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 = 3 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1 m wide sits in the wall of a tank of seawater, with its top edge 2 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 2 mH = 1 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
A pitot-static tube in a wind tunnel reads a difference of 750 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
p0 − p = 750 Paρ = 1.2 kg/m³
v = m/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 40 mm while rising 1 m. At section 1 the pressure is 380 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 40 mmp1 = 380 kPav1 = 1.75 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.5 m/s through a horizontal pipe, 400 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.5 m/sL = 400 mD = 100 mmf = 0.018ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 300 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 75 L/s?
Q = 75 L/sL = 100 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 36 m while its shaft takes 7.1 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 15 L/sH = 36 mPshaft = 7.1 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 25 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 200 m long with an inside diameter of 150 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 = 25 L/sΔz = 10 mL = 200 mD = 150 mmf = 0.022ΣK = 8
H = m
3Solutions
1.Hydrostatic force
Answer: F = 235.4 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 + 3/2 = 4 m
A = H·b = 3·2 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4·6 = 235.4 kN
Answer
F = 235.4 kN
2.Hydrostatic force
Answer: ycp = 2.533 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 = 2 + 1/2 = 2.5 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·2.5·1 = 25.14 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·1³/12 = 0.08333 m⁴
ycp = yc + Ic/(yc·A) = 2.5 + 0.08333/(2.5·1) = 2.533 m
Answer
ycp = 2.533 m
3.Continuity & Bernoulli
Answer: v = 35.36 m/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 to the stagnation point
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
p0 − p = ½ρv²
v = √(2·750/1.2) = 35.36 m/s
Answer
v = 35.36 m/s
4.Continuity & Bernoulli
Answer: p2 = 352.8 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 = 1.75·(75/40)² = 6.152 m/s
½ρ(v1² − v2²) = ½·1000·(1.75² − 6.152²)/1000 = −17.39 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 380 + (−17.39) − 9.81 = 352.8 kPa
Answer
p2 = 352.8 kPa
5.Pipe friction & head loss
Answer: Δp = 225 kPa
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.
Head loss
hf = f·(L/D)·V²/(2g) = 0.018·(400/0.1)·2.5²/(2·9.81) = 22.94 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·22.94 = 225000 Pa = 225 kPa
Answer
Δp = 225 kPa
6.Pipe friction & head loss
Answer: H = 1.016 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 = 0.075/0.07069 = 1.061 m/s
V²/(2g) = 0.05738 m
Loss coefficients
f·L/D = 0.015·100/0.3 = 5
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 + 5 + 11.7)·0.05738 = 1.016 m
Answer
H = 1.016 m
7.Pumps & pump power
Answer: η = 74.61 %
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 = 1000·9.81·0.015·36 = 5.297 kW
Efficiency
η = Ph/Pshaft = 5.297/7.1 = 74.61%
Answer
η = 74.61%
8.Pumps & pump power
Answer: H = 13.81 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 = 0.025/0.01767 = 1.415 m/s
V²/(2g) = 0.102 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (29.33 + 8)·0.102 = 3.808 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 10 + 3.808 = 13.81 m
Answer
H = 13.81 m
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