Applied Thermodynamics: Energy Analysis of Closed Systems

Analysis of Energy in Closed Systems and Moving Boundary Work

Moving boundary work is a fundamental form of mechanical work commonly encountered in technical practice, specifically related to the expansion or compression of a gas within a cylinder-piston device. During this process, the internal face of the piston acts as a moving boundary that oscillates. Consequently, this work is formally identified as moving boundary work or simply boundary work ($W_b$). In a typical starting state, a gas exerts an initial pressure $P$ within a total volume $V$, where the piston has a cross-sectional area $A$. If the piston is allowed to move a differential distance $ds$ while maintaining a state of quasi-equilibrium, the differential work performed is defined as $\delta W_b = F \, ds = P A \, ds = P \, dV$.

The total boundary work performed during a complete process as the piston moves from an initial state to a final state is obtained by summing (integrating) the differential work increments: $W_b = \int_1^2 P \, dV$. This integral can only be evaluated if the functional relationship between pressure and volume during the process is known, meaning the function $P = f(V)$ must be available. Historically and conceptually, the area under the process curve on a $P-V$ diagram represents the total boundary work. It is essential to recognize that boundary work is a path-dependent function, meaning it depends on the specific path followed as well as the initial and final states. For a system undergoing a cycle, the net work is the difference between the work done by the system and the work done on the system.

Boundary Work in Specific Thermodynamic Processes

For a process occurring at a constant volume (isochoric process), such as in a rigid container, the boundary work is always zero. This is mathematically evident because a rigid container does not allow for a change in volume ($dV = 0$), leading to an integral value of zero regardless of changes in pressure or temperature. For example, if a rigid container holds air at $500\,kPa$ and $150^\circ C$ and cools to $400\,kPa$ and $65^\circ C$, no boundary work is performed.

In a constant pressure (isobaric) process, the boundary work is calculated as $W_b = P_0(V_2 - V_1)$. Since total volume can be expressed as mass times specific volume ($V = m v$), the work can also be calculated as $W_b = m P_0(v_2 - v_1)$. An example of this is a friction-less cylinder-piston device containing $10\,lbm$ of steam at $60\,psia$ and $320^\circ F$ that is heated to $400^\circ F$. In this scenario, the pressure remains constant because the atmospheric pressure and piston weight are constant. Using superheated steam tables (Table A-6E), specific volumes are identified ($v_1 = 7.4863\,ft^3/lbm$ and $v_2 = 8.3548\,ft^3/lbm$) to solve for the work.

For the isothermal compression or expansion of an ideal gas, the temperature remains constant ($T_0$). Based on the ideal gas law ($PV = mRT_0$), the pressure can be expressed as $P = \frac{mRT_0}{V}$. Integrating this yields $W_b = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right)$. Consider a device containing $0.4\,m^3$ of air at $100\,kPa$ and $80^\circ C$ compressed to $0.1\,m^3$. Assuming quasi-equilibrium and ideal gas behavior, the work is calculated using the logarithmic relationship of the volume limits or the pressure ratio ($\frac{P_1}{P_2}$) since $P_1 V_1 = P_2 V_2$.

Polytropic Processes and Complex Boundary Work Scenarios

Real-world expansion and compression processes often follow a relationship where pressure and volume are related by the equation $PV^n = C$, where $n$ and $C$ are constants. This is known as a polytropic process. The pressure is expressed as $P = C V^{-n}$, where $C = P_1 V_1^n = P_2 V_2^n$. Substituting this into the boundary work integral results in $W_b = \frac{P_2 V_2 - P_1 V_1}{1 - n}$ for $n \neq 1$. For an ideal gas, this becomes $W_b = \frac{mR(T_2 - T_1)}{1 - n}$. In the special case where $n = 1$, the process becomes isothermal and follows the logarithmic work formula.

A more complex scenario involves the expansion of an ideal gas against a linear spring. Consider a device with $0.05\,m^3$ of gas at $200\,kPa$ where a spring (constant $k = 150\,kN/m$) just touches the piston. If heat is added until the volume doubles ($V_2 = 0.1\,m^3$) and the piston area is $0.25\,m^2$, the piston displacement is $\Delta x = \frac{V_2 - V_1}{A} = \frac{0.05\,m^3}{0.25\,m^2} = 0.2\,m$. The final spring force is $F = kx = (150\,kN/m)(0.2\,m) = 30\,kN$. This adds a pressure of $P_{spring} = \frac{F}{A} = \frac{30\,kN}{0.25\,m^2} = 120\,kPa$. The final pressure is thus $P_1 + P_{spring} = 320\,kPa$. The total work is the area of the trapezoid on the $P-V$ diagram: $W_{total} = \frac{P_1 + P_2}{2}(V_2 - V_1) = 13\,kJ$. Here, the work done against the atmosphere/piston (rectangular region I) is $10\,kJ$, and the work against the spring (triangular region II) is $3\,kJ$.

Energy Balance for Closed Systems

The conservation of energy principle for a closed system is expressed as $E_{in} - E_{out} = \Delta E_{system}$. In rate form, this is $\dot{E}_{in} - \dot{E}_{out} = \frac{dE_{system}}{dt}$. For a stationary process over a time interval $\Delta t$, total quantities are $Q = \dot{Q} \, \Delta t$ and $W = \dot{W} \, \Delta t$. On a per-unit-mass basis, it is expressed as $e_{in} - e_{out} = \Delta e_{system}$. For a system undergoing a cycle, the change in energy is zero ($\Delta E = 0$), which implies $Q = W$ (net heat equals net work).

The energy balance for a general closed system is often written as $Q_{net,in} - W_{net,out} = \Delta E$. For processes involving constant pressure, moving boundary work ($W_b$) and the change in internal energy ($\Delta U$) can be combined into a single term: Enthalpy ($H$). The balance becomes $Q - W_{other} = \Delta H$, where $W_{other}$ includes electrical or shaft work. For example, if $25\,g$ of saturated water vapor at $300\,kPa$ is heated by a resistance (source: $120\,V$, current: $0.2\,A$, time: $5\,min$) while losing $3.7\,kJ$ of heat, the final temperature is found by calculating the electrical work $W_e = V I \, \Delta t$ and using the enthalpy relation $Q - W_e = m(h_2 - h_1)$. Based on initial saturated state properties and final enthalpy, the final temperature is determined to be $200^\circ C$.

Unrestricted Expansion and Vacuum Interactions

In an unrestricted expansion (also known as expansion into a vacuum), a rigid container divided by a partition has one side with substance (e.g., $5\,kg$ of water at $200\,kPa$, $25^\circ C$) and the other side at vacuum. When the partition is removed, the water expands. Even though the water moves, the boundary of the system (the rigid tank walls) is fixed, meaning $W_b = 0$. In this scenario, the energy balance simplifies to $Q = m(u_2 - u_1)$. In a specific example where the volume doubles and the final temperature returns to $25^\circ C$, the water ends as a saturated mixture. Because the initial internal energy ($u_1 \approx u_f@25^\circ C = 104.83\,kJ/kg$) and final internal energy ($u_2 \approx 104.88\,kJ/kg$) are very close, the net heat transfer is minimal (calculated as $0.25\,kJ$).

Specific Heats: $C_v$ and $C_p$

Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree. Two types are critical: specific heat at constant volume ($C_v$) and specific heat at constant pressure ($C_p$). Formally, $C_v = \left(\frac{\partial u}{\partial T}\right)_v$ and $C_p = \left(\frac{\partial h}{\partial T}\right)_p$. For example, Helium requires different energy inputs depending on whether the volume or pressure is held constant. Specific heat is a measure of a substance's energy storage capacity and varies based on how the process is executed.

Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

For an ideal gas ($Pv = RT$), internal energy ($u$) is a function of temperature only ($u = f(T)$), as demonstrated by Joule in $1843$. Consequently, enthalpy ($h = u + Pv = u + RT$) is also purely temperature-dependent. Changes in these properties are expressed as $du = C_v(T) \, dT$ and $dh = C_p(T) \, dT$. To find the change between states, one must integrate these relations: $\Delta u = \int_1^2 C_v(T) \, dT$ and $\Delta h = \int_1^2 C_p(T) \, dT$.

There are three ways to determine these changes:

1. Using tabulated data (e.g., Table A-17 for air), which is the most accurate and simple method.

2. Integrating specific heat as a function of temperature (usually a polynomial: $C_p = a + bT + cT^2 + dT^3$), which is precise for computer calculations.

3. Using average specific heat values ($C_{v,avg}$ or $C_{p,avg}$) at the average temperature $T_{avg} = \frac{T_1 + T_2}{2}$, which is convenient for small temperature ranges.

A vital relationship for ideal gases is $C_p - C_v = R$. If expressed on a molar basis, the universal gas constant $R_u$ is used ($\overline{C_p} - \overline{C_v} = R_u$). The ratio of specific heats is defined as $k = \frac{C_p}{C_v}$.

Thermal Analysis of Stationary Systems

Consider an insulated rigid container with $1.5\,lbm$ of Helium at $80^\circ F$ and $50\,psia$. A paddle wheel (power input $0.02\,hp$) operates for $30\,min$. Since the container is rigid and insulated, $W_b = 0$ and $Q = 0$. The energy balance is $-W_{shaft,in} = \Delta U = m C_v(T_2 - T_1)$. For Helium, a monatomic gas, specific heat is constant ($C_v = 0.753\,Btu/lbm \cdot ^\circ F$). By calculating the shaft work input in $Btu$, the final temperature and subsequent final pressure (via the ideal gas law constant volume ratio) can be determined.

In another example, $0.5\,m^3$ of Nitrogen at $400\,kPa$ and $27^\circ C$ is heated by an electrical resistor ($2\,A$, $120\,V$, $5\,min$) while expanding at constant pressure and losing $2800\,J$ of heat. Here, the energy balance $W_{e,in} - Q_{out} = \Delta H = m C_p(T_2 - T_1)$ is used. Nitrogen is treated as an ideal gas with $C_p = 1.039\,kJ/kg \cdot K$ at room temperature. The mass is found using $m = \frac{P_1 V_1}{RT_1}$, and the final temperature is solved as $56.7^\circ C$.

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

Substances with constant specific volume or density are called incompressible substances. For solids and liquids, the specific volume remains essentially constant during processes, which implies that $C_p = C_v = C$. The energy changes are defined as $du = C(T) \, dT \implies \Delta u \approx C_{avg} \Delta T$.

Enthalpy changes for incompressible substances use the differential form $dh = du + v \, dP + P \, dv$. Since $dv = 0$, this becomes $dh = du + v \, dP$.

1. For solids, $v \, \Delta P$ is negligible, so $\Delta h \approx \Delta u \approx C_{avg} \Delta T$.

2. For liquids at constant pressure ($\Delta P = 0$), $\Delta h = \Delta u \approx C_{avg} \Delta T$.

3. For liquids at constant temperature ($\Delta T = 0$), such as in pumps, $\Delta h = v \, \Delta P$.

The enthalpy of a compressed liquid can be approximated using the saturated liquid property at the same temperature plus a correction factor for pressure: $h_{(P,T)} \approx h_{f@T} + v_{f@T}(P - P_{sat@T})$. At $100^\circ C$ and $15\,MPa$, a saturated liquid approximation yields $419.17\,kJ/kg$ (a $2.6\%$ error), while the corrected formula yields $434.07\,kJ/kg$, which is closer to the exact table value of $430.39\,kJ/kg$.

Thermal Equilibrium and Heat Transfer in Mixed Systems

When a $50\,kg$ iron block at $80^\circ C$ is submerged in an insulated tank of $0.5\,m^3$ of liquid water at $25^\circ C$, the system reaches thermal equilibrium. For such a stationary and insulated system ($Q=0, W=0$), the total change in internal energy is zero ($\Delta U_{total} = 0$). This is the sum of the internal energy changes of the components: $\Delta U_{iron} + \Delta U_{water} = 0$. This expands to $[m C(T_f - T_1)]_{iron} + [m C(T_f - T_1)]_{water} = 0$. Given the mass of water is $500\,kg$ (using $v_{water} \approx 0.001\,m^3/kg$) and specific heats ($C_{iron} = 0.45\,kJ/kg \cdot ^\circ C$; $C_{water} = 4.18\,kJ/kg \cdot ^\circ C$), the final equilibrium temperature can be solved.

Questions & Discussion

Is it possible to reach the same conclusions regarding the unrestricted expansion if the water alone is chosen as the system? Yes, it is possible. However, if only the water is considered the system, the boundaries become moving boundaries during the expansion. In that case, you must account for the moving boundary work performed by the water. However, since the water expands into a vacuum, there is no external pressure resisting this expansion ($P_{ext} = 0$), meaning the work performed against the surroundings is still zero ($W = \int P_{ext} \, dV = 0$). Thus, the energy balance result remains identical ($Q = \Delta U$).