Fluid Mechanics: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

d = 3 mH = 2 mb = 1.5 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 glycerin, 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 = 1.5 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal venturi meter with an inlet diameter of 80 mm and a throat diameter of 40 mm. The pressure at the inlet is 26 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 80 mmd₂ = 40 mmp₁ − p₂ = 26 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

A closed water tank has compressed air above the water at a gauge pressure of 170 kPa. Water leaves through an opening with a diameter of 50 mm, 2.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².

p_air = 170 kPa (gauge)h = 2.5 md = 50 mmρ = 1000 kg/m³g = 9.81 m/s²

v = m/s

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 2.1 m/s through a horizontal pipe, 200 m long, with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 2.1 m/sL = 200 mD = 50 mmf = 0.015ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 400 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 25 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 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 = 25 mL = 400 mD = 300 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 76 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 = 10 L/sH = 76 mη_pump = 82%η_motor = 95%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 20 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 400 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.015; 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 = 20 L/sΔz = 10 mL = 400 mD = 100 mmf = 0.015ΣK = 6η = 82%

P_shaft = kW

3Solutions

1.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 = 3 + 2/2 = 4 m

A = H·b = 2·1.5 = 3 m²

Resultant force

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

Answer

F = 117.7 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 = 2.5·1.5 = 3.75 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·5.25·3.75 = 243.3 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·2.5³/12 = 1.953 m⁴

y_cp = y_c + I_c/(y_c·A) = 5.25 + 1.953/(5.25·3.75) = 5.349 m

Answer

y_cp = 5.349 m

3.Continuity & Bernoulli

Answer: Q = 9.359 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·(40/80)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 = √(2·26000/(1000·(1 − 0.0625))) = 7.448 m/s

v₁ = 1.862 m/s

Flow rate

Q = A₂·v₂ = 0.001257·7.448 = 0.009359 m³/s = 9.359 L/s

Answer

Q = 9.359 L/s

4.Continuity & Bernoulli

Answer: v = 19.72 m/s

Bernoulli from the surface to the jet

v = √(2·(170000/1000 + 9.81·2.5)) = 19.72 m/s

Answer

v = 19.72 m/s

5.Pipe friction & head loss

Answer: Δp = 132.3 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) = 0.015·(200/0.05)·2.1²/(2·9.81) = 13.49 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 = 1000·9.81·13.49 = 132300 Pa = 132.3 kPa

Answer

Δp = 132.3 kPa

6.Pipe friction & head loss

Answer: Q = 280.9 L/s

Loss coefficients

f·L/D = 0.02·400/0.3 = 26.67

ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4

Energy equation

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

25 = (26.67 + 4.4)·V²/(2·9.81)

V = √(2·9.81·25/31.07) = 3.973 m/s

Flow rate

Q = V·π·D²/4 = 3.973·0.07069 = 0.2809 m³/s = 280.9 L/s

Answer

Q = 280.9 L/s

7.Pumps & pump power

Answer: P_el = 9.571 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 = 1000·9.81·0.01·76 = 7.456 kW

Shaft power

P_shaft = 7.456/0.82 = 9.092 kW

Electrical power

P_el = 9.092/0.95 = 9.571 kW

Answer

P_el = 9.571 kW

8.Pumps & pump power

Answer: P_shaft = 7.612 kW

Velocity

V = Q/A = 0.02/0.007854 = 2.546 m/s

V²/(2g) = 0.3305 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (60 + 6)·0.3305 = 21.81 m

Pump head

H = Δz + h_L = 10 + 21.81 = 31.81 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.02·31.81/0.82 = 7.612 kW

Answer

P_shaft = 7.612 kW

ExamProblems · examproblems.com/courses/fluid-mechanics