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

d = 1.5 mH = 1 mb = 1.5 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 water, 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 = 2.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water drains from a large open tank through a sharp-edged orifice with a diameter of 25 mm. The water level is kept 3.5 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².

d = 25 mmh = 3.5 mC_d = 0.62g = 9.81 m/s²

Q = L/s

4.Continuity & BernoulliHard

A horizontal venturi meter (inlet 200 mm, throat 150 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 190 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 = 200 mmd₂ = 150 mmΔh = 190 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, 400 m long, with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1 m/sL = 400 mD = 75 mmf = 0.02ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 5) 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 = 200 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 5K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump with an efficiency of 60% delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 50 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 100 L/sH = 50 mη = 60%g = 9.81 m/s²

P_shaft = kW

8.Pumps & pump powerHard

A pump moves 10 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 100 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 = 5 (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 350 kPa. Find the head H the pump must add.

Q = 10 L/sΔz = 5 mL = 100 mD = 100 mmf = 0.018ΣK = 5p_tank = 350 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 29.43 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 = 1.5 + 1/2 = 2 m

A = H·b = 1·1.5 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2·1.5 = 29.43 kN

Answer

F = 29.43 kN

2.Hydrostatic force

Answer: y_cp = 5.349 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 + 2.5/2 = 5.25 m

A = H·b = 2.5·3 = 7.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·5.25·7.5 = 386.3 kN

Centre of pressure

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

I_c = b·H³/12 = 3·2.5³/12 = 3.906 m⁴

y_cp = y_c + I_c/(y_c·A) = 5.25 + 3.906/(5.25·7.5) = 5.349 m

Answer

y_cp = 5.349 m

3.Continuity & Bernoulli

Answer: Q = 2.522 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.

Ideal jet velocity

√(2gh) = √(2·9.81·3.5) = 8.287 m/s

Actual flow rate

The discharge coefficient accounts for the contraction of the jet and friction at the edge.

Q = C_d·A·√(2gh) = 0.62·0.0004909·8.287 = 0.002522 m³/s = 2.522 L/s

Answer

Q = 2.522 L/s

4.Continuity & Bernoulli

Answer: Q = 142.1 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.19 = 23485 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 150/200 = 0.75

v₂ = √(2·23485/(1000·(1 − 0.3164))) = 8.289 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.97·0.01767·8.289 = 0.1421 m³/s = 142.1 L/s

Answer

Q = 142.1 L/s

5.Pipe friction & head loss

Answer: Δp = 53.33 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) = 0.02·(400/0.075)·1²/(2·9.81) = 5.437 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 = 1000·9.81·5.437 = 53333 Pa = 53.33 kPa

Answer

Δp = 53.33 kPa

6.Pipe friction & head loss

Answer: Δz = 9.786 m

Velocity

V = Q/A = 0.1/0.03142 = 3.183 m/s

V²/(2g) = 0.5164 m

Loss coefficients

f·L/D = 0.015·150/0.2 = 11.25

ΣK = 0.5 + 4·0.3 + 5 + 1.0 = 7.7

Total head loss

h_f = 11.25·0.5164 = 5.81 m

h_m = 7.7·0.5164 = 3.976 m

Δz = h_f + h_m = 9.786 m

Answer

Δz = 9.786 m

7.Pumps & pump power

Answer: P_shaft = 81.75 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 = 1000·9.81·0.1·50 = 49.05 kW

Shaft power

P_shaft = P_h/η = 49.05/0.6 = 81.75 kW

Answer

P_shaft = 81.75 kW

8.Pumps & pump power

Answer: H = 42.58 m

Velocity

V = Q/A = 0.01/0.007854 = 1.273 m/s

V²/(2g) = 0.08263 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (18 + 5)·0.08263 = 1.9 m

Pressure head

p/(ρ·g) = 350000/(1000·9.81) = 35.68 m

Pump head

Both surfaces are practically at rest, so there is no velocity head term.

H = Δz + p/(ρ·g) + h_L = 5 + 35.68 + 1.9 = 42.58 m

Answer

H = 42.58 m

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