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 1.5 m wide sits in the wall of a tank of water, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 2 m wide sits in the wall of a tank of oil, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1.5 mb = 2 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows at 24 L/s through a horizontal pipe that narrows from 200 mm to 100 mm. The pressure in the narrow section is 70 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 24 L/sd1 = 200 mmd2 = 100 mmp2 = 70 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 1 m. At section 1 the pressure is 250 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 75 mmp1 = 250 kPav1 = 1 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 4 L/s through a pipe 800 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
Q = 4 L/sL = 800 mD = 50 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 150 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.025) 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), 3 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 200 L/s?
Q = 200 L/sL = 150 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)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 62 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 20 L/sH = 62 mη = 85%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 120 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 600 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 120 L/sΔz = 10 mL = 600 mD = 250 mmf = 0.02ΣK = 5
H = m
3Solutions
1.Hydrostatic force
Answer: F = 82.77 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 = 3 + 1.5/2 = 3.75 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.75·2.25 = 82.77 kN
Answer
F = 82.77 kN
2.Hydrostatic force
Answer: F = 97.12 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 = 3 + 1.5/2 = 3.75 m
A = H·b = 1.5·2 = 3 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·3.75·3 = 97.12 kN
Answer
F = 97.12 kN
3.Continuity & Bernoulli
Answer: p1 = 74.38 kPa
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.
Velocities
v1 = Q/A1 = 0.024/0.03142 = 0.7639 m/s
v2 = Q/A2 = 0.024/0.007854 = 3.056 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 70 + ½·1000·(3.056² − 0.7639²)/1000 = 70 + 4.377 = 74.38 kPa
Answer
p1 = 74.38 kPa
4.Continuity & Bernoulli
Answer: p2 = 239.1 kPa
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 and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 1·(100/75)² = 1.778 m/s
½ρ(v1² − v2²) = ½·1000·(1² − 1.778²)/1000 = −1.08 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 250 + (−1.08) − 9.81 = 239.1 kPa
Answer
p2 = 239.1 kPa
5.Pipe friction & head loss
Answer: hf = 101.5 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 = 0.004/0.001963 = 2.037 m/s
Velocity head
V²/(2g) = 2.037²/(2·9.81) = 0.2115 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(800/0.05)·2.037²/(2·9.81) = 101.5 m
Answer
hf = 101.5 m
6.Pipe friction & head loss
Answer: H = 6.161 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 = 0.2/0.07069 = 2.829 m/s
V²/(2g) = 0.408 m
Loss coefficients
f·L/D = 0.025·150/0.3 = 12.5
ΣK = 0.5 + 3·0.3 + 0.2 = 1.6
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 12.5 + 1.6)·0.408 = 6.161 m
Answer
H = 6.161 m
7.Pumps & pump power
Answer: Pshaft = 14.31 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 = 1000·9.81·0.02·62 = 12.16 kW
Shaft power
Pshaft = Ph/η = 12.16/0.85 = 14.31 kW
Answer
Pshaft = 14.31 kW
8.Pumps & pump power
Answer: H = 26.14 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (48 + 5)·0.3046 = 16.14 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 10 + 16.14 = 26.14 m
Answer
H = 26.14 m
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