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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 1 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 2.5 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 200 mm to 100 mm. In the wide section the pressure is 210 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 200 mmd2 = 100 mmp1 = 210 kPav1 = 0.5 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 9 m. At section 1 the pressure is 230 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 75 mmp1 = 230 kPav1 = 2 m/sΔz = 9 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 5 L/s through a pipe 150 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
Q = 5 L/sL = 150 mD = 100 mmf = 0.022g = 9.81 m/s²
hf = m
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 = 0.022. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 2) 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 15 L/s?
Q = 15 L/sL = 100 mD = 100 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 78 m. The pump efficiency is 85% 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 = 78 mηpump = 85%ηmotor = 95%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 30 m higher. The pipe is 150 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.025; 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 250 kPa. Find the head H the pump must add.
Q = 50 L/sΔz = 30 mL = 150 mD = 150 mmf = 0.025ΣK = 8ptank = 250 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 = 0.5 + 1/2 = 1 m
A = H·b = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1·2 = 19.62 kN
Answer
F = 19.62 kN
2.Hydrostatic force
Answer: F = 145.7 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 = 1 + 2.5/2 = 2.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·2.25·7.5 = 145.7 kN
Answer
F = 145.7 kN
3.Continuity & Bernoulli
Answer: p2 = 208.1 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·(200/100)² = 2 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 2²)/1000 = −1.875 kPa
p2 = p1 + ½ρ(v1² − v2²) = 210 + (−1.875) = 208.1 kPa
Answer
p2 = 208.1 kPa
4.Continuity & Bernoulli
Answer: p2 = 111.7 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
ρgΔz = 1000·9.81·9/1000 = 88.29 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 230 + (−30) − 88.29 = 111.7 kPa
Answer
p2 = 111.7 kPa
5.Pipe friction & head loss
Answer: hf = 0.6817 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 = 0.005/0.007854 = 0.6366 m/s
Velocity head
V²/(2g) = 0.6366²/(2·9.81) = 0.02066 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(150/0.1)·0.6366²/(2·9.81) = 0.6817 m
Answer
hf = 0.6817 m
6.Pipe friction & head loss
Answer: Δz = 4.908 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 = 0.015/0.007854 = 1.91 m/s
V²/(2g) = 0.1859 m
Loss coefficients
f·L/D = 0.022·100/0.1 = 22
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Total head loss
hf = 22·0.1859 = 4.09 m
hm = 4.4·0.1859 = 0.818 m
Δz = hf + hm = 4.908 m
Answer
Δz = 4.908 m
7.Pumps & pump power
Answer: Pel = 56.86 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 = 1000·9.81·0.06·78 = 45.91 kW
Shaft power
Pshaft = 45.91/0.85 = 54.01 kW
Electrical power
Pel = 54.01/0.95 = 56.86 kW
Answer
Pel = 56.86 kW
8.Pumps & pump power
Answer: H = 68.95 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (25 + 8)·0.408 = 13.47 m
Pressure head
p/(ρ·g) = 250000/(1000·9.81) = 25.48 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 30 + 25.48 + 13.47 = 68.95 m
Answer
H = 68.95 m
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