
- Two open reservoirs
- Δz = (f·L/D + ΣK)·V²/(2g)
- Free outlet
- H = (1 + f·L/D + ΣK)·V²/(2g)
Pumps & pump power
- Hydraulic power
- Ph = ρ·g·Q·H = Q·Δp
- Shaft power
- Pshaft = Ph/ηpump
- Electrical power
- Pel = Pshaft/ηmotor
- Pump head
- H = Δz + Δp/(ρ·g) + hL
- System losses
- hL = (f·L/D + ΣK)·V²/(2g)
2Practice problems
1.Hydrostatic forceMedium
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 3 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 3 m wide sits in the wall of a tank of seawater, with its top edge 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1.5 mH = 1 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 60 mm. The pressure at the inlet is 56 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 100 mmd2 = 60 mmp1 − p2 = 56 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 70 kPa. Water leaves through an opening with a diameter of 25 mm, 1 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
pair = 70 kPa (gauge)h = 1 md = 25 mmρ = 1000 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 3 L/s through a pipe 300 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 3 L/sL = 300 mD = 50 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.022) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each) and a valve (K = 0.2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 150 L/s?
Q = 150 L/sL = 100 mD = 300 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 0.2
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 85% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 50 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 20 L/sH = 50 mη = 85%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 10 m higher. The pipe is 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 75 L/sΔz = 10 mL = 300 mD = 250 mmf = 0.022ΣK = 6
H = m
3Solutions
1.Hydrostatic force
Answer: F = 485.6 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 4 + 3/2 = 5.5 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.5·9 = 485.6 kN
Answer
F = 485.6 kN
2.Hydrostatic force
Answer: ycp = 2.042 m
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 1.5 + 1/2 = 2 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·2·3 = 60.33 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·1³/12 = 0.25 m⁴
ycp = yc + Ic/(yc·A) = 2 + 0.25/(2·3) = 2.042 m
Answer
ycp = 2.042 m
3.Continuity & Bernoulli
Answer: Q = 32.07 L/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Continuity
v1 = v2·(60/100)² = 0.36·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·56000/(1000·(1 − 0.1296))) = 11.34 m/s
v1 = 4.084 m/s
Flow rate
Q = A2·v2 = 0.002827·11.34 = 0.03207 m³/s = 32.07 L/s
Answer
Q = 32.07 L/s
4.Continuity & Bernoulli
Answer: Q = 6.202 L/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Bernoulli from the surface to the jet
v = √(2·(70000/1000 + 9.81·1)) = 12.63 m/s
Flow rate
Q = A·v = 0.0004909·12.63 = 0.006202 m³/s = 6.202 L/s
Answer
Q = 6.202 L/s
5.Pipe friction & head loss
Answer: hf = 12.85 m
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Mean velocity
A = π·0.05²/4 = 0.001963 m²
V = Q/A = 0.003/0.001963 = 1.528 m/s
Velocity head
V²/(2g) = 1.528²/(2·9.81) = 0.119 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(300/0.05)·1.528²/(2·9.81) = 12.85 m
Answer
hf = 12.85 m
6.Pipe friction & head loss
Answer: H = 2.211 m
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Velocity
V = Q/A = 0.15/0.07069 = 2.122 m/s
V²/(2g) = 0.2295 m
Loss coefficients
f·L/D = 0.022·100/0.3 = 7.333
ΣK = 0.5 + 2·0.3 + 0.2 = 1.3
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 7.333 + 1.3)·0.2295 = 2.211 m
Answer
H = 2.211 m
7.Pumps & pump power
Answer: Pshaft = 11.54 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Hydraulic power
Ph = ρ·g·Q·H = 1000·9.81·0.02·50 = 9.81 kW
Shaft power
Pshaft = Ph/η = 9.81/0.85 = 11.54 kW
Answer
Pshaft = 11.54 kW
8.Pumps & pump power
Answer: H = 13.86 m
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Velocity
V = Q/A = 0.075/0.04909 = 1.528 m/s
V²/(2g) = 0.119 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (26.4 + 6)·0.119 = 3.855 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 10 + 3.855 = 13.86 m
Answer
H = 13.86 m
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