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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 1.5 mb = 1 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 glycerin, 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 = 2.5 mb = 1 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 20 mm. The water level is kept 6 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 20 mmh = 6 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 40 mm while rising 3 m. At section 1 the pressure is 220 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 40 mmp1 = 220 kPav1 = 2.25 m/sΔz = 3 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.7 m/s through a pipe 150 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.7 m/sL = 150 mD = 50 mmf = 0.015g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 75 L/s?
Q = 75 L/sL = 200 mD = 200 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 60% delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 22 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 150 L/sH = 22 mη = 60%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 100 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 9 m higher. The pipe is 500 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The upper tank is closed and the air above its water is at a gauge pressure of 100 kPa. Find the head H the pump must add.
Q = 100 L/sΔz = 9 mL = 500 mD = 250 mmf = 0.015ΣK = 8ptank = 100 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 33.11 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 = 1.5 + 1.5/2 = 2.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·1.5 = 33.11 kN
Answer
F = 33.11 kN
2.Hydrostatic force
Answer: ycp = 2.939 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 + 2.5/2 = 2.75 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·2.75·2.5 = 84.98 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·2.5³/12 = 1.302 m⁴
ycp = yc + Ic/(yc·A) = 2.75 + 1.302/(2.75·2.5) = 2.939 m
Answer
ycp = 2.939 m
3.Continuity & Bernoulli
Answer: Q = 2.216 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) = √(2·9.81·6) = 10.85 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.0003142·10.85 = 0.002216 m³/s = 2.216 L/s
Answer
Q = 2.216 L/s
4.Continuity & Bernoulli
Answer: p2 = 94.22 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 = 2.25·(100/40)² = 14.06 m/s
½ρ(v1² − v2²) = ½·1000·(2.25² − 14.06²)/1000 = −96.35 kPa
ρgΔz = 1000·9.81·3/1000 = 29.43 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 220 + (−96.35) − 29.43 = 94.22 kPa
Answer
p2 = 94.22 kPa
5.Pipe friction & head loss
Answer: hf = 16.72 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) = 0.015·(150/0.05)·2.7²/(2·9.81) = 16.72 m
Answer
hf = 16.72 m
6.Pipe friction & head loss
Answer: Δz = 7.959 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Loss coefficients
f·L/D = 0.015·200/0.2 = 15
ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4
Total head loss
hf = 15·0.2905 = 4.357 m
hm = 12.4·0.2905 = 3.602 m
Δz = hf + hm = 7.959 m
Answer
Δz = 7.959 m
7.Pumps & pump power
Answer: Pshaft = 53.95 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 = 1000·9.81·0.15·22 = 32.37 kW
Shaft power
Pshaft = Ph/η = 32.37/0.6 = 53.95 kW
Answer
Pshaft = 53.95 kW
8.Pumps & pump power
Answer: H = 27.23 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (30 + 8)·0.2115 = 8.038 m
Pressure head
p/(ρ·g) = 100000/(1000·9.81) = 10.19 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 9 + 10.19 + 8.038 = 27.23 m
Answer
H = 27.23 m
ExamProblems · examproblems.com/courses/fluid-mechanics