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 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1 m wide sits in the wall of a tank of seawater, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 2 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows at 39 L/s through a horizontal pipe that narrows from 150 mm to 75 mm. The pressure in the narrow section is 100 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 39 L/sd1 = 150 mmd2 = 75 mmp2 = 100 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 6 m. At section 1 the pressure is 310 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 75 mmp1 = 310 kPav1 = 0.75 m/sΔz = 6 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.6 m/s through a horizontal pipe, 500 m long, with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 1.6 m/sL = 500 mD = 250 mmf = 0.03ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 300 m long with an inside diameter of 250 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), 3 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 = 0.015Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 5
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 12 m while its shaft takes 14.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 100 L/sH = 12 mPshaft = 14.4 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 20 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 9 m higher. The pipe is 150 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 6 (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 150 kPa. Find the head H the pump must add.
Q = 20 L/sΔz = 9 mL = 150 mD = 100 mmf = 0.025ΣK = 6ptank = 150 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 79.71 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 + 2.5/2 = 3.25 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·2.5 = 79.71 kN
Answer
F = 79.71 kN
2.Hydrostatic force
Answer: F = 30.17 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 + 2/2 = 1.5 m
A = H·b = 2·1 = 2 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·1.5·2 = 30.17 kN
Answer
F = 30.17 kN
3.Continuity & Bernoulli
Answer: p1 = 136.5 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.
Velocities
v1 = Q/A1 = 0.039/0.01767 = 2.207 m/s
v2 = Q/A2 = 0.039/0.004418 = 8.828 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 100 + ½·1000·(8.828² − 2.207²)/1000 = 100 + 36.53 = 136.5 kPa
Answer
p1 = 136.5 kPa
4.Continuity & Bernoulli
Answer: p2 = 246.9 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·(150/75)² = 3 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 3²)/1000 = −4.219 kPa
ρgΔz = 1000·9.81·6/1000 = 58.86 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 310 + (−4.219) − 58.86 = 246.9 kPa
Answer
p2 = 246.9 kPa
5.Pipe friction & head loss
Answer: Δp = 76.8 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) = 0.03·(500/0.25)·1.6²/(2·9.81) = 7.829 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 = 1000·9.81·7.829 = 76800 Pa = 76.8 kPa
Answer
Δp = 76.8 kPa
6.Pipe friction & head loss
Answer: H = 5.373 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 = 0.015·300/0.25 = 18
ΣK = 0.5 + 3·0.3 + 5 = 6.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 18 + 6.4)·0.2115 = 5.373 m
Answer
H = 5.373 m
7.Pumps & pump power
Answer: η = 81.75 %
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 = 1000·9.81·0.1·12 = 11.77 kW
Efficiency
η = Ph/Pshaft = 11.77/14.4 = 81.75%
Answer
η = 81.75%
8.Pumps & pump power
Answer: H = 38.67 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 = 0.02/0.007854 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (37.5 + 6)·0.3305 = 14.38 m
Pressure head
p/(ρ·g) = 150000/(1000·9.81) = 15.29 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 9 + 15.29 + 14.38 = 38.67 m
Answer
H = 38.67 m
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