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 3 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 = 3 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 3 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 3 mH = 2 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²

y_cp = 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 390 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 100 mmd₂ = 75 mmp₁ = 390 kPav₁ = 1.25 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 150 mm to 50 mm while rising 3 m. At section 1 the pressure is 360 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 150 mmd₂ = 50 mmp₁ = 360 kPav₁ = 0.5 m/sΔz = 3 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 2.3 m/s through a horizontal pipe, 500 m long, with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 2.3 m/sL = 500 mD = 150 mmf = 0.025ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018. The water surface of the upper reservoir is 40 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 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 = 40 mL = 300 mD = 200 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 72 m. The pump efficiency is 80% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 25 L/sH = 72 mη_pump = 80%η_motor = 90%g = 9.81 m/s²

P_el = 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 18 m higher. The pipe is 150 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 = 5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 75%. Find the shaft power P_shaft in kW.

Q = 120 L/sΔz = 18 mL = 150 mD = 250 mmf = 0.015ΣK = 5η = 75%

P_shaft = kW

3Solutions

1.Hydrostatic force

Answer: F = 132.4 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 = 4 + 1/2 = 4.5 m

A = H·b = 1·3 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.5·3 = 132.4 kN

Answer

F = 132.4 kN

2.Hydrostatic force

Answer: y_cp = 4.083 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 = 3 + 2/2 = 4 m

A = H·b = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·4·6 = 207.2 kN

Centre of pressure

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

I_c = b·H³/12 = 3·2³/12 = 2 m⁴

y_cp = y_c + I_c/(y_c·A) = 4 + 2/(4·6) = 4.083 m

Answer

y_cp = 4.083 m

3.Continuity & Bernoulli

Answer: p₂ = 388.3 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 = 1.25·(100/75)² = 2.222 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 2.222²)/1000 = −1.688 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 390 + (−1.688) = 388.3 kPa

Answer

p₂ = 388.3 kPa

4.Continuity & Bernoulli

Answer: p₂ = 320.6 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 0.5·(150/50)² = 4.5 m/s

½ρ(v₁² − v₂²) = ½·1000·(0.5² − 4.5²)/1000 = −10 kPa

ρgΔz = 1000·9.81·3/1000 = 29.43 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 360 + (−10) − 29.43 = 320.6 kPa

Answer

p₂ = 320.6 kPa

5.Pipe friction & head loss

Answer: Δp = 220.4 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 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.

Head loss

h_f = f·(L/D)·V²/(2g) = 0.025·(500/0.15)·2.3²/(2·9.81) = 22.47 m

Pressure drop

The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·h_f.

Δp = ρ·g·h_f = 1000·9.81·22.47 = 220417 Pa = 220.4 kPa

Answer

Δp = 220.4 kPa

6.Pipe friction & head loss

Answer: Q = 157.1 L/s

Loss coefficients

f·L/D = 0.018·300/0.2 = 27

ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4

Energy equation

Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.

40 = (27 + 4.4)·V²/(2·9.81)

V = √(2·9.81·40/31.4) = 4.999 m/s

Flow rate

Q = V·π·D²/4 = 4.999·0.03142 = 0.1571 m³/s = 157.1 L/s

Answer

Q = 157.1 L/s

7.Pumps & pump power

Answer: P_el = 24.53 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 = 1000·9.81·0.025·72 = 17.66 kW

Shaft power

P_shaft = 17.66/0.8 = 22.07 kW

Electrical power

P_el = 22.07/0.9 = 24.53 kW

Answer

P_el = 24.53 kW

8.Pumps & pump power

Answer: P_shaft = 34.95 kW

Velocity

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

V²/(2g) = 0.3046 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (9 + 5)·0.3046 = 4.264 m

Pump head

H = Δz + h_L = 18 + 4.264 = 22.26 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.12·22.26/0.75 = 34.95 kW

Answer

P_shaft = 34.95 kW

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