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 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 3 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 seawater, 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 = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 1 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 1 mCd = 0.62g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 1 m. At section 1 the pressure is 290 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 75 mmp1 = 290 kPav1 = 1.25 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.9 m/s through a horizontal pipe, 100 m long, with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.9 m/sL = 100 mD = 50 mmf = 0.03ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 500 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.015. The water surface of the upper reservoir is 29 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 5) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.
Δz = 29 mL = 500 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 60% delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 76 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 40 L/sH = 76 mη = 60%g = 9.81 m/s²
Pshaft = kW
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 9 m higher. The pipe is 300 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 = 4 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 50 L/sΔz = 9 mL = 300 mD = 150 mmf = 0.022ΣK = 4
H = m
3Solutions
1.Hydrostatic force
Answer: F = 323.7 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 + 3/2 = 5.5 m
A = H·b = 3·2 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.5·6 = 323.7 kN
Answer
F = 323.7 kN
2.Hydrostatic force
Answer: F = 181 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 = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·4·4.5 = 181 kN
Answer
F = 181 kN
3.Continuity & Bernoulli
Answer: Q = 5.392 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) = √(2·9.81·1) = 4.429 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.62·0.001963·4.429 = 0.005392 m³/s = 5.392 L/s
Answer
Q = 5.392 L/s
4.Continuity & Bernoulli
Answer: p2 = 268.5 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 = 1.25·(150/75)² = 5 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 5²)/1000 = −11.72 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 290 + (−11.72) − 9.81 = 268.5 kPa
Answer
p2 = 268.5 kPa
5.Pipe friction & head loss
Answer: Δp = 252.3 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) = 0.03·(100/0.05)·2.9²/(2·9.81) = 25.72 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 = 1000·9.81·25.72 = 252300 Pa = 252.3 kPa
Answer
Δp = 252.3 kPa
6.Pipe friction & head loss
Answer: Q = 297.6 L/s
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.
Loss coefficients
f·L/D = 0.015·500/0.3 = 25
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
29 = (25 + 7.1)·V²/(2·9.81)
V = √(2·9.81·29/32.1) = 4.21 m/s
Flow rate
Q = V·π·D²/4 = 4.21·0.07069 = 0.2976 m³/s = 297.6 L/s
Answer
Q = 297.6 L/s
7.Pumps & pump power
Answer: Pshaft = 49.7 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 = 1000·9.81·0.04·76 = 29.82 kW
Shaft power
Pshaft = Ph/η = 29.82/0.6 = 49.7 kW
Answer
Pshaft = 49.7 kW
8.Pumps & pump power
Answer: H = 28.59 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (44 + 4)·0.408 = 19.59 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 9 + 19.59 = 28.59 m
Answer
H = 28.59 m
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