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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 3 mb = 1.5 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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 3 mb = 2 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 110 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 75 mmp1 = 110 kPav1 = 2.25 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 8 m. At section 1 the pressure is 340 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 40 mmp1 = 340 kPav1 = 2.25 m/sΔz = 8 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.5 m/s through a horizontal pipe, 150 m long, with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 1.5 m/sL = 150 mD = 75 mmf = 0.015ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 150 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.018) 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 75 L/s?
Q = 75 L/sL = 150 mD = 200 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 5
H = m
7.Pumps & pump powerMedium
A pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 10 m. The pump efficiency is 65% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 60 L/sH = 10 mηpump = 65%ηmotor = 95%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 13 m higher. The pipe is 300 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 = 4 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 120 L/sΔz = 13 mL = 300 mD = 250 mmf = 0.025ΣK = 4
H = m
3Solutions
1.Hydrostatic force
Answer: F = 132.4 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 = 1.5 + 3/2 = 3 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·4.5 = 132.4 kN
Answer
F = 132.4 kN
2.Hydrostatic force
Answer: F = 271.5 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 + 3/2 = 4.5 m
A = H·b = 3·2 = 6 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·4.5·6 = 271.5 kN
Answer
F = 271.5 kN
3.Continuity & Bernoulli
Answer: p2 = 72.03 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 = 2.25·(150/75)² = 9 m/s
½ρ(v1² − v2²) = ½·1000·(2.25² − 9²)/1000 = −37.97 kPa
p2 = p1 + ½ρ(v1² − v2²) = 110 + (−37.97) = 72.03 kPa
Answer
p2 = 72.03 kPa
4.Continuity & Bernoulli
Answer: p2 = 165.2 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 = 2.25·(100/40)² = 14.06 m/s
½ρ(v1² − v2²) = ½·1000·(2.25² − 14.06²)/1000 = −96.35 kPa
ρgΔz = 1000·9.81·8/1000 = 78.48 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 340 + (−96.35) − 78.48 = 165.2 kPa
Answer
p2 = 165.2 kPa
5.Pipe friction & head loss
Answer: Δp = 33.75 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) = 0.015·(150/0.075)·1.5²/(2·9.81) = 3.44 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 = 1000·9.81·3.44 = 33750 Pa = 33.75 kPa
Answer
Δp = 33.75 kPa
6.Pipe friction & head loss
Answer: H = 6.071 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Loss coefficients
f·L/D = 0.018·150/0.2 = 13.5
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 + 13.5 + 6.4)·0.2905 = 6.071 m
Answer
H = 6.071 m
7.Pumps & pump power
Answer: Pel = 9.532 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 = 1000·9.81·0.06·10 = 5.886 kW
Shaft power
Pshaft = 5.886/0.65 = 9.055 kW
Electrical power
Pel = 9.055/0.95 = 9.532 kW
Answer
Pel = 9.532 kW
8.Pumps & pump power
Answer: H = 23.36 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) = (30 + 4)·0.3046 = 10.36 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 13 + 10.36 = 23.36 m
Answer
H = 23.36 m
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