Thermodynamics · Chapter 4
Thermodynamics · Chapter 4

Energy Analysis of Closed Systems

With properties in hand, we apply the first law to fixed masses. The key new ingredient is moving-boundary work, and the key new properties are the specific heats that connect heat input to temperature change.

Learning Objectives

  • Evaluate moving-boundary work for common quasi-equilibrium processes.
  • Write and solve the energy balance for a closed system.
  • Define the specific heats cv and cp and relate them.
  • Compute internal energy and enthalpy changes for ideal gases.
  • Treat solids and liquids as incompressible substances.

Model selection and property discipline

Begin by drawing a boundary around the fixed mass and identifying the complete process path. Heat is positive into the system and work positive out on this page. Absolute pressure and temperature are mandatory in equations of state. Determine phase before choosing an ideal-gas relation or pure-substance table; interpolate only between bracketing entries in the same region and record the source. Internal energy (u) belongs in a closed-system balance, whereas enthalpy \(h=u+Pv\) becomes especially convenient for constant-pressure heating.

Physical meaning

The first law accounts for quantity, while the process model determines how energy crosses the boundary. Boundary work is the signed area under the actual (P)-(V) path. Specific heats are slopes of properties: \(c_v=(\partial u/\partial T)_v\) and \(c_p=(\partial h/\partial T)_P\); they are not interchangeable labels for the same heating process.

Problem-solving strategy

  1. Recognize a sealed or fixed-mass device as a closed system and identify whether the description is a process or a complete cycle.
  2. Mark heat into and work out as positive, identify moving, electrical, shaft, or other work, and decide whether kinetic or potential energy matters.
  3. Locate each state by two independent intensive properties, establish phase, and cite ideal-gas data or an authoritative property table with interpolation.
  4. Write \(Q-W=\Delta U+\Delta KE+\Delta PE\) before deleting terms; then solve symbolically and iterate if properties depend on the unknown temperature.
  5. Audit mass, energy, units, direction, model limits, and—when entropy data are available—confirm \(S_{gen}\ge0\).
4–1

Moving Boundary Work

Boundary work

The work done as a system boundary moves, such as gas expanding against a piston. It equals the area under the process path on a P–V diagram, so it depends on the path.

Boundary work
\[W_b = \int_1^2 P\,dV\]
Isobaric (constant pressure)
\[W_b=P_0\,(V_2-V_1)\]
Isothermal ideal gas
\[W_b = P_1 V_1 \ln\frac{V_2}{V_1} = mRT\ln\frac{V_2}{V_1}\]
Polytropic process \(PV^n=\text{const}\)
\[W_b=\frac{P_2V_2-P_1V_1}{1-n}\quad(n\ne1)\]
vP12W = ∫P dv
Figure 4–1  Boundary work is the shaded area under the process curve on a P–v diagram.
4–2

Energy Balance for Closed Systems

Closed-system first law
\[Q_{\text{net,in}} - W_{\text{net,out}} = \Delta U + \Delta KE + \Delta PE\]
Stationary closed system
\[Q-W=\Delta U=m\,(u_2-u_1)\]
Key IdeaFor a stationary closed system the kinetic and potential terms vanish, so all net energy transfer goes into internal energy.
4–3

Specific Heats

Specific heat

The energy needed to raise the temperature of a unit mass by one degree. \(c_v\) applies at constant volume, \(c_p\) at constant pressure; \(c_p > c_v\) because a constant-pressure process also does expansion work.

Definitions
\[c_v=\left(\frac{\partial u}{\partial T}\right)_v,\qquad c_p=\left(\frac{\partial h}{\partial T}\right)_P\]
4–4

Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

Ideal-gas property dependence

For an ideal gas, internal energy and enthalpy depend on temperature alone. Their changes follow directly from the specific heats.

Ideal-gas energy changes
\[\Delta u = \int c_v\,dT \approx c_v\,\Delta T,\qquad \Delta h = \int c_p\,dT \approx c_p\,\Delta T\]
Enthalpy definition
\[h = u + Pv\]
Mayer's Relation (ideal gas)

For an ideal gas the constant-pressure and constant-volume specific heats differ exactly by the gas constant.

\(c_p-c_v=R,\qquad k=\frac{c_p}{c_v}\)

4–5

Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids

Incompressible substance

Solids and liquids change volume so little that their specific volume is treated as constant, which makes \(c_p\) and \(c_v\) essentially equal: \(c_p \approx c_v = c\).

Energy change of an incompressible substance
\[\Delta u \approx c_{\text{avg}}\,\Delta T,\qquad \Delta h \approx c_{\text{avg}}\,\Delta T + v\,\Delta P\]
EXAMPLE 1Weighted piston heating audit
A sealed piston–cylinder contains 0.80 kg of gas and expands at a constant absolute pressure of 250 kPa from 0.120 to 0.200 m³ while receiving 38.0 kJ of heat. Determine boundary work and the change in specific internal energy.

Recognition strategy

Sealed and piston identify a closed system with moving-boundary work. Constant pressure makes the path integral exact, and the first law then reveals the internal-energy change.

Given information

  • \(m=0.80\ \text{kg}\), \(P=250\ \text{kPa absolute}\), \(V_1=0.120\ \text{m}^3\), \(V_2=0.200\ \text{m}^3\), and \(Q_{in}=38.0\ \text{kJ}\).

Model and interpretation

The boundary encloses the fixed gas during one noncyclic quasi-equilibrium process. Heat in and work out are positive. This is a mass-basis analysis; pressures are absolute and any temperatures would be kelvin. No property table is required, the constant-pressure model is exact as given, and kinetic and potential energy changes are neglected.

Step-by-step solution
1. Select the path relation \(W_b=\int_1^2P\,dV=P(V_2-V_1)\); expansion predicts positive work out.
2. Evaluate \(W_b=(250\ \text{kPa})(0.200-0.120\ \text{m}^3)=20.0\ \text{kJ}\).
3. Apply closed-system energy conservation: \(\Delta U=Q-W_b=38.0-20.0=18.0\ \text{kJ}\).
4. Convert to the requested basis: \(\Delta u=\Delta U/m=18.0/0.80=22.5\ \text{kJ/kg}\).

Verification

Units: Pressure times volume is energy because \(1\ \text{kPa}\!\cdot\!\text{m}^3=1\ \text{kJ}\), and division by mass gives kJ/kg.

Signs and direction: Expansion work is positive out, but the larger heat input makes the internal-energy change positive, matching the computed direction.

Plausibility: The independent balance \(38.0=20.0+18.0\ \text{kJ}\) closes; fixed mass is conserved, and any realizable path must also satisfy nonnegative entropy generation.

Final result: \(W_b=20.0\ \text{kJ out}\) and \(\Delta u=+22.5\ \text{kJ/kg}\).

Engineering interpretation

About 53% of the supplied heat leaves as piston work. Release work would verify friction, leakage, pressure variation, and fluid property data rather than treating the ideal weighted piston as certified behavior.

EXAMPLE 2Rigid receiver electrical heating
A rigid insulated receiver contains 2.50 kg of nitrogen at 300 K and 180 kPa absolute. A 0.600 kW heater runs for 12.0 min. Using the supplied constant ideal-gas value \(c_v=0.743\ \text{kJ/(kg}\!\cdot\!\text{K)}\), determine the final temperature.

Recognition strategy

A sealed rigid vessel is closed and has no boundary work. Insulation removes heat transfer, but electrical work crosses the boundary, so internal energy rather than enthalpy determines the state change.

Given information

  • \(m=2.50\ \text{kg}\), \(T_1=300\ \text{K}\), \(P_1=180\ \text{kPa absolute}\), \(\dot W_{elec,in}=0.600\ \text{kW}\), \(t=720\ \text{s}\), and supplied \(c_v=0.743\ \text{kJ/(kg}\!\cdot\!\text{K)}\).

Model and interpretation

The boundary encloses the fixed nitrogen mass during a transient process, not a cycle. Work out is positive, so electrical input is negative work out. Temperature is absolute and the basis is mass. Nitrogen is modeled approximately as an ideal gas with constant supplied \(c_v\); heat, boundary work, kinetic energy, and potential energy changes are neglected.

Step-by-step solution
1. Compute electrical energy entering: \(W_{elec,in}=(0.600\ \text{kJ/s})(720\ \text{s})=432\ \text{kJ}\).
2. Reduce the first law with \(Q=0\) and \(W_b=0\), obtaining \(\Delta U=W_{elec,in}=432\ \text{kJ}\).
3. Use the approximate supplied property model \(\Delta U=mc_v(T_2-T_1)\), appropriate for a constant-\(c_v\) ideal gas.
4. Solve \(T_2=300+432/(2.50\times0.743)=532.6\ \text{K}\), which rounds to \(533\ \text{K}\).

Verification

Units: Kilowatts times seconds produce kJ, and kJ divided by \(mc_v\) leaves kelvin, so every term is dimensionally consistent.

Signs and direction: Electrical work enters the receiver and must raise internal energy; the positive temperature rise therefore has the correct sign and direction.

Plausibility: Back-substitution gives \(2.50(0.743)(532.6-300)=432\ \text{kJ}\); mass and energy close, and insulated irreversible heating has \(S_{gen}>0\).

Final result: \(T_2\approx533\ \text{K}\) on an absolute temperature basis.

Engineering interpretation

The receiver warms by about 233 K. For release work, authoritative variable-specific-heat data should replace the supplied constant value, with iteration on (u(T)) and a separate vessel pressure-rating review.

EXAMPLE 3Liquid thermal-store cooldown
A sealed 120 kg thermal store cools from 92°C to 44°C while delivering 9.60 MJ of useful boundary heat. Treat the liquid as incompressible with the explicitly supplied average \(c=3.90\ \text{kJ/(kg}\!\cdot\!\text{K)}\). Estimate parasitic heat loss to the room.

Recognition strategy

A sealed rigid liquid store is a closed-system sensible-energy problem. The difference between its internal-energy decrease and useful delivered heat is the parasitic heat transfer.

Given information

  • \(m=120\ \text{kg}\), \(T_1=365.15\ \text{K}\), \(T_2=317.15\ \text{K}\), useful heat out \(=9600\ \text{kJ}\), and supplied \(c=3.90\ \text{kJ/(kg}\!\cdot\!\text{K)}\).

Model and interpretation

The boundary encloses a fixed liquid mass for one cooldown, on a mass basis. Heat into is positive, so both useful delivery and loss are negative. The incompressible constant-property model is approximate; pressure is absolute but irrelevant here, no phase change or quality exists, and work, kinetic energy, and potential energy changes are neglected. The room is the reference environment for interpreting loss.

Step-by-step solution
1. Compute the absolute temperature difference, \(\Delta T=317.15-365.15=-48.00\ \text{K}\); kelvin and Celsius intervals are equal.
2. Estimate \(\Delta U=mc\Delta T=(120)(3.90)(-48.00)=-22464\ \text{kJ}\).
3. With no work, energy conservation gives total heat \(Q=\Delta U=-22464\ \text{kJ}\).
4. Separate useful and parasitic paths: \(Q_{loss,out}=22464-9600=12864\ \text{kJ}=12.864\ \text{MJ}\).

Verification

Units: The product kg times kJ/(kg·K) times K gives kJ, and the final conversion uses exactly \(1000\ \text{kJ}=1\ \text{MJ}\).

Signs and direction: Cooling makes \(\Delta U<0\); heat leaves the store, so both useful output and parasitic loss have the physically correct outward direction.

Plausibility: Useful heat plus loss equals \(9.600+12.864=22.464\ \text{MJ}\), independently closing energy; entropy generation for finite-temperature leakage cannot be negative.

Final result: Estimated parasitic heat loss is \(Q_{loss,out}=12.9\ \text{MJ}\).

Engineering interpretation

Only about 43% of the released sensible energy reaches the useful load. Design release requires measured temperature stratification, authoritative liquid properties, insulation testing, uncertainty bounds, and applicable pressure and safety review.

Chapter recap

  • Choose a fixed-mass boundary before writing the closed-system first law.
  • Evaluate boundary work from the process path and keep heat-in/work-out signs consistent.
  • Use \(u,c_v\) for internal-energy changes and \(h,c_p\) for enthalpy changes; identify phase and source properties.
  • Before moving on, be able to close mass and energy balances, explain approximations, and test units, signs, plausibility, uncertainty, and \(S_{gen}\ge0\).