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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 2.5 mb = 2 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 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 = 1 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows at 17 L/s through a horizontal pipe that narrows from 100 mm to 75 mm. The pressure in the narrow section is 300 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 17 L/sd1 = 100 mmd2 = 75 mmp2 = 300 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 6 m. At section 1 the pressure is 110 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 100 mmp1 = 110 kPav1 = 2.75 m/sΔz = 6 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 5 L/s through a pipe 50 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss hf.
Q = 5 L/sL = 50 mD = 100 mmf = 0.02g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 2 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 20 L/s?
Q = 20 L/sL = 300 mD = 150 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 75% delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 80 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 60 L/sH = 80 mη = 75%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 150 m long with an inside diameter of 250 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. The pump efficiency is 80%. Find the shaft power Pshaft in kW.
Q = 75 L/sΔz = 5 mL = 150 mD = 250 mmf = 0.022ΣK = 8η = 80%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 208.5 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 + 2.5/2 = 4.25 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.25·5 = 208.5 kN
Answer
F = 208.5 kN
2.Hydrostatic force
Answer: ycp = 1.556 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 + 1/2 = 1.5 m
A = H·b = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.5·1.5 = 22.07 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·1³/12 = 0.125 m⁴
ycp = yc + Ic/(yc·A) = 1.5 + 0.125/(1.5·1.5) = 1.556 m
Answer
ycp = 1.556 m
3.Continuity & Bernoulli
Answer: p1 = 305.1 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.
Velocities
v1 = Q/A1 = 0.017/0.007854 = 2.165 m/s
v2 = Q/A2 = 0.017/0.004418 = 3.848 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 300 + ½·1000·(3.848² − 2.165²)/1000 = 300 + 5.061 = 305.1 kPa
Answer
p1 = 305.1 kPa
4.Continuity & Bernoulli
Answer: p2 = 35.78 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 = 2.75·(150/100)² = 6.188 m/s
½ρ(v1² − v2²) = ½·1000·(2.75² − 6.188²)/1000 = −15.36 kPa
ρgΔz = 1000·9.81·6/1000 = 58.86 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 110 + (−15.36) − 58.86 = 35.78 kPa
Answer
p2 = 35.78 kPa
5.Pipe friction & head loss
Answer: hf = 0.2066 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·0.1²/4 = 0.007854 m²
V = Q/A = 0.005/0.007854 = 0.6366 m/s
Velocity head
V²/(2g) = 0.6366²/(2·9.81) = 0.02066 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.02·(50/0.1)·0.6366²/(2·9.81) = 0.2066 m
Answer
hf = 0.2066 m
6.Pipe friction & head loss
Answer: Δz = 2.226 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 = 0.02/0.01767 = 1.132 m/s
V²/(2g) = 0.06529 m
Loss coefficients
f·L/D = 0.015·300/0.15 = 30
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Total head loss
hf = 30·0.06529 = 1.959 m
hm = 4.1·0.06529 = 0.2677 m
Δz = hf + hm = 2.226 m
Answer
Δz = 2.226 m
7.Pumps & pump power
Answer: Pshaft = 62.78 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 = 1000·9.81·0.06·80 = 47.09 kW
Shaft power
Pshaft = Ph/η = 47.09/0.75 = 62.78 kW
Answer
Pshaft = 62.78 kW
8.Pumps & pump power
Answer: Pshaft = 6.918 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 = 0.075/0.04909 = 1.528 m/s
V²/(2g) = 0.119 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (13.2 + 8)·0.119 = 2.522 m
Pump head
H = Δz + hL = 5 + 2.522 = 7.522 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.075·7.522/0.8 = 6.918 kW
Answer
Pshaft = 6.918 kW
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