Fluid Mechanics: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

d = 0.5 mH = 1 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 100 mm to 40 mm. In the wide section the pressure is 320 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 100 mmd₂ = 40 mmp₁ = 320 kPav₁ = 1.25 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 75 mm to 25 mm while rising 1 m. At section 1 the pressure is 390 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 75 mmd₂ = 25 mmp₁ = 390 kPav₁ = 1 m/sΔz = 1 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 2.7 m/s through a horizontal pipe, 1000 m long, with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 2.7 m/sL = 1000 mD = 250 mmf = 0.022ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 150 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.025) 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 = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 120 L/s?

Q = 120 L/sL = 150 mD = 250 mmf = 0.025K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 5

H = m

7.Pumps & pump powerMedium

A pump with an efficiency of 70% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 30 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 20 L/sH = 30 mη = 70%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 19 m higher. The pipe is 400 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 65%. Find the shaft power P_shaft in kW.

Q = 40 L/sΔz = 19 mL = 400 mD = 150 mmf = 0.018ΣK = 8η = 65%

P_shaft = kW

3Solutions

1.Hydrostatic force

Answer: F = 71.74 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 + 1.5/2 = 3.25 m

A = H·b = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.25·2.25 = 71.74 kN

Answer

F = 71.74 kN

2.Hydrostatic force

Answer: y_cp = 1.083 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 = 0.5 + 1/2 = 1 m

A = H·b = 1·1 = 1 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·1·1 = 10.06 kN

Centre of pressure

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

I_c = b·H³/12 = 1·1³/12 = 0.08333 m⁴

y_cp = y_c + I_c/(y_c·A) = 1 + 0.08333/(1·1) = 1.083 m

Answer

y_cp = 1.083 m

3.Continuity & Bernoulli

Answer: p₂ = 290.3 kPa

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 and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1.25·(100/40)² = 7.813 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 7.813²)/1000 = −29.74 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 320 + (−29.74) = 290.3 kPa

Answer

p₂ = 290.3 kPa

4.Continuity & Bernoulli

Answer: p₂ = 340.2 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1·(75/25)² = 9 m/s

½ρ(v₁² − v₂²) = ½·1000·(1² − 9²)/1000 = −40 kPa

ρgΔz = 1000·9.81·1/1000 = 9.81 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 390 + (−40) − 9.81 = 340.2 kPa

Answer

p₂ = 340.2 kPa

5.Pipe friction & head loss

Answer: Δp = 320.8 kPa

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.

Head loss

h_f = f·(L/D)·V²/(2g) = 0.022·(1000/0.25)·2.7²/(2·9.81) = 32.7 m

Pressure drop

The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·h_f.

Δp = ρ·g·h_f = 1000·9.81·32.7 = 320760 Pa = 320.8 kPa

Answer

Δp = 320.8 kPa

6.Pipe friction & head loss

Answer: H = 6.823 m

Velocity

V = Q/A = 0.12/0.04909 = 2.445 m/s

V²/(2g) = 0.3046 m

Loss coefficients

f·L/D = 0.025·150/0.25 = 15

ΣK = 0.5 + 3·0.3 + 5 = 6.4

Energy equation

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

H = (1 + 15 + 6.4)·0.3046 = 6.823 m

Answer

H = 6.823 m

7.Pumps & pump power

Answer: P_shaft = 8.409 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 = 1000·9.81·0.02·30 = 5.886 kW

Shaft power

P_shaft = P_h/η = 5.886/0.7 = 8.409 kW

Answer

P_shaft = 8.409 kW

8.Pumps & pump power

Answer: P_shaft = 20.3 kW

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) = (48 + 8)·0.2611 = 14.62 m

Pump head

H = Δz + h_L = 19 + 14.62 = 33.62 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.04·33.62/0.65 = 20.3 kW

Answer

P_shaft = 20.3 kW

ExamProblems · examproblems.com/courses/fluid-mechanics