Cryogenic Milling Application

Reference ID: MET-9668 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

Cryogenic milling is a size-reduction process in which a feedstock—often a heat-sensitive spice or pharmaceutical powder—is embrittled by direct contact with liquid nitrogen (LN₂) and then milled. The engineering task is to size the cryogenic heat exchanger (usually a jacketed screw conveyor or coiled tube) so that the product is cooled from ambient to its embrittlement temperature while the nitrogen is consumed at the minimum economic rate. The worksheet therefore solves two coupled problems:

the steady-state heat duty required to sub-cool the solids, and
the required heat-transfer area (tube length) and LN₂ flow that satisfy that duty.

Results are used to specify mill throughput, LN₂ storage capacity, and venting systems in food, fine-chemical, and polymer industries.

Methodology & Formulas

1. Heat Duty

The sensible heat removed from the product is

\[ Q = \dot{m}_{\text{spice}}\,c_{p,\text{spice}}\,(T_{\text{in}}-T_{\text{out}}) \]

where \( \dot{m}_{\text{spice}} \) is the mass flow rate of solids and \( c_{p,\text{spice}} \) its specific heat.

2. LN₂ Consumption

The nitrogen absorbs the duty by a combination of sensible heating and latent boil-off:

\[ \dot{m}_{\text{LN}_2} = \frac{Q}{h_{\text{fg}}+c_{p,\text{LN}_2}(T_{\text{out}}-T_{\text{LN}_2})} \]

with \( h_{\text{fg}} \) the latent heat of vaporisation and \( c_{p,\text{LN}_2} \) the liquid specific heat.

3. Flow Regime and Velocity

A target Reynolds number is imposed to guarantee turbulent heat transfer on the nitrogen side:

\[ \text{Re} = \frac{\rho\,u\,D}{\mu} \quad\Rightarrow\quad u = \frac{\text{Re}\,\mu}{\rho\,D} \]

where \( \rho \) and \( \mu \) are the density and dynamic viscosity of LN₂ at operating temperature.

4. Friction Factor

The Haaland explicit correlation for smooth pipes gives the Darcy friction factor:

\[ f = \left(0.79\,\ln(\text{Re})-1.64\right)^{-2} \]

5. Gnielinski Nusselt Number

The turbulent Nusselt number for internal pipe flow is

\[ \text{Nu} = \frac{(f/8)(\text{Re}-1000)\text{Pr}}{1+12.7\,(f/8)^{1/2}(\text{Pr}^{2/3}-1)} \]

with \( \text{Pr} = \frac{c_{p,\text{LN}_2}\,\mu}{k_{\text{LN}_2}} \). The inner heat-transfer coefficient follows from

\[ h_{\text{i}} = \frac{\text{Nu}\,k_{\text{LN}_2}}{D} \]

Parameter	Validity Range
Reynolds number	3000 ≤ Re ≤ 5×10⁶
Prandtl number	0.5 ≤ Pr ≤ 2000
Length-to-diameter ratio	L/D ≥ 10

6. Overall Heat-Transfer Coefficient

Resistances are added in series: inner convection, conduction through the wall, boiling on the outer surface, and fouling:

\[ \frac{1}{U} = \frac{1}{h_{\text{i}}} + \frac{D_{\text{o}}}{k_{\text{wall}}\,\ln(D_{\text{o}}/D)} + \frac{1}{h_{\text{boil}}} + R_{\text{fouling}} \]

7. Log-Mean Temperature Difference

For counter-current cooling the driving force is

\[ \Delta T_{\text{lm}} = \frac{(T_{\text{in}}-T_{\text{LN}_2})-(T_{\text{out}}-T_{\text{LN}_2})}{\ln\left(\frac{T_{\text{in}}-T_{\text{LN}_2}}{T_{\text{out}}-T_{\text{LN}_2}}\right)} \]

8. Required Area and Length

The heat-exchanger area and tube length are obtained from

\[ A = \frac{Q}{U\,\Delta T_{\text{lm}}} \quad\text{and}\quad L = \frac{A}{\pi\,D_{\text{o}}} \]

The calculated length is compared with the physical length provided to ensure the duty can be met.

Cryogenic milling typically operates between −100 °C and −196 °C (LN₂ boiling point). At these temperatures many polymers become brittle, metals lose ductility, and waxes solidify, allowing fracture-based size reduction instead of smearing or melting. This prevents clogging, retains volatile components, and yields finer, more uniform particle size distributions.

Start with the heat-balance equation: Q = m·C_p·ΔT + m·ΔH_fus + losses. A quick rule-of-thumb for plastics is 1.2–1.8 kg LN₂ per kg of feed to reach −150 °C. For metals, double the value. Always add 15% extra for heat-in-leak from the mill housing and piping. Run a 5 kg pilot test and scale linearly while logging flowmeter data to refine the factor for your specific mill geometry.

Pin mills: good for friable organics, easy to jacket.
Hammer mills with tungsten-carbide liners: handle tough metals.
Attritor mills: effective for nano-range ceramics when LN₂ is injected through hollow shaft.
Always verify the spindle seal is rated for −196 °C and that the bearing grease is perfluoropolyether (PFPE) based.

Keep the receiver under dry nitrogen purge at +5 °C to +10 °C. Install a jacketed rotary valve with 5 rpm max speed and use a 100-mesh delumper right before the valve. Insulate the mill outlet with low-emissivity foil-faced foam and maintain a slight positive pressure (0.5 kPa) to block moist room air.

Oxygen analyzer with alarm at 19% and shut-down at 18%.
Low-temperature limit switch on mill housing to stop LN₂ solenoid at −190 °C to prevent embrittlement of seals.
Pressure relief valve set at 1.5 bar on all cryogenic lines.
Emergency pull-cord that closes LN₂ supply and vents the mill to atmosphere.

Worked Example: Sizing a Cryogenic Mill Cooling Coil

A spice processor needs to cool 500 kg h⁻¹ of milled cinnamon from 25 °C to −110 °C before cryogenic grinding. Liquid nitrogen (LN₂) at 5 bar and −196 °C will be circulated counter-currently through a stainless-steel coil immersed in the product. Determine the required tube length so that the LN₂ Reynolds number is 10,000 and the overall heat-transfer coefficient is acceptable.

Knowns

Product mass flow: 500 kg h⁻¹ (0.139 kg s⁻¹)
Product specific heat: 2.1 kJ kg⁻¹ K⁻¹ (2100 J kg⁻¹ K⁻¹)
Inlet/outlet product temperatures: 25 °C / −110 °C
LN₂ temperature: −196 °C
LN₂ pressure: 5 bar
Tube inner diameter: 25 mm (0.025 m)
Wall thickness: 2 mm (0.002 m)
Target Reynolds number: 10,000
LN₂ properties at −196 °C: ρ = 810 kg m⁻³, μ = 0.00017 Pa s, k = 0.13 W m⁻¹ K⁻¹, c_p = 1040 J kg⁻¹ K⁻¹, h_fg = 199 kJ kg⁻¹
Stainless-steel thermal conductivity: 16 W m⁻¹ K⁻¹
Boiling-side coefficient: 2000 W m⁻² K⁻¹
Fouling resistance: 0.0002 m² K W⁻¹

Step-by-Step Calculation

Calculate heat duty
Q = m c_p ΔT = 0.139 kg s⁻¹ × 2100 J kg⁻¹ K⁻¹ × (298.15 – 163.15) K = 39.4 kW
Determine LN₂ velocity for Re = 10,000
Re = ρ u D / μ → u = Re μ / (ρ D) = 10,000 × 0.00017 / (810 × 0.025) = 0.084 m s⁻¹
Compute Prandtl number
Pr = μ c_p / k = 0.00017 × 1040 / 0.13 = 1.36
Estimate Dittus–Boelter Nusselt number
Nu = 0.023 Re^0.8 Pr^0.3 = 0.023 × 10,000^0.8 × 1.36^0.3 = 40.8
Find inside heat-transfer coefficient
h_i = Nu k / D = 40.8 × 0.13 / 0.025 = 212 W m⁻² K⁻¹
Sum thermal resistances
r_i = 1 / h_i = 0.0047 m² K W⁻¹
r_wall = (D_o/D) ln(D_o/D) / (2 k_wall) = (0.029/0.025) ln(0.029/0.025) / (2 × 16) = 0.012 m² K W⁻¹
r_boil = 1 / h_boil = 0.0005 m² K W⁻¹
r_foul = 0.0002 m² K W⁻¹
ΣR = 0.0176 m² K W⁻¹
Overall heat-transfer coefficient
U = 1 / ΣR = 56.7 W m⁻² K⁻¹
Log-mean temperature difference
ΔT₁ = 25 – (−196) = 221 K
ΔT₂ = −110 – (−196) = 86 K
LMTD = (221 – 86) / ln(221 / 86) = 143 K
Required heat-transfer area
A = Q / (U LMTD) = 39,400 / (56.7 × 143) = 4.85 m²
Convert area to tube length
L = A / (π D) = 4.85 / (π × 0.025) = 61.8 m → use 62 m
LN₂ consumption
Assuming saturated liquid at inlet and negligible exit sub-cooling, latent heat removal dominates:
m_LN2 = Q / h_fg = 39,400 / 199,000 = 0.198 kg s⁻¹ ≈ 713 kg h⁻¹
(Actual flow will be slightly higher if exit vapor is superheated.)

Final Answer

A 62 m long, 25 mm ID stainless-steel tube coil will cool the spice stream under the specified conditions, consuming approximately 710 kg h⁻¹ of LN₂.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle