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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 2 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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1.5 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 200 mm and a throat diameter of 100 mm. The pressure at the inlet is 60 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 200 mmd2 = 100 mmp1 − p2 = 60 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 180 kPa. Water leaves through an opening with a diameter of 25 mm, 5 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
pair = 180 kPa (gauge)h = 5 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 1.5 m/s through a pipe 400 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.5 m/sL = 400 mD = 75 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.025) 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), 4 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 100 L/s?
Q = 100 L/sL = 200 mD = 200 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 5
H = m
7.Pumps & pump powerMedium
A pump delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 10 m. The pump efficiency is 75% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 40 L/sH = 10 mηpump = 75%ηmotor = 90%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 80 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 100 m long with an inside diameter of 250 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 = 80 L/sΔz = 7 mL = 100 mD = 250 mmf = 0.022ΣK = 4
H = m
3Solutions
1.Hydrostatic force
Answer: F = 235.4 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 + 2/2 = 4 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4·6 = 235.4 kN
Answer
F = 235.4 kN
2.Hydrostatic force
Answer: F = 49.02 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 + 1.5/2 = 3.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3.25·1.5 = 49.02 kN
Answer
F = 49.02 kN
3.Continuity & Bernoulli
Answer: Q = 88.86 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.
Continuity
v1 = v2·(100/200)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·60000/(1000·(1 − 0.0625))) = 11.31 m/s
v1 = 2.828 m/s
Flow rate
Q = A2·v2 = 0.007854·11.31 = 0.08886 m³/s = 88.86 L/s
Answer
Q = 88.86 L/s
4.Continuity & Bernoulli
Answer: Q = 10.51 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·(180000/1000 + 9.81·5)) = 21.4 m/s
Flow rate
Q = A·v = 0.0004909·21.4 = 0.01051 m³/s = 10.51 L/s
Answer
Q = 10.51 L/s
5.Pipe friction & head loss
Answer: hf = 15.29 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) = 1.5²/(2·9.81) = 0.1147 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(400/0.075)·1.5²/(2·9.81) = 15.29 m
Answer
hf = 15.29 m
6.Pipe friction & head loss
Answer: H = 16.89 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 = 0.1/0.03142 = 3.183 m/s
V²/(2g) = 0.5164 m
Loss coefficients
f·L/D = 0.025·200/0.2 = 25
ΣK = 0.5 + 4·0.3 + 5 = 6.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 25 + 6.7)·0.5164 = 16.89 m
Answer
H = 16.89 m
7.Pumps & pump power
Answer: Pel = 5.813 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 = 1000·9.81·0.04·10 = 3.924 kW
Shaft power
Pshaft = 3.924/0.75 = 5.232 kW
Electrical power
Pel = 5.232/0.9 = 5.813 kW
Answer
Pel = 5.813 kW
8.Pumps & pump power
Answer: H = 8.733 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (8.8 + 4)·0.1354 = 1.733 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 7 + 1.733 = 8.733 m
Answer
H = 8.733 m
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