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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 1 mb = 1 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 seawater, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 200 mm to 100 mm. In the wide section the pressure is 210 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 200 mmd2 = 100 mmp1 = 210 kPav1 = 1.25 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 75 mm and a gauge pressure of 60 kPa. The nozzle exit diameter is 40 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
d1 = 75 mmp1 = 60 kPa (gauge)d2 = 40 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at 40 L/s through a pipe 800 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 40 L/sL = 800 mD = 150 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 600 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.015) 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), 4 bends (K = 0.3 each) and a valve (K = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 100 L/s?
Q = 100 L/sL = 600 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 5
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 78% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 44 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 20 L/sH = 44 mη = 78%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 80 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 18 m higher. The pipe is 200 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 85%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 18 mL = 200 mD = 200 mmf = 0.025ΣK = 6η = 85%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 19.62 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 + 1/2 = 2 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·1 = 19.62 kN
Answer
F = 19.62 kN
2.Hydrostatic force
Answer: F = 211.2 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 + 2/2 = 3.5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3.5·6 = 211.2 kN
Answer
F = 211.2 kN
3.Continuity & Bernoulli
Answer: p2 = 198.3 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·(200/100)² = 5 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 5²)/1000 = −11.72 kPa
p2 = p1 + ½ρ(v1² − v2²) = 210 + (−11.72) = 198.3 kPa
Answer
p2 = 198.3 kPa
4.Continuity & Bernoulli
Answer: H = 6.655 m
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.
Exit velocity
β = 40/75 = 0.5333
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·60000/(1000·0.9191)) = 11.43 m/s
Height of the jet
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
H = v2²/(2g) = 11.43²/(2·9.81) = 6.655 m
Answer
H = 6.655 m
5.Pipe friction & head loss
Answer: hf = 25.07 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·0.15²/4 = 0.01767 m²
V = Q/A = 0.04/0.01767 = 2.264 m/s
Velocity head
V²/(2g) = 2.264²/(2·9.81) = 0.2611 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(800/0.15)·2.264²/(2·9.81) = 25.07 m
Answer
hf = 25.07 m
6.Pipe friction & head loss
Answer: H = 3.846 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 = 0.1/0.07069 = 1.415 m/s
V²/(2g) = 0.102 m
Loss coefficients
f·L/D = 0.015·600/0.3 = 30
ΣK = 0.5 + 4·0.3 + 5 = 6.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 30 + 6.7)·0.102 = 3.846 m
Answer
H = 3.846 m
7.Pumps & pump power
Answer: Pshaft = 11.07 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 = 1000·9.81·0.02·44 = 8.633 kW
Shaft power
Pshaft = Ph/η = 8.633/0.78 = 11.07 kW
Answer
Pshaft = 11.07 kW
8.Pumps & pump power
Answer: Pshaft = 26.08 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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (25 + 6)·0.3305 = 10.25 m
Pump head
H = Δz + hL = 18 + 10.25 = 28.25 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·28.25/0.85 = 26.08 kW
Answer
Pshaft = 26.08 kW
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