Thermodynamics · Chapter 1

Introduction and Basic Concepts

Thermodynamics begins with vocabulary. Before energy can be tracked or balanced, we need precise meanings for system, property, state, process, temperature, and pressure — the foundation everything later rests on.

Learning objectives

Classify closed, open, and isolated systems from a physical boundary.
Separate intensive, extensive, and specific properties with consistent units.
Identify a simple compressible state from independent intensive properties.
Convert temperature and pressure to the absolute bases required by property relations.
Apply density, ideal-gas, and hydrostatic relations within their stated limits.
Verify thermodynamic models through units, signs, conservation, and plausibility.

1–1

Thermodynamics and Energy

A boundary-centered science

Thermodynamics describes stored energy and energy crossing a chosen boundary. Energy is conserved, while entropy generation distinguishes real processes from reversible ideals. Rates require time-dependent transport models; equilibrium thermodynamics alone compares states.

General accounting statement

\[E_{in}-E_{out}=\Delta E_{system}\]

Physical meaning

The equation is bookkeeping: every joule entering either leaves or remains as increased storage. It does not by itself identify the transfer mechanism or guarantee that a proposed process satisfies the second law.

Figure 1–1 A system is the matter chosen for study; everything else is the surroundings, and the surface between them is the boundary.

1–2

Importance of Dimensions and Units

Dimensions expose bad models

A valid equation is dimensionally homogeneous. Keep conversion factors visible, distinguish mass from force, and use compound units such as \(\text{kJ}/(\text{kg}\!\cdot\!\text{K})\). Significant figures should reflect measurement uncertainty, not calculator display.

Force and weight

\[F=ma,\qquad W_{weight}=mg\]

EXAMPLE 1 · FOUNDATIONInventory in a pressurized calibration vessel

A rigid calibration vessel contains air at 180 kPa gage and 35 °C. Its internal volume is 0.120 m³, local atmospheric pressure is 98.0 kPa, and the supplied ideal-gas constant is \(R=0.287\ \text{kJ}/(\text{kg}\!\cdot\!\text{K})\). Determine the contained mass.

Recognition strategy

The words rigid vessel and contained mass identify a closed-system state calculation, not an energy balance. Pressure and temperature must first be converted to absolute values before the supplied ideal-gas model can identify mass.

Given information

Closed boundary: fixed air mass, no process or cycle is occurring.
\(P_{gage}=180\ \text{kPa}\), \(P_{atm}=98.0\ \text{kPa}\), \(T=35^\circ\text{C}\), and \(V=0.120\ \text{m}^3\).
Mass basis; supplied ideal-gas property relation and gas constant; kinetic and potential energies are irrelevant to this state inventory.

Model and interpretation

Take the vessel wall as the system boundary. Use absolute pressure and kelvins, with \(PV=mRT\). This is an approximate ideal-gas state model; no heat, work, entropy generation, exergy reference environment, or efficiency is evaluated because no process is specified.

Step-by-step solution

1. Convert the measured state: \(P_{abs}=180+98.0=278\ \text{kPa}\) and \(T=35+273.15=308.15\ \text{K}\).

2. Rearrange symbolically before inserting values: \(m=PV/(RT)\), which exposes every measured input and the property-model dependence.

3. Substitute compatible SI units: \(m=(278)(0.120)/[(0.287)(308.15)]=0.3772\ \text{kg}\).

4. Because this is one equilibrium state, mass is the inventory itself; energy, entropy, and exergy balances have no process transfer terms to reconcile.

Verification

Units: Using \(\text{kPa}\!\cdot\!\text{m}^3=\text{kJ}\), the energy units cancel against \(R T\) to leave kilograms.

Signs and direction: The mass is positive, as required; adding atmosphere rather than subtracting it gives the correct absolute-pressure direction.

Plausibility: The implied density is \(3.14\ \text{kg/m}^3\), about 2.6 times room-air density, reasonable for roughly 2.8 bar absolute.

Final result: \(m=0.377\ \text{kg}\) of air on the stated ideal-gas basis.

Engineering interpretation

The atmospheric-pressure uncertainty enters mass directly: a 1 kPa error changes this result by about 0.36%. A released inventory calculation should use calibrated absolute sensors and an approved real-gas property source if pressure increases enough for nonideality to matter.

1–3

Systems and Control Volumes

Choose what is being counted

A closed system follows fixed matter, although its boundary may move. A control volume encloses a region through which mass may flow. An isolated system transfers neither mass nor energy. Boundary choice determines which interactions appear in balances.

Figure 1–2 A control volume with one inlet and one exit; mass and energy may cross the dashed control surface.

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Properties of a System

Properties belong to states

Pressure and temperature are intensive; mass, volume, internal energy, enthalpy, entropy, and exergy are extensive. Dividing an extensive property by mass produces a specific property such as \(u=U/m\) or \(h=H/m\). Heat and work are transfers, never properties.

Physical meaning

Cutting a uniform system in half leaves its intensive state unchanged but halves each extensive inventory. This scaling test is a quick way to classify an unfamiliar quantity.

1–5

Density and Specific Gravity

Density, specific volume, and specific gravity

\[\rho=\frac{m}{V},\qquad v=\frac{1}{\rho},\qquad SG=\frac{\rho}{\rho_{ref}}\]

Reference conditions matter

Specific gravity is dimensionless only when numerator and reference density use compatible conditions and units. For precision work, report the reference fluid and temperature instead of silently treating water as exactly 1000 kg/m³.

1–6

State and Equilibrium

State postulate

Two independent intensive properties fix the equilibrium state of a simple compressible substance. In a two-phase region, saturation pressure and temperature are dependent, so quality or another independent property is needed.

Equilibrium and state identification

Thermal, mechanical, phase, and chemical driving forces must all vanish for complete equilibrium. A quasi-equilibrium path is an ideal sequence of nearly equilibrium states, not proof of reversibility.

1–7

Processes and Cycles

Path, process, and cycle

A process connects states; its path affects heat and work. A cycle returns every state property to its initial value, so \(\Delta U=\Delta H=\Delta S=0\) for the working fluid over the complete cycle even though net heat and work need not vanish.

Figure 1–3 A process drawn on a P–v diagram; the path matters only for quasi-equilibrium processes, where each point is an equilibrium state.

1–8

Temperature and the Zeroth Law

Absolute temperature conversions

\[T(\text{K})=T({}^\circ\text{C})+273.15,\qquad T(\text{R})=T({}^\circ\text{F})+459.67\]

Thermal equilibrium

The zeroth law makes temperature comparison possible. Temperature differences may share Celsius and kelvin increments, but equations of state, efficiency limits, and temperature ratios require absolute scales.

1–9

Pressure

Pressure references and hydrostatics

\[P_{abs}=P_{gage}+P_{atm},\qquad P_2-P_1=\rho g(z_1-z_2)\]

Pressure needs a datum

Absolute pressure is measured from vacuum and is required in property relations. Gage pressure is relative to the local atmosphere. The hydrostatic relation assumes a stationary fluid with approximately constant density over the elevation change.

EXAMPLE 2 · INTERMEDIATEOil-column diagnosis of vessel pressure

A service manometer connected to a nitrogen vessel shows the vessel side depressing an oil column by 0.450 m. The supplied oil specific gravity is 0.820, local atmosphere is 96.5 kPa absolute, and \(g=9.81\ \text{m/s}^2\). Find vessel gage and absolute pressures.

Recognition strategy

A static column height and known density indicate hydrostatics, not an ideal-gas calculation. The depressed vessel side establishes that vessel pressure exceeds atmospheric pressure, fixing the sign before arithmetic.

Given information

Open manometer fluid is the system for the pressure walk; nitrogen does not cross that diagnostic boundary.
\(SG=0.820\), \(h=0.450\ \text{m}\), \(P_{atm}=96.5\ \text{kPa absolute}\); supplied \(\rho_{water}=1000\ \text{kg/m}^3\).
Mass basis for density; static, incompressible oil; no temperature-dependent table correction is used.

Model and interpretation

Walk through the stationary oil from the atmosphere toward the lower vessel-side interface. Pressure rises downward. This exact hydrostatic balance within the constant-density approximation neglects capillarity; KE and PE changes of a moving stream, heat, work, entropy, exergy, and efficiencies do not apply to the static state.

Step-by-step solution

1. Convert specific gravity to density: \(\rho=SG\rho_{water}=0.820(1000)=820\ \text{kg/m}^3\).

2. Write the pressure difference symbolically as \(P_{gage}=\rho g h\), positive because the vessel pushes its side downward.

3. Evaluate \(P_{gage}=(820)(9.81)(0.450)=3620\ \text{Pa}=3.62\ \text{kPa}\).

4. Reference the answer to vacuum by adding atmosphere: \(P_{abs}=96.5+3.62=100.12\ \text{kPa}\).

Verification

Units: Density times acceleration times height gives \(\text{kg}/(\text{m}\!\cdot\!\text{s}^2)=\text{Pa}\); conversion to kPa is explicit.

Signs and direction: The observed lower level on the vessel side demands positive gage pressure, and the computed absolute pressure exceeds atmosphere.

Plausibility: A 0.45 m oil head producing only a few kilopascals is reasonable; a water-column estimate would be about 4.4 kPa.

Final result: \(P_{gage}=3.62\ \text{kPa}\) and \(P_{abs}=100.1\ \text{kPa}\).

Engineering interpretation

The result depends linearly on height and oil density. Field acceptance should include temperature-corrected certified manometer-fluid density, level-reading uncertainty, capillary correction when relevant, and a local atmospheric measurement rather than a standard-atmosphere assumption.

1–10

Pressure Measurement Devices

Trace a pressure path

For a manometer, move through each fluid column: pressure increases downward and decreases upward. At a shared elevation in one connected static fluid, pressure is equal. Barometers compare atmosphere with a near-vacuum reference.

Figure 1–4 A gas confined by a piston; the piston weight and any applied force set the gas pressure, which a manometer or gage can read.

1–11

Problem-Solving Technique

Recognition before algebra

Mark requested outputs, define a boundary, list states and property sources, then choose conservation laws and constitutive models. If properties depend on the unknown state, iterate with a stated tolerance and retain the physically valid branch.

EXAMPLE 3 · PROFESSIONALControl-surface air-flow inventory rate

Air crosses a commissioning duct station at 250 kPa absolute and 320 K with uniform speed 18.0 m/s through a 0.0150 m² area. Using the supplied ideal-gas constant \(R=0.287\ \text{kJ}/(\text{kg}\!\cdot\!\text{K})\), determine the steady mass flow rate.

Recognition strategy

The cue crosses a station makes this an open-system control-volume problem. First identify density from the absolute thermodynamic state, then combine density with normal velocity and area in the steady mass balance.

Given information

Control surface cuts the duct at the measurement station; steady one-dimensional inflow is assumed.
\(P=250\ \text{kPa absolute}\), \(T=320\ \text{K absolute}\), \(A=0.0150\ \text{m}^2\), and \(V_n=18.0\ \text{m/s}\).
Mass basis and supplied ideal-gas source; uniform profiles; no cycle and no heat or shaft-work transfer at the mathematical station.

Model and interpretation

Apply \(\rho=P/(RT)\) and \(\dot m=\rho A V_n\). The ideal-gas and one-dimensional profile models are approximate. KE is not neglected in defining the stream but no device energy balance is requested; elevation PE is irrelevant at one section. Positive flow is defined into the selected control volume.

Step-by-step solution

1. Find density from the stated absolute variables: \(\rho=250/[(0.287)(320)]=2.722\ \text{kg/m}^3\).

2. Compute volumetric flow at the section: \(\dot{\mathcal V}=A V_n=(0.0150)(18.0)=0.270\ \text{m}^3/\text{s}\).

3. Apply steady mass transport: \(\dot m=\rho\dot{\mathcal V}=(2.722)(0.270)=0.7350\ \text{kg/s}\).

4. For a one-inlet, one-exit steady duct, the downstream mass rate must equal this value; no mass accumulation is permitted.

Verification

Units: The product \(\text{kg/m}^3\times\text{m}^3/\text{s}\) reduces exactly to the required \(\text{kg/s}\).

Signs and direction: Velocity is defined inward, so mass flow is positive; reversing the surface normal would change its signed flux, not its magnitude.

Plausibility: The volume flow of 0.270 m³/s times a density above ambient gives less than 1 kg/s, a physically reasonable scale.

Final result: \(\dot m=0.735\ \text{kg/s}\) into the control volume.

Engineering interpretation

A commissioning value should correct for nonuniform velocity and temperature profiles and use a calibrated absolute-pressure instrument. Mass conservation supplies a strong independent downstream check; entropy generation and exergy destruction are nonnegative for a real duct but require inlet/outlet states and an environment reference to quantify.

Problem-solving strategy

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

Chapter recap

Boundary choice separates closed-system storage from control-volume transport.
State properties require absolute pressure and temperature where a property model demands them.
Specific and intensive quantities do not scale with mass; extensive inventories do.
Property data, assumptions, reference conditions, and uncertainty belong beside every result.

Before moving on

You should be able to sketch a boundary, classify the model, identify two independent state properties, convert pressure and temperature bases, and audit units, signs, conservation, and physical plausibility.

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