14.4 Ideal Gas Processes and Power/Refrigeration Cycles

14.4 Ideal Gas Processes, Power/Refrigeration Cycles, and Energy Balance

Core FE Exam Principle: Working fluids undergo thermodynamic cycles to convert thermal energy into mechanical power (power cycles) or to transfer thermal energy from cold to hot regions using work input (refrigeration cycles). On the FE exam, mastery of ideal gas relations, cycle efficiencies, and component energy balances is essential.

Ideal Gas Equation of State & Specific Heat Relations

An ideal gas is a hypothetical fluid whose molecules exert no intermolecular forces and occupy negligible volume.

Equation of State

$P V = m R T \quad \text{or} \quad P v = R T$

where (P) is absolute pressure ((\text{kPa})), (v) is specific volume ((\text{m}^3/\text{kg})), (T) is absolute temperature ((\text{K})), and (R) is the specific gas constant:

$R = \frac{\bar{R}}{M} \quad [\text{kJ/(kg}\cdot\text{K)}]$

where (\bar{R} = 8.314 \text{ kJ/(kmol}\cdot\text{K)}) is the universal gas constant and (M) is molecular weight (e.g., Air: (M = 28.97 \text{ kg/kmol}), (R_{air} = 0.287 \text{ kJ/(kg}\cdot\text{K)})).

Ideal Gas Specific Heats

For ideal gases, internal energy (u) and enthalpy (h) are functions of temperature only:

$du = c_v dT \implies \Delta u = c_v (T_2 - T_1)$ $dh = c_p dT \implies \Delta h = c_p (T_2 - T_1)$ $c_p - c_v = R$ $k = \frac{c_p}{c_v} \quad (\text{Ratio of specific heats; Air: } k = 1.4)$

Boundary Work & Fundamental Ideal Gas Processes

Boundary work done during a quasi-equilibrium process is evaluated as:

$W_b = \int_1^2 P dV$

Vapor Power Cycles: The Rankine Cycle

The Rankine cycle is the ideal model for steam power plants.

Four Standard Components & Processes

1. Process 1-2 (Pump): Isentropic compression of liquid in pump: $w_{pump,in} = h_2 - h_1 \approx v_1 (P_2 - P_1)$
2. Process 2-3 (Boiler): Constant-pressure heat addition in boiler: $q_{in} = h_3 - h_2$
3. Process 3-4 (Turbine): Isentropic expansion of steam in turbine: $w_{turb,out} = h_3 - h_4$
4. Process 4-1 (Condenser): Constant-pressure heat rejection in condenser: $q_{out} = h_4 - h_1$

Thermal Efficiency

$\eta_{th,Rankine} = \frac{w_{net}}{q_{in}} = \frac{w_{turb,out} - w_{pump,in}}{q_{in}} = \frac{(h_3 - h_4) - (h_2 - h_1)}{h_3 - h_2} = 1 - \frac{q_{out}}{q_{in}}$

Air-Standard Gas Power Cycles: Otto and Diesel Cycles

Air-standard cycles model internal combustion engines using air as an ideal gas working fluid.

The Otto Cycle (Ideal Spark-Ignition / Gasoline Engine)

Consists of four internally reversible processes:

• 1-2: Isentropic compression (Compression ratio (r = V_1 / V_2)).
• 2-3: Constant-volume (isochoric) heat addition: (q_{in} = c_v (T_3 - T_2)).
• 3-4: Isentropic expansion (power stroke).
• 4-1: Constant-volume (isochoric) heat rejection: (q_{out} = c_v (T_4 - T_1)).

$\text{Otto Cycle Thermal Efficiency: } \eta_{th,Otto} = 1 - \frac{1}{r^{k-1}}$

The Diesel Cycle (Ideal Compression-Ignition / Diesel Engine)

Differs from Otto cycle by having constant-pressure (isobaric) heat addition during Process 2-3 (Cutoff ratio (r_c = V_3 / V_2)):

$\eta_{th,Diesel} = 1 - \frac{1}{r^{k-1}} \left[ \frac{r_c^k - 1}{k(r_c - 1)} \right]$

Vapor-Compression Refrigeration Cycles

Refrigeration cycles operate in reverse of power cycles to remove heat from a low-temperature space.

Four Standard Components & Processes

1. Process 1-2 (Compressor): Isentropic compression of saturated vapor to superheated vapor: $w_{in} = h_2 - h_1$
2. Process 2-3 (Condenser): Constant-pressure heat rejection to high-temp surroundings: $q_H = h_2 - h_3$
3. Process 3-4 (Expansion Valve): Throttling expansion at constant enthalpy (Isenthalpic): $h_4 = h_3$
4. Process 4-1 (Evaporator): Constant-pressure heat absorption from cold refrigerated space: $q_L = h_1 - h_4$

Coefficient of Performance (COP)

$\text{Refrigerators: } COP_{ref} = \frac{\text{Desired Cooling Output}}{\text{Work Input}} = \frac{q_L}{w_{in}} = \frac{h_1 - h_4}{h_2 - h_1}$ $\text{Heat Pumps: } COP_{HP} = \frac{\text{Desired Heating Output}}{\text{Work Input}} = \frac{q_H}{w_{in}} = \frac{h_2 - h_3}{h_2 - h_1} = COP_{ref} + 1$

Comprehensive Worked Engineering Example

Problem Statement

Air enters an ideal air-standard Otto cycle compressor at state 1 where (P_1 = 100 \text{ kPa}) and (T_1 = 300 \text{ K}). The engine has a compression ratio of (r = 8.5). During the constant-volume heat addition process, (q_{in} = 800 \text{ kJ/kg}) of heat is added to the air. Assume constant specific heats for air at room temperature: (c_v = 0.718 \text{ kJ/(kg}\cdot\text{K)}), (c_p = 1.005 \text{ kJ/(kg}\cdot\text{K)}), (R = 0.287 \text{ kJ/(kg}\cdot\text{K)}), and (k = 1.4).

Calculate:

1. The temperature (T_2) and pressure (P_2) at the end of the isentropic compression stroke.
2. The peak temperature (T_3) reached during the cycle.
3. The thermal efficiency (\eta_{th,Otto}) and net work output per unit mass (w_{net}) of the cycle.

Step-by-Step Solution

Step 1: Evaluate State 2 Properties (End of Isentropic Compression)

Using the isentropic property relations for ideal gas with constant (k = 1.4): $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{k-1} = T_1 (r)^{k-1}$ $T_2 = 300 \text{ K} \times (8.5)^{1.4 - 1} = 300 \times (8.5)^{0.4} = 300 \times 2.3539 = 706.17 \text{ K}$

$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^k = P_1 (r)^k$ $P_2 = 100 \text{ kPa} \times (8.5)^{1.4} = 100 \times 20.008 = 2000.8 \text{ kPa} = 2.001 \text{ MPa}$

Step 2: Evaluate Peak Temperature (T_3) at End of Heat Addition

For constant-volume heat addition (Process 2-3): $q_{in} = c_v (T_3 - T_2) \implies T_3 = T_2 + \frac{q_{in}}{c_v}$ $T_3 = 706.17 \text{ K} + \frac{800 \text{ kJ/kg}}{0.718 \text{ kJ/(kg}\cdot\text{K)}} = 706.17 + 1114.21 = 1820.38 \text{ K}$

Step 3: Compute Thermal Efficiency and Net Work Output

• Thermal efficiency of Otto cycle: $\eta_{th,Otto} = 1 - \frac{1}{r^{k-1}} = 1 - \frac{1}{(8.5)^{0.4}} = 1 - \frac{1}{2.3539} = 1 - 0.4248 = 0.5752 \text{ or } 57.52\%$
• Net work output (w_{net}): $w_{net} = \eta_{th,Otto} \times q_{in} = 0.5752 \times 800 \text{ kJ/kg} = 460.16 \text{ kJ/kg}$

Final Answer: (T_2 = 706.2 \text{ K}), (P_2 = 2.00 \text{ MPa}), peak temperature (T_3 = 1820.4 \text{ K}), thermal efficiency (\eta_{th} = 57.5%), and net work (w_{net} = 460.2 \text{ kJ/kg}).