Fluid Mechanics · Unit 1 · Lesson 12 of 13
How Do Engineers Measure Everything In Fluids?
In 1999 NASA lost the Mars Climate Orbiter because one team's software produced results in pound-force seconds while another team expected newton seconds. How do engineers make sure that never happens to them?
1. Real-World Problem
A pump data sheet gives flow in litres per minute, a pressure gauge reads in bar, an oil data sheet lists viscosity in centipoise and a motor nameplate shows horsepower. Before any calculation, an engineer must put all of these into one consistent system.
2. Predict
Make your guess before you try the experiment. You will see the answer after the next step.
Predict
3. Experiment
In Build mode, pick a quantity such as pressure or viscosity and see it built from M, L and T. In Convert mode, turn 32 psi into kPa and bar.
In Check mode, decide which of the six equations balance, then see the dimensions of each side worked out.
4. Understand
A dimension is the kind of quantity (length, mass, time). A unit is the agreed size used to measure it (metre, kilogram, second). For the quantities covered in this unit, four fundamental dimensions are sufficient; everything else is built from them. (SI defines seven base quantities in total; the other three, electric current, amount of substance and luminous intensity, are not needed here.)
| Fundamental dimension | Symbol | SI base unit |
|---|---|---|
| Mass | M | kilogram (kg) |
| Length | L | metre (m) |
| Time | T | second (s) |
| Temperature | Θ | kelvin (K) |
Derived quantities used in this subject
| Quantity | Dimensions | SI unit | Other common units |
|---|---|---|---|
| Area | L² | m² | cm², mm² |
| Volume | L³ | m³ | 1 L = 0.001 m³ |
| Velocity | LT⁻¹ | m/s | km/h (1 m/s = 3.6 km/h) |
| Acceleration | LT⁻² | m/s² | g = 9.81 m/s² |
| Force | MLT⁻² | N = kg·m/s² | 1 kgf = 9.81 N |
| Pressure, stress | ML⁻¹T⁻² | Pa = N/m² | 1 bar = 100,000 Pa; 1 atm = 101,325 Pa; 1 psi = 6895 Pa; 1 mm Hg = 133.3 Pa |
| Density | ML⁻³ | kg/m³ | g/cm³ (1 g/cm³ = 1000 kg/m³) |
| Specific weight | ML⁻²T⁻² | N/m³ | |
| Dynamic viscosity | ML⁻¹T⁻¹ | Pa·s = N·s/m² | 1 poise = 0.1 Pa·s; 1 cP = 0.001 Pa·s |
| Kinematic viscosity | L²T⁻¹ | m²/s | 1 stokes = 0.0001 m²/s |
| Surface tension | MT⁻² | N/m | |
| Volume flow rate | L³T⁻¹ | m³/s | L/min, L/s |
| Mass flow rate | MT⁻¹ | kg/s | |
| Work, energy | ML²T⁻² | J = N·m | 1 kWh = 3.6 MJ |
| Torque | ML²T⁻² | N·m (never written as J) | |
| Power | ML²T⁻³ | W = J/s | 1 hp = 746 W (mechanical horsepower, the one on Indian pump and motor nameplates); 1 metric hp = 735.5 W |
| Bulk modulus | ML⁻¹T⁻² | Pa | GPa |
Same dimensions, different meaning: torque and energy. Both are ML²T⁻². Work (energy) is a force times the distance moved along that force: a scalar, measured in joules. Torque is a force times its perpendicular lever arm: a turning effect with a direction (a vector), written N·m and never J. Dimensions cannot tell them apart; their physical meaning can. They are linked through rotation: a torque T turning through an angle θ (radians) does work Tθ, and the power delivered by a shaft is P = Tω.
Some books use force instead of mass as a fundamental dimension (the FLT system). Then mass has dimensions FT²L⁻¹, from F = ma.
5. Derive
Step 1. Dimensional homogeneity: every term in a correct physical equation has the same dimensions. Check the hydrostatic equation:
p = \rho g h \;\Rightarrow\; [ML^{-1}T^{-2}] = [ML^{-3}][LT^{-2}][L] = [ML^{-1}T^{-2}]Both sides match, so the equation can be right. If they do not match, the equation is certainly wrong.
Step 2. Dimensions of viscosity from τ = μ du/dy:
[\mu] = \frac{[\tau]}{[du/dy]} = \frac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}Step 3. Shaft power from torque and rotational speed (ω in rad/s, N in rpm):
P = T\,\omega \qquad \omega = \frac{2\pi N}{60}6. Calculate
- Tyre at 32 psi: 32 × 6895 = 220,640 Pa = 220.6 kPa = 2.21 bar.
- Engine oil at 150 cP: 150 × 0.001 = 0.15 Pa·s.
- Pump flow 60 L/min: 60 ÷ 1000 ÷ 60 = 0.001 m³/s.
- A motor gives a torque of 10 N·m at 1440 rpm: ω = 2π × 1440 ÷ 60 = 150.8 rad/s, so P = 10 × 150.8 = 1508 W, about 1.5 kW. The 10 N·m is a torque, the 1508 J each second is energy: same dimensions, different quantities.
7. Apply
- Tyre pressure in psi, gas cylinders in bar, weather in hPa: all convertible to Pa.
- Engine oil viscosity data sheets in mPa·s (same as cP).
- Pump flow in litres per minute, converted to m³/s before any calculation.
- Motor nameplates in kW or hp; shaft torque in N·m.
8. Validate
In Convert mode enter 32 psi: it must show 220.6 kPa and 2.21 bar. In Check mode, p = ρgh must show Balanced and Manning's equation must show Not balanced when written V = (1/n)R_h^(2/3)S^(1/2) with n as a plain number, and Balanced once k = 1 m^(1/3)/s is included (or once n is given units of s/m^(1/3)). Where dimensional checks do not help: a dimensionally correct equation can still be wrong (p = 2ρgh balances), and dimensions cannot tell torque from energy.
Easy To Mix Up
- Kilogram is a unit of mass, not weight. Weight is a force, in newtons.
- Gauge and absolute pressure use the same unit but different zero points.
- Dimensionally correct does not mean correct. p = 2ρgh also balances, but is wrong.
- Torque and energy share dimensions but are different quantities. Write torque in N·m, energy in J.
9. Quiz
Four quick questions: a concept, a calculation and an interpretation. Each answer comes with an explanation. Score 3 or more to mark this lesson completed.
Check Your Understanding
Four questions: a concept, a calculation and an interpretation. Every answer comes with an explanation. Score 3 of 4 to mark the lesson completed.
Now you know the language. Where does a mechanical engineer actually use fluid mechanics at work?
Go to Lesson 13: Where Do Mechanical Engineers Use Fluid Mechanics? →Read More
- Cengel and Cimbala, "Introduction and Basic Concepts" (dimensions and units) and "Dimensional Analysis and Modeling"; BIPM, The International System of Units (SI Brochure, 9th edition).
- NPTEL, Introduction to Fluid Mechanics (Prof. Suman Chakraborty, IIT Kharagpur): Week 1.