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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 3 mb = 3 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 4 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 4 mH = 2 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal venturi meter with an inlet diameter of 80 mm and a throat diameter of 40 mm. The pressure at the inlet is 33 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 80 mmd₂ = 40 mmp₁ − p₂ = 33 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

A closed water tank has compressed air above the water at a gauge pressure of 40 kPa. Water leaves through an opening with a diameter of 40 mm, 2.5 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².

p_air = 40 kPa (gauge)h = 2.5 md = 40 mmρ = 1000 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 1.1 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 = 1.1 m/sL = 500 mD = 150 mmf = 0.018g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 0.2) 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 100 L/s?

Q = 100 L/sL = 100 mD = 200 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 0.2K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 64 m. The pump efficiency is 82% and the electric motor driving it has an efficiency of 92%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 50 L/sH = 64 mη_pump = 82%η_motor = 92%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 12 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 500 m long with an inside diameter of 100 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 300 kPa. Find the head H the pump must add.

Q = 12 L/sΔz = 8 mL = 500 mD = 100 mmf = 0.015ΣK = 8p_tank = 300 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 353.2 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 + 3/2 = 4 m

A = H·b = 3·3 = 9 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4·9 = 353.2 kN

Answer

F = 353.2 kN

2.Hydrostatic force

Answer: y_cp = 5.067 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 = 4 + 2/2 = 5 m

A = H·b = 2·2 = 4 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·5·4 = 196.2 kN

Centre of pressure

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

I_c = b·H³/12 = 2·2³/12 = 1.333 m⁴

y_cp = y_c + I_c/(y_c·A) = 5 + 1.333/(5·4) = 5.067 m

Answer

y_cp = 5.067 m

3.Continuity & Bernoulli

Answer: Q = 10.54 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·(40/80)² = 0.25·v₂

Bernoulli between inlet and throat

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

p₁ − p₂ = ½ρv₂²(1 − β⁴)

v₂ = √(2·33000/(1000·(1 − 0.0625))) = 8.39 m/s

v₁ = 2.098 m/s

Flow rate

Q = A₂·v₂ = 0.001257·8.39 = 0.01054 m³/s = 10.54 L/s

Answer

Q = 10.54 L/s

4.Continuity & Bernoulli

Answer: Q = 14.28 L/s

Bernoulli from the surface to the jet

v = √(2·(40000/1000 + 9.81·2.5)) = 11.36 m/s

Flow rate

Q = A·v = 0.001257·11.36 = 0.01428 m³/s = 14.28 L/s

Answer

Q = 14.28 L/s

5.Pipe friction & head loss

Answer: h_f = 3.7 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) = 1.1²/(2·9.81) = 0.06167 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.018·(500/0.15)·1.1²/(2·9.81) = 3.7 m

Answer

h_f = 3.7 m

6.Pipe friction & head loss

Answer: Δz = 6.352 m

Velocity

V = Q/A = 0.1/0.03142 = 3.183 m/s

V²/(2g) = 0.5164 m

Loss coefficients

f·L/D = 0.02·100/0.2 = 10

ΣK = 0.5 + 2·0.3 + 0.2 + 1.0 = 2.3

Total head loss

h_f = 10·0.5164 = 5.164 m

h_m = 2.3·0.5164 = 1.188 m

Δz = h_f + h_m = 6.352 m

Answer

Δz = 6.352 m

7.Pumps & pump power

Answer: P_el = 41.61 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 = 1000·9.81·0.05·64 = 31.39 kW

Shaft power

P_shaft = 31.39/0.82 = 38.28 kW

Electrical power

P_el = 38.28/0.92 = 41.61 kW

Answer

P_el = 41.61 kW

8.Pumps & pump power

Answer: H = 48.46 m

Velocity

V = Q/A = 0.012/0.007854 = 1.528 m/s

V²/(2g) = 0.119 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (75 + 8)·0.119 = 9.876 m

Pressure head

p/(ρ·g) = 300000/(1000·9.81) = 30.58 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 8 + 30.58 + 9.876 = 48.46 m

Answer

H = 48.46 m

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