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 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 water, with its top edge 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1.5 mH = 1.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 250 mm and a throat diameter of 125 mm. The pressure at the inlet is 50 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 250 mmd2 = 125 mmp1 − p2 = 50 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 5 m. At section 1 the pressure is 120 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 100 mmp1 = 120 kPav1 = 1.75 m/sΔz = 5 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 15 L/s through a pipe 50 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 friction head loss hf.
Q = 15 L/sL = 50 mD = 75 mmf = 0.015g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 4 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 20 L/s?
Q = 20 L/sL = 150 mD = 150 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×4)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 66 m. The pump efficiency is 80% 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 = 10 L/sH = 66 mηpump = 80%ηmotor = 95%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 35 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 78%. Find the shaft power Pshaft in kW.
Q = 75 L/sΔz = 35 mL = 150 mD = 200 mmf = 0.022ΣK = 8η = 78%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 36.79 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 + 1/2 = 2.5 m
A = H·b = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·1.5 = 36.79 kN
Answer
F = 36.79 kN
2.Hydrostatic force
Answer: ycp = 2.333 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 = 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
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·1.5³/12 = 0.4219 m⁴
ycp = yc + Ic/(yc·A) = 2.25 + 0.4219/(2.25·2.25) = 2.333 m
Answer
ycp = 2.333 m
3.Continuity & Bernoulli
Answer: Q = 126.7 L/s
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
v1 = v2·(125/250)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·50000/(1000·(1 − 0.0625))) = 10.33 m/s
v1 = 2.582 m/s
Flow rate
Q = A2·v2 = 0.01227·10.33 = 0.1267 m³/s = 126.7 L/s
Answer
Q = 126.7 L/s
4.Continuity & Bernoulli
Answer: p2 = 64.73 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
ρgΔz = 1000·9.81·5/1000 = 49.05 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 120 + (−6.221) − 49.05 = 64.73 kPa
Answer
p2 = 64.73 kPa
5.Pipe friction & head loss
Answer: hf = 5.876 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·0.075²/4 = 0.004418 m²
V = Q/A = 0.015/0.004418 = 3.395 m/s
Velocity head
V²/(2g) = 3.395²/(2·9.81) = 0.5876 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.015·(50/0.075)·3.395²/(2·9.81) = 5.876 m
Answer
hf = 5.876 m
6.Pipe friction & head loss
Answer: Δz = 1.678 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 = 0.02/0.01767 = 1.132 m/s
V²/(2g) = 0.06529 m
Loss coefficients
f·L/D = 0.018·150/0.15 = 18
ΣK = 0.5 + 4·0.3 + 5 + 1.0 = 7.7
Total head loss
hf = 18·0.06529 = 1.175 m
hm = 7.7·0.06529 = 0.5027 m
Δz = hf + hm = 1.678 m
Answer
Δz = 1.678 m
7.Pumps & pump power
Answer: Pel = 8.519 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 = 1000·9.81·0.01·66 = 6.475 kW
Shaft power
Pshaft = 6.475/0.8 = 8.093 kW
Electrical power
Pel = 8.093/0.95 = 8.519 kW
Answer
Pel = 8.519 kW
8.Pumps & pump power
Answer: Pshaft = 39.73 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.
Velocity
V = Q/A = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (16.5 + 8)·0.2905 = 7.117 m
Pump head
H = Δz + hL = 35 + 7.117 = 42.12 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.075·42.12/0.78 = 39.73 kW
Answer
Pshaft = 39.73 kW
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