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 3 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 = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 2 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 = 2 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 80 mm and a throat diameter of 40 mm. The pressure at the inlet is 44 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 80 mmd2 = 40 mmp1 − p2 = 44 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 8 m. At section 1 the pressure is 370 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 75 mmp1 = 370 kPav1 = 1.25 m/sΔz = 8 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 4 L/s through a pipe 500 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
Q = 4 L/sL = 500 mD = 100 mmf = 0.03g = 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), 3 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 80 L/s?
Q = 80 L/sL = 600 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 5
H = m
7.Pumps & pump powerMedium
A pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 34 m. The pump efficiency is 75% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 50 L/sH = 34 mηpump = 75%ηmotor = 90%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 10 m higher. The pipe is 500 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 4 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 72%. Find the shaft power Pshaft in kW.
Q = 120 L/sΔz = 10 mL = 500 mD = 250 mmf = 0.018ΣK = 4η = 72%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 147.2 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 = 1.5 + 2/2 = 2.5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·6 = 147.2 kN
Answer
F = 147.2 kN
2.Hydrostatic force
Answer: F = 98.1 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 + 2/2 = 2.5 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·4 = 98.1 kN
Answer
F = 98.1 kN
3.Continuity & Bernoulli
Answer: Q = 12.17 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.
Continuity
v1 = v2·(40/80)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·44000/(1000·(1 − 0.0625))) = 9.688 m/s
v1 = 2.422 m/s
Flow rate
Q = A2·v2 = 0.001257·9.688 = 0.01217 m³/s = 12.17 L/s
Answer
Q = 12.17 L/s
4.Continuity & Bernoulli
Answer: p2 = 289.8 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·(100/75)² = 2.222 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 2.222²)/1000 = −1.688 kPa
ρgΔz = 1000·9.81·8/1000 = 78.48 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 370 + (−1.688) − 78.48 = 289.8 kPa
Answer
p2 = 289.8 kPa
5.Pipe friction & head loss
Answer: hf = 1.983 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 = 0.004/0.007854 = 0.5093 m/s
Velocity head
V²/(2g) = 0.5093²/(2·9.81) = 0.01322 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(500/0.1)·0.5093²/(2·9.81) = 1.983 m
Answer
hf = 1.983 m
6.Pipe friction & head loss
Answer: H = 2.442 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 = 0.08/0.07069 = 1.132 m/s
V²/(2g) = 0.06529 m
Loss coefficients
f·L/D = 0.015·600/0.3 = 30
ΣK = 0.5 + 3·0.3 + 5 = 6.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 30 + 6.4)·0.06529 = 2.442 m
Answer
H = 2.442 m
7.Pumps & pump power
Answer: Pel = 24.71 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 = 1000·9.81·0.05·34 = 16.68 kW
Shaft power
Pshaft = 16.68/0.75 = 22.24 kW
Electrical power
Pel = 22.24/0.9 = 24.71 kW
Answer
Pel = 24.71 kW
8.Pumps & pump power
Answer: Pshaft = 36.27 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (36 + 4)·0.3046 = 12.18 m
Pump head
H = Δz + hL = 10 + 12.18 = 22.18 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.12·22.18/0.72 = 36.27 kW
Answer
Pshaft = 36.27 kW
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