Fluid Mechanics: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Hydrostatic force

Pressure at depth: p = ρgh
Force on a vertical wall: F = ρg·h_c·A
Centre of pressure: y_p = y_c + I_c/(y_cA)
Rectangle: I_c = bH³/12

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: h_f = f·(L/D)·V²/(2g)
Minor losses: h_m = ΣK·V²/(2g)
Typical K values: entrance 0.5, exit 1.0, 90° bend ≈ 0.3
Pressure drop: Δp = ρ·g·h_f
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: P_h = ρ·g·Q·H = Q·Δp
Shaft power: P_shaft = P_h/η_pump
Electrical power: P_el = P_shaft/η_motor
Pump head: H = Δz + Δp/(ρ·g) + h_L
System losses: h_L = (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 3 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 3 mH = 1.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2.5 mH = 1.5 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 75 mm to 40 mm. In the wide section the pressure is 100 kPa and the velocity is 2.5 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 75 mmd₂ = 40 mmp₁ = 100 kPav₁ = 2.5 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 150 mm to 40 mm while rising 3 m. At section 1 the pressure is 350 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 150 mmd₂ = 40 mmp₁ = 350 kPav₁ = 0.5 m/sΔz = 3 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at 40 L/s through a pipe 100 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 40 L/sL = 100 mD = 150 mmf = 0.015g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.025. The minor losses are a square-edged entrance (K = 0.5), 4 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 120 L/s?

Q = 120 L/sL = 150 mD = 250 mmf = 0.025K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 10K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 66 m. The pump efficiency is 82% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 150 L/sH = 66 mη_pump = 82%η_motor = 95%g = 9.81 m/s²

P_el = kW

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 100 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 4 (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 200 kPa. Find the head H the pump must add.

Q = 50 L/sΔz = 5 mL = 100 mD = 150 mmf = 0.02ΣK = 4p_tank = 200 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 55.18 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 3 + 1.5/2 = 3.75 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.75·1.5 = 55.18 kN

Answer

F = 55.18 kN

2.Hydrostatic force

Answer: y_cp = 3.308 m

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 2.5 + 1.5/2 = 3.25 m

A = H·b = 1.5·2 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.25·3 = 95.65 kN

Centre of pressure

Pressure grows with depth, so the resultant acts below the centroid.

I_c = b·H³/12 = 2·1.5³/12 = 0.5625 m⁴

y_cp = y_c + I_c/(y_c·A) = 3.25 + 0.5625/(3.25·3) = 3.308 m

Answer

y_cp = 3.308 m

3.Continuity & Bernoulli

Answer: p₂ = 64.5 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 = 2.5·(75/40)² = 8.789 m/s

½ρ(v₁² − v₂²) = ½·1000·(2.5² − 8.789²)/1000 = −35.5 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 100 + (−35.5) = 64.5 kPa

Answer

p₂ = 64.5 kPa

4.Continuity & Bernoulli

Answer: p₂ = 296 kPa

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v₂ = v₁·(d₁/d₂)² = 0.5·(150/40)² = 7.031 m/s

½ρ(v₁² − v₂²) = ½·1000·(0.5² − 7.031²)/1000 = −24.59 kPa

ρgΔz = 1000·9.81·3/1000 = 29.43 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 350 + (−24.59) − 29.43 = 296 kPa

Answer

p₂ = 296 kPa

5.Pipe friction & head loss

Answer: h_f = 2.611 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 h_f = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses h_m = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = h_f + h_m, with the exit loss K = 1 included.

Mean velocity

A = π·0.15²/4 = 0.01767 m²

V = Q/A = 0.04/0.01767 = 2.264 m/s

Velocity head

V²/(2g) = 2.264²/(2·9.81) = 0.2611 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.015·(100/0.15)·2.264²/(2·9.81) = 2.611 m

Answer

h_f = 2.611 m

6.Pipe friction & head loss

Answer: Δz = 8.437 m

Velocity

V = Q/A = 0.12/0.04909 = 2.445 m/s

V²/(2g) = 0.3046 m

Loss coefficients

f·L/D = 0.025·150/0.25 = 15

ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7

Total head loss

h_f = 15·0.3046 = 4.569 m

h_m = 12.7·0.3046 = 3.868 m

Δz = h_f + h_m = 8.437 m

Answer

Δz = 8.437 m

7.Pumps & pump power

Answer: P_el = 124.7 kW

Pump power: the power given to the water is the hydraulic power P_h = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: P_shaft = P_h/η_pump, and the motor draws P_el = P_shaft/η_motor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Hydraulic power

P_h = ρ·g·Q·H = 1000·9.81·0.15·66 = 97.12 kW

Shaft power

P_shaft = 97.12/0.82 = 118.4 kW

Electrical power

P_el = 118.4/0.95 = 124.7 kW

Answer

P_el = 124.7 kW

8.Pumps & pump power

Answer: H = 32.46 m

Velocity

V = Q/A = 0.05/0.01767 = 2.829 m/s

V²/(2g) = 0.408 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (13.33 + 4)·0.408 = 7.073 m

Pressure head

p/(ρ·g) = 200000/(1000·9.81) = 20.39 m

Pump head

Both surfaces are practically at rest, so there is no velocity head term.

H = Δz + p/(ρ·g) + h_L = 5 + 20.39 + 7.073 = 32.46 m

Answer

H = 32.46 m

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