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 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 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 oil, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 200 mm and a throat diameter of 100 mm. The pressure at the inlet is 40 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 200 mmd2 = 100 mmp1 − p2 = 40 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 8 m. At section 1 the pressure is 120 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 50 mmp1 = 120 kPav1 = 2 m/sΔz = 8 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 4 L/s through a pipe 150 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss hf.
Q = 4 L/sL = 150 mD = 50 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.025) 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 200 L/s?
Q = 200 L/sL = 200 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 5
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 28 m while its shaft takes 13.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 40 L/sH = 28 mPshaft = 13.4 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 10 m higher. The pipe is 150 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.015; 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 350 kPa. Find the head H the pump must add.
Q = 20 L/sΔz = 10 mL = 150 mD = 100 mmf = 0.015ΣK = 8ptank = 350 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 159.4 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 + 2.5/2 = 3.25 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·5 = 159.4 kN
Answer
F = 159.4 kN
2.Hydrostatic force
Answer: F = 72.84 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 + 1.5/2 = 3.75 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·3.75·2.25 = 72.84 kN
Answer
F = 72.84 kN
3.Continuity & Bernoulli
Answer: Q = 72.55 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·(100/200)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 = √(2·40000/(1000·(1 − 0.0625))) = 9.238 m/s
v1 = 2.309 m/s
Flow rate
Q = A2·v2 = 0.007854·9.238 = 0.07255 m³/s = 72.55 L/s
Answer
Q = 72.55 L/s
4.Continuity & Bernoulli
Answer: p2 = 33.39 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·(75/50)² = 4.5 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 4.5²)/1000 = −8.125 kPa
ρgΔz = 1000·9.81·8/1000 = 78.48 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 120 + (−8.125) − 78.48 = 33.39 kPa
Answer
p2 = 33.39 kPa
5.Pipe friction & head loss
Answer: hf = 15.86 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·0.05²/4 = 0.001963 m²
V = Q/A = 0.004/0.001963 = 2.037 m/s
Velocity head
V²/(2g) = 2.037²/(2·9.81) = 0.2115 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(150/0.05)·2.037²/(2·9.81) = 15.86 m
Answer
hf = 15.86 m
6.Pipe friction & head loss
Answer: H = 9.942 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 = 0.2/0.07069 = 2.829 m/s
V²/(2g) = 0.408 m
Loss coefficients
f·L/D = 0.025·200/0.3 = 16.67
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 + 16.67 + 6.7)·0.408 = 9.942 m
Answer
H = 9.942 m
7.Pumps & pump power
Answer: η = 81.99 %
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 = 1000·9.81·0.04·28 = 10.99 kW
Efficiency
η = Ph/Pshaft = 10.99/13.4 = 81.99%
Answer
η = 81.99%
8.Pumps & pump power
Answer: H = 55.76 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) = (22.5 + 8)·0.3305 = 10.08 m
Pressure head
p/(ρ·g) = 350000/(1000·9.81) = 35.68 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 10 + 35.68 + 10.08 = 55.76 m
Answer
H = 55.76 m
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