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 = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1 m wide sits in the wall of a tank of glycerin, 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 = 1.5 mb = 1 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a wind tunnel reads a difference of 100 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.

p₀ − p = 100 Paρ = 1.2 kg/m³

v = m/s

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 100 mm to 40 mm while rising 9 m. At section 1 the pressure is 390 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 100 mmd₂ = 40 mmp₁ = 390 kPav₁ = 1.5 m/sΔz = 9 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at 3 L/s through a pipe 150 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 3 L/sL = 150 mD = 50 mmf = 0.015g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 400 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 2 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 120 L/s?

Q = 120 L/sL = 400 mD = 300 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 2K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump with an efficiency of 65% delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 58 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 25 L/sH = 58 mη = 65%g = 9.81 m/s²

P_shaft = kW

8.Pumps & pump powerHard

A pump moves 40 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 20 m higher. The pipe is 100 m long with an inside diameter of 150 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 upper tank is closed and the air above its water is at a gauge pressure of 100 kPa. Find the head H the pump must add.

Q = 40 L/sΔz = 20 mL = 100 mD = 150 mmf = 0.02ΣK = 8p_tank = 100 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 206 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 = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.5·6 = 206 kN

Answer

F = 206 kN

2.Hydrostatic force

Answer: y_cp = 1.857 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 + 1.5/2 = 1.75 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·1.75·1.5 = 32.45 kN

Centre of pressure

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

I_c = b·H³/12 = 1·1.5³/12 = 0.2813 m⁴

y_cp = y_c + I_c/(y_c·A) = 1.75 + 0.2813/(1.75·1.5) = 1.857 m

Answer

y_cp = 1.857 m

3.Continuity & Bernoulli

Answer: v = 12.91 m/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.

Bernoulli to the stagnation point

The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.

p₀ − p = ½ρv²

v = √(2·100/1.2) = 12.91 m/s

Answer

v = 12.91 m/s

4.Continuity & Bernoulli

Answer: p₂ = 258.9 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1.5·(100/40)² = 9.375 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.5² − 9.375²)/1000 = −42.82 kPa

ρgΔz = 1000·9.81·9/1000 = 88.29 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 390 + (−42.82) − 88.29 = 258.9 kPa

Answer

p₂ = 258.9 kPa

5.Pipe friction & head loss

Answer: h_f = 5.354 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·0.05²/4 = 0.001963 m²

V = Q/A = 0.003/0.001963 = 1.528 m/s

Velocity head

V²/(2g) = 1.528²/(2·9.81) = 0.119 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.015·(150/0.05)·1.528²/(2·9.81) = 5.354 m

Answer

h_f = 5.354 m

6.Pipe friction & head loss

Answer: Δz = 3.54 m

Velocity

V = Q/A = 0.12/0.07069 = 1.698 m/s

V²/(2g) = 0.1469 m

Loss coefficients

f·L/D = 0.015·400/0.3 = 20

ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1

Total head loss

h_f = 20·0.1469 = 2.938 m

h_m = 4.1·0.1469 = 0.6023 m

Δz = h_f + h_m = 3.54 m

Answer

Δz = 3.54 m

7.Pumps & pump power

Answer: P_shaft = 21.88 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 = 1000·9.81·0.025·58 = 14.22 kW

Shaft power

P_shaft = P_h/η = 14.22/0.65 = 21.88 kW

Answer

P_shaft = 21.88 kW

8.Pumps & pump power

Answer: H = 35.76 m

Velocity

V = Q/A = 0.04/0.01767 = 2.264 m/s

V²/(2g) = 0.2611 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (13.33 + 8)·0.2611 = 5.571 m

Pressure head

p/(ρ·g) = 100000/(1000·9.81) = 10.19 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 20 + 10.19 + 5.571 = 35.76 m

Answer

H = 35.76 m

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