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 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 3 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 glycerin, with its top edge 0.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 0.5 mH = 1 mb = 1.5 mρ = 1260 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 40 mm. In the wide section the pressure is 390 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 40 mmp1 = 390 kPav1 = 0.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 90 kPa. The nozzle exit diameter is 20 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
d1 = 50 mmp1 = 90 kPa (gauge)d2 = 20 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at 25 L/s through a pipe 100 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss hf.
Q = 25 L/sL = 100 mD = 250 mmf = 0.02g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 23 m above that of the lower one. 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. Find the flow rate Q in L/s.
Δz = 23 mL = 100 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 5Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 80% delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 78 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 100 L/sH = 78 mη = 80%g = 9.81 m/s²
Pshaft = kW
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 13 m higher. The pipe is 300 m long with an inside diameter of 100 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. Find the head H the pump must add.
Q = 20 L/sΔz = 13 mL = 300 mD = 100 mmf = 0.02ΣK = 6
H = m
3Solutions
1.Hydrostatic force
Answer: F = 485.6 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 = 4 + 3/2 = 5.5 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.5·9 = 485.6 kN
Answer
F = 485.6 kN
2.Hydrostatic force
Answer: ycp = 1.083 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 = 0.5 + 1/2 = 1 m
A = H·b = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·1·1.5 = 18.54 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·1³/12 = 0.125 m⁴
ycp = yc + Ic/(yc·A) = 1 + 0.125/(1·1.5) = 1.083 m
Answer
ycp = 1.083 m
3.Continuity & Bernoulli
Answer: p2 = 334.7 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·(150/40)² = 10.55 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 10.55²)/1000 = −55.34 kPa
p2 = p1 + ½ρ(v1² − v2²) = 390 + (−55.34) = 334.7 kPa
Answer
p2 = 334.7 kPa
4.Continuity & Bernoulli
Answer: H = 9.415 m
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.
Exit velocity
β = 20/50 = 0.4
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·90000/(1000·0.9744)) = 13.59 m/s
Height of the jet
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
H = v2²/(2g) = 13.59²/(2·9.81) = 9.415 m
Answer
H = 9.415 m
5.Pipe friction & head loss
Answer: hf = 0.1058 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·0.25²/4 = 0.04909 m²
V = Q/A = 0.025/0.04909 = 0.5093 m/s
Velocity head
V²/(2g) = 0.5093²/(2·9.81) = 0.01322 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.02·(100/0.25)·0.5093²/(2·9.81) = 0.1058 m
Answer
hf = 0.1058 m
6.Pipe friction & head loss
Answer: Q = 396.2 L/s
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.
Loss coefficients
f·L/D = 0.02·100/0.3 = 6.667
ΣK = 0.5 + 4·0.3 + 5 + 1.0 = 7.7
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
23 = (6.667 + 7.7)·V²/(2·9.81)
V = √(2·9.81·23/14.37) = 5.604 m/s
Flow rate
Q = V·π·D²/4 = 5.604·0.07069 = 0.3962 m³/s = 396.2 L/s
Answer
Q = 396.2 L/s
7.Pumps & pump power
Answer: Pshaft = 95.65 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 = 1000·9.81·0.1·78 = 76.52 kW
Shaft power
Pshaft = Ph/η = 76.52/0.8 = 95.65 kW
Answer
Pshaft = 95.65 kW
8.Pumps & pump power
Answer: H = 34.81 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) = (60 + 6)·0.3305 = 21.81 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 13 + 21.81 = 34.81 m
Answer
H = 34.81 m
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