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

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

d = 1.5 mH = 1 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 250 mm and a throat diameter of 125 mm. The pressure at the inlet is 10 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 250 mmd₂ = 125 mmp₁ − p₂ = 10 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

A hose with an inner diameter of 50 mm siphons water from an open tank. The highest point C of the hose is 1.5 m above the water surface, and the outlet (open to the air) is 4.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.

d = 50 mmz_C = +1.5 mz₂ = −4.5 mg = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at 40 L/s through a pipe 150 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 40 L/sL = 150 mD = 300 mmf = 0.028g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 200 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 = 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 50 L/s?

Q = 50 L/sL = 200 mD = 200 mmf = 0.025K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 2K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 38 m. The pump efficiency is 78% 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 = 25 L/sH = 38 mη_pump = 78%η_motor = 92%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 20 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 10 m higher. The pipe is 500 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 80%. Find the shaft power P_shaft in kW.

Q = 20 L/sΔz = 10 mL = 500 mD = 100 mmf = 0.02ΣK = 8η = 80%

P_shaft = kW

3Solutions

1.Hydrostatic force

Answer: F = 40.47 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 + 1.5/2 = 2.75 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.75·1.5 = 40.47 kN

Answer

F = 40.47 kN

2.Hydrostatic force

Answer: y_cp = 2.042 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 = 1.5 + 1/2 = 2 m

A = H·b = 1·2 = 2 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2·2 = 39.24 kN

Centre of pressure

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

I_c = b·H³/12 = 2·1³/12 = 0.1667 m⁴

y_cp = y_c + I_c/(y_c·A) = 2 + 0.1667/(2·2) = 2.042 m

Answer

y_cp = 2.042 m

3.Continuity & Bernoulli

Answer: Q = 56.68 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·(125/250)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 = √(2·10000/(1000·(1 − 0.0625))) = 4.619 m/s

v₁ = 1.155 m/s

Flow rate

Q = A₂·v₂ = 0.01227·4.619 = 0.05668 m³/s = 56.68 L/s

Answer

Q = 56.68 L/s

4.Continuity & Bernoulli

Answer: Q = 18.45 L/s

Velocity in the hose

Surface and outlet are both open to the air, so the pressure terms cancel.

v = √(2g·4.5) = √(2·9.81·4.5) = 9.396 m/s

Flow rate

Q = A·v = 0.001963·9.396 = 0.01845 m³/s = 18.45 L/s

Answer

Q = 18.45 L/s

5.Pipe friction & head loss

Answer: h_f = 0.2285 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·0.3²/4 = 0.07069 m²

V = Q/A = 0.04/0.07069 = 0.5659 m/s

Velocity head

V²/(2g) = 0.5659²/(2·9.81) = 0.01632 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.028·(150/0.3)·0.5659²/(2·9.81) = 0.2285 m

Answer

h_f = 0.2285 m

6.Pipe friction & head loss

Answer: Δz = 3.834 m

Velocity

V = Q/A = 0.05/0.03142 = 1.592 m/s

V²/(2g) = 0.1291 m

Loss coefficients

f·L/D = 0.025·200/0.2 = 25

ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7

Total head loss

h_f = 25·0.1291 = 3.228 m

h_m = 4.7·0.1291 = 0.6068 m

Δz = h_f + h_m = 3.834 m

Answer

Δz = 3.834 m

7.Pumps & pump power

Answer: P_el = 12.99 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 = 1000·9.81·0.025·38 = 9.319 kW

Shaft power

P_shaft = 9.319/0.78 = 11.95 kW

Electrical power

P_el = 11.95/0.92 = 12.99 kW

Answer

P_el = 12.99 kW

8.Pumps & pump power

Answer: P_shaft = 11.21 kW

Velocity

V = Q/A = 0.02/0.007854 = 2.546 m/s

V²/(2g) = 0.3305 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (100 + 8)·0.3305 = 35.69 m

Pump head

H = Δz + h_L = 10 + 35.69 = 45.69 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.02·45.69/0.8 = 11.21 kW

Answer

P_shaft = 11.21 kW

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