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 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 mb = 1 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 water, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1 mH = 2.5 mb = 3 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 200 mm and a throat diameter of 150 mm. The pressure at the inlet is 37 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 200 mmd2 = 150 mmp1 − p2 = 37 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 200 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 330 mm. The discharge coefficient is 0.99. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
d1 = 200 mmd2 = 100 mmΔh = 330 mmCd = 0.99ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.7 m/s through a pipe 50 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.7 m/sL = 50 mD = 300 mmf = 0.022g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 600 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.015) 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), 2 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 150 L/s?
Q = 150 L/sL = 600 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 75% delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 40 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 15 L/sH = 40 mη = 75%g = 9.81 m/s²
Pshaft = 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 13 m higher. The pipe is 600 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 65%. Find the shaft power Pshaft in kW.
Q = 50 L/sΔz = 13 mL = 600 mD = 150 mmf = 0.02ΣK = 6η = 65%
Pshaft = kW
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 = 1.5 + 1/2 = 2 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·1 = 19.62 kN
Answer
F = 19.62 kN
2.Hydrostatic force
Answer: ycp = 2.481 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 + 2.5/2 = 2.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·7.5 = 165.5 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·2.5³/12 = 3.906 m⁴
ycp = yc + Ic/(yc·A) = 2.25 + 3.906/(2.25·7.5) = 2.481 m
Answer
ycp = 2.481 m
3.Continuity & Bernoulli
Answer: Q = 183.9 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·(150/200)² = 0.5625·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·37000/(1000·(1 − 0.3164))) = 10.4 m/s
v1 = 5.852 m/s
Flow rate
Q = A2·v2 = 0.01767·10.4 = 0.1839 m³/s = 183.9 L/s
Answer
Q = 183.9 L/s
4.Continuity & Bernoulli
Answer: Q = 72.53 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.
Pressure difference from the manometer
Δp = (13600 − 1000)·9.81·0.33 = 40790 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 100/200 = 0.5
v2 = √(2·40790/(1000·(1 − 0.0625))) = 9.328 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.007854·9.328 = 0.07253 m³/s = 72.53 L/s
Answer
Q = 72.53 L/s
5.Pipe friction & head loss
Answer: hf = 1.362 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 head
V²/(2g) = 2.7²/(2·9.81) = 0.3716 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(50/0.3)·2.7²/(2·9.81) = 1.362 m
Answer
hf = 1.362 m
6.Pipe friction & head loss
Answer: H = 8.515 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 = 0.15/0.07069 = 2.122 m/s
V²/(2g) = 0.2295 m
Loss coefficients
f·L/D = 0.015·600/0.3 = 30
ΣK = 0.5 + 2·0.3 + 5 = 6.1
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 30 + 6.1)·0.2295 = 8.515 m
Answer
H = 8.515 m
7.Pumps & pump power
Answer: Pshaft = 7.848 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 = 1000·9.81·0.015·40 = 5.886 kW
Shaft power
Pshaft = Ph/η = 5.886/0.75 = 7.848 kW
Answer
Pshaft = 7.848 kW
8.Pumps & pump power
Answer: Pshaft = 36.29 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (80 + 6)·0.408 = 35.09 m
Pump head
H = Δz + hL = 13 + 35.09 = 48.09 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.05·48.09/0.65 = 36.29 kW
Answer
Pshaft = 36.29 kW
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