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 2 m wide sits in the wall of a tank of seawater, with its top edge 4 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 4 mH = 1 mb = 2 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 190 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 75 mmp1 = 190 kPav1 = 2 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 100 mm while rising 3 m. At section 1 the pressure is 220 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 100 mmp1 = 220 kPav1 = 2.75 m/sΔz = 3 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 25 L/s through a pipe 300 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 = 25 L/sL = 300 mD = 100 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 150 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), 2 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 25 L/s?
Q = 25 L/sL = 100 mD = 150 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 16 m while its shaft takes 24.1 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 100 L/sH = 16 mPshaft = 24.1 kWg = 9.81 m/s²
η = %
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 9 m higher. The pipe is 400 m long with an inside diameter of 250 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 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 = 120 L/sΔz = 9 mL = 400 mD = 250 mmf = 0.025ΣK = 6ptank = 300 kPa
H = m
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: ycp = 4.519 m
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 + 1/2 = 4.5 m
A = H·b = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·4.5·2 = 90.5 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·1³/12 = 0.1667 m⁴
ycp = yc + Ic/(yc·A) = 4.5 + 0.1667/(4.5·2) = 4.519 m
Answer
ycp = 4.519 m
3.Continuity & Bernoulli
Answer: p2 = 160 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·(150/75)² = 8 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 8²)/1000 = −30 kPa
p2 = p1 + ½ρ(v1² − v2²) = 190 + (−30) = 160 kPa
Answer
p2 = 160 kPa
4.Continuity & Bernoulli
Answer: p2 = 133.9 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·(200/100)² = 11 m/s
½ρ(v1² − v2²) = ½·1000·(2.75² − 11²)/1000 = −56.72 kPa
ρgΔz = 1000·9.81·3/1000 = 29.43 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 220 + (−56.72) − 29.43 = 133.9 kPa
Answer
p2 = 133.9 kPa
5.Pipe friction & head loss
Answer: hf = 46.48 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 = 0.025/0.007854 = 3.183 m/s
Velocity head
V²/(2g) = 3.183²/(2·9.81) = 0.5164 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(300/0.1)·3.183²/(2·9.81) = 46.48 m
Answer
hf = 46.48 m
6.Pipe friction & head loss
Answer: H = 1.744 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 = 0.025/0.01767 = 1.415 m/s
V²/(2g) = 0.102 m
Loss coefficients
f·L/D = 0.015·100/0.15 = 10
ΣK = 0.5 + 2·0.3 + 5 = 6.1
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 10 + 6.1)·0.102 = 1.744 m
Answer
H = 1.744 m
7.Pumps & pump power
Answer: η = 65.13 %
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 = 1000·9.81·0.1·16 = 15.7 kW
Efficiency
η = Ph/Pshaft = 15.7/24.1 = 65.13%
Answer
η = 65.13%
8.Pumps & pump power
Answer: H = 53.59 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (40 + 6)·0.3046 = 14.01 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 = 9 + 30.58 + 14.01 = 53.59 m
Answer
H = 53.59 m
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