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 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2 mb = 3 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 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1 mH = 2.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 5.5 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 5.5 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 50 mm while rising 10 m. At section 1 the pressure is 320 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 50 mmp1 = 320 kPav1 = 0.75 m/sΔz = 10 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.9 m/s through a horizontal pipe, 50 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 1.9 m/sL = 50 mD = 200 mmf = 0.025ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 4 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 12 L/s?
Q = 12 L/sL = 200 mD = 100 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 34 m while its shaft takes 13.9 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 34 mPshaft = 13.9 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 100 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 150 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 100 L/sΔz = 8 mL = 150 mD = 250 mmf = 0.022ΣK = 5
H = m
3Solutions
1.Hydrostatic force
Answer: F = 176.6 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 + 2/2 = 3 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·6 = 176.6 kN
Answer
F = 176.6 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 = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·3.75 = 82.77 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·2.5³/12 = 1.953 m⁴
ycp = yc + Ic/(yc·A) = 2.25 + 1.953/(2.25·3.75) = 2.481 m
Answer
ycp = 2.481 m
3.Continuity & Bernoulli
Answer: Q = 13.26 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.
Ideal jet velocity
√(2gh) = √(2·9.81·5.5) = 10.39 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.001963·10.39 = 0.01326 m³/s = 13.26 L/s
Answer
Q = 13.26 L/s
4.Continuity & Bernoulli
Answer: p2 = 150.2 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 = 0.75·(200/50)² = 12 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 12²)/1000 = −71.72 kPa
ρgΔz = 1000·9.81·10/1000 = 98.1 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 320 + (−71.72) − 98.1 = 150.2 kPa
Answer
p2 = 150.2 kPa
5.Pipe friction & head loss
Answer: Δp = 11.28 kPa
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.
Head loss
hf = f·(L/D)·V²/(2g) = 0.025·(50/0.2)·1.9²/(2·9.81) = 1.15 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·1.15 = 11281 Pa = 11.28 kPa
Answer
Δp = 11.28 kPa
6.Pipe friction & head loss
Answer: Δz = 4.129 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 = 0.012/0.007854 = 1.528 m/s
V²/(2g) = 0.119 m
Loss coefficients
f·L/D = 0.015·200/0.1 = 30
ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7
Total head loss
hf = 30·0.119 = 3.569 m
hm = 4.7·0.119 = 0.5592 m
Δz = hf + hm = 4.129 m
Answer
Δz = 4.129 m
7.Pumps & pump power
Answer: η = 59.99 %
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 = 1000·9.81·0.025·34 = 8.338 kW
Efficiency
η = Ph/Pshaft = 8.338/13.9 = 59.99%
Answer
η = 59.99%
8.Pumps & pump power
Answer: H = 11.85 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (13.2 + 5)·0.2115 = 3.85 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 8 + 3.85 = 11.85 m
Answer
H = 11.85 m
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