Basic Thermodynamics Cheatsheet

Cheatsheet Content

Module 1: Basic Concepts & Temperature Measurement Problem 1: Constant Volume Gas Thermometer Given: $P_i = 1000 \, \text{mm Hg}$ at ice point ($0^\circ C$), $P_s = 1366 \, \text{mm Hg}$ at steam point ($100^\circ C$). Linear relationship: $t = a + \beta P$. At ice point: $0 = a + \beta (1000)$ At steam point: $100 = a + \beta (1366)$ Solving for $a$ and $\beta$: $\beta = \frac{100 - 0}{1366 - 1000} = \frac{100}{366} \approx 0.2732$ $a = -1000 \beta = -1000 \times \frac{100}{366} \approx -273.22$ Relation: $t = 0.2732 P - 273.22$ When $P = 1074 \, \text{mm Hg}$: $t = 0.2732 (1074) - 273.22 \approx 293.75 - 273.22 = 20.53^\circ C$ Problem 2: Fahrenheit and Centigrade Comparison Given: $R_F = 2 R_C$ (Fahrenheit reading is twice Centigrade reading). Conversion formulas: $R_F = \frac{9}{5} R_C + 32$ $R_K = R_C + 273.15$ Substitute $R_F = 2 R_C$: $2 R_C = \frac{9}{5} R_C + 32$ $2 R_C - \frac{9}{5} R_C = 32 \Rightarrow \frac{10 - 9}{5} R_C = 32 \Rightarrow \frac{1}{5} R_C = 32$ $R_C = 160^\circ C$ $R_F = 2 \times 160 = 320^\circ F$ $R_K = 160 + 273.15 = 433.15 \, K$ Problem 3: Celsius Thermometric Scale with Property p Given: $p = e^{(t-B)/A}$. Ice point ($0^\circ C$), steam point ($100^\circ C$). At $0^\circ C$, $p = 1.86$. At $100^\circ C$, $p = 6.81$. $1.86 = e^{(0-B)/A} \Rightarrow \ln(1.86) = -B/A \Rightarrow 0.6259 = -B/A$ $6.81 = e^{(100-B)/A} \Rightarrow \ln(6.81) = (100-B)/A \Rightarrow 1.9184 = (100-B)/A$ Subtracting the two equations: $1.9184 - 0.6259 = (100-B)/A - (-B/A) = 100/A$ $1.2925 = 100/A \Rightarrow A = 100/1.2925 \approx 77.377$ $B = -A \times 0.6259 = -77.377 \times 0.6259 \approx -48.43$ Relation: $t = A \ln(p) + B = 77.377 \ln(p) - 48.43$ For $p = 2.5$: $t = 77.377 \ln(2.5) - 48.43 \approx 77.377 \times 0.9163 - 48.43 \approx 70.92 - 48.43 = 22.49^\circ C$ Module 2: Energy, Work, and Heat Problem 4: System Energy Change Heat received $Q = 42 \, \text{kJ}$. Volume change $\Delta V = 0.123 \, \text{m}^3$. Atmospheric pressure $P_{atm} = 12 \, \text{N/m}^2$. Work done by system on atmosphere $W_{atm} = P_{atm} \Delta V = 12 \times 0.123 = 1.476 \, \text{J} = 0.001476 \, \text{kJ}$. Mass lifted $m = 80 \, \text{kg}$, distance $h = 6 \, \text{m}$. Work done by system on surroundings (lifting mass) $W_{lift} = mgh = 80 \times 9.81 \times 6 = 4708.8 \, \text{J} = 4.7088 \, \text{kJ}$. Total work done by system $W = W_{atm} + W_{lift} = 0.001476 + 4.7088 = 4.710276 \, \text{kJ}$. i) Change in energy of the system ($\Delta U_1$): First Law of Thermodynamics: $\Delta U = Q - W$. $\Delta U_1 = 42 - 4.710276 = 37.289724 \, \text{kJ}$. ii) System returned to initial volume by adiabatic process, work required $100 \, \text{kJ}$. Adiabatic process: $Q = 0$. Work done ON system $W_{on} = 100 \, \text{kJ}$. So $W = -100 \, \text{kJ}$. Change in energy of the system for this process ($\Delta U_2$): $\Delta U_2 = Q - W = 0 - (-100) = 100 \, \text{kJ}$. iii) Total change in energy of the system: $\Delta U_{total} = \Delta U_1 + \Delta U_2 = 37.289724 + 100 = 137.289724 \, \text{kJ}$. Problem 5: Centrifugal Air Compressor Inlet: $P_1 = 100 \, \text{kPa}$, $T_1 = 300 \, \text{K}$. Outlet: $P_2 = 400 \, \text{kPa}$, $T_2 = 500 \, \text{K}$. Outlet velocity $V_2 = 100 \, \text{m/s}$. Inlet velocity $V_1 \approx 0$. Mass flow rate $\dot{m} = 15 \, \text{kg/s}$. Specific heat $C_p = 1 \, \text{kJ/kg-K}$. No heat transfer $Q = 0$. Steady Flow Energy Equation (SFEE): $\dot{Q} - \dot{W} = \dot{m} \left[ (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1) \right]$. Neglecting potential energy change $(z_2 - z_1 = 0)$. $\dot{Q} = 0$. So, $-\dot{W} = \dot{m} \left[ C_p(T_2 - T_1) + \frac{V_2^2 - V_1^2}{2} \right]$. $C_p(T_2 - T_1) = 1 \, \text{kJ/kg-K} \times (500 - 300) \, \text{K} = 200 \, \text{kJ/kg}$. $\frac{V_2^2 - V_1^2}{2} = \frac{(100 \, \text{m/s})^2 - 0^2}{2} = \frac{10000}{2} = 5000 \, \text{J/kg} = 5 \, \text{kJ/kg}$. $-\dot{W} = 15 \, \text{kg/s} \times (200 + 5) \, \text{kJ/kg} = 15 \times 205 = 3075 \, \text{kW}$. Power required to drive compressor $\dot{W} = -3075 \, \text{kW}$ (negative sign indicates work input). Problem 6: Water-Cooled Compressor Mass flow rate of air $\dot{m} = 0.5 \, \text{kg/s}$. Shaft input $\dot{W}_{in} = 60 \, \text{kW}$. Heat lost to cooling water $\dot{Q}_{loss,water} = 0.30 \times \dot{W}_{in} = 0.30 \times 60 = 18 \, \text{kW}$. Heat lost due to bearings/friction $\dot{Q}_{loss,friction} = 0.10 \times \dot{W}_{in} = 0.10 \times 60 = 6 \, \text{kW}$. Inlet: $P_1 = 1 \, \text{bar}$, $T_1 = 20^\circ C = 293.15 \, \text{K}$. Specific heat $C_p = 1 \, \text{kJ/kg}^\circ C = 1 \, \text{kJ/kg-K}$. SFEE: $\dot{Q} - \dot{W} = \dot{m} C_p (T_2 - T_1)$. (Neglecting KE & PE changes). Net heat transfer from system $\dot{Q}_{net} = \dot{Q}_{loss,water} + \dot{Q}_{loss,friction} = 18 + 6 = 24 \, \text{kW}$. (This is heat leaving the system, so $Q = -24 \, \text{kW}$). Work input $\dot{W} = -60 \, \text{kW}$. $-24 - (-60) = 0.5 \times 1 \times (T_2 - 293.15)$. $36 = 0.5 (T_2 - 293.15)$. $T_2 - 293.15 = 36 / 0.5 = 72$. $T_2 = 293.15 + 72 = 365.15 \, \text{K} = 92^\circ C$. Module 3: Refrigeration and Heat Engines Problem 7: Refrigerator Performance Heat extracted from cold compartment $Q_L = 1500 \, \text{kJ/min} = 1500/60 = 25 \, \text{kJ/s} = 25 \, \text{kW}$. Cold compartment temperature $T_L = -5^\circ C = 268.15 \, \text{K}$. Room temperature $T_H = 25^\circ C = 298.15 \, \text{K}$. Power input $W_{in} = 1 \, \text{kW}$. i) Actual COP of the refrigerator: $COP_{actual} = \frac{Q_L}{W_{in}} = \frac{25 \, \text{kW}}{1 \, \text{kW}} = 25$. ii) Maximum possible COP (Carnot COP): $COP_{Carnot} = \frac{T_L}{T_H - T_L} = \frac{268.15}{298.15 - 268.15} = \frac{268.15}{30} \approx 8.938$. Note: The calculated actual COP is much higher than Carnot COP. This indicates an error in the problem statement or units (e.g., $1500 \, \text{kJ/min}$ is a very large heat extraction for $1 \, \text{kW}$ input). Assuming the given numbers are correct. Problem 8: Heat Engine Performance Heat received from high-temperature reservoir $Q_H = 2000 \, \text{kJ}$. Heat rejected to sink $Q_L = 800 \, \text{kJ}$. i) Net-work output: $W_{net} = Q_H - Q_L = 2000 - 800 = 1200 \, \text{kJ}$. ii) Thermal efficiency of the engine: $\eta_{th} = \frac{W_{net}}{Q_H} = \frac{1200}{2000} = 0.6 = 60\%$. Problem 9: Minimum Power Input for Refrigerator Heat extracted from cold body $Q_L = 500 \, \text{kJ/min} = 500/60 = 8.333 \, \text{kW}$. Cold body temperature $T_L = -10^\circ C = 263.15 \, \text{K}$. Room temperature $T_H = 30^\circ C = 303.15 \, \text{K}$. Minimum power input corresponds to Carnot refrigerator. $COP_{Carnot} = \frac{T_L}{T_H - T_L} = \frac{263.15}{303.15 - 263.15} = \frac{263.15}{40} = 6.57875$. $W_{in,min} = \frac{Q_L}{COP_{Carnot}} = \frac{8.333 \, \text{kW}}{6.57875} \approx 1.266 \, \text{kW}$. Module 4: Real Gases and Entropy Problem 10: Pressure of CO Gas (Generalized Chart) This problem requires a generalized compressibility chart, which cannot be directly represented or "solved" in a text-based cheatsheet. The methodology involves: 1. Find critical properties ($T_c$, $P_c$) for CO from tables. 2. Calculate reduced temperature ($T_r = T/T_c$) and reduced specific volume ($v_r = v/v_c$ or $v_r' = v P_c / (R T_c)$). 3. Use these reduced properties to find the compressibility factor ($Z$) from the generalized chart. 4. Calculate pressure using $P = Z R T / v$. For the second part (volume reduced to 80%), repeat steps 2-4 with the new specific volume and the same pressure, iterating to find the new temperature. Typical values for CO: $T_c = 132.9 \, \text{K}$, $P_c = 3.5 \, \text{MPa}$. Gas constant $R = 0.2968 \, \text{kJ/kg-K}$. Problem 11: Entropy Change for Mixing Gases Cylinder volume $V_{total} = 2 \times 0.1 = 0.2 \, \text{m}^3$. Initial temperature $T = 20^\circ C = 293.15 \, \text{K}$. Compartment 1: $V_1 = 0.1 \, \text{m}^3$, $P_1 = 2.5 \, \text{MPa} = 2500 \, \text{kPa}$. Compartment 2: $V_2 = 0.1 \, \text{m}^3$, $P_2 = 1 \, \text{MPa} = 1000 \, \text{kPa}$. Assume air is an ideal gas: $R_{air} = 0.287 \, \text{kJ/kg-K}$. Mass in compartment 1: $m_1 = \frac{P_1 V_1}{R T} = \frac{2500 \times 0.1}{0.287 \times 293.15} \approx 2.972 \, \text{kg}$. Mass in compartment 2: $m_2 = \frac{P_2 V_2}{R T} = \frac{1000 \times 0.1}{0.287 \times 293.15} \approx 1.189 \, \text{kg}$. Total mass $m_{total} = m_1 + m_2 = 2.972 + 1.189 = 4.161 \, \text{kg}$. Final pressure $P_f = \frac{m_{total} R T}{V_{total}} = \frac{4.161 \times 0.287 \times 293.15}{0.2} \approx 1749.9 \, \text{kPa} = 1.75 \, \text{MPa}$. Change in entropy for mixing ideal gases (at constant temperature): $\Delta S = - R \sum_{i} m_i \ln(y_i)$, where $y_i$ is mole fraction. Or, more simply, $\Delta S = m_1 R \ln\left(\frac{P_1}{P_f}\right) + m_2 R \ln\left(\frac{P_2}{P_f}\right)$. (This is for a single gas expanding into a new volume, not mixing two different gases). For mixing of a single gas initially at different pressures in insulated compartments: $\Delta S = m_1 R \ln(V_f/V_1) + m_2 R \ln(V_f/V_2)$ where $V_f$ is final volume for each mass. This is not correct for mixing of same substance. For two bodies of the same ideal gas at different pressures and same temperature mixing in an insulated container, the total entropy change is given by: $\Delta S = m_1 R \ln\left(\frac{P_1}{P_f}\right) + m_2 R \ln\left(\frac{P_2}{P_f}\right)$. This formula is for a constant volume process where pressures equalize. In this case, the gas from compartment 1 expands to $V_{total}$, and gas from compartment 2 expands to $V_{total}$. $\Delta S_1 = m_1 R \ln(V_{total}/V_1) = 2.972 \times 0.287 \times \ln(0.2/0.1) = 2.972 \times 0.287 \times \ln(2) \approx 0.590 \, \text{kJ/K}$. $\Delta S_2 = m_2 R \ln(V_{total}/V_2) = 1.189 \times 0.287 \times \ln(0.2/0.1) = 1.189 \times 0.287 \times \ln(2) \approx 0.236 \, \text{kJ/K}$. Total $\Delta S = \Delta S_1 + \Delta S_2 = 0.590 + 0.236 = 0.826 \, \text{kJ/K}$. Problem 12: CO2 Pressure (Van der Waals vs. Ideal Gas) Specific volume $v = 1.2 \, \text{m}^3/\text{kg}$. Temperature $T = 120^\circ C = 393.15 \, \text{K}$. CO2 molar mass $M = 44.01 \, \text{kg/kmol}$. Gas constant $R_{CO2} = R_u/M = 8.314/44.01 = 0.1889 \, \text{kJ/kg-K}$. Van der Waals constants for CO2: $a = 0.364 \, \text{Pa} \cdot \text{m}^6/\text{mol}^2 = 0.364 \times (M^{-1})^2 \text{ MPa} \cdot (\text{m}^3/\text{kg})^2 = 0.364 \times (44.01)^{-2} \times 10^6 \, \text{Pa} \cdot (\text{m}^3/\text{kg})^2 \approx 188.6 \, \text{Pa} \cdot (\text{m}^3/\text{kg})^2$. (Using $a=364000 \text{ N} \cdot \text{m}^4/\text{kmol}^2$ and $b=0.0427 \text{ m}^3/\text{kmol}$ from common tables). To convert $a$ and $b$ to per unit mass: $a_{mass} = a_{molar} / M^2 = 364000 / (44.01)^2 = 188.58 \, \text{N} \cdot \text{m}^4/\text{kg}^2$. $b_{mass} = b_{molar} / M = 0.0427 / 44.01 = 0.000970 \, \text{m}^3/\text{kg}$. Van der Waals Equation: $(P + \frac{a}{v^2})(v - b) = R T$ $P = \frac{R T}{v - b} - \frac{a}{v^2}$ $P = \frac{0.1889 \times 393.15}{1.2 - 0.000970} - \frac{188.58}{(1.2)^2}$ $P = \frac{74.28}{1.19903} - \frac{188.58}{1.44} \approx 61.95 - 130.96 = -69.01 \, \text{kPa}$. (This negative pressure indicates an issue with constants or calculation. Let's re-check $a$ and $b$ for CO2 in correct units, often $a$ is in $\text{Pa} \cdot \text{m}^6/\text{mol}^2$ and $b$ in $\text{m}^3/\text{mol}$). Using $a = 0.364 \, \text{Pa} \cdot \text{m}^6/\text{mol}^2$ and $b = 4.27 \times 10^{-5} \, \text{m}^3/\text{mol}$. For $R = 8.314 \, \text{J/mol-K}$, $P = \frac{R T}{v_m - b} - \frac{a}{v_m^2}$, where $v_m = v \times M_{CO2} = 1.2 \times 44.01 = 52.812 \, \text{m}^3/\text{kmol}$. $P = \frac{8314 \times 393.15}{52.812 - 0.0427} - \frac{0.364 \times 10^6}{(52.812)^2}$ (converting $a$ to $\text{Pa} \cdot \text{m}^6/\text{kmol}^2$) $P = \frac{3269387.1}{52.7693} - \frac{364000}{(52.812)^2} \approx 61958 - 130.3 = 61827.7 \, \text{Pa} = 61.8 \, \text{kPa}$. Ideal Gas Equation: $P = \frac{R_{CO2} T}{v}$ $P = \frac{0.1889 \, \text{kJ/kg-K} \times 393.15 \, \text{K}}{1.2 \, \text{m}^3/\text{kg}} = \frac{74.28}{1.2} = 61.9 \, \text{kPa}$. Comparison: For this specific volume and temperature, CO2 behaves very close to an ideal gas. The Van der Waals pressure is $61.8 \, \text{kPa}$ and ideal gas pressure is $61.9 \, \text{kPa}$.