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 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 1 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 2 m wide sits in the wall of a tank of glycerin, with its top edge 2.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 2.5 mH = 2 mb = 2 mρ = 1260 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 3 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 40 mmh = 3 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 130 kPa. Water leaves through an opening with a diameter of 25 mm, 3 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
pair = 130 kPa (gauge)h = 3 md = 25 mmρ = 1000 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.7 m/s through a pipe 800 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.7 m/sL = 800 mD = 100 mmf = 0.02g = 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 200 mm and a Darcy friction factor f = 0.025. The water surface of the upper reservoir is 15 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 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 = 15 mL = 150 mD = 200 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 20 m while its shaft takes 5.8 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 20 mPshaft = 5.8 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 15 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 75%. Find the shaft power Pshaft in kW.
Q = 50 L/sΔz = 15 mL = 150 mD = 200 mmf = 0.022ΣK = 8η = 75%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 24.52 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 + 1/2 = 2.5 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·1 = 24.52 kN
Answer
F = 24.52 kN
2.Hydrostatic force
Answer: ycp = 3.595 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 = 2.5 + 2/2 = 3.5 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3.5·4 = 173 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·2³/12 = 1.333 m⁴
ycp = yc + Ic/(yc·A) = 3.5 + 1.333/(3.5·4) = 3.595 m
Answer
ycp = 3.595 m
3.Continuity & Bernoulli
Answer: Q = 6.267 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) = √(2·9.81·3) = 7.672 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.001257·7.672 = 0.006267 m³/s = 6.267 L/s
Answer
Q = 6.267 L/s
4.Continuity & Bernoulli
Answer: Q = 8.765 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.
Bernoulli from the surface to the jet
v = √(2·(130000/1000 + 9.81·3)) = 17.86 m/s
Flow rate
Q = A·v = 0.0004909·17.86 = 0.008765 m³/s = 8.765 L/s
Answer
Q = 8.765 L/s
5.Pipe friction & head loss
Answer: hf = 59.45 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) = 2.7²/(2·9.81) = 0.3716 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.02·(800/0.1)·2.7²/(2·9.81) = 59.45 m
Answer
hf = 59.45 m
6.Pipe friction & head loss
Answer: Q = 112 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 = 0.025·150/0.2 = 18.75
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
15 = (18.75 + 4.4)·V²/(2·9.81)
V = √(2·9.81·15/23.15) = 3.565 m/s
Flow rate
Q = V·π·D²/4 = 3.565·0.03142 = 0.112 m³/s = 112 L/s
Answer
Q = 112 L/s
7.Pumps & pump power
Answer: η = 84.57 %
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 = 1000·9.81·0.025·20 = 4.905 kW
Efficiency
η = Ph/Pshaft = 4.905/5.8 = 84.57%
Answer
η = 84.57%
8.Pumps & pump power
Answer: Pshaft = 11.88 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.
Velocity
V = Q/A = 0.05/0.03142 = 1.592 m/s
V²/(2g) = 0.1291 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (16.5 + 8)·0.1291 = 3.163 m
Pump head
H = Δz + hL = 15 + 3.163 = 18.16 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.05·18.16/0.75 = 11.88 kW
Answer
Pshaft = 11.88 kW
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