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 1.5 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 1 mb = 1.5 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 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 mρ = 880 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 75 mm. In the wide section the pressure is 290 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 75 mmp1 = 290 kPav1 = 1 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 160 kPa. Water leaves through an opening with a diameter of 40 mm, 1 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
pair = 160 kPa (gauge)h = 1 md = 40 mmρ = 1000 kg/m³g = 9.81 m/s²
v = m/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.3 m/s through a horizontal pipe, 100 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.3 m/sL = 100 mD = 200 mmf = 0.03ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 400 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 30 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 = 30 mL = 400 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 80% delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 54 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 25 L/sH = 54 mη = 80%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 6 m higher. The pipe is 600 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018; 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 = 120 L/sΔz = 6 mL = 600 mD = 250 mmf = 0.018ΣK = 4
H = m
3Solutions
1.Hydrostatic force
Answer: F = 66.22 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 = 4 + 1/2 = 4.5 m
A = H·b = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.5·1.5 = 66.22 kN
Answer
F = 66.22 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 = 1 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·2·1 = 17.27 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·1³/12 = 0.08333 m⁴
ycp = yc + Ic/(yc·A) = 2 + 0.08333/(2·1) = 2.042 m
Answer
ycp = 2.042 m
3.Continuity & Bernoulli
Answer: p2 = 288.9 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.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 1·(100/75)² = 1.778 m/s
½ρ(v1² − v2²) = ½·1000·(1² − 1.778²)/1000 = −1.08 kPa
p2 = p1 + ½ρ(v1² − v2²) = 290 + (−1.08) = 288.9 kPa
Answer
p2 = 288.9 kPa
4.Continuity & Bernoulli
Answer: v = 18.43 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·(160000/1000 + 9.81·1)) = 18.43 m/s
Answer
v = 18.43 m/s
5.Pipe friction & head loss
Answer: Δp = 39.67 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) = 0.03·(100/0.2)·2.3²/(2·9.81) = 4.044 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 = 1000·9.81·4.044 = 39675 Pa = 39.67 kPa
Answer
Δp = 39.67 kPa
6.Pipe friction & head loss
Answer: Q = 309.2 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 = 0.02·400/0.3 = 26.67
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.
30 = (26.67 + 4.1)·V²/(2·9.81)
V = √(2·9.81·30/30.77) = 4.374 m/s
Flow rate
Q = V·π·D²/4 = 4.374·0.07069 = 0.3092 m³/s = 309.2 L/s
Answer
Q = 309.2 L/s
7.Pumps & pump power
Answer: Pshaft = 16.55 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 = 1000·9.81·0.025·54 = 13.24 kW
Shaft power
Pshaft = Ph/η = 13.24/0.8 = 16.55 kW
Answer
Pshaft = 16.55 kW
8.Pumps & pump power
Answer: H = 20.38 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (43.2 + 4)·0.3046 = 14.38 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 6 + 14.38 = 20.38 m
Answer
H = 20.38 m
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