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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1 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 seawater, with its top edge 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1.5 mH = 2.5 mb = 3 mρ = 1025 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 40 mm. The water level is kept 5 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
d = 40 mmh = 5 mCd = 0.63g = 9.81 m/s²
Q = L/s
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 150 kPa. The nozzle exit diameter is 25 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 = 150 kPa (gauge)d2 = 25 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.7 m/s through a pipe 500 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.7 m/sL = 500 mD = 75 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 0.2) 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 50 L/s?
Q = 50 L/sL = 150 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 0.2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 36 m while its shaft takes 5.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 10 L/sH = 36 mPshaft = 5.4 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 7 m higher. The pipe is 150 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 4 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 50 L/sΔz = 7 mL = 150 mD = 150 mmf = 0.022ΣK = 4
H = m
3Solutions
1.Hydrostatic force
Answer: F = 34.34 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 = 3 + 1/2 = 3.5 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.5·1 = 34.34 kN
Answer
F = 34.34 kN
2.Hydrostatic force
Answer: ycp = 2.939 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 = 1.5 + 2.5/2 = 2.75 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·2.75·7.5 = 207.4 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·2.5³/12 = 3.906 m⁴
ycp = yc + Ic/(yc·A) = 2.75 + 3.906/(2.75·7.5) = 2.939 m
Answer
ycp = 2.939 m
3.Continuity & Bernoulli
Answer: Q = 7.841 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) = 9.905 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.63·0.001257·9.905 = 0.007841 m³/s = 7.841 L/s
Answer
Q = 7.841 L/s
4.Continuity & Bernoulli
Answer: H = 16.31 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 = 25/50 = 0.5
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·150000/(1000·0.9375)) = 17.89 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) = 17.89²/(2·9.81) = 16.31 m
Answer
H = 16.31 m
5.Pipe friction & head loss
Answer: hf = 29.46 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 head
V²/(2g) = 1.7²/(2·9.81) = 0.1473 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(500/0.075)·1.7²/(2·9.81) = 29.46 m
Answer
hf = 29.46 m
6.Pipe friction & head loss
Answer: Δz = 0.8355 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 = 0.05/0.04909 = 1.019 m/s
V²/(2g) = 0.05288 m
Loss coefficients
f·L/D = 0.022·150/0.25 = 13.2
ΣK = 0.5 + 3·0.3 + 0.2 + 1.0 = 2.6
Total head loss
hf = 13.2·0.05288 = 0.698 m
hm = 2.6·0.05288 = 0.1375 m
Δz = hf + hm = 0.8355 m
Answer
Δz = 0.8355 m
7.Pumps & pump power
Answer: η = 65.4 %
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 = 1000·9.81·0.01·36 = 3.532 kW
Efficiency
η = Ph/Pshaft = 3.532/5.4 = 65.4%
Answer
η = 65.4%
8.Pumps & pump power
Answer: H = 17.61 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (22 + 4)·0.408 = 10.61 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 7 + 10.61 = 17.61 m
Answer
H = 17.61 m
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