4.3 Process Energy Balance Applications

4.3 Process Energy Balance Applications

Quick Answer: For steady heaters/coolers with no shaft work, Q̇ ≈ ṁ Cp ΔT (sensible) plus ṁ λ (latent) as needed. For mixers, solve mass balances first, then Σ ṁ_in ĥ_in = ṁ_out ĥ_out if adiabatic. Always close material balance before energy.

This section is the applied payoff of 4.1–4.2 for UPDA/MMUP Chemical Domain A: compute duties and outlet temperatures on everyday units. Depth matches what a 25-question professional exam can test—clean setups, consistent units, and no skipped streams—not full rigorous simulator thermo.

General Steady Open-System Pattern

For equipment at steady state, neglect KE/PE unless given:

Σ ṁ_in ĥ_in − Σ ṁ_out ĥ_out + Q̇ + Ẇ_s ≈ 0

Sensible Heat: m·Cp·ΔT

When a stream changes temperature without phase change, and Cp is roughly constant:

Δĥ ≈ Cp ΔT

Q̇ ≈ ṁ Cp (T_out − T_in) (single stream, no W_s, no phase change)

Mean Cp: If Cp varies, use Cp at average T or tabulated enthalpy differences. On MCQs, a single Cp is usually supplied.

Solids, liquids, gases: Liquids often have higher Cp per mass than gases; do not use water’s Cp for hydrocarbon gas without data.

Worked Example A — Liquid Heater Duty

Problem. A desalination preheat train warms brackish water (treat as liquid water for Cp):

• ṁ = 3.0 kg/s
• T_in = 28 °C
• T_out = 52 °C
• Cp = 4.18 kJ/(kg·°C)
• Steady, no phase change, no shaft work, neglect KE/PE

Find heat duty added to the water.

Solution.

Q̇ = ṁ Cp (T_out − T_in) = 3.0 × 4.18 × (52 − 28) = 3.0 × 4.18 × 24 = 3.0 × 100.32 = 300.96 kW ≈ 301 kW

Exam check: Order of magnitude—about 0.3 MW for 3 kg/s and 24 °C rise on water is reasonable.

Latent Heat

Phase change at constant temperature (idealized pure component at fixed P):

Q̇_latent ≈ ṁ λ

where λ is latent heat (kJ/kg) of vaporization, fusion, etc.

Combined sensible + latent (e.g., cool vapor to dew point, condense fully, subcool liquid):

Q̇ ≈ ṁ [ Cp_v (T_dew − T_in) + (−λ) + Cp_l (T_out − T_dew) ]

with careful signs: condensation releases heat from the process stream (negative contribution to Q into the stream).

Worked Example B — Condenser Duty

Problem. Pure organic vapor enters a condenser:

• ṁ = 0.50 kg/s
• Inlet: saturated vapor at 80 °C (already at dew point)
• Fully condensed to saturated liquid at 80 °C
• λ = 380 kJ/kg at 80 °C
• Then subcooled liquid to 45 °C with Cp_l = 2.2 kJ/(kg·°C)
• Steady, no shaft work

Find total heat removal rate from the process fluid.

Solution.

Condensation: heat leaving fluid = ṁ λ = 0.50 × 380 = 190 kW

Subcooling: heat leaving fluid = ṁ Cp_l (80 − 45) = 0.50 × 2.2 × 35 = 38.5 kW

Total heat removed = 190 + 38.5 = 228.5 kW

(If Q̇ is heat into the organic stream: Q̇ = −228.5 kW.)

Latent heat dominates—a common qualitative exam point.

Mixers (Adiabatic and Non-adiabatic)

Mass balance first (Section 3.1):

ṁ₁ + ṁ₂ = ṁ_out

Energy (steady, no W_s):

ṁ₁ ĥ₁ + ṁ₂ ĥ₂ + Q̇ = ṁ_out ĥ_out

Adiabatic mixer (Q̇ = 0):

ĥ_out = (ṁ₁ ĥ₁ + ṁ₂ ĥ₂) / ṁ_out

If streams are the same liquid with constant Cp and no mixing heat beyond sensible:

T_out ≈ (ṁ₁ T₁ + ṁ₂ T₂) / ṁ_out

Caution: Mixing concentrated acid and water can release significant heat of mixing—then simple Cp blending under-predicts T_out. Unless the stem mentions heat of mixing, UPDA-style problems usually assume ideal sensible mixing.

Worked Example C — Two-Stream Temperature Blend

Problem. Adiabatic mixing of two liquid streams (same Cp, ideal mixing):

• Stream 1: 2.0 kg/s at 20 °C
• Stream 2: 1.0 kg/s at 80 °C
• Cp identical and constant; no phase change

Find T_out.

Solution.

ṁ_out = 3.0 kg/s

T_out = (2.0 × 20 + 1.0 × 80) / 3.0 = (40 + 80) / 3.0 = 40 °C

If the mixer is not adiabatic and a coil removes 50 kW, with Cp = 4.0 kJ/(kg·°C):

Energy: ṁ₁ Cp T₁ + ṁ₂ Cp T₂ + Q̇ = ṁ_out Cp T_out

Q̇ = −50 kW (into system)

3.0 × 4.0 × T_out = 2.0×4.0×20 + 1.0×4.0×80 − 50 = 160 + 320 − 50 = 430

T_out = 430 / 12 = 35.8 °C

Cooling lowers the outlet temperature below the adiabatic 40 °C—as expected.

Combined Mass and Energy Workflow

1. Flowsheet sketch and system boundary around the unit.
2. Material balances → all unknown ṁ and compositions.
3. State each stream (T, phase) → choose Cp, λ, or ĥ data.
4. Energy balance → Q̇, Ẇ_s, or unknown T.
5. Sanity check: energy units; exo/endo consistency; outlet T between feed temperatures for adiabatic same-phase mixing of similar fluids.

Worked Example D — Heater with Unknown Outlet Temperature

Problem. A gas stream is heated with Q̇ = 120 kW electric duty (all absorbed by the gas):

• ṁ = 1.2 kg/s
• T_in = 40 °C
• Cp = 1.05 kJ/(kg·°C)
• No phase change, no shaft work, steady

Find T_out.

Solution.

Q̇ = ṁ Cp (T_out − T_in)

120 = 1.2 × 1.05 × (T_out − 40)

120 = 1.26 (T_out − 40)

T_out − 40 = 120 / 1.26 ≈ 95.24

T_out ≈ 135.2 °C

Reacting Systems (Brief Tie-In)

When reaction occurs, either:

• Include ΔH_rxn × extent with sensible terms, or
• Use absolute/formation-based enthalpies for each species

Do not double-count by adding ΔH_rxn on top of formation-based ĥ values. Section 4.2 covered the chemistry side; here remember that reactor energy balances are still open-system first-law statements with mass-closed extents.

Unit Consistency Checklist

UPDA Exam Strategy

• Domain A is about 20% of this guide's planning allocation—energy items often reward one clean equation more than exotic thermo.
• If numbers are given for ṁ, Cp, and two temperatures, compute Q̇ directly.
• If outlet T is asked and Q̇ is given, rearrange Q̇ = ṁ Cp ΔT.
• For multi-stream problems, write mass lines first on scratch paper even when the stem looks “only thermal.”
• Watch exothermic reactor + cooler wording: total jacket duty must remove reaction heat and any sensible cooling required.

Common Traps

• Using ΔT in absolute K incorrectly as a ratio (only differences matter for CpΔT).
• Applying liquid Cp to a two-phase condenser without splitting latent and sensible pieces.
• Adiabatic mixer: expecting T_out outside the feed temperature range for simple same-phase blending (usually wrong without heat of mixing or reaction).
• Solving energy before mass when an unknown flow still exists.

Mastering these applications completes Chapter 4’s energy triad: forms and first law (4.1), reaction heats (4.2), and unit duties with sensible/latent heat (4.3). Thermodynamics chapters that follow deepen state properties; transport chapters reuse the same duty concepts for heat exchangers with U and LMTD.