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 3 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1.5 mb = 3 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 water, with its top edge 4 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 4 mH = 3 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 40 mm. In the wide section the pressure is 240 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 40 mmp1 = 240 kPav1 = 1 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 200 mm, throat 150 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 160 mm. The discharge coefficient is 0.96. 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 = 150 mmΔh = 160 mmCd = 0.96ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 200 L/s through a pipe 500 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss hf.
Q = 200 L/sL = 500 mD = 300 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 300 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.025) 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 100 L/s?
Q = 100 L/sL = 300 mD = 250 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 10 m while its shaft takes 3.1 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 10 mPshaft = 3.1 kWg = 9.81 m/s²
η = %
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 14 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. Find the head H the pump must add.
Q = 50 L/sΔz = 14 mL = 150 mD = 200 mmf = 0.022ΣK = 8
H = m
3Solutions
1.Hydrostatic force
Answer: F = 77.25 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 + 1.5/2 = 1.75 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·4.5 = 77.25 kN
Answer
F = 77.25 kN
2.Hydrostatic force
Answer: ycp = 5.636 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 = 4 + 3/2 = 5.5 m
A = H·b = 3·2 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.5·6 = 323.7 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·3³/12 = 4.5 m⁴
ycp = yc + Ic/(yc·A) = 5.5 + 4.5/(5.5·6) = 5.636 m
Answer
ycp = 5.636 m
3.Continuity & Bernoulli
Answer: p2 = 221 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·(100/40)² = 6.25 m/s
½ρ(v1² − v2²) = ½·1000·(1² − 6.25²)/1000 = −19.03 kPa
p2 = p1 + ½ρ(v1² − v2²) = 240 + (−19.03) = 221 kPa
Answer
p2 = 221 kPa
4.Continuity & Bernoulli
Answer: Q = 129 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)·9.81·0.16 = 19777 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 150/200 = 0.75
v2 = √(2·19777/(1000·(1 − 0.3164))) = 7.607 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.96·0.01767·7.607 = 0.129 m³/s = 129 L/s
Answer
Q = 129 L/s
5.Pipe friction & head loss
Answer: hf = 17 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·0.3²/4 = 0.07069 m²
V = Q/A = 0.2/0.07069 = 2.829 m/s
Velocity head
V²/(2g) = 2.829²/(2·9.81) = 0.408 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(500/0.3)·2.829²/(2·9.81) = 17 m
Answer
hf = 17 m
6.Pipe friction & head loss
Answer: H = 7.848 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Loss coefficients
f·L/D = 0.025·300/0.25 = 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)·0.2115 = 7.848 m
Answer
H = 7.848 m
7.Pumps & pump power
Answer: η = 79.11 %
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 = 1000·9.81·0.025·10 = 2.453 kW
Efficiency
η = Ph/Pshaft = 2.453/3.1 = 79.11%
Answer
η = 79.11%
8.Pumps & pump power
Answer: H = 17.16 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 = 0.05/0.03142 = 1.592 m/s
V²/(2g) = 0.1291 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (16.5 + 8)·0.1291 = 3.163 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 14 + 3.163 = 17.16 m
Answer
H = 17.16 m
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