CHE-200 · Lesson 11 of 12 · 24 min

Energy balances on nonreactive processes

Most of the steam at Kingfisher Bioenergy is used without any chemical reaction: to heat the beer, to boil the bottoms of the columns, to evaporate water from the thin stillage, and to vaporize and superheat ethanol for the molecular sieves. The plant also recovers heat by passing hot stillage against cold beer in a preheater. Ingrid asks Theo to calculate the duty of the sieve vaporizer and to check the preheater. Both need enthalpy changes for heating, cooling and changing phase when steam tables are not available, which is the subject of this lesson.

Hypothetical process paths

Enthalpy is a state property, so the change between two states does not depend on how the change actually happens. That lets us replace any real process with a convenient imaginary path made of simple steps whose enthalpy changes are known: heating or cooling at constant pressure, changing phase at a temperature where the latent heat is known, changing pressure, and mixing. The real process might heat a liquid while it boils along a tube; the hypothetical path heats the liquid to its normal boiling point, vaporizes it there, and then heats the vapour. Both give the same total \(\Delta\hat{H}\).

Choosing the path is a matter of where data are available. Heats of vaporization are usually tabulated at the normal boiling point, so paths are built to change phase there. Heat capacities are known for each phase, so the sensible heating steps use them. For an ideal gas, enthalpy does not depend on pressure at all, and for liquids and solids the effect of pressure is small, \(\Delta\hat{H} \approx \hat{V}\Delta P\), which is often negligible.

Hypothetical path on a temperature against enthalpy diagram for ethanol. Step 1 rises along a liquid line from 25 degrees to the normal boiling point of 78.4 degrees. Step 2 is a horizontal line at 78.4 degrees representing vaporization, much longer than the other steps. Step 3 rises along a vapour line from 78.4 to 150 degrees. A dashed arrow labelled actual process goes directly from the start to the end.
A hypothetical path for heating and vaporizing ethanol. The latent heat step dominates the total enthalpy change.

Sensible heat and heat capacity

Heat that changes a substance's temperature without changing its phase is called sensible heat. The constant-pressure heat capacity \(C_p\) is the rate at which enthalpy changes with temperature, \(C_p = (\partial\hat{H}/\partial T)_P\). For ideal gases, and to a good approximation for liquids and solids, the enthalpy change for heating from \(T_1\) to \(T_2\) is the integral of \(C_p\) over that range.

\[\Delta\hat{H} = \int_{T_1}^{T_2} C_p(T)\,dT \qquad C_p = a + bT + cT^2 + dT^3\]

Heat capacities of gases rise with temperature, so tables give polynomial fits with constants valid over a stated range. Liquids and solids vary less, and a mean value is often good enough. Water's liquid heat capacity is about 4.18 kJ/(kg·K), much higher than that of most other liquids; organic liquids are typically 1.8 to 2.6 kJ/(kg·K). For gas mixtures and ideal liquid mixtures, the mixture heat capacity is the mole-fraction or mass-fraction weighted sum of the component values.

Approximate heat capacity data used in this lesson.
SubstancePhaseMean or polynomial C_pNotes
Waterliquid75.4 J/(mol·K), 4.18 kJ/(kg·K)Nearly constant 0 to 100 °C
Watervapour33.46 + 0.00688T + 7.604 × 10⁻⁶T² − 3.593 × 10⁻⁹T³ J/(mol·K)T in °C
Ethanolliquidabout 112 J/(mol·K), 2.43 kJ/(kg·K)Mean value 20 to 80 °C
Ethanolvapour61.34 + 0.1572T − 8.749 × 10⁻⁵T² + 1.983 × 10⁻⁸T³ J/(mol·K)T in °C
Airgasabout 29.2 J/(mol·K) near room temperatureRises slowly with T

Ideal gas heat capacities at constant volume and constant pressure differ by the gas constant, \(C_p = C_v + R\). For liquids and solids the two are nearly equal. Heat capacity data in US units are often in Btu/(lbm·°F); because a degree Fahrenheit is smaller than a kelvin by a factor of 1.8 and a Btu per pound-mass is 2.326 kJ/kg, the number in Btu/(lbm·°F) equals the number in cal/(g·°C), so water is 1.0 in both systems.

Phase changes and latent heats

Changing phase at constant temperature and pressure takes or releases latent heat. The heat of vaporization \(\Delta\hat{H}_v\) is the enthalpy change for turning a liquid into vapour at its boiling point; the heat of fusion \(\Delta\hat{H}_m\) is for melting a solid. Condensation and freezing release the same amounts. Latent heats are large: vaporizing water takes 40.66 kJ/mol at 100 °C, the same as heating liquid water through more than 500 K if it stayed liquid.

When a measured value is not available, estimates help. Trouton's rule says that for many nonpolar liquids the heat of vaporization at the normal boiling point is about 88 J/(mol·K) times the boiling point in kelvins. Water and alcohols, with hydrogen bonding, are higher: about 109 J/(mol·K) times \(T_b\). Watson's correlation converts a known heat of vaporization at one temperature to another, falling to zero at the critical point.

\[\Delta\hat{H}_v(T_2) = \Delta\hat{H}_v(T_1)\Big(\frac{T_c - T_2}{T_c - T_1}\Big)^{0.38}\]

The Watson form uses absolute temperatures, and the exponent 0.38 fits a wide range of liquids. For ethanol, with a heat of vaporization of 38.58 kJ/mol at its normal boiling point of 78.4 °C and a critical temperature of 513.9 K, Watson's correlation gives about 43.0 kJ/mol at 25 °C, close to the measured value of about 42.3 kJ/mol.

Worked problem

Duty of the sieve vaporizer

A start-up vaporizer takes 10 000 kg/h of liquid 92.5 wt% ethanol (the rest water) at 25 °C and produces vapour at 150 °C for the molecular sieve. Treating the components as an ideal mixture and ignoring pressure effects, find the heat duty using the data in the table and heats of vaporization of 38.58 kJ/mol (ethanol, 78.4 °C) and 40.66 kJ/mol (water, 100 °C).

Show solution
  1. Molar flows: ethanol \(= 9250/46.07 = 200.8\ \text{kmol/h}\); water \(= 750/18.02 = 41.6\ \text{kmol/h}\).
  2. Ethanol path: liquid \(0.112(78.4 - 25) = 5.98\); vaporize 38.58; vapour \(\int_{78.4}^{150} C_p\,dT = 5.60\ \text{kJ/mol}\); total 50.16 kJ/mol.
  3. Water path: liquid \(0.0754(100 - 25) = 5.65\); vaporize 40.66; vapour \(\int_{100}^{150} C_p\,dT = 1.72\ \text{kJ/mol}\); total 48.04 kJ/mol.
  4. Duty \(= 200.8 \times 50.16 + 41.6 \times 48.04 = 10\,070 + 2000 = 12\,070\ \text{MJ/h} = 3.35\ \text{MW}\).

Answer: More than three quarters of the duty is latent heat. At 10 bar, condensing steam supplies about \(2777.7 - 762.6 = 2015\ \text{kJ/kg}\), so the vaporizer needs about 6000 kg/h of steam. In normal operation the sieve takes vapour straight from the rectifier, and this vaporizer is used only at start-up, which is why plants integrate the sieve with the distillation columns.

Hypothetical-path enthalpy changes for heating and vaporizing 92.5 wt% ethanol, with Watson's correlation for the heat of vaporization at other temperatures.
# Heating and vaporizing 92.5 wt% ethanol: a hypothetical-path enthalpy calculation
# Heat capacity polynomials for the vapours, kJ/(mol C) with T in C (approximate)
CP = {'ethanol': (61.34e-3, 15.72e-5, -8.749e-8, 19.83e-12),
      'water':   (33.46e-3, 0.6880e-5, 0.7604e-8, -3.593e-12)}
CP_LIQ = {'ethanol': 0.112, 'water': 0.0754}       # kJ/(mol C), mean liquid values
TB = {'ethanol': 78.4, 'water': 100.0}               # normal boiling points, C
HV = {'ethanol': 38.58, 'water': 40.66}              # kJ/mol at the normal boiling point
TC = {'ethanol': 513.9, 'water': 647.1}              # critical temperatures, K
M = {'ethanol': 46.07, 'water': 18.02}

def int_cp(sp, T1, T2):
    a, b, c, d = CP[sp]
    f = lambda T: a * T + b * T ** 2 / 2 + c * T ** 3 / 3 + d * T ** 4 / 4
    return f(T2) - f(T1)

def watson(sp, T):
    """Heat of vaporization at T (C) from the value at the normal boiling point."""
    Tr1, Tr2 = (TB[sp] + 273.15) / TC[sp], (T + 273.15) / TC[sp]
    return HV[sp] * ((1 - Tr2) / (1 - Tr1)) ** 0.38

feed = 10_000.0                                       # kg/h of 92.5 wt% ethanol
n = {'ethanol': feed * 0.925 / M['ethanol'], 'water': feed * 0.075 / M['water']}  # kmol/h
T_in, T_out = 25.0, 150.0
total = 0.0                                           # MJ/h (kmol/h times kJ/mol)
for sp in n:
    h1 = CP_LIQ[sp] * (TB[sp] - T_in)               # heat liquid to its boiling point
    h2 = HV[sp]                                     # vaporize
    h3 = int_cp(sp, TB[sp], T_out)                  # heat vapour to the outlet
    dh = h1 + h2 + h3
    total += n[sp] * dh
    print(f"{sp:8s} {n[sp]:7.1f} kmol/h: liquid {h1:6.2f} + vaporize {h2:6.2f} + vapour {h3:6.2f} = {dh:6.2f} kJ/mol")
print(f"Vaporizer and superheater duty {total / 3600:.2f} MW ({total:.0f} MJ/h)")

print("\nHeat of vaporization of ethanol by Watson's correlation")
for T in (25, 50, 78.4, 120, 150):
    print(f"{T:6.1f} C  {watson('ethanol', T):6.2f} kJ/mol")
Output
ethanol    200.8 kmol/h: liquid   5.98 + vaporize  38.58 + vapour   5.60 =  50.16 kJ/mol
water       41.6 kmol/h: liquid   5.65 + vaporize  40.66 + vapour   1.72 =  48.04 kJ/mol
Vaporizer and superheater duty 3.35 MW (12070 MJ/h)

Heat of vaporization of ethanol by Watson's correlation
  25.0 C   42.98 kJ/mol
  50.0 C   41.02 kJ/mol
  78.4 C   38.58 kJ/mol
 120.0 C   34.48 kJ/mol
 150.0 C   30.93 kJ/mol

Heat exchangers

A heat exchanger transfers heat from a hot stream to a cold one without mixing them. For the whole exchanger with negligible heat loss, the heat lost by the hot stream equals the heat gained by the cold stream. Each stream's enthalpy change comes from its heat capacity, or from latent heat if it condenses or boils. The energy balance fixes the outlet temperatures, but not the area of the exchanger, which depends on heat transfer rates and is the subject of a heat transfer course.

\[\dot{m}_h\,C_{p,h}(T_{h,\text{in}} - T_{h,\text{out}}) = \dot{m}_c\,C_{p,c}(T_{c,\text{out}} - T_{c,\text{in}})\]

Worked problem

The beer preheater

Beer at 280 000 kg/h is heated from 32 °C to 65 °C in a preheater by whole stillage from the beer column bottoms, which enters at 108 °C at 215 000 kg/h. Mean heat capacities are 4.05 kJ/(kg·K) for the beer and 3.95 kJ/(kg·K) for the stillage. Find the heat transferred and the stillage outlet temperature.

Show solution
  1. Heat gained by the beer \(= 280\,000 \times 4.05 \times (65 - 32) = 37.42\ \text{GJ/h} = 10.4\ \text{MW}\).
  2. Stillage temperature drop \(= 37.42 \times 10^6/(215\,000 \times 3.95) = 44.1\ \text{K}\).
  3. Stillage outlet \(= 108 - 44.1 = 63.9\ °\text{C}\).

Answer: The preheater recovers 10.4 MW that would otherwise need about 18.6 t/h of steam at 10 bar in the beer column. The beer leaves at 65 °C while the stillage leaves at 63.9 °C, which is possible only with countercurrent flow, where the streams pass in opposite directions. Pinch analysis, taught in process design courses, finds the most heat that can be recovered across a whole plant this way.

Reusing latent heat: multiple-effect evaporation

The evaporator in the stillage train shows how much latent heat matters to plant economics. Lesson 4 found that it removes about 77 t/h of water. In a single-effect evaporator, every kilogram of water boiled off needs roughly a kilogram of steam, because the latent heat of the condensing steam is about equal to the latent heat of the boiling water. The plant's evaporator has four effects in series, each running at a lower pressure than the one before. Vapour boiled off in the first effect condenses in the second effect's heating tubes, where the lower pressure means the liquid boils at a lower temperature; the second effect's vapour heats the third, and so on. Only the first effect needs fresh steam.

An energy balance on each effect, condensing vapour in and boiling vapour out with small sensible-heat corrections, shows that four effects remove about 3.5 kg of water per kilogram of steam rather than one. That saving, around 55 t/h of steam at this plant, is far larger than any other single heat recovery measure. Many plants go further and drive one stage with vapour from the distillation columns, or compress the evaporator vapour mechanically so that electricity rather than steam supplies the latent heat. All of these are designed from the same energy balance you have just learned.

Mixing and solutions

Mixing two liquids or dissolving a solid can release or absorb heat, the heat of mixing or heat of solution. For ideal mixtures it is zero, and the enthalpy of a mixture is the sum of the component enthalpies. For some systems it is large: diluting concentrated sulphuric acid releases enough heat to boil water, which is why acid is always added to water and never the reverse. Ethanol and water have a small exothermic heat of mixing, a few hundred joules per mole at most, which is usually neglected in plant balances but appears in accurate simulator models. Heats of solution are tabulated for common salts, acids and bases, and enthalpy-concentration charts combine all the effects for systems such as sulphuric acid and water or ammonia and water.

Humidification and the psychrometric chart

Air-water systems, such as the grain dryers and the cooling towers, combine sensible heat and latent heat. The psychrometric chart plots humidity against temperature at atmospheric pressure with lines of relative humidity, specific volume, enthalpy and wet-bulb temperature. Adiabatic humidification, in which air passes over water and evaporates some of it without any external heat, follows a line of nearly constant wet-bulb temperature: the air cools as it picks up moisture. Cooling towers use this to reject the heat from the condensers, and evaporative coolers use it for comfort cooling in dry climates.

Common mistakes

  • MistakeUsing a heat of vaporization at the wrong temperature.

    InsteadBuild the path to change phase where the latent heat is known, or correct it with Watson's correlation.

  • MistakeTreating C_p as constant over a wide range for a gas.

    InsteadIntegrate the polynomial; gas heat capacities rise noticeably with temperature.

  • MistakeUsing degrees Fahrenheit differences with kJ/(kg·K) data.

    InsteadConvert the temperature difference, or convert C_p to Btu/(lbm·°F).

  • MistakeAssuming any outlet temperatures are possible in a heat exchanger.

    InsteadThe hot outlet must stay above the cold inlet and vice versa; check feasibility.

  • MistakeIgnoring heats of mixing for acids and bases.

    InsteadStrongly nonideal solutions need heats of solution or enthalpy-concentration charts.

Summary and key terms

  • Enthalpy changes can be calculated along any convenient hypothetical path.
  • Sensible heat: ΔH = ∫C_p dT, with polynomial C_p for gases and mean values for liquids.
  • Latent heats are large; Trouton's rule estimates them and Watson's correlation shifts them in temperature.
  • In a heat exchanger, the hot stream's enthalpy loss equals the cold stream's gain.
  • Mixing and solution heats are zero for ideal mixtures but large for acids and bases in water.
  • Air-water systems combine sensible and latent heat and are analysed with psychrometric charts.
Sensible heat
Heat that changes temperature without changing phase.
Latent heat
The enthalpy change accompanying a phase change at constant temperature.
Heat capacity
The rate of change of enthalpy (C_p) or internal energy (C_v) with temperature.
Hypothetical process path
An imaginary sequence of simple steps used to calculate a state property change.
Trouton's rule
An estimate of the heat of vaporization as about 88 J/(mol·K) times the normal boiling point.
Heat of solution
The enthalpy change when a substance dissolves in a solvent.
Countercurrent flow
Heat exchanger flow in which the two streams move in opposite directions.
Wet-bulb temperature
The temperature reached by water evaporating into air under adiabatic conditions.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. How much heat raises 1000 kg of water from 20 to 80 °C?

    Show answer

    1000 × 4.18 × 60 = 250 800 kJ, about 251 MJ.

  2. Estimate the heat of vaporization of n-hexane (T_b = 341.9 K) by Trouton's rule.

    Show answer

    88 × 341.9 = 30 100 J/mol, about 30.1 kJ/mol (measured about 28.9).

  3. Using the program's Watson output, what is ethanol's heat of vaporization at 120 °C?

    Show answer

    About 34.5 kJ/mol.

  4. In the preheater example, what stillage outlet temperature results if the beer is heated to 70 °C?

    Show answer

    Q = 280 000 × 4.05 × 38 = 43.09 GJ/h; ΔT = 50.7 K; outlet 57.3 °C.

  5. Convert 2.43 kJ/(kg·K) to Btu/(lbm·°F).

    Show answer

    2.43/4.187 = 0.580 Btu/(lbm·°F).

  6. Why is the latent heat step the largest part of the vaporizer duty?

    Show answer

    Vaporization breaks intermolecular attractions, which takes far more energy than raising the temperature of either phase by tens of degrees.

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