Units, dimensions and unit conversions in process work

SI base and derived units, FPS and CGS conversions, gauge vs absolute pressure, temperature scales, dimensional homogeneity and converting empirical correlations.

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Why it matters

Every process calculation – a pump size, a reactor feed rate, a heat duty – is a number with a unit, and plant data arrive in a mix of SI, CGS and FPS (British) units. A single wrong conversion factor (gauge vs absolute pressure, US vs imperial gallon, °C vs K) can make a design off by orders of magnitude. Reliable unit handling and dimensional checks are the first habit a chemical engineer must build.

Key ideas

Dimensions vs units. A dimension is the physical nature of a quantity: mass [M], length [L], time [T], temperature [θ], amount of substance [N]. A unit is the agreed size used to measure it: kg, m, s, K, mol. Velocity always has dimension [L T⁻¹], whether written in m/s or ft/h.

SI base units. The seven SI base units are metre (m), kilogram (kg), second (s), kelvin (K), mole (mol), ampere (A) and candela (cd). Everything else is derived:

  • force: newton, 1 N = 1 kg·m/s²
  • pressure: pascal, 1 Pa = 1 N/m² = 1 kg/(m·s²)
  • energy: joule, 1 J = 1 N·m = 1 kg·m²/s²
  • power: watt, 1 W = 1 J/s

Note that the litre and °C are accepted for use with SI but are not base units; bar (10⁵ Pa) and atm (101 325 Pa) are common non-SI pressure units.

Other systems you will meet. CGS (g, cm, s; dyne, erg, calorie) and FPS / engineering units (lbm, ft, h, °F, Btu, psi). In the FPS engineering system the lbf is defined so that 1 lbf = 32.174 lbm·ft/s², which introduces the conversion constant g_c = 32.174 lbm·ft/(lbf·s²). In SI, g_c = 1 kg·m/(N·s²) and simply disappears.

Conversion factors as "unity". A conversion factor such as (1000 m / 1 km) equals one, so multiplying by it changes the unit but not the quantity. Chain factors so that unwanted units cancel; write the units in every step so cancellation can be seen.

Dimensional homogeneity. Every additive term in a physically derived equation must have the same dimensions, and arguments of exp, log and sin must be dimensionless. This is the fastest check of any formula you derive or recall.

Empirical (dimensional) equations. Many design correlations were fitted to data in particular units, e.g. h = 0.026 G^0.8 / D^0.2 with h in Btu/(h·ft²·°F). Such an equation is not homogeneous: its constant carries hidden units. You must either use the original units or convert the constant – a frequent GATE task.

Temperature. Absolute temperature T(K) = t(°C) + 273.15; T(°R) = t(°F) + 459.67. A temperature difference converts as 1 K = 1 °C = 1.8 °F = 1.8 °R. Gas laws, radiation and equilibrium constants need absolute temperature; sensible-heat formulas need only differences.

Pressure. Absolute pressure = gauge pressure + atmospheric pressure. Gauges read relative to the local atmosphere; gas-law calculations need absolute pressure.

Formulas

  • 1 N = 1 kg·m/s² ; 1 Pa = 1 N/m² ; 1 J = 1 N·m ; 1 W = 1 J/s
  • 1 atm = 101.325 kPa = 1.01325 bar = 760 mmHg = 14.696 psi
  • 1 bar = 10⁵ Pa ; 1 psi = 6.8948 kPa
  • 1 lbm = 0.45359 kg ; 1 ft = 0.3048 m ; 1 in = 2.54 cm ; 1 US gal = 3.7854 L
  • 1 cal = 4.184 J ; 1 Btu = 1055.06 J ; 1 Btu/(h·ft²·°F) = 5.6783 W/(m²·K)
  • T(K) = t(°C) + 273.15 ; T(°R) = t(°F) + 459.67 ; ΔT(K) = ΔT(°F) / 1.8
  • P_abs = P_gauge + P_atm
  • F = m·a / g_c – F force (N or lbf), m mass (kg or lbm), a acceleration (m/s² or ft/s²); g_c = 1 kg·m/(N·s²) in SI and 32.174 lbm·ft/(lbf·s²) in FPS.
  • R = 8.314 J/(mol·K) = 0.08206 L·atm/(mol·K) = 8.314 kPa·m³/(kmol·K) = 1.987 cal/(mol·K)

Worked examples

Example 1 (standard). A pump transfers 250 US gal/min of an oil of specific gravity 0.85. Find the mass flow rate in kg/s.

  1. Volumetric flow: Q = 250 gal/min × 3.7854 L/gal × (1 min / 60 s) = 15.77 L/s = 0.01577 m³/s.
  2. Density: ρ = SG × ρ_water = 0.85 × 1000 kg/m³ = 850 kg/m³.
  3. Mass flow: ṁ = ρ·Q = 850 kg/m³ × 0.01577 m³/s = 13.41 kg/s.

Answer: ṁ ≈ 13.4 kg/s

Example 2 (GATE level). A heat-transfer correlation was fitted in FPS units: h = 0.026 G^0.8 / D^0.2, with h in Btu/(h·ft²·°F), G in lbm/(h·ft²) and D in ft. Find the constant when h is in W/(m²·K), G in kg/(m²·s) and D in m.

  1. Unit relations: 1 Btu/(h·ft²·°F) = 5.6783 W/(m²·K); 1 lbm/(h·ft²) = 0.45359 kg / (3600 s × 0.092903 m²) = 1.3562 × 10⁻³ kg/(m²·s); 1 ft = 0.3048 m.
  2. Express the FPS variables through SI ones: h_FPS = h_SI / 5.6783; G_FPS = G_SI / (1.3562 × 10⁻³); D_FPS = D_SI / 0.3048.
  3. Substitute: h_SI / 5.6783 = 0.026 × (G_SI / (1.3562 × 10⁻³))^0.8 / (D_SI / 0.3048)^0.2.
  4. Collect the constant: C = 5.6783 × 0.026 × (1 / (1.3562 × 10⁻³))^0.8 × 0.3048^0.2 = 0.14764 × 196.9 × 0.7885 ≈ 22.9.
  5. Check with numbers: G = 10 kg/(m²·s), D = 0.05 m gives h = 263 W/(m²·K) from both forms.

Answer: h = 22.9 G^0.8 / D^0.2 (SI units)

Common mistakes

  • Using gauge pressure in PV = nRT; always add atmospheric pressure first.
  • Converting a temperature difference with the +273.15 offset (a 10 °C rise is a 10 K rise, not 283 K).
  • Mixing the US gallon (3.785 L) with the imperial gallon (4.546 L).
  • Squaring or cubing a length but not its conversion factor: 1 ft² = 0.0929 m², 1 ft³ = 0.02832 m³.
  • Treating lbm and lbf as interchangeable without g_c.
  • Plugging SI values into an empirical correlation fitted in FPS units without converting its constant.

For GATE CH

Expect direct conversion questions folded into larger numericals (pressure, energy and flow units, gauge vs absolute), MCQs on dimensions of derived quantities (viscosity, diffusivity, heat-transfer coefficient), and converting the constant of an empirical correlation between unit systems. Practise writing every conversion as a chain of unity factors and checking the dimensions of the result before computing.

Quick check

  1. Convert 5 L to m³.
  2. What are the SI base units of the pascal?
  3. A gauge reads 2.5 bar at a site where the barometer reads 1.0 bar. What is the absolute pressure in kPa?
  4. A liquid is heated by 36 °F. What is the temperature rise in K?
  5. What are the dimensions of dynamic viscosity?

Answers: 1. 0.005 m³; 2. kg/(m·s²); 3. 350 kPa; 4. 20 K; 5. M L⁻¹ T⁻¹.

See it move

All Chemical animations
Unit Conversion Interactive

Use the slider to convert speed from kilometers per hour (km/h) to meters per second (m/s). Observe how the conversion factor changes the speed value.

Equations used
  • v_{m/s} = v_{km/h} × \frac{1000}{3600} — v_{m/s} speed in meters per second, v_{km/h} speed in kilometers per hour

Interview questions

All Process Calculations interview questions

Try answering each one aloud before you open it.

  1. 1.What is the difference between units and dimensions in process calculations?Concept

    Units are the specific measures used to quantify dimensions, such as meters for length or kilograms for mass. Dimensions are the physical quantities themselves, like length, mass, time, etc. In process calculations, dimensions provide the framework for understanding the physical world, while units provide the scale for measurement.

  2. 2.Explain why unit conversions are important in chemical engineering processes.Concept

    Unit conversions are crucial in chemical engineering because they ensure consistency and accuracy in calculations. Different parts of a process may use different units, and converting them to a common unit system helps avoid errors. It also facilitates communication and understanding among engineers who may be using different unit systems.

  3. 3.How do you convert pressure from bar to pascal?Numerical

    To convert pressure from bar to pascal, multiply the value in bar by 100,000, since 1 bar is equal to 100,000 pascals. For example, if the pressure is 2 bar, it would be 2 × 100,000 = 200,000 pascals.

  4. 4.Why is the SI unit system preferred in chemical engineering calculations?Application

    SI is coherent: derived units are built from base units with a factor of one (1 N = 1 kg·m/s², 1 J = 1 N·m, 1 W = 1 J/s), so no conversion constant such as g_c = 32.174 lbm·ft/(lbf·s²) appears in the equations. That removes a whole class of errors in force, energy and power balances. It is also the system of most modern data books and standards, so data can be combined without conversion.

  5. 5.What happens if you use inconsistent units in a process calculation?Application

    The arithmetic still produces a number, but it is wrong by the missing conversion factor – often by 10³, 3600 or 101.325, which can mean an undersized pump, wrong reactor volume or an unsafe relief valve. Typical culprits are gauge instead of absolute pressure in PV = nRT, °C instead of K in a ratio, or mixing kmol with mol. Carrying units through every step and checking dimensions of the result catches these errors.

  6. 6.Explain the concept of dimensional homogeneity in process calculations.Concept

    Dimensional homogeneity means that all terms in a physical equation must have the same dimensions. This ensures that the equation is physically meaningful and consistent. In process calculations, checking for dimensional homogeneity helps verify that equations are set up correctly and can prevent errors.

  7. 7.Convert a flow rate of 500 liters per hour to cubic meters per second.Numerical

    First, convert liters to cubic meters: 500 liters = 0.5 cubic meters. Then, convert hours to seconds: 1 hour = 3600 seconds. Finally, divide the volume by time: 0.5 cubic meters / 3600 seconds = 0.0001389 cubic meters per second.

  8. 8.Why might a chemical engineer need to convert temperature from Celsius to Kelvin?Application

    A chemical engineer might need to convert temperature from Celsius to Kelvin because many thermodynamic equations require absolute temperature, which is measured in Kelvin. Kelvin is the SI unit for temperature and ensures that calculations involving temperature differences or ratios are accurate.

  9. 9.Explain how dimensional analysis can be used to check the correctness of a process calculation.Concept

    Dimensional analysis involves checking that all terms in an equation have consistent dimensions. By ensuring that the dimensions on both sides of an equation match, engineers can verify that the equation is set up correctly. This method helps identify errors in unit conversions or in the formulation of the equation itself.

  10. 10.What are the SI base units, and why does it help to reduce derived units to them?Application

    The seven SI base units are m, kg, s, K, mol, A and cd. Reducing a derived unit to base units (for example Pa = kg/(m·s²), J = kg·m²/s²) lets you check whether an equation or an answer is dimensionally consistent and spot a missing factor. It also shows quickly that, say, Pa·m³ is the same as J, which is why PV has units of energy.

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