Fluid Mechanics: free practice, theory and problems

1. Hydrostatics

What is it about?

Hydrostatics deals with fluids at rest. Even though nothing is flowing, the fluid still exerts pressure on everything it touches – the bottom and walls of a tank, a dam, a submerged body or the piston in a hydraulic cylinder. Pressure increases with depth because every column of fluid must support the weight of everything above it.

For an engineer this is the foundation for sizing tanks, dams, diving equipment, hydraulic systems and manometers.

Key quantities and formulas

• Hydrostatic pressure: $p = p_0 + \rho gh$, where $p_0$ is the pressure at the surface (often atmospheric) and $h$ is the depth.
• Gauge pressure is measured relative to the atmosphere; absolute pressure is $p_{abs} = p_{atm} + p_{gauge}$, with $p_{atm} \approx 101.3$ kPa.
• Archimedes' principle: the buoyant force on a submerged body is $F_B = \rho_{fluid}gV_{displaced}$.
• A floating body displaces exactly its own weight in fluid: $\rho_{body}V_{body}g = \rho_{fluid}V_{displaced}g$, so the submerged volume fraction is $\rho_{body}/\rho_{fluid}$.
• Hydraulic press (Pascal's principle): the pressure is equal on both pistons, $F_2 = F_1A_2/A_1$.
• U-tube manometer: the pressure difference corresponds to a height difference in a fluid column, $\Delta p = \rho g\,\Delta h$.

How to solve the problems

1. Find which fluid it is (density $\rho$) and which depth or height difference $h$ applies.
2. Decide whether you need gauge pressure (just $\rho gh$) or absolute pressure (add $p_{atm}$).
3. For buoyancy: use the entire submerged volume. For floating: set the weight equal to the buoyant force.
4. For hydraulics and manometers: set up the pressure balance and solve for the unknown.

Example

An iceberg ($\rho_{ice} = 917$ kg/m³) floats in seawater ($\rho_{sea} = 1025$ kg/m³). What fraction of its volume sticks up above the surface?

1. Floating: the weight of the iceberg equals the buoyant force, $\rho_{ice}Vg = \rho_{sea}V_{under}g$.
2. Fraction below water: $V_{under}/V = \rho_{ice}/\rho_{sea} = 917/1025 \approx 0.895$.
3. Fraction above water: $1 - 0.895 = 0.105$, i.e. about 10.5%.

Answer: Only about 10.5% of the iceberg is visible above the water surface.

Common mistakes

• Believing that the pressure depends on the shape of the container or the total volume of fluid (the hydrostatic paradox says it does not).
• Mixing up gauge pressure and absolute pressure.
• Forgetting to multiply by $g$ in $\rho gh$, or using $\rho$ in g/cm³ instead of kg/m³.
• Believing a floating body has a buoyant force equal to its entire weight displaced in air; only the submerged part counts.

Concepts in this part

Practise hydrostatics in the app →

2. Continuity and Bernoulli

What is it about?

When a fluid flows through a pipe whose cross-section changes, it must speed up where the pipe is narrow, so that the same volume passes every point each second (continuity). Bernoulli's equation extends this to an energy balance along a streamline: pressure, speed and height can convert into one another, but their sum stays constant when there are no losses.

These two relationships are the foundation for sizing piping systems, understanding flow meters (venturi meters, Pitot tubes) and analyzing everything from water supply to aircraft wings.

Key quantities and formulas

• Volumetric flow rate: $Q = Av$ (m³/s), constant along a pipe with no branches: $A_1v_1 = A_2v_2$.
• Mass flow rate: $\dot m = \rho Av$ (kg/s), constant even if the fluid is compressible.
• Bernoulli's equation (steady, frictionless, incompressible flow along a streamline): $p + \tfrac12\rho v^2 + \rho gz = \text{constant}$.
• Static pressure $p$, dynamic pressure $\tfrac12\rho v^2$ and stagnation pressure $p + \tfrac12\rho v^2$ (the pressure where the flow is brought to rest).
• Torricelli's law (outflow from a tank): $v = \sqrt{2gh}$.
• Pitot tube: measures the speed from the pressure difference between the stagnation and static pressure, $v = \sqrt{2\Delta p/\rho}$.
• The extended energy equation adds pump work and losses to account for real piping systems.

How to solve the problems

1. Mark point 1 and point 2 along the streamline, and write down what you know about area, speed, pressure and height at each one.
2. Use continuity to relate the speeds at the two points if the area changes.
3. Set up Bernoulli's equation between the points, and cancel terms that are equal or zero (e.g. the same height).
4. Solve for the unknown quantity, and check that the pressure does not become unrealistically negative.

Example

Water flows in a horizontal pipe with a diameter of 100 mm and a speed of 2 m/s. The pipe narrows to 60 mm. What is the pressure drop across the constriction?

1. Continuity: $v_2 = v_1(d_1/d_2)^2 = 2\cdot(100/60)^2 \approx 5.56$ m/s.
2. Bernoulli (same height): $p_1 - p_2 = \tfrac12\rho(v_2^2 - v_1^2) = \tfrac12\cdot 1000\cdot(5.56^2 - 2^2) \approx 13.4$ kPa.

Answer: The pressure drops by about 13.4 kPa across the constriction.

Common mistakes

• Applying Bernoulli's equation across a point with losses (a valve, a bend) or a pump without adding those terms.
• Forgetting the elevation term $\rho gz$ when the height of the pipe changes.
• Mixing up static pressure and stagnation pressure.
• Using the diameter instead of the area (remember $A = \pi d^2/4$) in the continuity equation.

Concepts in this part

Practise continuity and Bernoulli in the app →

3. Pipe flow and dimensionless numbers

What is it about?

In a real pipe there is friction between the fluid and the wall, and the flow can be smooth and orderly (laminar) or chaotic with eddies (turbulent). The Reynolds number tells you which type of flow you have, and the friction factor $f$ tells you how much pressure you lose to friction along the pipe.

For an engineer this is crucial for correctly sizing pumps and pipelines: too small a diameter or too long a pipe run gives large pressure losses and requires more pump power than necessary.

Key quantities and formulas

• Reynolds number: $Re = \rho vD/\mu$ (dimensionless). Laminar flow below about 2300, turbulent above about 4000, with a transitional range in between.
• No-slip condition: the fluid has the same speed as the wall where it is in contact with it, which creates a boundary layer.
• Darcy–Weisbach equation for friction loss: $h_f = f\dfrac{L}{D}\dfrac{v^2}{2g}$ (m), where $f$ depends on $Re$ and the relative roughness.
• For laminar flow: $f = 64/Re$. For turbulent flow you find $f$ from the Moody chart or the Colebrook equation; a rougher pipe gives a higher $f$.
• Minor (local) losses in valves, bends and constrictions: $h_{minor} = K\dfrac{v^2}{2g}$, with the loss coefficient $K$ from a table.
• Pump power to overcome a loss: $P = \rho gQh_f$ (W), with the volumetric flow rate $Q$ in m³/s.

How to solve the problems

1. Compute the Reynolds number to decide whether the flow is laminar or turbulent.
2. Find the friction factor: $64/Re$ if laminar, otherwise from a given value or the Moody chart.
3. Substitute into Darcy–Weisbach for the friction loss, and add any minor losses.
4. For pump power: multiply the total loss (in meters) by $\rho g$ and the volumetric flow rate.

Example

Water ($\rho = 1000$ kg/m³, $\mu = 10^{-3}$ Pa·s) flows at 1.5 m/s in a pipe with a diameter of 30 mm and a length of 20 m. The friction factor is $f = 0.027$. How large is the friction loss?

1. Reynolds number (as a check): $Re = \rho vD/\mu = 1000\cdot 1.5\cdot 0.03/10^{-3} = 45000$ – turbulent, so a given $f$ makes sense.
2. Darcy–Weisbach: $h_f = f\dfrac{L}{D}\dfrac{v^2}{2g} = 0.027\cdot\dfrac{20}{0.03}\cdot\dfrac{1.5^2}{2\cdot 9.81} \approx 2.06$ m.

Answer: The friction loss is about 2.06 m of fluid column.

Common mistakes

• Using $f = 64/Re$ for turbulent flow; that formula only applies to laminar flow.
• Mixing up diameter and radius in $Re$ and Darcy–Weisbach.
• Forgetting the minor losses and only accounting for friction in the pipe itself.
• Using the wrong unit for $\mu$ (Pa·s, not cP, without converting).

Concepts in this part

Practise pipe flow and dimensionless numbers in the app →

4. Pumps and pipe systems

What is it about?

Pumps move liquid through pipes: water in buildings, cooling water in machines, oil in hydraulics. To choose the right pump you need to know how high the liquid must be lifted, how much must flow per second, and how much power that takes.

Concepts and formulas

• The head $H$ (m) is the pressure rise expressed as a column of water:

• The hydraulic power is the power that actually goes into the liquid, and the shaft power is what the motor must deliver:

• The system curve shows the head the pipe system requires: a static part (height difference) plus friction losses that grow with the square of the flow, $H_{sys} = H_{st} + kQ^2$.
• The operating point is where the pump curve crosses the system curve.
• The affinity laws when the speed $n$ changes:

• Cavitation: if the pressure on the suction side drops below the vapour pressure, vapour bubbles form and collapse, damaging the impeller.

How to solve the problems

1. Convert units: L/s to m³/s (divide by 1000), kPa to Pa (multiply by 1000).
2. Use $P_h = \rho g Q H$ with $\rho = 1000$ kg/m³ for water.
3. Divide by the efficiency to find the shaft power.
4. For a change of speed: use the affinity laws (1st, 2nd and 3rd power).

Example

A pump lifts 10 L/s of water 20 m and has an efficiency of 0.7.

1. $Q = 0.01$ m³/s.
2. $P_h = 1000\cdot 9.81\cdot 0.01\cdot 20 = 1962$ W.
3. $P_{shaft} = 1962/0.7 \approx 2803$ W, about 2.8 kW.

Common mistakes

• Using L/s directly in the formula. It needs m³/s.
• Multiplying by the efficiency instead of dividing. The motor must deliver more than the liquid receives.
• Thinking the power grows linearly with speed. It grows with the cube: 10 % more speed gives 33 % more power.

Concepts in this part

Practise pumps and pipe systems in the app →