2.1 Enthalpy Accounting, Sensible Heat, and Phase Changes

Enthalpy Accounting, Sensible Heat, and Phase Changes

Energy balance calculations sit inside Mass/Energy Balances, which the NCEES PE Chemical CBT specification (effective January 1, 2020) weights at 12-18 of the 80 exam questions, with 6-9 of those drawn specifically from the Energy Balances subtopic. Mastering open-system energy conservation requires absolute clarity regarding enthalpy accounting, reference state conventions, sensible heat path integrals, and latent heat scaling.

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1. Open-System First Law Formulation

For a continuous, steady-state open chemical process unit with multiple entering and exiting streams, the First Law of Thermodynamics is stated as:

$$\dot{Q} - \dot{W}_s = \sum_{out} \dot{m}_{out} \left( \hat{h} + \frac{u^2}{2} + gz \right)_{out} - \sum_{in} \dot{m}_{in} \left( \hat{h} + \frac{u^2}{2} + gz \right)_{in}$$

Where:

• $\dot{Q}$ = rate of heat transfer across the system boundary ($\text{kW}$ or $\text{Btu/h}$), where $\dot{Q} > 0$ denotes heat added to the system.
• $\dot{W}_s$ = shaft work delivered by the system to the surroundings ($\text{kW}$ or $\text{hp}$), where $\dot{W}_s > 0$ denotes work produced (e.g., by a turbine or expander) and $\dot{W}_s < 0$ denotes work consumed (e.g., by a pump or compressor).
• $\hat{h}$ = specific flowing enthalpy ($\text{kJ/kg}$ or $\text{Btu/lb}$), defined fundamentally as $\hat{h} = \hat{u} + P\hat{v}$.
• $u^2 / 2$ = specific kinetic energy ($\text{m}^2/\text{s}^2 = \text{J/kg}$).
• $gz$ = specific potential energy ($\text{J/kg}$).

In the vast majority of chemical process units (heat exchangers, distillation columns, reboilers, condensers, reactors, and flash vessels), kinetic and potential energy variations between process inlets and outlets are negligible compared to thermal enthalpy changes ($\Delta \hat{h} \gg \Delta \hat{e}_k, \Delta \hat{e}_p$). The energy balance simplifies to:

$$\dot{Q} - \dot{W}_s = \Delta \dot{H} = \sum_{out} \dot{m}_{out} \hat{h}_{out} - \sum_{in} \dot{m}_{in} \hat{h}_{in}$$

When no moving shafts penetrate the control volume (e.g., heat exchangers, flash drums), $\dot{W}_s = 0$, yielding $\dot{Q} = \Delta \dot{H}$.

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2. Reference States and Hypothetical Process Paths

Enthalpy is a relative state function; absolute enthalpy cannot be measured. Only differences in enthalpy ($\Delta \hat{h}$) between defined thermodynamic states have physical meaning. To assign numerical enthalpy values to process streams, chemical engineers define an arbitrary reference state $(T_{ref}, P_{ref}, \text{phase})$.

Hypothetical Process Path Construction

Because enthalpy is a state function independent of the actual mechanism or physical path taken, the enthalpy change between State 1 $(T_1, P_1, \text{liquid})$ and State 2 $(T_2, P_2, \text{vapor})$ is evaluated by constructing a convenient hypothetical multi-step path composed of individual steps where physical property data are readily available:

1. Step 1: Isothermal Pressure Change on Liquid (often neglected for liquids below $30\text{ bar}$): $$\Delta \hat{h}_1 = \int_{P_1}^{P_{sat}(T_1)} \left[ \hat{v} - T \left(\frac{\partial \hat{v}}{\partial T}\right)_P \right] dP \approx \hat{v}_L (P_{sat} - P_1)$$
2. Step 2: Isobaric Sensible Heating of Liquid from $T_1$ to normal boiling point $T_b$: $$\Delta \hat{h}_2 = \int_{T_1}^{T_b} C_{p,L}(T) dT$$
3. Step 3: Isothermal, Isobaric Phase Transition (Vaporization) at $T_b$ and $P_{sat}(T_b)$: $$\Delta \hat{h}_3 = \Delta \hat{h}_{vap}(T_b)$$
4. Step 4: Isobaric Sensible Heating of Vapor from $T_b$ to $T_2$: $$\Delta \hat{h}_4 = \int_{T_b}^{T_2} C_{p,V}(T) dT$$
5. Step 5: Isothermal Pressure Change on Vapor from $P_{sat}$ to $P_2$: For ideal gases, $(\partial \hat{h} / \partial P)_T = 0$, so $\Delta \hat{h}_5 = 0$. For real gases at high pressure, enthalpy departure functions must be incorporated.

The overall enthalpy change is the algebraic sum of all steps:

$$\Delta \hat{h} = \Delta \hat{h}_1 + \Delta \hat{h}_2 + \Delta \hat{h}_3 + \Delta \hat{h}_4 + \Delta \hat{h}_5$$

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3. Sensible Heat and Heat Capacity Integrations

Sensible heat refers to energy exchanged that produces a temperature change without inducing a phase transition. The fundamental differential relation is:

$$d\hat{h} = C_p(T) dT + \left[ \hat{v} - T \left(\frac{\partial \hat{v}}{\partial T}\right)_P \right] dP$$

Temperature-Dependent Heat Capacities

For ideal gases and real fluids at moderate pressures, the pressure differential vanishes, leaving $d\hat{h} = C_p(T) dT$. In the NCEES PE Chemical Reference Handbook, molar heat capacities are expressed as polynomials:

$$C_p(T) = a + bT + cT^2 + dT^3$$

Integrating between temperature limits $T_1$ and $T_2$ (where $T$ must be in absolute Kelvin or Rankine as specified by the correlation coefficients):

$$\Delta \hat{H} = \int_{T_1}^{T_2} C_p(T) dT = a(T_2 - T_1) + \frac{b}{2}(T_2^2 - T_1^2) + \frac{c}{3}(T_2^3 - T_1^3) + \frac{d}{4}(T_2^4 - T_1^4)$$

Alternatively, a mean heat capacity $\bar{C}_p$ over the operating range can be evaluated:

$$\bar{C}_p = \frac{\int_{T_1}^{T_2} C_p(T) dT}{T_2 - T_1} \implies \Delta \hat{H} = \bar{C}_p (T_2 - T_1)$$

Incompressible Liquids and Mechanical Enthalpy

For liquids modeled as incompressible (constant specific volume $\hat{v}$ and thermal expansion coefficient $\beta = \frac{1}{\hat{v}}(\partial \hat{v}/\partial T)_P \approx 0$):

$$\Delta \hat{h} = \int_{T_1}^{T_2} C_p dT + \hat{v}(P_2 - P_1)$$

In low-pressure liquid operations, $\hat{v}\Delta P$ is negligible. For example, water with $\hat{v} = 0.001\text{ m}^3/\text{kg}$ experiencing a $100\text{ kPa}$ ($1\text{ bar}$) pressure rise exhibits $\hat{v}\Delta P = 0.1\text{ kJ/kg}$. By comparison, heating water by just $1^\circ\text{C}$ requires $C_p \Delta T \approx 4.184\text{ kJ/kg}$—over 40 times greater. However, in boiler feed pumps operating up to $15\text{ MPa}$ ($150\text{ bar}$), $\hat{v}\Delta P = 0.001\text{ m}^3/\text{kg} \times 15,000\text{ kPa} = 15.0\text{ kJ/kg}$, which represents significant shaft work.

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4. Latent Heat of Phase Changes

Phase transitions occur at constant temperature and pressure for pure components. Latent heat is the enthalpy required to alter the molecular arrangement between phases:

• Latent Heat of Vaporization ($\Delta H_{vap}$): Liquid to vapor transition.
• Latent Heat of Fusion ($\Delta H_{fus}$): Solid to liquid transition.
• Latent Heat of Sublimation ($\Delta H_{sub}$): Solid directly to vapor transition, where $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$ at the triple point.

Temperature Dependence: The Watson Correlation

When latent heat of vaporization is tabulated only at the normal boiling point $T_{b1}$, but vaporization occurs at an elevated or reduced system pressure corresponding to saturation temperature $T_2$, the Watson correlation must be applied:

$$\frac{\Delta H_{vap}(T_2)}{\Delta H_{vap}(T_1)} = \left( \frac{1 - T_{r2}}{1 - T_{r1}} \right)^{0.38} = \left( \frac{T_c - T_2}{T_c - T_1} \right)^{0.38}$$

Where:

• $T_c$ = critical temperature of the substance (absolute temperature, $\text{K}$ or $^\circ\text{R}$).
• $T_{r} = T / T_c$ = reduced temperature.

As temperature approaches the critical point ($T \to T_c$), reduced temperature approaches unity ($T_r \to 1$), and the latent heat of vaporization vanishes identically to zero ($\Delta H_{vap} \to 0$).

Two-Phase Quality Calculations

For a two-phase liquid-vapor mixture (e.g., wet steam exiting a boiler or flash drum), the stream enthalpy is determined by the quality $x$ (mass fraction of vapor):

$$\hat{h} = (1 - x)\hat{h}_f + x\hat{h}_g = \hat{h}_f + x\hat{h}_{fg}$$

Where $\hat{h}_f$ is saturated liquid enthalpy, $\hat{h}_g$ is saturated vapor enthalpy, and $\hat{h}_{fg} = \hat{h}_g - \hat{h}_f = \Delta \hat{h}_{vap}$.

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5. Summary Table: Enthalpy Accounting Across Regimes

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6. Worked Numerical Example: Multistage Condenser-Subcooler Sizing

Problem Statement

An organic recovery unit processes a superheated pure ethanol vapor stream at a rate of $\dot{m} = 1,800\text{ kg/h}$ ($0.500\text{ kg/s}$). The vapor enters a shell-and-tube condenser at $T_1 = 120.0^\circ\text{C}$ and $P = 101.325\text{ kPa}$ ($1\text{ atm}$). The ethanol is completely condensed and subcooled to $T_2 = 30.0^\circ\text{C}$ before entering an atmospheric storage vessel.

Cooling water enters the tube side at $T_{cw,in} = 20.0^\circ\text{C}$ with a permitted maximum temperature rise of $\Delta T_{cw} = 15.0^\circ\text{C}$ (exiting at $35.0^\circ\text{C}$).

Physical Property Data for Ethanol ($MW = 46.07\text{ g/mol}$):

• Normal boiling point: $T_b = 78.3^\circ\text{C}$
• Average vapor heat capacity: $C_{p,V} = 1.750\text{ kJ/(kg}\cdot\text{K)}$
• Latent heat of vaporization at $78.3^\circ\text{C}$: $\Delta \hat{h}_{vap} = 846.0\text{ kJ/kg}$
• Average liquid heat capacity: $C_{p,L} = 2.570\text{ kJ/(kg}\cdot\text{K)}$
• Cooling water heat capacity: $C_{p,w} = 4.184\text{ kJ/(kg}\cdot\text{K)}$

Calculate:

1. The desuperheating duty ($\dot{Q}_1$).
2. The condensation duty ($\dot{Q}_2$).
3. The subcooling duty ($\dot{Q}_3$).
4. The total heat removal rate ($\dot{Q}_{total}$).
5. The required cooling water mass flow rate ($\dot{m}_{cw}$).

Step-by-Step Solution

Step 1: Construct the Hypothetical Path for Ethanol

The process operates at steady state with no shaft work ($\dot{W}_s = 0$). Hence, $\dot{Q} = \Delta \dot{H}$. The path breaks into three sequential steps:

$$\Delta \dot{H} = \Delta \dot{H}_1 + \Delta \dot{H}_2 + \Delta \dot{H}_3$$

Step 2: Zone 1 — Desuperheating the Vapor ($120.0^\circ\text{C} \to 78.3^\circ\text{C}$)

$$\Delta \dot{H}_1 = \dot{m} \int_{120.0}^{78.3} C_{p,V} dT = \dot{m} C_{p,V} (T_b - T_1)$$ $$\Delta \dot{H}_1 = (0.500\text{ kg/s}) \times (1.750\text{ kJ/(kg}\cdot\text{K)}) \times (78.3 - 120.0)\text{ K}$$ $$\Delta \dot{H}_1 = (0.500) \times (1.750) \times (-41.7) = -36.488\text{ kW}$$

Step 3: Zone 2 — Condensing Saturated Vapor to Saturated Liquid ($78.3^\circ\text{C}$)

Vaporization is endothermic ($+\Delta \hat{h}_{vap}$); condensation is exothermic ($-\Delta \hat{h}_{vap}$):

$$\Delta \dot{H}_2 = -\dot{m} \Delta \hat{h}_{vap}$$ $$\Delta \dot{H}_2 = -(0.500\text{ kg/s}) \times (846.0\text{ kJ/kg}) = -423.000\text{ kW}$$

Step 4: Zone 3 — Subcooling the Liquid ($78.3^\circ\text{C} \to 30.0^\circ\text{C}$)

$$\Delta \dot{H}_3 = \dot{m} \int_{78.3}^{30.0} C_{p,L} dT = \dot{m} C_{p,L} (T_2 - T_b)$$ $$\Delta \dot{H}_3 = (0.500\text{ kg/s}) \times (2.570\text{ kJ/(kg}\cdot\text{K)}) \times (30.0 - 78.3)\text{ K}$$ $$\Delta \dot{H}_3 = (0.500) \times (2.570) \times (-48.3) = -62.066\text{ kW}$$

Step 5: Total Heat Duty

$$\dot{Q}_{total} = \Delta \dot{H}_1 + \Delta \dot{H}_2 + \Delta \dot{H}_3 = -36.488 - 423.000 - 62.066 = -521.554\text{ kW} \approx -521.6\text{ kW}$$

The cooling duty required is $521.6\text{ kW}$ removed from the ethanol stream. Condensation accounts for $423.0 / 521.6 = 81.1\%$ of the total duty.

Step 6: Cooling Water Flow Rate

An energy balance on the tube side yields $\dot{Q}_{absorbed} = -\dot{Q}_{total} = 521.554\text{ kW}$:

$$\dot{m}_{cw} = \frac{\dot{Q}_{absorbed}}{C_{p,w} \Delta T_{cw}} = \frac{521.554\text{ kJ/s}}{(4.184\text{ kJ/(kg}\cdot\text{K)}) \times (15.0\text{ K})} = \frac{521.554}{62.76} = 8.310\text{ kg/s}$$

In hourly mass units: $8.310\text{ kg/s} \times 3,600\text{ s/h} = 29,916\text{ kg/h}$.

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7. Critical PE Exam Traps & Pitfalls

Trap 1: Misinterpreting Reference States Across Sources
Steam tables set liquid internal energy and enthalpy to zero at the triple point of water ($0.01^\circ\text{C}$, $0.00611\text{ bar}$). Chemical thermodynamic tables (like Perry's or NIST) set the elemental standard enthalpy of formation to zero at $25^\circ\text{C}$ ($298.15\text{ K}$) and $1\text{ bar}$. If you calculate $\Delta H$ for water by subtracting a steam table enthalpy from a NIST enthalpy, your answer will be off by approximately $105\text{ kJ/kg}$! Always compute stream differences using the same reference source.

Trap 2: Ignoring Temperature Variations in Latent Heat
The latent heat of vaporization decreases significantly as temperature increases. Using the normal boiling point latent heat $\Delta H_{vap}(T_b)$ for a high-pressure reboiler running at elevated saturation temperature severely overpredicts the available heat duty and undersizes heat transfer areas. Always scale with Watson's correlation when working at off-normal pressures.

Trap 3: Sign Confusion on Shaft Work
The standard NCEES open-system formulation writes $\dot{Q} - \dot{W}_s = \Delta \dot{H}$. When solving for a pump or compressor that consumes energy, $\dot{W}_s$ is negative. Therefore, $\Delta \dot{H} = \dot{Q} - (-\dot{W}_{pump}) = \dot{Q} + \dot{W}_{pump}$. The enthalpy of the fluid increases due to mechanical work input.