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 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1.5 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 water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 80 mm and a throat diameter of 40 mm. The pressure at the inlet is 52 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 80 mmd2 = 40 mmp1 − p2 = 52 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 100 kPa. Water leaves through an opening with a diameter of 50 mm, 4.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
pair = 100 kPa (gauge)h = 4.5 md = 50 mmρ = 1000 kg/m³g = 9.81 m/s²
v = m/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.3 m/s through a horizontal pipe, 1000 m long, with an inside diameter of 250 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 = 1.3 m/sL = 1000 mD = 250 mmf = 0.018ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 25 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 2 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 = 25 mL = 150 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 40 m while its shaft takes 14.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 30 L/sH = 40 mPshaft = 14.4 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 150 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 = 2.5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 75 L/sΔz = 8 mL = 150 mD = 200 mmf = 0.018ΣK = 2.5
H = m
3Solutions
1.Hydrostatic force
Answer: F = 77.25 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 + 1.5/2 = 1.75 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·4.5 = 77.25 kN
Answer
F = 77.25 kN
2.Hydrostatic force
Answer: F = 79.71 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 + 2.5/2 = 3.25 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·2.5 = 79.71 kN
Answer
F = 79.71 kN
3.Continuity & Bernoulli
Answer: Q = 13.24 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·(40/80)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 = √(2·52000/(1000·(1 − 0.0625))) = 10.53 m/s
v1 = 2.633 m/s
Flow rate
Q = A2·v2 = 0.001257·10.53 = 0.01324 m³/s = 13.24 L/s
Answer
Q = 13.24 L/s
4.Continuity & Bernoulli
Answer: v = 16.98 m/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·(100000/1000 + 9.81·4.5)) = 16.98 m/s
Answer
v = 16.98 m/s
5.Pipe friction & head loss
Answer: Δp = 60.84 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·(1000/0.25)·1.3²/(2·9.81) = 6.202 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 = 1000·9.81·6.202 = 60840 Pa = 60.84 kPa
Answer
Δp = 60.84 kPa
6.Pipe friction & head loss
Answer: Q = 416.9 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 = 0.02·150/0.3 = 10
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
25 = (10 + 4.1)·V²/(2·9.81)
V = √(2·9.81·25/14.1) = 5.898 m/s
Flow rate
Q = V·π·D²/4 = 5.898·0.07069 = 0.4169 m³/s = 416.9 L/s
Answer
Q = 416.9 L/s
7.Pumps & pump power
Answer: η = 81.75 %
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 = 1000·9.81·0.03·40 = 11.77 kW
Efficiency
η = Ph/Pshaft = 11.77/14.4 = 81.75%
Answer
η = 81.75%
8.Pumps & pump power
Answer: H = 12.65 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (13.5 + 2.5)·0.2905 = 4.648 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 8 + 4.648 = 12.65 m
Answer
H = 12.65 m
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