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 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 = 1.5 mb = 1 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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 2.5 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 350 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 50 mmp1 = 350 kPav1 = 2.25 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 7 m. At section 1 the pressure is 250 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 75 mmp1 = 250 kPav1 = 0.5 m/sΔz = 7 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.1 m/s through a pipe 400 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.1 m/sL = 400 mD = 200 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.02) 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), 3 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 40 L/s?
Q = 40 L/sL = 200 mD = 200 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 74 m while its shaft takes 51.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 60 L/sH = 74 mPshaft = 51.2 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 20 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 15 m higher. The pipe is 500 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The upper tank is closed and the air above its water is at a gauge pressure of 300 kPa. Find the head H the pump must add.
Q = 20 L/sΔz = 15 mL = 500 mD = 100 mmf = 0.018ΣK = 8ptank = 300 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 47.82 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 + 1.5/2 = 3.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·1.5 = 47.82 kN
Answer
F = 47.82 kN
2.Hydrostatic force
Answer: F = 106.8 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 = 3 + 2.5/2 = 4.25 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·4.25·2.5 = 106.8 kN
Answer
F = 106.8 kN
3.Continuity & Bernoulli
Answer: p2 = 339.7 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 = 2.25·(75/50)² = 5.063 m/s
½ρ(v1² − v2²) = ½·1000·(2.25² − 5.063²)/1000 = −10.28 kPa
p2 = p1 + ½ρ(v1² − v2²) = 350 + (−10.28) = 339.7 kPa
Answer
p2 = 339.7 kPa
4.Continuity & Bernoulli
Answer: p2 = 175.1 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.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 0.5·(200/75)² = 3.556 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 3.556²)/1000 = −6.196 kPa
ρgΔz = 1000·9.81·7/1000 = 68.67 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 250 + (−6.196) − 68.67 = 175.1 kPa
Answer
p2 = 175.1 kPa
5.Pipe friction & head loss
Answer: hf = 2.22 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) = 1.1²/(2·9.81) = 0.06167 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(400/0.2)·1.1²/(2·9.81) = 2.22 m
Answer
hf = 2.22 m
6.Pipe friction & head loss
Answer: H = 2.677 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 = 0.04/0.03142 = 1.273 m/s
V²/(2g) = 0.08263 m
Loss coefficients
f·L/D = 0.02·200/0.2 = 20
ΣK = 0.5 + 3·0.3 + 10 = 11.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 20 + 11.4)·0.08263 = 2.677 m
Answer
H = 2.677 m
7.Pumps & pump power
Answer: η = 85.07 %
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 = 1000·9.81·0.06·74 = 43.56 kW
Efficiency
η = Ph/Pshaft = 43.56/51.2 = 85.07%
Answer
η = 85.07%
8.Pumps & pump power
Answer: H = 77.97 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 = 0.02/0.007854 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (90 + 8)·0.3305 = 32.39 m
Pressure head
p/(ρ·g) = 300000/(1000·9.81) = 30.58 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 15 + 30.58 + 32.39 = 77.97 m
Answer
H = 77.97 m
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