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 2 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1 mb = 2 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 water, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1 mH = 2 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 40 mm. The water level is kept 5 m above the orifice and the discharge coefficient is 0.64. Find the flow rate Q. Use g = 9.81 m/s².
d = 40 mmh = 5 mCd = 0.64g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A hose with an inner diameter of 32 mm siphons water from an open tank. The highest point C of the hose is 2.5 m above the water surface, and the outlet (open to the air) is 3.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
d = 32 mmzC = +2.5 mz2 = −3.5 mg = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.8 m/s through a pipe 100 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.8 m/sL = 100 mD = 100 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 150 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.015) 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), 4 bends (K = 0.3 each) and a valve (K = 10). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 80 L/s?
Q = 80 L/sL = 150 mD = 200 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 80% delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 36 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 25 L/sH = 36 mη = 80%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 20 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 11 m higher. The pipe is 400 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 20 L/sΔz = 11 mL = 400 mD = 100 mmf = 0.022ΣK = 8
H = m
3Solutions
1.Hydrostatic force
Answer: F = 58.86 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 = 2.5 + 1/2 = 3 m
A = H·b = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·2 = 58.86 kN
Answer
F = 58.86 kN
2.Hydrostatic force
Answer: ycp = 2.167 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 + 2/2 = 2 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·4 = 78.48 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·2³/12 = 1.333 m⁴
ycp = yc + Ic/(yc·A) = 2 + 1.333/(2·4) = 2.167 m
Answer
ycp = 2.167 m
3.Continuity & Bernoulli
Answer: Q = 7.966 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.
Ideal jet velocity
√(2gh) = √(2·9.81·5) = 9.905 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.64·0.001257·9.905 = 0.007966 m³/s = 7.966 L/s
Answer
Q = 7.966 L/s
4.Continuity & Bernoulli
Answer: Q = 6.665 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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·3.5) = √(2·9.81·3.5) = 8.287 m/s
Flow rate
Q = A·v = 0.0008042·8.287 = 0.006665 m³/s = 6.665 L/s
Answer
Q = 6.665 L/s
5.Pipe friction & head loss
Answer: hf = 8.791 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 head
V²/(2g) = 2.8²/(2·9.81) = 0.3996 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(100/0.1)·2.8²/(2·9.81) = 8.791 m
Answer
hf = 8.791 m
6.Pipe friction & head loss
Answer: H = 7.916 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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Loss coefficients
f·L/D = 0.015·150/0.2 = 11.25
ΣK = 0.5 + 4·0.3 + 10 = 11.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 11.25 + 11.7)·0.3305 = 7.916 m
Answer
H = 7.916 m
7.Pumps & pump power
Answer: Pshaft = 11.04 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 = 1000·9.81·0.025·36 = 8.829 kW
Shaft power
Pshaft = Ph/η = 8.829/0.8 = 11.04 kW
Answer
Pshaft = 11.04 kW
8.Pumps & pump power
Answer: H = 42.73 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 = 0.02/0.007854 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (88 + 8)·0.3305 = 31.73 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 11 + 31.73 = 42.73 m
Answer
H = 42.73 m
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