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 1.5 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 2 mb = 1.5 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 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2 mH = 2 mb = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 60 mm. The pressure at the inlet is 43 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 100 mmd₂ = 60 mmp₁ − p₂ = 43 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 100 mm and a gauge pressure of 150 kPa. The nozzle exit diameter is 40 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².

d₁ = 100 mmp₁ = 150 kPa (gauge)d₂ = 40 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 1.5 m/s through a pipe 500 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss h_f.

V = 1.5 m/sL = 500 mD = 150 mmf = 0.018g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 300 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), 3 bends (K = 0.3 each) and a valve (K = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 75 L/s?

Q = 75 L/sL = 300 mD = 300 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2

H = m

7.Pumps & pump powerMedium

A pump delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 26 m. The pump efficiency is 85% 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 = 40 L/sH = 26 mη_pump = 85%η_motor = 95%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 15 m higher. The pipe is 400 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 = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 72%. Find the shaft power P_shaft in kW.

Q = 120 L/sΔz = 15 mL = 400 mD = 250 mmf = 0.018ΣK = 6η = 72%

P_shaft = kW

3Solutions

1.Hydrostatic force

Answer: F = 103 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 = 2.5 + 2/2 = 3.5 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.5·3 = 103 kN

Answer

F = 103 kN

2.Hydrostatic force

Answer: y_cp = 3.111 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 + 2/2 = 3 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·3·3 = 90.5 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·2³/12 = 1 m⁴

y_cp = y_c + I_c/(y_c·A) = 3 + 1/(3·3) = 3.111 m

Answer

y_cp = 3.111 m

3.Continuity & Bernoulli

Answer: Q = 28.1 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.

Continuity

v₁ = v₂·(60/100)² = 0.36·v₂

Bernoulli between inlet and throat

The pipe is horizontal, so only pressure and velocity change.

p₁ − p₂ = ½ρv₂²(1 − β⁴)

v₂ = √(2·43000/(1000·(1 − 0.1296))) = 9.94 m/s

v₁ = 3.578 m/s

Flow rate

Q = A₂·v₂ = 0.002827·9.94 = 0.0281 m³/s = 28.1 L/s

Answer

Q = 28.1 L/s

4.Continuity & Bernoulli

Answer: H = 15.69 m

Exit velocity

β = 40/100 = 0.4

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·150000/(1000·0.9744)) = 17.55 m/s

Height of the jet

Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.

H = v₂²/(2g) = 17.55²/(2·9.81) = 15.69 m

Answer

H = 15.69 m

5.Pipe friction & head loss

Answer: h_f = 6.881 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.

Velocity head

V²/(2g) = 1.5²/(2·9.81) = 0.1147 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.018·(500/0.15)·1.5²/(2·9.81) = 6.881 m

Answer

h_f = 6.881 m

6.Pipe friction & head loss

Answer: H = 1.285 m

Velocity

V = Q/A = 0.075/0.07069 = 1.061 m/s

V²/(2g) = 0.05738 m

Loss coefficients

f·L/D = 0.018·300/0.3 = 18

ΣK = 0.5 + 3·0.3 + 2 = 3.4

Energy equation

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

H = (1 + 18 + 3.4)·0.05738 = 1.285 m

Answer

H = 1.285 m

7.Pumps & pump power

Answer: P_el = 12.63 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 = 1000·9.81·0.04·26 = 10.2 kW

Shaft power

P_shaft = 10.2/0.85 = 12 kW

Electrical power

P_el = 12/0.95 = 12.63 kW

Answer

P_el = 12.63 kW

8.Pumps & pump power

Answer: P_shaft = 41.86 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) = (28.8 + 6)·0.3046 = 10.6 m

Pump head

H = Δz + h_L = 15 + 10.6 = 25.6 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.12·25.6/0.72 = 41.86 kW

Answer

P_shaft = 41.86 kW

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