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 2 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 = 2 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1.5 m wide sits in the wall of a tank of seawater, 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 = 3 mb = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows at 33 L/s through a horizontal pipe that narrows from 200 mm to 100 mm. The pressure in the narrow section is 290 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 33 L/sd₁ = 200 mmd₂ = 100 mmp₂ = 290 kPaρ = 1000 kg/m³

p₁ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 75 mm to 25 mm while rising 7 m. At section 1 the pressure is 260 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 75 mmd₂ = 25 mmp₁ = 260 kPav₁ = 1.25 m/sΔz = 7 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1 m/s through a horizontal pipe, 500 m long, with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1 m/sL = 500 mD = 300 mmf = 0.015ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 500 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.018) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each) and a valve (K = 10). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 200 L/s?

Q = 200 L/sL = 500 mD = 300 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 10

H = m

7.Pumps & pump powerMedium

A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 46 m. The pump efficiency is 75% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 10 L/sH = 46 mη_pump = 75%η_motor = 94%g = 9.81 m/s²

P_el = kW

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 9 m higher. The pipe is 400 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; 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 = 75 L/sΔz = 9 mL = 400 mD = 200 mmf = 0.015ΣK = 4

H = m

3Solutions

1.Hydrostatic force

Answer: F = 78.48 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 = 1 + 2/2 = 2 m

A = H·b = 2·2 = 4 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2·4 = 78.48 kN

Answer

F = 78.48 kN

2.Hydrostatic force

Answer: y_cp = 4.188 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 + 3/2 = 4 m

A = H·b = 3·1.5 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·4·4.5 = 181 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·3³/12 = 3.375 m⁴

y_cp = y_c + I_c/(y_c·A) = 4 + 3.375/(4·4.5) = 4.188 m

Answer

y_cp = 4.188 m

3.Continuity & Bernoulli

Answer: p₁ = 298.3 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.

Velocities

v₁ = Q/A₁ = 0.033/0.03142 = 1.05 m/s

v₂ = Q/A₂ = 0.033/0.007854 = 4.202 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 290 + ½·1000·(4.202² − 1.05²)/1000 = 290 + 8.275 = 298.3 kPa

Answer

p₁ = 298.3 kPa

4.Continuity & Bernoulli

Answer: p₂ = 128.8 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1.25·(75/25)² = 11.25 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 11.25²)/1000 = −62.5 kPa

ρgΔz = 1000·9.81·7/1000 = 68.67 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 260 + (−62.5) − 68.67 = 128.8 kPa

Answer

p₂ = 128.8 kPa

5.Pipe friction & head loss

Answer: Δp = 12.5 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) = 0.015·(500/0.3)·1²/(2·9.81) = 1.274 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 = 1000·9.81·1.274 = 12500 Pa = 12.5 kPa

Answer

Δp = 12.5 kPa

6.Pipe friction & head loss

Answer: H = 17.18 m

Velocity

V = Q/A = 0.2/0.07069 = 2.829 m/s

V²/(2g) = 0.408 m

Loss coefficients

f·L/D = 0.018·500/0.3 = 30

ΣK = 0.5 + 2·0.3 + 10 = 11.1

Energy equation

The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.

H = (1 + 30 + 11.1)·0.408 = 17.18 m

Answer

H = 17.18 m

7.Pumps & pump power

Answer: P_el = 6.401 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 = 1000·9.81·0.01·46 = 4.513 kW

Shaft power

P_shaft = 4.513/0.75 = 6.017 kW

Electrical power

P_el = 6.017/0.94 = 6.401 kW

Answer

P_el = 6.401 kW

8.Pumps & pump power

Answer: H = 18.88 m

Velocity

V = Q/A = 0.075/0.03142 = 2.387 m/s

V²/(2g) = 0.2905 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (30 + 4)·0.2905 = 9.876 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + h_L = 9 + 9.876 = 18.88 m

Answer

H = 18.88 m

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