Two ways to remove water

Fresh ginger is about nine-tenths water; fresh cranberries and Dendrobium stems are not far behind. To keep them, the water must go. There are two broad ways to do it.

Hot-air drying blows warm, dry air over the product. Heat flows in, liquid water moves to the surface and evaporates, and the vapour is carried away. It is cheap and fast, but the product is warm and wet for hours, which are the conditions under which heat-sensitive molecules react, as Chapter 3 showed for gingerol. The surface dries first and shrinks, cells collapse, and the dried piece is dense and slow to rehydrate.

Freeze-drying (lyophilisation) freezes the product, lowers the pressure, and supplies just enough heat for the ice to turn directly into vapour. Because the water leaves as vapour from ice, the solid skeleton stays in place and the dried product is porous, light and quick to rehydrate. Because the product stays cold, reactions are slow. A review of drying for high-value foods describes freeze-drying as giving the best quality of the common methods, at the highest cost (Ratti 2001).

The triple point

Ice can turn straight into vapour only below a certain pressure. On a phase diagram of water, the lines separating solid, liquid and vapour meet at the triple point, where all three coexist. For ordinary water it lies at 273.16 K (0.01 °C) and 611.657 Pa (Wagner and Pruß 2002), about 0.6 % of atmospheric pressure. Below that pressure, liquid water cannot exist at equilibrium; warming ice makes it sublime.

In a freeze-dryer the chamber is held at a few pascals to a few tens of pascals, and a condenser at −50 °C or colder traps the vapour as ice. Water moves from the product to the condenser because the vapour pressure over the product’s ice is higher than the pressure in the chamber.

The Clausius–Clapeyron equation

How fast ice sublimes depends on its vapour pressure, which rises steeply with temperature. The Clausius–Clapeyron equation describes the slope:

$$\frac{d \ln p}{dT} = \frac{\Delta H_{\text{sub}}}{R T^2}$$

where $\Delta H_\text{sub}$ is the enthalpy of sublimation. If $\Delta H_\text{sub}$ is roughly constant, a plot of $\ln p$ against $1/T$ is a straight line. Accurate work uses fitted formulas; one widely used formula for ice is that of Murphy and Koop (Murphy and Koop 2005).

import math

def p_ice(t_c):
    """Vapour pressure over ice, Pa (Murphy and Koop 2005)."""
    T = t_c + 273.15
    return math.exp(9.550426 - 5723.265 / T + 3.53068 * math.log(T) - 0.00728332 * T)

print(f"at the triple point (0.01 °C): {p_ice(0.01):.1f} Pa (accepted value 611.657 Pa)")
for t in (-50, -40, -30, -20, -10):
    print(f"{t:4d} °C  {p_ice(t):7.2f} Pa")

# Clausius–Clapeyron: ln p = const − ΔH_sub / (R T). Slope between −40 and −10 °C:
R = 8.314
T1, T2 = 233.15, 263.15
dH = R * math.log(p_ice(-10) / p_ice(-40)) / (1 / T1 - 1 / T2)
print(f"enthalpy of sublimation from the slope: {dH / 1000:.1f} kJ/mol "
      f"= {dH / 18.015:.0f} J/g")
print(f"each 10 °C warmer at the ice front: × {p_ice(-20) / p_ice(-30):.2f} "
      f"(−30 → −20 °C)")
# at the triple point (0.01 °C): 611.7 Pa (accepted value 611.657 Pa)
#  -50 °C     3.94 Pa
#  -40 °C    12.84 Pa
#  -30 °C    38.01 Pa
#  -20 °C   103.25 Pa
#  -10 °C   259.89 Pa
# enthalpy of sublimation from the slope: 51.1 kJ/mol = 2838 J/g
# each 10 °C warmer at the ice front: × 2.72 (−30 → −20 °C)

Two numbers in the output run the whole process. Every gram of ice needs about 2.8 kJ of heat to sublime, more than the roughly 2.45 kJ needed to evaporate a gram of liquid water at room temperature, because the energy of melting is added to that of evaporation. And each 10 °C rise in the temperature of the ice almost triples its vapour pressure, and so the driving force for drying.

A retreating ice front

Primary drying, the removal of the frozen water, is usually modelled by assuming that a sharp boundary between frozen and dried material moves into the product. This is the uniformly retreating ice front model, first applied to the freeze-drying of foods in the 1960s (Sandall et al. 1967). Above the front lies a porous dried layer through which the vapour must escape. The thicker that layer, the more it resists the flow. In pharmaceutical freeze-drying the resistance is measured and written as $R_p = R_0 + R_1 x$, where $x$ is the thickness of the dried layer (Pikal 1985).

If the flux of vapour per unit area is $J = \Delta p / R_p$, and the front advances as the ice beneath it is removed, then for a slab with $m$ grams of ice per cubic centimetre the time to dry a thickness $L$ is

$$t = \frac{m\,(R_0 L + R_1 L^2 / 2)}{\Delta p}$$

The $L^2$ term is the important one: double the thickness and, once the dried layer dominates the resistance, the time roughly quadruples.

import math

def p_ice(t_c):                                  # Pa, Murphy and Koop (2005)
    T = t_c + 273.15
    return math.exp(9.550426 - 5723.265 / T + 3.53068 * math.log(T) - 0.00728332 * T)

TORR = 133.322                                   # Pa per Torr
DH_SUB = 2838                                    # J per g of ice

def primary_drying_hours(L_cm, t_front, p_chamber_pa=10.0,
                         ice_g_per_cm3=0.80, R0=2.0, R1=30.0):
    """Uniformly retreating ice front in a slab dried from one face.
    The dried layer resists vapour flow: R = R0 + R1·x (cm²·Torr·h/g, x in cm).
    Flux J = Δp / R, and the front moves dx/dt = J / ice_g_per_cm3, so
    t = ice · (R0·L + R1·L²/2) / Δp. All parameters are illustrative."""
    dp = (p_ice(t_front) - p_chamber_pa) / TORR
    return ice_g_per_cm3 * (R0 * L_cm + R1 * L_cm ** 2 / 2) / dp, dp

print("slice   front    hours   heat needed at start")
for L in (0.3, 0.6, 1.0):
    for tf in (-30, -20):
        h, dp = primary_drying_hours(L, tf)
        j0 = dp / 2.0                                    # g/(h·cm²) when x = 0
        watts = j0 * DH_SUB / 3600 * 1e4                 # W per m²
        print(f"{10 * L:3.0f} mm  {tf:4d} °C  {h:6.1f}   {watts:5.0f} W/m²")
# slice   front    hours   heat needed at start
#   3 mm   -30 °C     7.4     828 W/m²
#   3 mm   -20 °C     2.2    2757 W/m²
#   6 mm   -30 °C    25.1     828 W/m²
#   6 mm   -20 °C     7.5    2757 W/m²
#  10 mm   -30 °C    64.7     828 W/m²
#  10 mm   -20 °C    19.4    2757 W/m²

Thin slices dry far faster than thick ones, and a warmer front dries much faster than a colder one. That is why operators push the temperature up, and why freeze-drying has a limit: if the ice front or the dried layer just above it becomes too warm, the freeze-concentrated sugars between the ice crystals soften and the structure collapses, closing the pores. Products rich in sugars, such as berries, have low collapse temperatures and must be dried colder and more slowly than lean products.

What the heat does

The difference in quality between the two methods comes mostly from temperature and time with water present.

import math

R = 8.314

def retained(steps, k_ref=0.04, t_ref=60.0, Ea=75_000):
    """Share of a heat-sensitive compound left after a schedule of
    (temperature °C, hours) steps, first-order loss with Arrhenius rate.
    k_ref (per hour at t_ref) and Ea are illustrative, not measured values
    for any one compound; moisture effects are ignored."""
    total = 0.0
    for t, h in steps:
        k = k_ref * math.exp(Ea / R * (1 / (t_ref + 273.15) - 1 / (t + 273.15)))
        total += k * h
    return math.exp(-total)

schedules = {
    "hot air 70 °C, 7 h":                [(70, 7)],
    "hot air 60 °C, 10 h":               [(60, 10)],
    "hot air 50 °C, 16 h":               [(50, 16)],
    "freeze-dry −25 °C 20 h + 25 °C 8 h": [(-25, 20), (25, 8)],
}
for name, steps in schedules.items():
    print(f"{name:36s} {100 * retained(steps):5.1f} % left")
# hot air 70 °C, 7 h                    54.0 % left
# hot air 60 °C, 10 h                   67.0 % left
# hot air 50 °C, 16 h                   75.8 % left
# freeze-dry −25 °C 20 h + 25 °C 8 h    98.7 % left

Hot-air drying at lower temperatures keeps more, but takes longer, and the gain flattens; freeze-drying keeps nearly everything in this model because the product spends its long drying time cold. Real products are more complicated. Water content falls during drying and slows many reactions; oxygen exposure, enzymes and light matter; and some compounds are lost by evaporation rather than reaction. Freeze-drying also loses some volatile aroma compounds, though usually fewer than hot-air drying does.

Porosity and rehydration

The porous structure that freeze-drying leaves behind has practical consequences. It rehydrates quickly, which is useful for powders and for whole pieces that will be soaked. It also has a large internal surface exposed to air, which makes the dried product hygroscopic and, if not packed well, vulnerable to oxidation. A freeze-dried product needs packaging with low permeability to water vapour and oxygen. Chapter 6 shows why its water activity has to be kept low.