Thermodynamics · Chapter 2
Thermodynamics · Chapter 2

Energy, Energy Transfer, and General Energy Analysis

Energy is the central currency of thermodynamics. This chapter catalogs the forms energy takes, the two ways it crosses a boundary — heat and work — and the bookkeeping rule that ties them together: the first law.

Learning objectives

  • Distinguish stored energy from heat, work, and mass-carried transfer.
  • Select internal energy for closed storage and enthalpy for flowing streams.
  • Apply consistent heat and work signs to process and cycle balances.
  • Estimate kinetic and potential contributions before neglecting them.
  • Compute actual conversion efficiency from clearly defined desired output and input.
  • Audit first-law closure and second-law admissibility.
2–1

Introduction

Energy accounting begins at the boundary

Name the system before naming heat or work. Energy is a property stored as internal, kinetic, potential, chemical, and other forms; heat and work describe transfer across a boundary and therefore depend on the path.

Physical meaning

An energy balance can close perfectly yet describe the wrong equipment if its boundary omits a motor, cooling jacket, or flow stream. Boundary sketches are part of the calculation, not decoration.

2–2

Forms of Energy

Total energy
\[E=U+KE+PE=U+\frac{mV^2}{2}+mgz\]

Microscopic and macroscopic storage

Internal energy \(U\) is microscopic and independent of the external coordinate frame. Kinetic energy depends on velocity relative to a selected frame; potential energy depends on an elevation datum. State these references whenever their changes matter.

EXAMPLE 1Energy storage in a lifted test capsule
A 4.00 kg sealed test capsule receives 90.0 kJ of heat and 35.0 kJ of paddle work while rising 12.0 m and accelerating from 2.0 to 8.0 m/s. With \(g=9.81\ \text{m/s}^2\), find its internal-energy change.

Recognition strategy

A sealed object is a closed system. Heat and shaft work enter, while changes in kinetic and gravitational potential energy consume part of the net transfer; the remainder changes internal energy.

Given information

  • Fixed 4.00 kg system undergoing a process, not a cycle; \(Q_{in}=90.0\ \text{kJ}\) and \(W_{shaft,in}=35.0\ \text{kJ}\).
  • \(V_1=2.0\ \text{m/s}\), \(V_2=8.0\ \text{m/s}\), and \(z_2-z_1=12.0\ \text{m}\).
  • No property table is needed; mass basis; heat into and work out are positive in \(Q-W=\Delta E\), so paddle input means \(W=-35.0\ \text{kJ}\).

Model and interpretation

The system boundary follows the capsule. Use the exact closed-system first law with measured transfers, neglect deformation energy, and reference PE to any common elevation datum. Temperature and pressure are not used; absolute bases and an exergy environment are therefore unnecessary. No mass crosses the boundary.

Step-by-step solution
1. Write \(Q-W=\Delta U+\Delta KE+\Delta PE\), retaining both macroscopic terms rather than assuming them negligible.
2. Calculate \(\Delta KE=\tfrac12(4)[8^2-2^2]=120\ \text{J}=0.120\ \text{kJ}\).
3. Calculate \(\Delta PE=mg\Delta z=(4)(9.81)(12)=471\ \text{J}=0.471\ \text{kJ}\).
4. Resolve storage: \(\Delta U=90-(-35)-0.120-0.471=124.409\ \text{kJ}\).

Verification

Units: Every first-law term is energy in kJ after explicit conversion of mechanical joules, so the balance is dimensionally closed.

Signs and direction: Both transfers enter the capsule and increase storage; positive elevation and speed changes correctly reduce the portion assigned to internal energy.

Plausibility: Macroscopic storage is only 0.591 kJ of a 125 kJ input, so a roughly 124 kJ internal rise is reasonable and independently closes energy.

Final result: \(\Delta U=+124.4\ \text{kJ}\) for the sealed capsule.

Engineering interpretation

The calculation shows why KE and PE should be estimated before being discarded. Entropy generation cannot be quantified without boundary temperatures and state entropy data, but it must be nonnegative; a claimed negative value would expose inconsistent data or signs.

2–3

Energy Transfer by Heat

Heat requires a temperature difference

Conduction, convection, and radiation are modes of heat transfer. Positive \(Q\) denotes heat entering the system in the convention used here. Adiabatic means \(Q=0\), not constant temperature and not automatically reversible or isentropic.

Heat-transfer rate and amount
\[Q=\int_{t_1}^{t_2}\dot Q\,dt\]
2–4

Energy Transfer by Work

Work is organized boundary transfer

Shaft rotation, electrical current, springs, and moving boundaries can transfer work. With \(Q-W=\Delta E\), \(W>0\) means work done by the system; work input is negative in that signed equation. Declare alternatives explicitly rather than switching midway.

Mechanical work and power
\[W=\int F\,ds,\qquad \dot W=\frac{dW}{dt}\]
2–5

Mechanical Forms of Work

Common mechanical work models
\[W_b=\int_{V_1}^{V_2}P_b\,dV,\qquad \dot W_{shaft}=\tau\omega\]

Applicability of work models

Boundary work requires a moving boundary and a known external boundary pressure path. Shaft power uses torque and angular speed at the same location. Spring energy requires a valid force–deflection law; none of these should be inserted merely because their variables appear.

EXAMPLE 2Electrical demand for a transfer skid
A transfer skid must deliver 14.0 kW of mechanical energy to a liquid stream. The supplied vendor efficiencies at the operating point are 78.0% for the pump and 91.0% for the motor. Determine electrical input and heat loss to the surroundings.

Recognition strategy

Two devices in series signal an efficiency-chain calculation. Desired fluid power is downstream of both conversion losses, so divide by both supplied real-component efficiencies rather than multiplying the output.

Given information

  • Steady control volume surrounds motor and pump together; one inlet and outlet have equal steady mass flow.
  • Desired stream mechanical-energy increase is 14.0 kW; vendor givens are \(\eta_p=0.780\) and \(\eta_m=0.910\).
  • Rate basis; work entering is positive for reporting. KE, PE, and fluid enthalpy details are already included in the specified desired stream-power increase.

Model and interpretation

Use actual vendor efficiencies, not reversible or isentropic efficiencies. Assume steady operation, no energy accumulation, and that all conversion loss ultimately leaves as heat. Pressure and temperature bases, property tables, phase quality, and a reference environment are not required for this aggregate power audit.

Step-by-step solution
1. Combine the serial efficiencies symbolically: \(\eta_{overall}=\eta_m\eta_p=(0.910)(0.780)=0.7098\).
2. Relate required input to desired output: \(\dot W_{elec,in}=\dot E_{fluid}/\eta_{overall}\).
3. Substitute \(\dot W_{elec,in}=14.0/0.7098=19.724\ \text{kW}\).
4. At steady state, loss is \(\dot Q_{out}=19.724-14.0=5.724\ \text{kW}\).

Verification

Units: Efficiencies are dimensionless, leaving kW for both input and loss; input minus output also has power dimensions.

Signs and direction: Required input exceeds desired output and heat leaves the skid, signs that agree with real conversion losses and the selected direction convention.

Plausibility: An overall efficiency near 71% makes the 19.7 kW demand and 5.72 kW loss reasonable; their sum independently recovers input.

Final result: \(\dot W_{elec,in}=19.7\ \text{kW}\) and \(\dot Q_{out}=5.72\ \text{kW}\).

Engineering interpretation

Heat rejection is a thermal-management load, not missing energy. Actual selection must use the vendor efficiency map at the operating point, tolerances, duty cycle, and motor service factor; entropy generation is positive because irreversible losses degrade work to heat.

2–6

The First Law of Thermodynamics

Energy conservation

For a closed process, \(Q-W=\Delta U+\Delta KE+\Delta PE\). For a steady control volume, flowing streams carry \(h+V^2/2+gz\). Over a complete cycle every state-property change is zero, so net heat equals net work with consistent directions.

Steady-flow energy balance
\[\dot Q-\dot W=\sum_{out}\dot m\left(h+\frac{V^2}{2}+gz\right)-\sum_{in}\dot m\left(h+\frac{V^2}{2}+gz\right)\]

Physical meaning

Enthalpy \(h=u+Pv\) packages internal energy and the flow work needed to push mass across a control surface. It is a property, not an additional form of energy to add beside \(u+Pv\).

EXAMPLE 3Energy and entropy audit of a process-air heater
A steady air heater handles 0.800 kg/s. Supplied inlet and outlet properties are \(h_1=305\ \text{kJ/kg}\), \(s_1=1.710\ \text{kJ/(kg\!\cdot\!K)}\), \(h_2=365\ \text{kJ/kg}\), and \(s_2=1.860\ \text{kJ/(kg\!\cdot\!K)}\). It receives electrical work only. Find power and entropy generation.

Recognition strategy

A continuous stream identifies a steady control volume, so enthalpy carries internal energy plus flow work. The entropy rise with no heat transfer gives entropy generation directly and tests second-law admissibility.

Given information

  • Steady one-inlet, one-exit heater; \(\dot m=0.800\ \text{kg/s}\); electrically insulated element and adiabatic outer boundary.
  • Supplied property values are givens from an authoritative air-property calculation; states are absolute although their \(P,T\) coordinates are not needed here.
  • Mass basis; electrical work enters. Inlet and outlet speeds and elevations are equal, so KE and PE changes are neglected.

Model and interpretation

Apply steady mass, energy, and entropy balances to the whole heater. This is an open, adiabatic, irreversible process—not a cycle and not isentropic. Enthalpy, rather than internal energy, is the flow property. No environment reference is needed unless converting entropy generation to exergy destruction.

Step-by-step solution
1. Mass conservation gives \(\dot m_{out}=\dot m_{in}=0.800\ \text{kg/s}\), with zero accumulation.
2. Energy conservation gives \(\dot W_{elec,in}=\dot m(h_2-h_1)\) after the stated KE, PE, and heat terms vanish.
3. Evaluate \(\dot W_{elec,in}=0.800(365-305)=48.0\ \text{kW}\).
4. Entropy balance gives \(\dot S_{gen}=\dot m(s_2-s_1)=0.800(1.860-1.710)=0.120\ \text{kW/K}\).

Verification

Units: Mass rate times kJ/kg gives kJ/s or kW; mass rate times kJ/(kg·K) gives kW/K as required.

Signs and direction: Positive work input accompanies increasing enthalpy, and positive entropy generation satisfies the required second-law direction.

Plausibility: The 60 kJ/kg enthalpy rise at 0.8 kg/s independently predicts 48 kW; nonnegative entropy generation confirms possibility.

Final result: \(\dot W_{elec,in}=48.0\ \text{kW}\), with \(\dot S_{gen}=0.120\ \text{kW/K}\).

Engineering interpretation

At an environment temperature \(T_0\), exergy destruction would be \(T_0\dot S_{gen}\); \(T_0\) must be measured or specified, never assumed silently. Release analysis requires traceable air properties, heat-leak measurements, and uncertainty on flow rate.

2–7

Energy Conversion Efficiencies

Defined performance ratio
\[\eta=\frac{\text{desired output}}{\text{required input}}\]

Name both sides of the ratio

A motor efficiency, pump efficiency, thermal efficiency, COP, and isentropic efficiency compare different quantities. COP may exceed one because it compares moved heat with work; thermal efficiency cannot exceed one. Ideal component efficiencies compare actual performance with an isentropic benchmark and require state properties.

2–8

Energy and Environment

Conservation is not quality

Energy is conserved, but irreversibility destroys exergy and generates entropy. For a specified environment, exergy destruction is \(X_{dest}=T_0S_{gen}\ge0\). Efficiency improvements can reduce fuel and emissions, but environmental claims require lifecycle boundaries and measured data.

Problem-solving strategy

  1. Draw the boundary first and classify a fixed mass, control volume, process, or cycle.
  2. Record state variables on an absolute basis and identify the property source before selecting equations.
  3. Declare mass or mole basis, heat and work signs, reference datum, and every approximation.
  4. Count unknowns, solve symbolically, then iterate only when properties and state depend on the result.
  5. Audit mass, energy, entropy generation, units, signs, phase validity, and limiting behavior.

Chapter recap

  • Systems store energy; heat, work, and mass transport energy across boundaries.
  • Use \(u\) for nonflow storage and \(h=u+Pv\) for flowing-stream bookkeeping.
  • Estimate KE and PE before omission, and state every reference datum.
  • Efficiency definitions must identify desired output, required input, and ideal or actual basis.
  • Close energy and mass balances and require \(S_{gen}\ge0\).

Before moving on

You should be able to draw the boundary, choose a sign convention, distinguish heat from work, write closed and steady-flow balances, estimate neglected terms, and interpret conservation residuals without confusing them with model validation.

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