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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 2.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of oil, 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 = 1 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows at 14 L/s through a horizontal pipe that narrows from 80 mm to 40 mm. The pressure in the narrow section is 110 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 14 L/sd1 = 80 mmd2 = 40 mmp2 = 110 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 150 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 240 mm. The discharge coefficient is 0.95. 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 = 150 mmd2 = 100 mmΔh = 240 mmCd = 0.95ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 50 L/s through a pipe 200 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 50 L/sL = 200 mD = 150 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 10 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 = 10 mL = 300 mD = 300 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 36 m while its shaft takes 43.5 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 80 L/sH = 36 mPshaft = 43.5 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 5 m higher. The pipe is 400 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. Find the head H the pump must add.
Q = 50 L/sΔz = 5 mL = 400 mD = 200 mmf = 0.018ΣK = 6
H = m
3Solutions
1.Hydrostatic force
Answer: F = 128.8 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 + 2.5/2 = 1.75 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·7.5 = 128.8 kN
Answer
F = 128.8 kN
2.Hydrostatic force
Answer: ycp = 2.042 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 + 1/2 = 2 m
A = H·b = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·2·1.5 = 25.9 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·1³/12 = 0.125 m⁴
ycp = yc + Ic/(yc·A) = 2 + 0.125/(2·1.5) = 2.042 m
Answer
ycp = 2.042 m
3.Continuity & Bernoulli
Answer: p1 = 168.2 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.
Velocities
v1 = Q/A1 = 0.014/0.005027 = 2.785 m/s
v2 = Q/A2 = 0.014/0.001257 = 11.14 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 110 + ½·1000·(11.14² − 2.785²)/1000 = 110 + 58.18 = 168.2 kPa
Answer
p1 = 168.2 kPa
4.Continuity & Bernoulli
Answer: Q = 64.16 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)·9.81·0.24 = 29665 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 100/150 = 0.6667
v2 = √(2·29665/(1000·(1 − 0.1975))) = 8.599 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.95·0.007854·8.599 = 0.06416 m³/s = 64.16 L/s
Answer
Q = 64.16 L/s
5.Pipe friction & head loss
Answer: hf = 9.793 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·0.15²/4 = 0.01767 m²
V = Q/A = 0.05/0.01767 = 2.829 m/s
Velocity head
V²/(2g) = 2.829²/(2·9.81) = 0.408 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(200/0.15)·2.829²/(2·9.81) = 9.793 m
Answer
hf = 9.793 m
6.Pipe friction & head loss
Answer: Q = 192.7 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 = 0.022·300/0.3 = 22
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.
10 = (22 + 4.4)·V²/(2·9.81)
V = √(2·9.81·10/26.4) = 2.726 m/s
Flow rate
Q = V·π·D²/4 = 2.726·0.07069 = 0.1927 m³/s = 192.7 L/s
Answer
Q = 192.7 L/s
7.Pumps & pump power
Answer: η = 64.95 %
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 = 1000·9.81·0.08·36 = 28.25 kW
Efficiency
η = Ph/Pshaft = 28.25/43.5 = 64.95%
Answer
η = 64.95%
8.Pumps & pump power
Answer: H = 10.42 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 = 0.05/0.03142 = 1.592 m/s
V²/(2g) = 0.1291 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (36 + 6)·0.1291 = 5.422 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 5 + 5.422 = 10.42 m
Answer
H = 10.42 m
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