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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1 mb = 3 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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 3 mb = 1 mρ = 1025 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 4.5 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 = 4.5 mCd = 0.63g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A hose with an inner diameter of 40 mm siphons water from an open tank. The highest point C of the hose is 2.5 m above the water surface, and the outlet (open to the air) is 2 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure pC at point C. The atmospheric pressure is 101.3 kPa.
d = 40 mmzC = +2.5 mz2 = −2 mg = 9.81 m/s²patm = 101.3 kPa
pC = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.8 m/s through a horizontal pipe, 800 m long, with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.8 m/sL = 800 mD = 50 mmf = 0.018ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 10) 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 80 L/s?
Q = 80 L/sL = 100 mD = 200 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 72% delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 28 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 50 L/sH = 28 mη = 72%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 7 m higher. The pipe is 200 m long with an inside diameter of 100 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 250 kPa. Find the head H the pump must add.
Q = 20 L/sΔz = 7 mL = 200 mD = 100 mmf = 0.018ΣK = 6ptank = 250 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 88.29 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 + 1/2 = 3 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·3 = 88.29 kN
Answer
F = 88.29 kN
2.Hydrostatic force
Answer: F = 90.5 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 = 1.5 + 3/2 = 3 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3·3 = 90.5 kN
Answer
F = 90.5 kN
3.Continuity & Bernoulli
Answer: Q = 7.439 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) = √(2·9.81·4.5) = 9.396 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.63·0.001257·9.396 = 0.007439 m³/s = 7.439 L/s
Answer
Q = 7.439 L/s
4.Continuity & Bernoulli
Answer: pC = 57.15 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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·2) = √(2·9.81·2) = 6.264 m/s
Bernoulli from the surface to C
patm = pC + ½ρv² + ρgzC
pC = 101.3 − 1000·(6.264²/2 + 9.81·2.5)/1000 = 101.3 − 44.15 = 57.15 kPa
Answer
pC = 57.15 kPa
If this pressure falls to the vapour pressure of the water, the siphon stops working.
5.Pipe friction & head loss
Answer: Δp = 1129 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) = 0.018·(800/0.05)·2.8²/(2·9.81) = 115.1 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 = 1000·9.81·115.1 = 1.129 × 10⁶ Pa = 1129 kPa
Answer
Δp = 1129 kPa
6.Pipe friction & head loss
Answer: Δz = 7.172 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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Loss coefficients
f·L/D = 0.018·100/0.2 = 9
ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7
Total head loss
hf = 9·0.3305 = 2.975 m
hm = 12.7·0.3305 = 4.197 m
Δz = hf + hm = 7.172 m
Answer
Δz = 7.172 m
7.Pumps & pump power
Answer: Pshaft = 19.07 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 = 1000·9.81·0.05·28 = 13.73 kW
Shaft power
Pshaft = Ph/η = 13.73/0.72 = 19.07 kW
Answer
Pshaft = 19.07 kW
8.Pumps & pump power
Answer: H = 46.37 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) = (36 + 6)·0.3305 = 13.88 m
Pressure head
p/(ρ·g) = 250000/(1000·9.81) = 25.48 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 7 + 25.48 + 13.88 = 46.37 m
Answer
H = 46.37 m
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