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 1.5 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 1 mH = 2.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

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

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

F = kN

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 75 mm to 25 mm. In the wide section the pressure is 400 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 75 mmd₂ = 25 mmp₁ = 400 kPav₁ = 1.25 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

A hose with an inner diameter of 32 mm siphons water from an open tank. The highest point C of the hose is 3 m above the water surface, and the outlet (open to the air) is 2.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure p_C at point C. The atmospheric pressure is 101.3 kPa.

d = 32 mmz_C = +3 mz₂ = −2.5 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1.6 m/s through a horizontal pipe, 200 m long, with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1.6 m/sL = 200 mD = 250 mmf = 0.025ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 29 m above that of the lower one. 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. Find the flow rate Q in L/s.

Δz = 29 mL = 300 mD = 250 mmf = 0.022K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 70 m. The pump efficiency is 82% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 150 L/sH = 70 mη_pump = 82%η_motor = 95%g = 9.81 m/s²

P_el = 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 6 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 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 = 6 mL = 400 mD = 150 mmf = 0.018ΣK = 8p_tank = 100 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 82.77 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 = 1 + 2.5/2 = 2.25 m

A = H·b = 2.5·1.5 = 3.75 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.25·3.75 = 82.77 kN

Answer

F = 82.77 kN

2.Hydrostatic force

Answer: F = 117.7 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 + 2/2 = 3 m

A = H·b = 2·2 = 4 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3·4 = 117.7 kN

Answer

F = 117.7 kN

3.Continuity & Bernoulli

Answer: p₂ = 337.5 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·(75/25)² = 11.25 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 11.25²)/1000 = −62.5 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 400 + (−62.5) = 337.5 kPa

Answer

p₂ = 337.5 kPa

4.Continuity & Bernoulli

Answer: p_C = 47.34 kPa

Velocity in the hose

Surface and outlet are both open to the air, so the pressure terms cancel.

v = √(2g·2.5) = √(2·9.81·2.5) = 7.004 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(7.004²/2 + 9.81·3)/1000 = 101.3 − 53.96 = 47.34 kPa

Answer

p_C = 47.34 kPa

If this pressure falls to the vapour pressure of the water, the siphon stops working.

5.Pipe friction & head loss

Answer: Δp = 25.6 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) = 0.025·(200/0.25)·1.6²/(2·9.81) = 2.61 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 = 1000·9.81·2.61 = 25600 Pa = 25.6 kPa

Answer

Δp = 25.6 kPa

6.Pipe friction & head loss

Answer: Q = 212 L/s

Loss coefficients

f·L/D = 0.022·300/0.25 = 26.4

ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1

Energy equation

Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.

29 = (26.4 + 4.1)·V²/(2·9.81)

V = √(2·9.81·29/30.5) = 4.319 m/s

Flow rate

Q = V·π·D²/4 = 4.319·0.04909 = 0.212 m³/s = 212 L/s

Answer

Q = 212 L/s

7.Pumps & pump power

Answer: P_el = 132.2 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 = 1000·9.81·0.15·70 = 103 kW

Shaft power

P_shaft = 103/0.82 = 125.6 kW

Electrical power

P_el = 125.6/0.95 = 132.2 kW

Answer

P_el = 132.2 kW

8.Pumps & pump power

Answer: H = 30.82 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) = (48 + 8)·0.2611 = 14.62 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 = 6 + 10.19 + 14.62 = 30.82 m

Answer

H = 30.82 m

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