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 2 m wide sits in the wall of a tank of water, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.

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

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of seawater, 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 = 3 mb = 3 mρ = 1025 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 49 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 80 mmd₂ = 40 mmp₁ − p₂ = 49 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

A horizontal venturi meter (inlet 100 mm, throat 60 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 230 mm. The discharge coefficient is 0.97. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².

d₁ = 100 mmd₂ = 60 mmΔh = 230 mmC_d = 0.97ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at 15 L/s through a pipe 800 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 15 L/sL = 800 mD = 100 mmf = 0.022g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 200 m long with an inside diameter of 150 mm (Darcy friction factor f = 0.022) 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), 4 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 60 L/s?

Q = 60 L/sL = 200 mD = 150 mmf = 0.022K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 5

H = m

7.Pumps & pump powerMedium

A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 58 m. The pump efficiency is 60% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 25 L/sH = 58 mη_pump = 60%η_motor = 94%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 15 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 200 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 = 3 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 15 L/sΔz = 5 mL = 200 mD = 100 mmf = 0.015ΣK = 3

H = m

3Solutions

1.Hydrostatic force

Answer: F = 264.9 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 = 3 + 3/2 = 4.5 m

A = H·b = 3·2 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.5·6 = 264.9 kN

Answer

F = 264.9 kN

2.Hydrostatic force

Answer: y_cp = 2.8 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 + 3/2 = 2.5 m

A = H·b = 3·3 = 9 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·2.5·9 = 226.2 kN

Centre of pressure

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

I_c = b·H³/12 = 3·3³/12 = 6.75 m⁴

y_cp = y_c + I_c/(y_c·A) = 2.5 + 6.75/(2.5·9) = 2.8 m

Answer

y_cp = 2.8 m

3.Continuity & Bernoulli

Answer: Q = 12.85 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 = √(2·49000/(1000·(1 − 0.0625))) = 10.22 m/s

v₁ = 2.556 m/s

Flow rate

Q = A₂·v₂ = 0.001257·10.22 = 0.01285 m³/s = 12.85 L/s

Answer

Q = 12.85 L/s

4.Continuity & Bernoulli

Answer: Q = 22.17 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.23 = 28429 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 60/100 = 0.6

v₂ = √(2·28429/(1000·(1 − 0.1296))) = 8.082 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.97·0.002827·8.082 = 0.02217 m³/s = 22.17 L/s

Answer

Q = 22.17 L/s

5.Pipe friction & head loss

Answer: h_f = 32.72 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·0.1²/4 = 0.007854 m²

V = Q/A = 0.015/0.007854 = 1.91 m/s

Velocity head

V²/(2g) = 1.91²/(2·9.81) = 0.1859 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.022·(800/0.1)·1.91²/(2·9.81) = 32.72 m

Answer

h_f = 32.72 m

6.Pipe friction & head loss

Answer: H = 21.76 m

Velocity

V = Q/A = 0.06/0.01767 = 3.395 m/s

V²/(2g) = 0.5876 m

Loss coefficients

f·L/D = 0.022·200/0.15 = 29.33

ΣK = 0.5 + 4·0.3 + 5 = 6.7

Energy equation

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

H = (1 + 29.33 + 6.7)·0.5876 = 21.76 m

Answer

H = 21.76 m

7.Pumps & pump power

Answer: P_el = 25.22 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 = 14.22/0.6 = 23.71 kW

Electrical power

P_el = 23.71/0.94 = 25.22 kW

Answer

P_el = 25.22 kW

8.Pumps & pump power

Answer: H = 11.14 m

Velocity

V = Q/A = 0.015/0.007854 = 1.91 m/s

V²/(2g) = 0.1859 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (30 + 3)·0.1859 = 6.135 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + h_L = 5 + 6.135 = 11.14 m

Answer

H = 11.14 m

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