2.1 First and Second Laws of Thermodynamics & Open/Closed System Energy Balances

2.1 First and Second Laws of Thermodynamics & Open/Closed System Energy Balances

Thermodynamic energy balances form the core of every calculation on the PE Mechanical: HVAC and Refrigeration exam. Whether analyzing a 500-ton centrifugal chiller, a steam heating coil, an air handler economizer, or a thermal expansion valve, engineers must systematically define control volumes, track energy and mass flows, apply property relations, and account for second-law irreversibilities.

---

1. Thermodynamic System Definitions & Sign Conventions

Before writing an energy equation, clearly classify the system boundary:

1. Closed System (Control Mass): A fixed quantity of matter. No mass crosses the boundary ($m = \text{constant}$), but energy crosses as heat ($Q$) and work ($W$). Examples include a sealed refrigerant storage cylinder or a rigid refrigerant charging cylinder.
2. Open System (Control Volume): A designated region in space through which mass flows. Mass, heat, and work can all cross the control surface. Examples include compressors, evaporators, cooling towers, fans, pumps, and mixing boxes.
3. Isolated System: Neither mass nor energy crosses the boundary (e.g., an adiabatic rigid container).

Standard Sign Convention (NCEES Reference Handbook)

In standard mechanical engineering thermodynamics:

• Heat Transfer ($Q$, $\dot{Q}$): Positive ($+$) when heat is transferred INTO the system from the surroundings; Negative ($-$) when heat is transferred OUT OF the system to the surroundings.
• Work Transfer ($W$, $\dot{W}$): Positive ($+$) when work is done BY the system on the surroundings (e.g., turbine shaft work output); Negative ($-$) when work is done ON the system by the surroundings (e.g., compressor or pump power input).

$\Delta E_{system} = Q_{in} - W_{out} = Q_{net,in} - W_{net,out}$

$\frac{dE_{cv}}{dt} = \dot{Q} - \dot{W} + \sum \dot{m}_{in} \left(h + \frac{V^2}{2} + gz\right)_{in} - \sum \dot{m}_{out} \left(h + \frac{V^2}{2} + gz\right)_{out}$

---

2. Closed System Energy Balances & Polytropic Processes

For a stationary closed system (where changes in macroscopic kinetic and potential energy are zero, $\Delta KE = \Delta PE = 0$), the First Law reduces to:

$Q - W = \Delta U = m (u_2 - u_1)$

Where $u$ is specific internal energy ($\text{Btu/lb}_m$ or $\text{kJ/kg}$). For an ideal gas with constant specific heats, $\Delta u = c_v (T_2 - T_1)$ and $\Delta h = c_p (T_2 - T_1)$.

Boundary Work ($P,dV$ Work)

Boundary work done during a quasi-equilibrium expansion or compression process is:

$W_b = \int_1^2 P \, dV$

Polytropic processes follow the pressure-volume relationship $P V^n = C$, where $n$ is the polytropic index:

NCEES Exam Strategy — Polytropic Index Sign Traps: When computing $W_b = \frac{P_2 V_2 - P_1 V_1}{1 - n}$, ensure consistent units. In US Customary units, $P$ is often in $\text{psia}$ (or $\text{lb}_f/\text{in}^2$) and $V$ is in $\text{ft}^3$. You must multiply by $144 \text{ in}^2/\text{ft}^2$ to convert pressure to $\text{lb}_f/\text{ft}^2$ before evaluating work in $\text{ft}\cdot\text{lb}_f$, and then divide by $778.17 \text{ ft}\cdot\text{lb}_f/\text{Btu}$ if the question asks for work in $\text{Btu}$!

---

3. Open System Steady-State Steady-Flow (SS-SF) Energy Balances

Most HVAC and refrigeration equipment operates under steady-state steady-flow (SS-SF) conditions, meaning fluid properties at any spatial point within the control volume do not change over time ($\frac{dm_{cv}}{dt} = 0$, $\frac{dE_{cv}}{dt} = 0$).

Conservation of Mass

$\sum \dot{m}_{in} = \sum \dot{m}_{out}$

$\dot{m} = \rho A V = \frac{A V}{v} = \frac{\dot{V}}{v}$

Where $\dot{m}$ is mass flow rate ($\text{lb}_m/\text{s}$ or $\text{lb}_m/\text{hr}$), $\dot{V}$ is volumetric flow rate ($\text{CFM}$ or $\text{ft}^3/\text{min}$), $v$ is specific volume ($\text{ft}^3/\text{lb}_m$), $A$ is cross-sectional area ($\text{ft}^2$), and $V$ is velocity ($\text{ft/s}$).

Conservation of Energy (First Law for Open Systems)

$\dot{Q} - \dot{W}_{cv} = \sum \dot{m}_{out} \left(h_{out} + \frac{V_{out}^2}{2 g_c} + \frac{g z_{out}}{g_c}\right) - \sum \dot{m}_{in} \left(h_{in} + \frac{V_{in}^2}{2 g_c} + \frac{g z_{in}}{g_c}\right)$

In US Customary units, kinetic energy $\frac{V^2}{2 g_c}$ with $V$ in $\text{ft/s}$ has units of $\text{ft}\cdot\text{lb}_f/\text{lb}_m$. To convert to $\text{Btu/lb}_m$, divide by $778.17 \text{ ft}\cdot\text{lb}_f/\text{Btu}$:

$\text{Specific Kinetic Energy (Btu/lb}_m\text{)} = \frac{V^2}{2 \times 32.174 \times 778.17} = \frac{V^2}{50,073}$

In most HVAC components (pumps, compressors, coils, valves), kinetic and potential energy changes ($\Delta ke$ and $\Delta pe$) are negligible compared to enthalpy changes ($\Delta h$), reducing the energy equation for single-inlet single-outlet devices to:

$\dot{q} - w_{cv} = h_e - h_i \implies \dot{Q} - \dot{W}_{cv} = \dot{m} (h_e - h_i)$

Energy Balances for Primary HVAC/R Equipment

+-----------------------+------------------------+---------------------------------------------+
| Component             | Simplifying Assumption | Governing Energy Balance                    |
+-----------------------+------------------------+---------------------------------------------+
| Compressor / Pump     | Adiabatic (\dot{Q}=0)  | \dot{W}_{in} = \dot{m}(h_{out} - h_{in})    |
| Turbine / Expander    | Adiabatic (\dot{Q}=0)  | \dot{W}_{out} = \dot{m}(h_{in} - h_{out})   |
| Heat Exchanger (Coil) | No Shaft Work (\dot{W}=0)| \dot{Q} = \dot{m}(h_{out} - h_{in})       |
| Throttling Device (TXV)| Adiabatic & Rigid      | h_{out} = h_{in}  (Isenthalpic)             |
| Mixing Chamber / Box  | Adiabatic & Rigid      | \sum \dot{m}_{in} h_{in} = \dot{m}_{out} h_{out} |
| Nozzle / Diffuser     | Adiabatic (\dot{Q}=\dot{W}=0)| h_{in} + \frac{V_{in}^2}{2} = h_{out} + \frac{V_{out}^2}{2} |
+-----------------------+------------------------+---------------------------------------------+

---

4. Second Law of Thermodynamics, Entropy & Isentropic Efficiency

The First Law establishes that energy is conserved; the Second Law dictates the direction of spontaneous processes and establishes theoretical limits on efficiency and performance.

Classical Statements of the Second Law

1. Kelvin-Planck Statement: It is impossible for any device operating on a thermodynamic cycle to receive heat from a single thermal reservoir and produce a net amount of work (i.e., thermal efficiency $\eta_{th} < 100%$).
2. Clausius Statement: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body without external work input (i.e., refrigeration COP is finite and requires compressor power).

Entropy and the T-ds Equations

Entropy ($S$) is a thermodynamic state property defined for a reversible process by $dS = \left(\frac{\delta Q}{T}\right)_{rev}$. For any real (irreversible) process:

$dS = \frac{\delta Q}{T} + \delta S_{gen} \quad \text{where } \delta S_{gen} > 0$

For a closed system undergoing an adiabatic process ($\delta Q = 0$), $\Delta S = S_{gen} \ge 0$. Entropy can be generated, but it can never be destroyed.

The fundamental differential property relations (Gibbs equations) hold for all processes (reversible or irreversible):

$T\,ds = du + P\,dv$

$T\,ds = dh - v\,dP$

For an ideal gas with constant specific heats:

$s_2 - s_1 = c_v \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{v_2}{v_1}\right)$

$s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)$

For an isentropic process ($s_2 = s_1$, reversible and adiabatic) of an ideal gas:

$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = \left(\frac{v_1}{v_2}\right)^{k-1}$

Isentropic Efficiency

In actual turbomachinery, fluid friction, turbulence, and heat leakage generate entropy, causing real expansion and compression paths to deviate from the ideal vertical line on a $T-s$ or $h-s$ (Mollier) diagram.

$\text{Turbine Isentropic Efficiency: } \eta_t = \frac{\text{Actual Work}}{\text{Isentropic Work}} = \frac{w_a}{w_s} = \frac{h_1 - h_{2a}}{h_1 - h_{2s}}$

$\text{Compressor / Pump Isentropic Efficiency: } \eta_c = \frac{\text{Isentropic Work}}{\text{Actual Work}} = \frac{w_s}{w_a} = \frac{h_{2s} - h_1}{h_{2a} - h_1}$

Enthalpy (h) / Temperature (T)
      ^
      |                  2a (Actual: higher h, s2a > s1)
      |                 / 
      |         2s     /  (Compressor Path: w_actual = h2a - h1 > w_s = h2s - h1)
      |          |    /   eta_c = (h2s - h1) / (h2a - h1)
 P2 --+----------+---/----
      |          |  /
      |          | /
      |          1 (Inlet)
 P1 --+----------+---------------------> Entropy (s)
                 s1 = s2s

---

5. Worked Example: Compressor Power with Heat Loss and Isentropic Efficiency

Problem: An industrial R-134a refrigeration compressor receives saturated vapor at $20 \text{ psia}$ ($h_1 = 105.30 \text{ Btu/lb}_m$, $s_1 = 0.2248 \text{ Btu/lb}m\cdot^\circ\text{R}$) and compresses it to a discharge pressure of $140 \text{ psia}$. At $140 \text{ psia}$, an isentropic compression to $s{2s} = 0.2248 \text{ Btu/lb}m\cdot^\circ\text{R}$ gives an ideal enthalpy of $h{2s} = 124.80 \text{ Btu/lb}m$. The compressor has an isentropic efficiency of $\eta_c = 0.78$. During operation, the compressor loses heat to the ambient air at a rate of $\dot{Q}{loss} = 1.80 \text{ Btu/lb}_m$ of refrigerant circulated. The mass flow rate of refrigerant is $\dot{m} = 120 \text{ lb}_m/\text{min}$.

Find:

1. The actual specific work input ($w_{actual}$) required by the compressor in $\text{Btu/lb}_m$.
2. The actual discharge enthalpy ($h_{2a}$) of the refrigerant exiting the compressor in $\text{Btu/lb}_m$.
3. The total electrical/brake horsepower ($HP$) required if the mechanical-electrical drive train efficiency is $\eta_{mech} = 0.92$.

Step-by-Step Solution:

Step 1: Calculate isentropic work and actual work. $w_s = h_{2s} - h_1 = 124.80 - 105.30 = 19.50 \text{ Btu/lb}_m$

By definition of compressor isentropic efficiency: $w_a = \frac{w_s}{\eta_c} = \frac{19.50 \text{ Btu/lb}_m}{0.78} = 25.00 \text{ Btu/lb}_m$

Step 2: Apply the open system First Law energy balance to find actual discharge enthalpy $h_{2a}$. $\dot{q}_{net,in} - w_a = h_{2a} - h_1$ Here, heat is lost from the control volume, so $\dot{q}{net,in} = -\dot{q}{loss} = -1.80 \text{ Btu/lb}_m$. Work is done on the fluid, so $w_a = -25.00 \text{ Btu/lb}_m$ under the sign convention $\dot{q} - w = \Delta h$, or directly writing an energy balance: $h_1 + w_a - q_{loss} = h_{2a}$ $h_{2a} = 105.30 + 25.00 - 1.80 = 128.50 \text{ Btu/lb}_m$

Step 3: Calculate the total power requirement in Horsepower. $\dot{W}_{fluid} = \dot{m} \times w_a = \left(120 \frac{\text{lb}_m}{\text{min}}\right) \times \left(25.00 \frac{\text{Btu}}{\text{lb}_m}\right) = 3,000 \frac{\text{Btu}}{\text{min}}$

Convert $\text{Btu/min}$ to Horsepower ($1 \text{ HP} = 42.418 \text{ Btu/min} = 2,544.4 \text{ Btu/hr} = 0.7457 \text{ kW}$): $\dot{W}_{fluid, HP} = \frac{3,000 \text{ Btu/min}}{42.418 \text{ Btu/min}\cdot\text{HP}} = 70.72 \text{ HP}$

Accounting for mechanical drive train efficiency ($\eta_{mech} = 0.92$): $\dot{W}_{brake} = \frac{\dot{W}_{fluid, HP}}{\eta_{mech}} = \frac{70.72 \text{ HP}}{0.92} = 76.87 \text{ HP} \approx 76.9 \text{ HP}$

---

6. NCEES Reference Handbook Navigation & Exam Traps

• Section 1.3: Thermodynamics: Find the general steady-state control volume equation, isentropic efficiency formulas, ideal gas relations, and polytropic work expressions.
• Unit Conversion Trap ($P,dV$): Never multiply $\text{psia}$ by $\text{ft}^3$ without multiplying by $144 \text{ in}^2/\text{ft}^2$. $\text{psia} \times \text{ft}^3 \ne \text{ft}\cdot\text{lb}_f$.
• Isentropic Efficiency Inversion: Candidates frequently invert the numerator and denominator. Remember that real compressors always consume more work than ideal ($w_a = w_s / \eta_c > w_s$), while real turbines always produce less work than ideal ($w_a = \eta_t w_s < w_s$). Both efficiencies must be $\le 1.0$.