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 3 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 = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 250 mm and a throat diameter of 125 mm. The pressure at the inlet is 56 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 250 mmd2 = 125 mmp1 − p2 = 56 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 9 m. At section 1 the pressure is 380 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 100 mmp1 = 380 kPav1 = 2 m/sΔz = 9 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.7 m/s through a horizontal pipe, 1000 m long, with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.7 m/sL = 1000 mD = 75 mmf = 0.025ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 0.2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 120 L/s?
Q = 120 L/sL = 300 mD = 250 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 0.2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 72 m while its shaft takes 68.9 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 80 L/sH = 72 mPshaft = 68.9 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 500 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; 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 = 50 L/sΔz = 8 mL = 500 mD = 200 mmf = 0.015ΣK = 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: 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 = 3 + 2/2 = 4 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4·6 = 235.4 kN
Answer
F = 235.4 kN
3.Continuity & Bernoulli
Answer: Q = 134.1 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·(125/250)² = 0.25·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 − 0.0625))) = 10.93 m/s
v1 = 2.733 m/s
Flow rate
Q = A2·v2 = 0.01227·10.93 = 0.1341 m³/s = 134.1 L/s
Answer
Q = 134.1 L/s
4.Continuity & Bernoulli
Answer: p2 = 283.6 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 = 2·(150/100)² = 4.5 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 4.5²)/1000 = −8.125 kPa
ρgΔz = 1000·9.81·9/1000 = 88.29 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 380 + (−8.125) − 88.29 = 283.6 kPa
Answer
p2 = 283.6 kPa
5.Pipe friction & head loss
Answer: Δp = 1215 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) = 0.025·(1000/0.075)·2.7²/(2·9.81) = 123.9 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 = 1000·9.81·123.9 = 1.215 × 10⁶ Pa = 1215 kPa
Answer
Δp = 1215 kPa
6.Pipe friction & head loss
Answer: Δz = 7.463 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Loss coefficients
f·L/D = 0.018·300/0.25 = 21.6
ΣK = 0.5 + 4·0.3 + 0.2 + 1.0 = 2.9
Total head loss
hf = 21.6·0.3046 = 6.579 m
hm = 2.9·0.3046 = 0.8833 m
Δz = hf + hm = 7.463 m
Answer
Δz = 7.463 m
7.Pumps & pump power
Answer: η = 82.01 %
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 = 1000·9.81·0.08·72 = 56.51 kW
Efficiency
η = Ph/Pshaft = 56.51/68.9 = 82.01%
Answer
η = 82.01%
8.Pumps & pump power
Answer: H = 13.87 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 = 0.05/0.03142 = 1.592 m/s
V²/(2g) = 0.1291 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (37.5 + 8)·0.1291 = 5.874 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 8 + 5.874 = 13.87 m
Answer
H = 13.87 m
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