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

d = 0.5 mH = 1.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1.5 m wide sits in the wall of a tank of seawater, with its top edge 2 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2 mH = 2 mb = 1.5 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 250 mm and a throat diameter of 125 mm. The pressure at the inlet is 50 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 250 mmd₂ = 125 mmp₁ − p₂ = 50 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

A horizontal venturi meter (inlet 300 mm, throat 150 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 0.95. 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 = 300 mmd₂ = 150 mmΔh = 230 mmC_d = 0.95ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 1.7 m/s through a pipe 500 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss h_f.

V = 1.7 m/sL = 500 mD = 150 mmf = 0.028g = 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 250 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 3 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. What difference in water level Δz between the reservoirs gives a flow rate of 100 L/s?

Q = 100 L/sL = 150 mD = 250 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 0.2K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump with an efficiency of 80% delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 14 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 60 L/sH = 14 mη = 80%g = 9.81 m/s²

P_shaft = kW

8.Pumps & pump powerHard

A pump moves 25 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 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 = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 82%. Find the shaft power P_shaft in kW.

Q = 25 L/sΔz = 10 mL = 100 mD = 150 mmf = 0.02ΣK = 6η = 82%

P_shaft = kW

3Solutions

1.Hydrostatic force

Answer: F = 55.18 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 = 1.5·3 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.25·4.5 = 55.18 kN

Answer

F = 55.18 kN

2.Hydrostatic force

Answer: y_cp = 3.111 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 = 2 + 2/2 = 3 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·3·3 = 90.5 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·2³/12 = 1 m⁴

y_cp = y_c + I_c/(y_c·A) = 3 + 1/(3·3) = 3.111 m

Answer

y_cp = 3.111 m

3.Continuity & Bernoulli

Answer: Q = 126.7 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 = √(2·50000/(1000·(1 − 0.0625))) = 10.33 m/s

v₁ = 2.582 m/s

Flow rate

Q = A₂·v₂ = 0.01227·10.33 = 0.1267 m³/s = 126.7 L/s

Answer

Q = 126.7 L/s

4.Continuity & Bernoulli

Answer: Q = 130.7 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 = 150/300 = 0.5

v₂ = √(2·28429/(1000·(1 − 0.0625))) = 7.788 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.95·0.01767·7.788 = 0.1307 m³/s = 130.7 L/s

Answer

Q = 130.7 L/s

5.Pipe friction & head loss

Answer: h_f = 13.75 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.

Velocity head

V²/(2g) = 1.7²/(2·9.81) = 0.1473 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.028·(500/0.15)·1.7²/(2·9.81) = 13.75 m

Answer

h_f = 13.75 m

6.Pipe friction & head loss

Answer: Δz = 2.834 m

Velocity

V = Q/A = 0.1/0.04909 = 2.037 m/s

V²/(2g) = 0.2115 m

Loss coefficients

f·L/D = 0.018·150/0.25 = 10.8

ΣK = 0.5 + 3·0.3 + 0.2 + 1.0 = 2.6

Total head loss

h_f = 10.8·0.2115 = 2.284 m

h_m = 2.6·0.2115 = 0.55 m

Δz = h_f + h_m = 2.834 m

Answer

Δz = 2.834 m

7.Pumps & pump power

Answer: P_shaft = 10.3 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 = 1000·9.81·0.06·14 = 8.24 kW

Shaft power

P_shaft = P_h/η = 8.24/0.8 = 10.3 kW

Answer

P_shaft = 10.3 kW

8.Pumps & pump power

Answer: P_shaft = 3.581 kW

Velocity

V = Q/A = 0.025/0.01767 = 1.415 m/s

V²/(2g) = 0.102 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (13.33 + 6)·0.102 = 1.972 m

Pump head

H = Δz + h_L = 10 + 1.972 = 11.97 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.025·11.97/0.82 = 3.581 kW

Answer

P_shaft = 3.581 kW

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