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

d = 3 mH = 2 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 4 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 4 mH = 2 mb = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 370 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 150 mmd₂ = 75 mmp₁ = 370 kPav₁ = 1.25 m/sρ = 1000 kg/m³

p₂ = kPa

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 340 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 = 340 mmC_d = 0.97ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1 m/s through a horizontal pipe, 500 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1 m/sL = 500 mD = 100 mmf = 0.028ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 4 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 40 L/s?

Q = 40 L/sL = 200 mD = 150 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 2K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump with an efficiency of 82% delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 48 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 50 L/sH = 48 mη = 82%g = 9.81 m/s²

P_shaft = 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 10 m higher. The pipe is 200 m long with an inside diameter of 100 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 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 = 12 L/sΔz = 10 mL = 200 mD = 100 mmf = 0.018ΣK = 8p_tank = 100 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 235.4 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 + 2/2 = 4 m

A = H·b = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4·6 = 235.4 kN

Answer

F = 235.4 kN

2.Hydrostatic force

Answer: y_cp = 5.067 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 = 4 + 2/2 = 5 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·5·3 = 150.8 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) = 5 + 1/(5·3) = 5.067 m

Answer

y_cp = 5.067 m

3.Continuity & Bernoulli

Answer: p₂ = 358.3 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·(150/75)² = 5 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 5²)/1000 = −11.72 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 370 + (−11.72) = 358.3 kPa

Answer

p₂ = 358.3 kPa

4.Continuity & Bernoulli

Answer: Q = 26.95 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.34 = 42026 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 60/100 = 0.6

v₂ = √(2·42026/(1000·(1 − 0.1296))) = 9.827 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.97·0.002827·9.827 = 0.02695 m³/s = 26.95 L/s

Answer

Q = 26.95 L/s

5.Pipe friction & head loss

Answer: Δp = 70 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) = 0.028·(500/0.1)·1²/(2·9.81) = 7.136 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 = 1000·9.81·7.136 = 70000 Pa = 70 kPa

Answer

Δp = 70 kPa

6.Pipe friction & head loss

Answer: Δz = 7.495 m

Velocity

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

V²/(2g) = 0.2611 m

Loss coefficients

f·L/D = 0.018·200/0.15 = 24

ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7

Total head loss

h_f = 24·0.2611 = 6.267 m

h_m = 4.7·0.2611 = 1.227 m

Δz = h_f + h_m = 7.495 m

Answer

Δz = 7.495 m

7.Pumps & pump power

Answer: P_shaft = 28.71 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 = 1000·9.81·0.05·48 = 23.54 kW

Shaft power

P_shaft = P_h/η = 23.54/0.82 = 28.71 kW

Answer

P_shaft = 28.71 kW

8.Pumps & pump power

Answer: H = 25.43 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) = (36 + 8)·0.119 = 5.235 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 = 10 + 10.19 + 5.235 = 25.43 m

Answer

H = 25.43 m

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