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DSC Degree of Crystallinity Percentage Formula

Determines the weight percentage degree of crystallinity (Xc) in semi-crystalline polymers from Differential Scanning Calorimetry (DSC) endothermic melting enthalpy and cold crystallization enthalpy.

Mathematical Canonical Relation
Xc=ΔHm−ΔHccΔHm0⋅(1−wf)×100X_c = \frac{\Delta H_m - \Delta H_{cc}}{\Delta H_m^0 \cdot (1 - w_f)} \times 100
Plain Formula: Xc = [ (Delta Hm - Delta Hcc) / (Delta Hm0 * (1 - wf)) ] * 100

Live Engineering Parameter Solver

Input your production or test values below for instant, verified output.

Verified Engine
dH_m
Range: 10 – 250 J/g
dH_cc
Range: 0 – 80 J/g
dH_zero
Range: 50 – 300 J/g
w_filler
Range: 0 – 0.5 fraction
Calculated Engineering Result:
Xc = 47.5%
Engineering Verification:Net heat of fusion: 98.4 J/g.

When & Where to Apply

Use when analyzing polymer morphology, injection molding cooling rates, annealing effects, and barrier/mechanical stiffness properties.

Common Engineering Pitfalls

  • Ignoring cold crystallization enthalpy Delta Hcc in PET or PLA
  • Forgetting to subtract inorganic filler weight fraction (1 - wf) in glass-filled compounds

Physical Assumptions & Validity Limits

  • Known theoretical 100% crystalline enthalpy Delta Hm0 from literature
  • Linear or sigmoidal baseline subtraction across melting transition peak
  • Accurately determined inorganic filler weight fraction wf

Variable Symbols, SI Units & Dimensional Specifications

SymbolVariable MeaningSI UnitTypical Industrial Range
X_cDegree of crystallinity%—
\Delta H_mMeasured melting enthalpy peak areaJ/g95.5
\Delta H_{cc}Cold crystallization enthalpy (if present)J/g0.0
\Delta H_m^0Melting enthalpy of 100% crystalline polymerJ/g207.1
w_fWeight fraction of non-crystallizable filler/reinforcementfraction0.0

Worked Industrial Numerical Example: Isotactic Polypropylene (iPP) DSC Crystallinity Measurement

Problem Statement:

A moulded Polypropylene sample displays a DSC melting peak enthalpy Delta Hm = 98.4 J/g. Given that 100% crystalline iPP has Delta Hm0 = 207.1 J/g and the sample has no filler (wf = 0), calculate Xc.

Step-by-Step Numerical Solution:
  1. Apply formula: Xc = (Delta Hm - Delta Hcc) / (Delta Hm0 * (1 - wf)) * 100
  2. Substitute values: Xc = (98.4 - 0) / (207.1 * 1.0) * 100
  3. Calculate: 98.4 / 207.1 * 100 = 47.51%
Final Answer: Xc = 47.5%
💡 Engineering Interpretation: The moulded PP part has 47.5% crystalline lamellae, providing the expected balance of flexural modulus (>1300 MPa) and impact toughness.

Authoritative Technical Reference & Source Provenance

Thermal Analysis of Polymers: Fundamentals and Applications by Joseph D. Menczel (Wiley, 2009), 1st Edition, Page 112-118.
Connected Curriculum Lesson

DSC & TGA Crystallinity & Thermal Degradation

Study Full Lesson