4.1 Energy Forms and the First Law for Processes

4.1 Energy Forms and the First Law for Processes

Quick Answer: The first law for processes is an energy balance: energy is conserved. Track U, H = U + PV, KE, PE, heat Q, and work W (including shaft work). Closed systems exchange Q and W only; open systems also transport enthalpy with mass. At steady flow with small KE/PE changes, ΔH ≈ Q + W_s once your sign convention is fixed.

Domain A on the UPDA/MMUP Chemical exam pairs material balances with energy balances. After you can close mass and moles (Chapters 2–3), the next step is accounting for heat duties, work, and temperature changes on heaters, coolers, compressors, and reactors. This section builds the energy vocabulary and the first-law forms you will reuse in every later Domain A energy item.

Why Energy Balances Matter on UPDA Chemical

Qatar’s process industries—LNG trains, gas processing, refining, petrochemicals, utilities—run on equipment whose duty (kW or MJ/h) is set by energy balances, not by material balances alone. Exam stems may ask for:

• Heat required to raise a stream temperature
• Cooling water load for a condenser
• Shaft power for a pump or compressor (conceptual or simplified numerical)
• Whether a process is endothermic or exothermic and what that means for utility demand

You do not need graduate-level thermodynamics here (that deepens in Chapters 5–6). You need correct first-law bookkeeping on a clear system.

Energy Forms Used in Process Calculations

Enthalpy is not “heat inside the fluid.” It is a state function defined so that open-system energy balances stay compact: the Pv term accounts for the work needed to push fluid into and out of the control volume.

Specific vs Extensive

• Extensive: U, H, KE, PE scale with mass or moles (kJ, MJ).
• Specific (per mass): û, ĥ, often kJ/kg.
• Molar: Ū, H̄, often kJ/kmol or kJ/mol.

Always match the basis of your material balance (kg/h vs kmol/h) to the energy units.

The First Law as a Process Balance

Write energy the same way you write mass:

In − Out + Generation − Consumption = Accumulation

For total energy, ordinary chemical processes do not “generate energy from nothing.” What people call “heat of reaction” is a change in the enthalpy (or internal energy) of the mixture relative to a reference—handled by evaluating H at inlet and outlet states, or by adding ΔH_rxn explicitly in a formation-enthalpy framework (Section 4.2).

Closed System (No Mass Crossing the Boundary)

A closed system has fixed mass. Energy crosses only as heat and work:

ΔU + ΔKE + ΔPE = Q + W

(with your chosen signs for Q and W). If the system is stationary (no bulk KE/PE change):

ΔU = Q + W

Batch reactor or sealed tank heating problems often start here. If volume is constant and only PV-type work is considered, W may be zero and Q = ΔU. If pressure is constant and the only work is expansion work, Q = ΔH for many simple cases—know which constraint the stem states.

Open System (Mass Crosses the Boundary)

An open system (control volume) has streams entering and leaving. Each stream carries:

ĥ + ½v² + g z

plus any energy transferred as Q and W to the control volume.

Rate form (continuous process):

Σ ṁ_in (ĥ + ½v² + gz)_in − Σ ṁ_out (ĥ + ½v² + gz)_out + Q̇ + Ẇ = dE_cv/dt

At steady state, accumulation dE_cv/dt = 0, and often KE and PE changes are dropped:

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

or, for a single inlet and single outlet with ṁ_in = ṁ_out = ṁ:

ṁ (ĥ_out − ĥ_in) ≈ Q̇ + Ẇ_s

which is the workhorse steady-flow energy balance for heaters, coolers, and many single-stream exchangers (when shaft work is zero, Q̇ ≈ ṁ Δĥ).

Sign Conventions Candidates Must Track

Textbooks and exam prep materials do not all use the same signs. Two common conventions:

Rule for UPDA-style MCQs:

1. Read whether the option set treats heat duty as a positive number “required” (magnitude) or as a signed term.
2. State one convention in your scratch work and never mix mid-problem.
3. For “how much heat must be added?”, compute the enthalpy rise and report a positive duty if the process needs heat—even if your equation had Q as a signed variable.

Shaft work: pumps and compressors do work on the fluid (energy of the stream rises if heat is negligible). Turbines extract work from the fluid.

When KE and PE Matter (and When They Do Not)

If a stem never gives velocities or heights, neglect KE and PE unless the unit is clearly a nozzle or hydroelectric-style device.

Worked Example: Closed vs Open Framing

Scenario. A horizontal tank holds 500 kg of liquid at rest. An electric heater adds 120 MJ of energy as heat; no mass enters or leaves; no stirring work; tank volume is essentially constant; kinetic and potential energy unchanged.

Closed-system balance:

ΔU = Q + W → ΔU = 120 MJ + 0 = 120 MJ

If the liquid’s effective heat capacity is such that ΔU ≈ m û(T) change, temperature rises accordingly. The point for the exam: no enthalpy of flow appears because no mass crossed the boundary.

Same energy as a heater on a flowing stream. Suppose instead a continuous heater raises a liquid stream:

• ṁ = 2.0 kg/s
• ĥ_out − ĥ_in = 60 kJ/kg
• Negligible KE/PE, no shaft work

Then Q̇ ≈ ṁ Δĥ = 2.0 × 60 = 120 kW (120 kJ/s).

Same numerical “120” energy rate idea, different system class: open steady flow uses enthalpy, closed tank used internal energy.

Worked Numerical Example: Steady Single-Stream Cooler

Problem. A process gas is cooled in a heat exchanger. Steady flow, one inlet, one outlet, no shaft work, neglect KE and PE.

• ṁ = 1.5 kg/s
• ĥ_in = 820 kJ/kg
• ĥ_out = 540 kJ/kg

Find the heat transfer rate and state whether heat is removed from the gas.

Solution.

ṁ (ĥ_out − ĥ_in) = Q̇ + Ẇ_s

Ẇ_s = 0

Q̇ = 1.5 × (540 − 820) = 1.5 × (−280) = −420 kW

Using the “Q into the system is positive” convention, Q̇ is negative → heat leaves the gas at 420 kW. The cooler’s process-side duty is 420 kW of heat removal (utility side picks that heat up, subject to losses).

Linking Energy to Material Balances

Always close mass first (Chapter 3). Energy balances need stream flow rates. For multi-stream units:

1. Draw the boundary.
2. Write total and component mass balances → all ṁ_i.
3. Assign state (T, phase, composition) → look up or estimate ĥ_i.
4. Write the steady energy balance for Q̇ and/or Ẇ_s.

Missing a stream flow is the fastest way to get a wrong duty.

Common Traps

• Using ΔU for a continuous heater when the right property is ΔH (open system).
• Forgetting shaft work on a compressor while including a huge invented KE term.
• Mixing kW (rate) with kJ (amount) without a time basis.
• Treating Q as a state function (“the heat of the outlet stream”).
• Inconsistent signs: adding heat duty as positive in one equation and negative in the next.

Exam Workflow

1. Classify: closed vs open; steady vs unsteady; single- vs multi-stream.
2. List forms: which of U, H, KE, PE, Q, W matter?
3. Write the first-law equation with an explicit sign convention.
4. Drop KE/PE only with justification.
5. Solve for the unknown duty or outlet enthalpy, then convert to temperature if Cp is given (Section 4.3).

Master these forms before heats of reaction: Section 4.2 folds chemistry into the same enthalpy framework, and Section 4.3 applies the balances to heaters, coolers, and mixers with sensible and latent heat.