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 = 1.5 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 glycerin, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 2 mb = 2 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 260 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 50 mmp1 = 260 kPav1 = 1.5 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 50 mm while rising 9 m. At section 1 the pressure is 160 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 50 mmp1 = 160 kPav1 = 0.5 m/sΔz = 9 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 25 L/s through a pipe 400 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss hf.
Q = 25 L/sL = 400 mD = 250 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.02. The minor losses are a square-edged entrance (K = 0.5), 2 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. What difference in water level Δz between the reservoirs gives a flow rate of 80 L/s?
Q = 80 L/sL = 300 mD = 250 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 60 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 92%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 10 L/sH = 60 mηpump = 72%ηmotor = 92%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 15 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 200 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 = 4 (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 100 kPa. Find the head H the pump must add.
Q = 15 L/sΔz = 8 mL = 200 mD = 100 mmf = 0.015ΣK = 4ptank = 100 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 49.66 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 + 1.5/2 = 2.25 m
A = H·b = 1.5·1.5 = 2.25 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·2.25 = 49.66 kN
Answer
F = 49.66 kN
2.Hydrostatic force
Answer: F = 98.88 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 + 2/2 = 2 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·2·4 = 98.88 kN
Answer
F = 98.88 kN
3.Continuity & Bernoulli
Answer: p2 = 255.4 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 = 1.5·(75/50)² = 3.375 m/s
½ρ(v1² − v2²) = ½·1000·(1.5² − 3.375²)/1000 = −4.57 kPa
p2 = p1 + ½ρ(v1² − v2²) = 260 + (−4.57) = 255.4 kPa
Answer
p2 = 255.4 kPa
4.Continuity & Bernoulli
Answer: p2 = 61.71 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 = 0.5·(150/50)² = 4.5 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 4.5²)/1000 = −10 kPa
ρgΔz = 1000·9.81·9/1000 = 88.29 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 160 + (−10) − 88.29 = 61.71 kPa
Answer
p2 = 61.71 kPa
5.Pipe friction & head loss
Answer: hf = 0.5288 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·0.25²/4 = 0.04909 m²
V = Q/A = 0.025/0.04909 = 0.5093 m/s
Velocity head
V²/(2g) = 0.5093²/(2·9.81) = 0.01322 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(400/0.25)·0.5093²/(2·9.81) = 0.5288 m
Answer
hf = 0.5288 m
6.Pipe friction & head loss
Answer: Δz = 4.21 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Loss coefficients
f·L/D = 0.02·300/0.25 = 24
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Total head loss
hf = 24·0.1354 = 3.249 m
hm = 7.1·0.1354 = 0.9612 m
Δz = hf + hm = 4.21 m
Answer
Δz = 4.21 m
7.Pumps & pump power
Answer: Pel = 8.886 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 = 1000·9.81·0.01·60 = 5.886 kW
Shaft power
Pshaft = 5.886/0.72 = 8.175 kW
Electrical power
Pel = 8.175/0.92 = 8.886 kW
Answer
Pel = 8.886 kW
8.Pumps & pump power
Answer: H = 24.51 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 = 0.015/0.007854 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (30 + 4)·0.1859 = 6.321 m
Pressure head
p/(ρ·g) = 100000/(1000·9.81) = 10.19 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 8 + 10.19 + 6.321 = 24.51 m
Answer
H = 24.51 m
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