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

d = 4 mH = 1 mb = 3 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 1 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 1 mH = 1.5 mb = 1 mρ = 1025 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a water pipe reads a difference of 10 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.

p₀ − p = 10 kPaρ = 1000 kg/m³

v = m/s

4.Continuity & BernoulliHard

A hose with an inner diameter of 50 mm siphons water from an open tank. The highest point C of the hose is 1.5 m above the water surface, and the outlet (open to the air) is 3 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 = 50 mmz_C = +1.5 mz₂ = −3 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 0.9 m/s through a horizontal pipe, 800 m long, with an inside diameter of 150 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 = 0.9 m/sL = 800 mD = 150 mmf = 0.022ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 100 m long with an inside diameter of 100 mm (Darcy friction factor f = 0.015) 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 = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 15 L/s?

Q = 15 L/sL = 100 mD = 100 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2

H = m

7.Pumps & pump powerMedium

In a test, a pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 78 m while its shaft takes 49.1 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.

Q = 50 L/sH = 78 mP_shaft = 49.1 kWg = 9.81 m/s²

η = %

8.Pumps & pump powerHard

A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 17 m higher. The pipe is 100 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 4 (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 = 120 L/sΔz = 17 mL = 100 mD = 250 mmf = 0.015ΣK = 4p_tank = 100 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 132.4 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 = 4 + 1/2 = 4.5 m

A = H·b = 1·3 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.5·3 = 132.4 kN

Answer

F = 132.4 kN

2.Hydrostatic force

Answer: y_cp = 1.857 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 + 1.5/2 = 1.75 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·1.75·1.5 = 26.4 kN

Centre of pressure

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

I_c = b·H³/12 = 1·1.5³/12 = 0.2813 m⁴

y_cp = y_c + I_c/(y_c·A) = 1.75 + 0.2813/(1.75·1.5) = 1.857 m

Answer

y_cp = 1.857 m

3.Continuity & Bernoulli

Answer: v = 4.472 m/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.

Bernoulli to the stagnation point

The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.

p₀ − p = ½ρv²

v = √(2·10000/1000) = 4.472 m/s

Answer

v = 4.472 m/s

4.Continuity & Bernoulli

Answer: p_C = 57.15 kPa

Velocity in the hose

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

v = √(2g·3) = √(2·9.81·3) = 7.672 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(7.672²/2 + 9.81·1.5)/1000 = 101.3 − 44.15 = 57.15 kPa

Answer

p_C = 57.15 kPa

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

5.Pipe friction & head loss

Answer: Δp = 47.52 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·(800/0.15)·0.9²/(2·9.81) = 4.844 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 = 1000·9.81·4.844 = 47520 Pa = 47.52 kPa

Answer

Δp = 47.52 kPa

6.Pipe friction & head loss

Answer: H = 3.607 m

Velocity

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

V²/(2g) = 0.1859 m

Loss coefficients

f·L/D = 0.015·100/0.1 = 15

ΣK = 0.5 + 3·0.3 + 2 = 3.4

Energy equation

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

H = (1 + 15 + 3.4)·0.1859 = 3.607 m

Answer

H = 3.607 m

7.Pumps & pump power

Answer: η = 77.92 %

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 = 1000·9.81·0.05·78 = 38.26 kW

Efficiency

η = P_h/P_shaft = 38.26/49.1 = 77.92%

Answer

η = 77.92%

8.Pumps & pump power

Answer: H = 30.24 m

Velocity

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

V²/(2g) = 0.3046 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (6 + 4)·0.3046 = 3.046 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 = 17 + 10.19 + 3.046 = 30.24 m

Answer

H = 30.24 m

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