Thermodynamics Property Evaluations and Internal Energy Calculations

Comprehensive vocabulary flashcards covering thermodynamic properties, ideal gas processes, polytropic work, steam turbine applications, Brayton cycles, refrigeration cycles, and pressure-enthalpy diagrams.

Specific Internal Energy ($u$)

The internal energy stored per unit mass of a substance, tabulated in thermodynamic tables such as Table A1 (saturated water) and Table A2 (superheated vapor). For example, water at $20^\circ\text{C}$ and $300\,\text{kPa}$ has $u = 83,91\,\text{kJ/kg}$, while saturated water vapour at $5^\circ\text{C}$ has $u = 2381,78\,\text{kJ/kg}$.

Boiler-to-Turbine Internal Energy Drop

The drop in specific internal energy when superheated steam flows between a boiler at $20\,\text{bar}$ and $350^\circ\text{C}$ ($u = 2860,45\,\text{kJ/kg}$) and a turbine inlet manifold at $18\,\text{bar}$ and $300^\circ\text{C}$ ($u = 2785,67\,\text{kJ/kg}$), resulting in a drop of $74,78\,\text{kJ/kg}$.

Absolute Pressure ($P_{abs}$)

The total pressure relative to a complete vacuum. To convert a pressure measurement reading from a boiler pressure gauge into absolute pressure, local atmospheric pressure must be added to the gauge reading ($P_{abs} = P_{gauge} + P_{atm}$).

Constant-Volume Heat Addition in a Rigid Tank

The thermal energy supplied to a sealed, rigid container with zero boundary work ($W = 0$), calculated by $Q = \Delta U = m \cdot C_v \cdot (T_2 - T_1)$. For example, heating $1,2\,\text{kg}$ of an ideal gas ($C_v = 1,6\,\text{kJ/kgK}$) from $320\,\text{K}$ to $400\,\text{K}$ requires $153,6\,\text{kJ}$ of heat transfer.

Isobaric Heat Transfer in a Piston-Cylinder

The amount of heat transferred per unit mass at constant pressure ($q = \Delta h$). For steam heated isobarically at $3\,\text{bar}$ from $200^\circ\text{C}$ to $250^\circ\text{C}$, the required specific heat transfer is $101,98\,\text{kJ/kg}$.

Air Compressor Shaft Power Input

The mechanical power input required on the shaft of an air compressor operating on an ideal gas. For a mass flow of $20\,\text{g/s}$ entering at $1,0\,\text{bar}$ and $25^\circ\text{C}$ and exiting at $8,0\,\text{bar}$ and $120^\circ\text{C}$, the needed power input is $1,9\,\text{kW}$.

Internal Energy Change of an Ideal Gas ($\Delta U$)

The change in stored thermal energy of an ideal gas governed by $\Delta U = m \cdot C_v \cdot \Delta T$ or $PV = mRT$. For instance, heating $3100\,\text{g}$ of air from $20^\circ\text{C}$ to $50^\circ\text{C}$ produces $\Delta U = 66,74\,\text{kJ}$, while heating a $1000\,\text{liter}$ rigid tank of nitrogen at $2\,\text{bar}$ from $10^\circ\text{C}$ to $50^\circ\text{C}$ produces $\Delta U = 70,73\,\text{kJ}$.

Polytropic Boundary Work Comparison ($n = 1,3$ vs. $n = 1$)

The work evaluated via $P_1 v_1^n = P_2 v_2^n$ between $P_1 = 200\,\text{kPa}$, $v_1 = 0,05\,\text{m}^3\text{/kg}$ and $v_2 = 0,1\,\text{m}^3\text{/kg}$. The polytropic process with $n = 1,3$ yields a specific boundary work of $6,258\,\text{kJ/kg}$, whereas the isothermal process ($n = 1$) produces a larger specific boundary work of $6,931\,\text{kJ/kg}$.

Isothermal Specific Boundary Work

The boundary work produced or absorbed per unit mass during constant-temperature compression or expansion. Compressing air isothermally at $60^\circ\text{C}$ from $0,7\,\text{bar}$ to $8\,\text{bar}$ requires a specific boundary work of $-232,927\,\text{kJ}$ done by the air to the surroundings.

Saturated Liquid Specific Enthalpy ($h_f$)

The specific enthalpy of a pure liquid at its boiling point for a given pressure or temperature. From thermodynamic table Appendix A1 for saturated water at $25^\circ\text{C}$, $h_f = 104,83\,\text{kJ/kg}$.

Compressed Liquid Water Properties

Thermodynamic state parameters for water below its saturation temperature at a specified pressure. From Appendix A3 at $5\,\text{bar}$ and $60^\circ\text{C}$, water has a specific volume of $v = 0,001017\,\text{m}^3\text{/kg}$ and a specific enthalpy of $h = 251,58\,\text{kJ/kg}$.

Refrigerant Saturation Temperature

The boiling/condensation phase-change temperature corresponding to a specific pressure. From Appendix C2 for superheated/saturated R134a, the saturation temperature at $1,5\,\text{bar}$ is $-17,13^\circ\text{C}$.

Polytropic Exponent ($n$)

The exponent describing the relationship $P_1 v_1^n = P_2 v_2^n$ during non-isobaric and non-adiabatic expansion or compression, calculated from two known states as $n = \frac{\ln(P_2 / P_1)}{\ln(v_1 / v_2)}$.

Polytropic Specific Heat Transfer ($q$)

The net heat transferred per unit mass in a polytropic process, defined as $q = \Delta u + w$, where the boundary work is $w = \frac{P_2 v_2 - P_1 v_1}{1 - n}$. For ammonia expanding from $20^\circ\text{C}$, $5\,\text{bar}$ to $30^\circ\text{C}$, $3\,\text{bar}$, the specific heat transfer is $108,4\,\text{kJ}$.

Back Work Ratio ($bwr$)

The fraction of mechanical work produced by a turbine that is consumed to drive the compressor in a gas turbine cycle, defined as $bwr = \frac{w_{comp}}{w_{turbine}} = \frac{C_p(T_2 - T_1)}{C_p(T_3 - T_4)} = \frac{T_2 - T_1}{T_3 - T_4}$.

Cold Air-Standard Brayton Cycle Efficiency ($\eta_{th}$)

The thermal efficiency of an ideal Brayton gas turbine cycle, calculated using the compressor pressure ratio $r_p = \frac{p_2}{p_1}$ and the specific heat ratio $k = \frac{C_p}{C_v}$ as $\eta_{th} = 1 - \frac{T_1}{T_2} = 1 - \left(\frac{p_1}{p_2}\right)^{\frac{k-1}{k}} = 1 - r_p^{\frac{1-k}{k}}$.

Net Power Output of an Ideal Brayton Cycle ($\dot{W}_{net}$)

The net rate of mechanical work output produced by a gas turbine system, computed via $\dot{W}_{net} = \dot{m} \cdot C_p \cdot (T_3 - T_4)(1 - bwr)$. For an ideal cold air-standard cycle with $\dot{m} = 6\,\text{kg/s}$, $r_p = 10$, $k = 1.4$, $T_1 = 300\,\text{K}$, and $T_3 = 1400\,\text{K}$, the net power is $2390\,\text{kW}$.

Isobaric Boundary Work ($_{1}W_2$)

The work transferred across a moving system boundary during constant-pressure expansion or cooling ($W = P(V_2 - V_1)$). For $100\,\text{g}$ of water vapor cooled isobarically at $3\,\text{bar}$ from $500^\circ\text{C}$ to $300^\circ\text{C}$, the boundary work is $-9,34\,\text{kJ}$.

Steam Compressor Power Input

The power required to compress superheated steam entering at $8\,\text{bar}$ and $300^\circ\text{C}$ with a volume flow rate of $500\,\text{lit/s}$ to an exit state of $1,0\,\text{MPa}$ and $350^\circ\text{C}$, which equals $156,36\,\text{kW}$.

Steam Turbine Shaft Power with Kinetic Energy

The mechanical shaft power delivered by expanding steam ($3,542\,\text{MW}$) calculated by accounting for the flow kinetic energy where steam enters at $4000\,\text{kPa}$, $500^\circ\text{C}$, $h_i = 3445,2\,\text{kJ/kg}$, an inlet velocity of $200\,\text{m/s}$, and an inlet diameter of $50\,\text{mm}$, and exits at $80\,\text{kPa}$, $h_e = 2665,7\,\text{kJ/kg}$, and an outlet diameter of $250\,\text{mm}$.

Turbine Efficiency and Mechanical Loss

The factors reducing shaft work output in a real steam turbine. For a steam turbine with a mass flow rate of $50\,000\,\text{kg/h}$, an isentropic efficiency of $40\%$, an additional mechanical loss of $8\,\text{kW}$, and enthalpies $h_i = 806,5\,\text{kJ/kg}$ and $h_e = 677,0\,\text{kJ/kg}$, the net delivered shaft power is $711\,\text{kW}$.

Coefficient of Performance for Refrigeration (COPR)

The efficiency metric of a refrigeration cycle, defined as the ratio of the cooling rate absorbed at the evaporator to the compressor power input ($\text{COPR} = \frac{\dot{Q}<em>L}{\dot{W}</em>{in}}$). For an ammonia cycle with $\dot{Q}<em>L = 109,44\,\text{kW}$ and $\dot{W}</em>{in} = 33,66\,\text{kW}$, $\text{COPR} = 3,25$.

Rankine Steam Power Plant Layout

A vapor power cycle consisting of an adiabatic turbine ($\dot{W}_{turbine} = 8\,\text{MW}$), a condenser with cooling water ($\dot{Q}_{cond} = 17.6\,\text{MW}$), a feedwater pump ($\dot{W}_{pump} = 80\,\text{kW}$), and a boiler ($\dot{Q}_{boiler} = 25,7\,\text{MW}$), resulting in a cycle thermal efficiency of $\eta_{th} = 31\%$.

Vapor-Compression Refrigeration Cycle Components (R-134a)

A four-component closed loop using refrigerant R-134a consisting of an adiabatic compressor (state 1 to 2), a condenser rejecting heat $q_{cond}$ (state 2 to 3), an adiabatic throttling valve (state 3 to 4), and an evaporator absorbing heat $q_{evap}$ (state 4 to 1).

R-134a Refrigeration Cycle on a Pressure-Enthalpy ($P\text{-}h$) Diagram

The thermodynamic trajectory of R-134a shown on a $P\text{-}h$ diagram: state 1 is saturated vapor at $-20^\circ\text{C}$; state 2 is superheated vapor at $1\,\text{MPa}$ and $70^\circ\text{C}$; state 3 is subcooled liquid at $1\,\text{MPa}$ and $30^\circ\text{C}$; and state 4 is a two-phase mixture after isenthalpic expansion ($h_4 = h_3$).

Steam Power Cycle on a Pressure-Enthalpy ($P\text{-}h$) Diagram

The thermodynamic representation of a Rankine steam cycle plotted on a water $P\text{-}h$ diagram, where boiler heat input is $q_{boiler} = h_1 - h_4$, turbine work output is $w_{turbine} = h_1 - h_2$, and thermal efficiency is evaluated as $\eta_{th} = \frac{w_{turbine}}{q_{boiler}}$.

Specific Internal Energy of Compressed/Saturated Liquid Water ($20^\circ\text{C}$, $300\,\text{kPa}$)

The specific internal energy property obtained from saturated water Table A1, equal to $83{,}91\,\text{kJ/kg}$.

Specific Internal Energy of Superheated Water Vapor ($50^\circ\text{C}$, $10\,\text{kPa}$)

The specific internal energy property obtained from superheated vapor Table A2, equal to $2443.30\,\text{kJ/kg}$.

Specific Internal Energy of Saturated Water Vapour ($5^\circ\text{C}$)

The specific internal energy value obtained from saturated water Table A1, equal to $2381{,}78\,\text{kJ/kg}$.

Internal Energy Drop from Boiler to Turbine

The reduction in internal energy between superheated steam produced by a boiler at $20\,\text{bar}$ and $350^\circ\text{C}$ ($u = 2860{,}45\,\text{kJ/kg}$) and steam at the turbine inlet manifold at $18\,\text{bar}$ and $300^\circ\text{C}$ ($u = 2785{,}67\,\text{kJ/kg}$), calculated as $2860{,}45\,\text{kJ/kg} - 2785{,}67\,\text{kJ/kg} = 74{,}78\,\text{kJ/kg}$.

First Law Heat Transfer for a Rigid Tank

The energy balance equation $Q = \Delta U + W$ where boundary work $W = 0$ because volume is constant, leading to $Q = \Delta U = m \cdot C_v \cdot (T_2 - T_1)$, which equals $153{,}6\,\text{kJ}$ for $1{,}2\,\text{kg}$ of ideal gas with $C_v = 1{,}6\,\text{kJ/kg}\cdot\text{K}$ heated from $320\,\text{K}$ to $400\,\text{K}$.

Absolute Pressure ($P_{\text{abs}}$) Conversion

The actual thermodynamic pressure utilized in property tables and diagrams, determined from a boiler pressure gauge reading by adding atmospheric pressure to the measured gauge pressure.

Isobaric Specific Heat Transfer

The heat transferred per unit mass during a constant-pressure process; for steam heated isobarically in a piston-cylinder device at $3\,\text{bar}$ from $200^\circ\text{C}$ to $250^\circ\text{C}$, the required heat transfer is $101{,}98\,\text{kJ/kg}$.

Air Compressor Shaft Power Input ($20\,\text{g/s}$)

The power required on the compressor shaft to process a mass flow rate of $20\,\text{g/s}$ of air (ideal gas) from $1{,}0\,\text{bar}$ and $25^\circ\text{C}$ to $8{,}0\,\text{bar}$ and $120^\circ\text{C}$, equal to $1{,}9\,\text{kW}$.

Internal Energy Change of Nitrogen in a Sealed Tank

The change in internal energy ($\Delta U$) when nitrogen in a sealed rigid $1000\,\text{liter}$ tank at $10^\circ\text{C}$ and $2\,\text{bar}$ is heated to $50^\circ\text{C}$, calculated via $PV = mRT$ to be $70{,}73\,\text{kJ}$.

Polytropic Process Specific Boundary Work ($w$) Comparison

The boundary work per unit mass for expansion governed by $P_1 v_1^n = P_2 v_2^n$ between $P_1 = 200\,\text{kPa}$, $v_1 = 0{,}05\,\text{m}^3/\text{kg}$ and $v_2 = 0{,}1\,\text{m}^3/\text{kg}$, yielding $6{,}258\,\text{kJ/kg}$ for $n = 1{,}3$ and a larger value of $6{,}931\,\text{kJ/kg}$ for an isothermal process ($n = 1$).

Isothermal Specific Boundary Work of Compression

The work per unit mass done by air during constant-temperature compression from $0{,}7\,\text{bar}$ and $60^\circ\text{C}$ to $8\,\text{bar}$, equal to $-232{,}927\,\text{kJ}$.

Saturated Liquid Specific Enthalpy ($h_f$)

The specific enthalpy of water at the saturated liquid phase boundary, which from Appendix Table A1 at $25^\circ\text{C}$ is $104{,}83\,\text{kJ/kg}$.

Compressed Liquid Water Properties (Table A3)

The state properties of subcooled water at $5\,\text{bar}$ and $60^\circ\text{C}$, consisting of a specific volume $v = 0{,}001017\,\text{m}^3/\text{kg}$ and a specific enthalpy $h = 251{,}58\,\text{kJ/kg}$.

Saturation Temperature of R134a at $1{,}5\,\text{bar}$

The phase change temperature for refrigerant R134a at a pressure of $1{,}5\,\text{bar}$, listed in Appendix Table C2 as $-17{,}13\,^\circ\text{C}$.

Polytropic Exponent ($n$) Formula

The mathematical exponent defining a polytropic path between two thermodynamic states, expressed as $n = \frac{\ln(P_2 / P_1)}{\ln(v_1 / v_2)}$.

Polytropic Specific Boundary Work Formula

The work equation per unit mass for a polytropic process, given by $w = \frac{P_2 v_2 - P_1 v_1}{1 - n}$, which alongside $q = \Delta u + w$ gives a specific heat transfer of $108{,}4\,\text{kJ}$ for ammonia expanding from $20^\circ\text{C}$, $5\,\text{bar}$ to $30^\circ\text{C}$, $3\,\text{bar}$.

Back Work Ratio ($bwr$) Formula

The ratio of compressor work to turbine work in a Brayton gas turbine cycle, defined as $bwr = \frac{w_{\text{comp}}}{w_{\text{turbine}}} = \frac{C_p(T_2 - T_1)}{C_p(T_3 - T_4)} = \frac{T_2 - T_1}{T_3 - T_4}$.

Brayton Cycle Thermal Efficiency ($\eta_{\text{therm}}$) Formula

The thermal efficiency of an ideal air-standard Brayton cycle expressed in terms of pressure ratio $r_p = \frac{p_2}{p_1}$ and specific heat ratio $k = \frac{C_p}{C_v}$ as $\eta_{\text{therm}} = 1 - \frac{T_1}{T_2} = 1 - \left(\frac{p_1}{p_2}\right)^{\frac{k-1}{k}} = 1 - r_p^{\frac{1-k}{k}}$.

Net Power Output of an Ideal Brayton Cycle ($\dot{W}_{\text{net}}$)

The net power produced by a gas turbine cycle calculated as $\dot{W}_{\text{net}} = \dot{m} \cdot C_p \cdot (T_3 - T_4)(1 - bwr)$, resulting in $2390\,\text{kW}$ for a cycle with $\dot{m} = 6\,\text{kg/s}$, $r_p = 10$, $T_1 = 300\,\text{K}$, $T_3 = 1400\,\text{K}$, and $k = 1.4$.

Constant-Pressure Boundary Work (${}_1W_2$) of Water Vapor

The boundary work done during constant-pressure cooling of $100\,\text{g}$ of superheated water vapor at $3\,\text{bar}$ from $500^\circ\text{C}$ to $300^\circ\text{C}$, equal to $-9{,}34\,\text{kJ}$.

Polytropic Expansion Final Specific Volume Calculation

The specific volume computed using the ideal gas law and $P_1 v_1^n = P_2 v_2^n$ for air changing from $4{,}9\,\text{bar}$ and $41^\circ\text{C}$ to $1{,}46\,\text{bar}$ with $n = 1{,}29$, equal to $0{,}4787\,\text{m}^3/\text{kg}$.

Steam Compressor Power Input ($500\,\text{L/s}$)

The shaft power input required to compress superheated steam at a volume flow rate of $500\,\text{L/s}$ from $8\,\text{bar}$ and $300^\circ\text{C}$ to $1{,}0\,\text{MPa}$ and $350^\circ\text{C}$, equal to $156{,}36\,\text{kW}$.

Turbine Shaft Power Accounting for Kinetic Energy

The power output of a steam turbine taking into account inlet velocity ($200\,\text{m/s}$) and nozzle dimensions ($D_i = 50\,\text{mm}$, $D_e = 250\,\text{mm}$) between $4000\,\text{kPa}$, $500^\circ\text{C}$ ($h_i = 3445{,}2\,\text{kJ/kg}$) and $80\,\text{kPa}$ ($h_e = 2665{,}7\,\text{kJ/kg}$), equal to $3{,}542\,\text{MW}$.

Turbine Shaft Power with Mechanical Loss

The shaft power developed by a turbine with $40\%$ efficiency and an additional $8\,\text{kW}$ mechanical loss, with mass flow rate $50\,000\,\text{kg/h}$, $h_i = 806{,}5\,\text{kJ/kg}$, and $h_e = 677{,}0\,\text{kJ/kg}$, equal to $711\,\text{kW}$.

Coefficient of Performance for Refrigeration ($\text{COP}_R$)

The efficiency metric of a refrigeration cycle defined as $\text{COP}<em>R = \frac{\dot{Q}_L}{\dot{W}</em>{\text{in}}}$, which equals $3{,}25$ for an ammonia cycle with cooling load $\dot{Q}<em>L = 109{,}44\,\text{kW}$ and compressor power $\dot{W}</em>{\text{in}} = 33{,}66\,\text{kW}$.

Steam Power Plant Cycle Schematic

A closed Rankine cycle system consisting of a boiler supplying $\dot{Q}<em>{\text{boiler}} = 25{,}7\,\text{MW}$, an adiabatic turbine delivering $\dot{W}</em>{\text{turbine}} = 8\,\text{MW}$, a condenser rejecting $\dot{Q}<em>{\text{cond}} = 17{,}6\,\text{MW}$, and a feedwater pump consuming $\dot{W}</em>{\text{pump}} = 80\,\text{kW}$ with an overall thermal efficiency $\eta_{\text{th}} = 31\%$.

R-134a Refrigeration Cycle on a P-h Diagram

A plotted thermodynamic refrigeration loop showing: (1-2) adiabatic compression of saturated vapor at $-20^\circ\text{C}$ to superheated vapor at $1\,\text{MPa}$ and $70^\circ\text{C}$, (2-3) isobaric condensation to liquid at $30^\circ\text{C}$, (3-4) isenthalpic throttling expansion, and (4-1) constant-pressure evaporation at $-20^\circ\text{C}$.

Steam Power Cycle on a Water P-h Diagram

A pressure-enthalpy diagram illustrating four cycle processes: constant-pressure boiler heating ($q_{\text{boiler}} = h_1 - h_4$), adiabatic turbine expansion ($w_{\text{turbine}} = h_1 - h_2$), constant-pressure condenser heat rejection, and feedwater pump compression, with cycle efficiency defined as $\eta_{\text{th}} = \frac{w_{\text{turbine}}}{q_{\text{boiler}}}$.