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 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 1.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of glycerin, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 3 mb = 1.5 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 75 mm. In the wide section the pressure is 250 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 75 mmp1 = 250 kPav1 = 0.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 75 mm while rising 2 m. At section 1 the pressure is 150 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 75 mmp1 = 150 kPav1 = 1 m/sΔz = 2 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.7 m/s through a pipe 200 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 = 2.7 m/sL = 200 mD = 100 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 400 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 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 75 L/s?
Q = 75 L/sL = 400 mD = 300 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 28 m while its shaft takes 48.5 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 150 L/sH = 28 mPshaft = 48.5 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 100 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 14 m higher. The pipe is 100 m long with an inside diameter of 250 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. The pump efficiency is 82%. Find the shaft power Pshaft in kW.
Q = 100 L/sΔz = 14 mL = 100 mD = 250 mmf = 0.015ΣK = 8η = 82%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 104.8 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 + 1.5/2 = 4.75 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.75·2.25 = 104.8 kN
Answer
F = 104.8 kN
2.Hydrostatic force
Answer: F = 166.9 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 = 1.5 + 3/2 = 3 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3·4.5 = 166.9 kN
Answer
F = 166.9 kN
3.Continuity & Bernoulli
Answer: p2 = 249.4 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 = 0.75·(100/75)² = 1.333 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 1.333²)/1000 = −0.6076 kPa
p2 = p1 + ½ρ(v1² − v2²) = 250 + (−0.6076) = 249.4 kPa
Answer
p2 = 249.4 kPa
4.Continuity & Bernoulli
Answer: p2 = 105.6 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 = 1·(200/75)² = 7.111 m/s
½ρ(v1² − v2²) = ½·1000·(1² − 7.111²)/1000 = −24.78 kPa
ρgΔz = 1000·9.81·2/1000 = 19.62 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 150 + (−24.78) − 19.62 = 105.6 kPa
Answer
p2 = 105.6 kPa
5.Pipe friction & head loss
Answer: hf = 16.35 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) = 2.7²/(2·9.81) = 0.3716 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(200/0.1)·2.7²/(2·9.81) = 16.35 m
Answer
hf = 16.35 m
6.Pipe friction & head loss
Answer: Δz = 1.63 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 = 0.018·400/0.3 = 24
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Total head loss
hf = 24·0.05738 = 1.377 m
hm = 4.4·0.05738 = 0.2525 m
Δz = hf + hm = 1.63 m
Answer
Δz = 1.63 m
7.Pumps & pump power
Answer: η = 84.95 %
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 = 1000·9.81·0.15·28 = 41.2 kW
Efficiency
η = Ph/Pshaft = 41.2/48.5 = 84.95%
Answer
η = 84.95%
8.Pumps & pump power
Answer: Pshaft = 20.29 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.
Velocity
V = Q/A = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (6 + 8)·0.2115 = 2.961 m
Pump head
H = Δz + hL = 14 + 2.961 = 16.96 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.1·16.96/0.82 = 20.29 kW
Answer
Pshaft = 20.29 kW
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