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 1.5 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 = 1.5 mb = 1.5 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 0.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 0.5 mH = 2.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 100 mm. In the wide section the pressure is 400 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 100 mmp1 = 400 kPav1 = 1.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 40 mm while rising 3 m. At section 1 the pressure is 260 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 40 mmp1 = 260 kPav1 = 2 m/sΔz = 3 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 0.9 m/s through a horizontal pipe, 200 m long, with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 0.9 m/sL = 200 mD = 50 mmf = 0.03ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.015. The water surface of the upper reservoir is 11 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 5) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.
Δz = 11 mL = 100 mD = 100 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 5Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 42 m while its shaft takes 10 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 20 L/sH = 42 mPshaft = 10 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 30 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 400 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 30 L/sΔz = 5 mL = 400 mD = 150 mmf = 0.022ΣK = 6
H = m
3Solutions
1.Hydrostatic force
Answer: F = 60.7 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 + 1.5/2 = 2.75 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.75·2.25 = 60.7 kN
Answer
F = 60.7 kN
2.Hydrostatic force
Answer: ycp = 2.048 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 = 0.5 + 2.5/2 = 1.75 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·1.75·3.75 = 56.65 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·2.5³/12 = 1.953 m⁴
ycp = yc + Ic/(yc·A) = 1.75 + 1.953/(1.75·3.75) = 2.048 m
Answer
ycp = 2.048 m
3.Continuity & Bernoulli
Answer: p2 = 393.8 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 = 1.75·(150/100)² = 3.938 m/s
½ρ(v1² − v2²) = ½·1000·(1.75² − 3.938²)/1000 = −6.221 kPa
p2 = p1 + ½ρ(v1² − v2²) = 400 + (−6.221) = 393.8 kPa
Answer
p2 = 393.8 kPa
4.Continuity & Bernoulli
Answer: p2 = 154.4 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·(100/40)² = 12.5 m/s
½ρ(v1² − v2²) = ½·1000·(2² − 12.5²)/1000 = −76.13 kPa
ρgΔz = 1000·9.81·3/1000 = 29.43 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 260 + (−76.13) − 29.43 = 154.4 kPa
Answer
p2 = 154.4 kPa
5.Pipe friction & head loss
Answer: Δp = 48.6 kPa
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.
Head loss
hf = f·(L/D)·V²/(2g) = 0.03·(200/0.05)·0.9²/(2·9.81) = 4.954 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·4.954 = 48600 Pa = 48.6 kPa
Answer
Δp = 48.6 kPa
6.Pipe friction & head loss
Answer: Q = 24.38 L/s
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.
Loss coefficients
f·L/D = 0.015·100/0.1 = 15
ΣK = 0.5 + 3·0.3 + 5 + 1.0 = 7.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
11 = (15 + 7.4)·V²/(2·9.81)
V = √(2·9.81·11/22.4) = 3.104 m/s
Flow rate
Q = V·π·D²/4 = 3.104·0.007854 = 0.02438 m³/s = 24.38 L/s
Answer
Q = 24.38 L/s
7.Pumps & pump power
Answer: η = 82.4 %
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 = 1000·9.81·0.02·42 = 8.24 kW
Efficiency
η = Ph/Pshaft = 8.24/10 = 82.4%
Answer
η = 82.4%
8.Pumps & pump power
Answer: H = 14.5 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 = 0.03/0.01767 = 1.698 m/s
V²/(2g) = 0.1469 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (58.67 + 6)·0.1469 = 9.499 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 5 + 9.499 = 14.5 m
Answer
H = 14.5 m
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