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 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1.5 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 glycerin, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2.5 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 40 mm. The water level is kept 1.5 m above the orifice and the discharge coefficient is 0.6. Find the flow rate Q. Use g = 9.81 m/s².
d = 40 mmh = 1.5 mCd = 0.6g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 2 m. At section 1 the pressure is 120 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 100 mmp1 = 120 kPav1 = 1.5 m/sΔz = 2 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 80 L/s through a pipe 150 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
Q = 80 L/sL = 150 mD = 300 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 600 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.015. The water surface of the upper reservoir is 21 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 2 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. Find the flow rate Q in L/s.
Δz = 21 mL = 600 mD = 300 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 65% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 54 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 20 L/sH = 54 mη = 65%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 19 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; 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 = 75 L/sΔz = 19 mL = 150 mD = 200 mmf = 0.015ΣK = 8
H = m
3Solutions
1.Hydrostatic force
Answer: F = 25.75 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 = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·1.5 = 25.75 kN
Answer
F = 25.75 kN
2.Hydrostatic force
Answer: F = 347.6 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 = 2.5 + 2.5/2 = 3.75 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3.75·7.5 = 347.6 kN
Answer
F = 347.6 kN
3.Continuity & Bernoulli
Answer: Q = 4.09 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) = √(2·9.81·1.5) = 5.425 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.6·0.001257·5.425 = 0.00409 m³/s = 4.09 L/s
Answer
Q = 4.09 L/s
4.Continuity & Bernoulli
Answer: p2 = 95.81 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 = 1.5·(150/100)² = 3.375 m/s
½ρ(v1² − v2²) = ½·1000·(1.5² − 3.375²)/1000 = −4.57 kPa
ρgΔz = 1000·9.81·2/1000 = 19.62 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 120 + (−4.57) − 19.62 = 95.81 kPa
Answer
p2 = 95.81 kPa
5.Pipe friction & head loss
Answer: hf = 0.9793 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 = 0.08/0.07069 = 1.132 m/s
Velocity head
V²/(2g) = 1.132²/(2·9.81) = 0.06529 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(150/0.3)·1.132²/(2·9.81) = 0.9793 m
Answer
hf = 0.9793 m
6.Pipe friction & head loss
Answer: Q = 245.7 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 = 0.015·600/0.3 = 30
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
21 = (30 + 4.1)·V²/(2·9.81)
V = √(2·9.81·21/34.1) = 3.476 m/s
Flow rate
Q = V·π·D²/4 = 3.476·0.07069 = 0.2457 m³/s = 245.7 L/s
Answer
Q = 245.7 L/s
7.Pumps & pump power
Answer: Pshaft = 16.3 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 = 1000·9.81·0.02·54 = 10.59 kW
Shaft power
Pshaft = Ph/η = 10.59/0.65 = 16.3 kW
Answer
Pshaft = 16.3 kW
8.Pumps & pump power
Answer: H = 24.59 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (11.25 + 8)·0.2905 = 5.592 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 19 + 5.592 = 24.59 m
Answer
H = 24.59 m
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