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

d = 0.5 mH = 1.5 mb = 1.5 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 oil, with its top edge 1 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 1 mH = 2.5 mb = 2 mρ = 880 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 56 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 250 mmd₂ = 125 mmp₁ − p₂ = 56 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 120 kPa. The nozzle exit diameter is 25 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 = 120 kPa (gauge)d₂ = 25 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water flows at 25 L/s through a pipe 800 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 25 L/sL = 800 mD = 200 mmf = 0.03g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 9 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 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. Find the flow rate Q in L/s.

Δz = 9 mL = 150 mD = 300 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 0.2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 28 m. The pump efficiency is 75% and the electric motor driving it has an efficiency of 88%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 100 L/sH = 28 mη_pump = 75%η_motor = 88%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 6 m higher. The pipe is 300 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 = 6 (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 200 kPa. Find the head H the pump must add.

Q = 12 L/sΔz = 6 mL = 300 mD = 100 mmf = 0.015ΣK = 6p_tank = 200 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 27.59 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 = 0.5 + 1.5/2 = 1.25 m

A = H·b = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.25·2.25 = 27.59 kN

Answer

F = 27.59 kN

2.Hydrostatic force

Answer: y_cp = 2.481 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 + 2.5/2 = 2.25 m

A = H·b = 2.5·2 = 5 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·2.25·5 = 97.12 kN

Centre of pressure

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

I_c = b·H³/12 = 2·2.5³/12 = 2.604 m⁴

y_cp = y_c + I_c/(y_c·A) = 2.25 + 2.604/(2.25·5) = 2.481 m

Answer

y_cp = 2.481 m

3.Continuity & Bernoulli

Answer: Q = 134.1 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·(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 = √(2·56000/(1000·(1 − 0.0625))) = 10.93 m/s

v₁ = 2.733 m/s

Flow rate

Q = A₂·v₂ = 0.01227·10.93 = 0.1341 m³/s = 134.1 L/s

Answer

Q = 134.1 L/s

4.Continuity & Bernoulli

Answer: H = 12.28 m

Exit velocity

β = 25/100 = 0.25

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·120000/(1000·0.9961)) = 15.52 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) = 15.52²/(2·9.81) = 12.28 m

Answer

H = 12.28 m

5.Pipe friction & head loss

Answer: h_f = 3.873 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·0.2²/4 = 0.03142 m²

V = Q/A = 0.025/0.03142 = 0.7958 m/s

Velocity head

V²/(2g) = 0.7958²/(2·9.81) = 0.03228 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.03·(800/0.2)·0.7958²/(2·9.81) = 3.873 m

Answer

h_f = 3.873 m

6.Pipe friction & head loss

Answer: Q = 261.5 L/s

Loss coefficients

f·L/D = 0.02·150/0.3 = 10

ΣK = 0.5 + 4·0.3 + 0.2 + 1.0 = 2.9

Energy equation

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

9 = (10 + 2.9)·V²/(2·9.81)

V = √(2·9.81·9/12.9) = 3.7 m/s

Flow rate

Q = V·π·D²/4 = 3.7·0.07069 = 0.2615 m³/s = 261.5 L/s

Answer

Q = 261.5 L/s

7.Pumps & pump power

Answer: P_el = 41.62 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 = 1000·9.81·0.1·28 = 27.47 kW

Shaft power

P_shaft = 27.47/0.75 = 36.62 kW

Electrical power

P_el = 36.62/0.88 = 41.62 kW

Answer

P_el = 41.62 kW

8.Pumps & pump power

Answer: H = 32.46 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) = (45 + 6)·0.119 = 6.068 m

Pressure head

p/(ρ·g) = 200000/(1000·9.81) = 20.39 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 6 + 20.39 + 6.068 = 32.46 m

Answer

H = 32.46 m

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