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 2 m wide sits in the wall of a tank of water, 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 = 2 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 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 1 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 6 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².
d = 40 mmh = 6 mCd = 0.62g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 200 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 230 mm. The discharge coefficient is 0.99. 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 = 100 mmΔh = 230 mmCd = 0.99ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 75 L/s through a pipe 800 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 75 L/sL = 800 mD = 250 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 150 m long with an inside diameter of 150 mm (Darcy friction factor f = 0.022) 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 30 L/s?
Q = 30 L/sL = 150 mD = 150 mmf = 0.022Kentrance = 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 56 m while its shaft takes 17.6 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 56 mPshaft = 17.6 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 60 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 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018; 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 100 kPa. Find the head H the pump must add.
Q = 60 L/sΔz = 13 mL = 100 mD = 200 mmf = 0.018ΣK = 6ptank = 100 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 58.86 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 = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.5·4 = 58.86 kN
Answer
F = 58.86 kN
2.Hydrostatic force
Answer: F = 166.9 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 + 1/2 = 4.5 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·4.5·3 = 166.9 kN
Answer
F = 166.9 kN
3.Continuity & Bernoulli
Answer: Q = 8.453 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) = √(2·9.81·6) = 10.85 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.62·0.001257·10.85 = 0.008453 m³/s = 8.453 L/s
Answer
Q = 8.453 L/s
4.Continuity & Bernoulli
Answer: Q = 60.55 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)·9.81·0.23 = 28429 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 100/200 = 0.5
v2 = √(2·28429/(1000·(1 − 0.0625))) = 7.788 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.007854·7.788 = 0.06055 m³/s = 60.55 L/s
Answer
Q = 60.55 L/s
5.Pipe friction & head loss
Answer: hf = 6.853 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 = 0.075/0.04909 = 1.528 m/s
Velocity head
V²/(2g) = 1.528²/(2·9.81) = 0.119 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(800/0.25)·1.528²/(2·9.81) = 6.853 m
Answer
hf = 6.853 m
6.Pipe friction & head loss
Answer: H = 4.275 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 = 0.03/0.01767 = 1.698 m/s
V²/(2g) = 0.1469 m
Loss coefficients
f·L/D = 0.022·150/0.15 = 22
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 + 22 + 6.1)·0.1469 = 4.275 m
Answer
H = 4.275 m
7.Pumps & pump power
Answer: η = 78.03 %
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 = 1000·9.81·0.025·56 = 13.73 kW
Efficiency
η = Ph/Pshaft = 13.73/17.6 = 78.03%
Answer
η = 78.03%
8.Pumps & pump power
Answer: H = 25.98 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 = 0.06/0.03142 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (9 + 6)·0.1859 = 2.789 m
Pressure head
p/(ρ·g) = 100000/(1000·9.81) = 10.19 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 13 + 10.19 + 2.789 = 25.98 m
Answer
H = 25.98 m
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