Thermodynamics Zeroth Law: • Thermal equilibrium is transitive First Law: • Energy is conserved, its form can be converted Second Law: • Energies can flow, equilibrate Third Law: • “Driving Force” for equilibration uniquely defined • Boltzmann’s Relation: S = k ln p Where: k = 1.3806 x 1023 J/K Systems of Units Newton’s Second Law: πΉ = gc = 1 %+ ⋅(# CGS: π( = 1 -./0 ⋅ 12 SI: π( = 1 3%+ ⋅# 4 ⋅ 12 156%⋅78 English: π( = 1 59 : ⋅ 12 #$ %& gc ¹ 1 % ⋅(# CGS: π( = 980.66 %+ ⋅ 12 Weight: πΉ = #%P %& : 3% ⋅# SI: π( = 9.8066 3% +⋅ 12 : 59 ⋅78 English: π( = 32.174 59 +⋅ 12 At the earth’s sea level, the local acc. due to the gravity (πQ ) is: 980.66 cm/ s2 : 9.8066 m/s2 Poundal ππ# ⋅ ππ‘ π( = 1 πππ’ππππ ⋅ π O 32.174 ft/s2 Conversion Temperature 1 slug = 32.174 lbm Celsius à Fahrenheit 1 kgm = 2.205 lbm t(°C) = S [t(°F) – 32] 1 ft = 0.3048 m = 30.48 cm Celsius à Kelvin 1 N = 100,000 dynes t(K) = t(°C) + 273 1 lbf = 32.174 poundal = 4.4484 N Fahrenheit à Rankine 1 kgf = 9.8066 N t(R) = t(°F) + 460 1 lbm = 453.592 gm R Density ρ= # T Mass density of water at sea level at standard condition (Standard temperature of 39.2°F or 4°C and standard pressure of 1 atm) 62.4 lbm / ft3 Specific Volume π£= T # V =W 1000 kgm / m3 1 kgm / L Mass density of air at sea level at standard conditions (Standard temperature of 70°F or 21.11°C and standard pressure of 1 atm) Conversion for Volume 0.075 lbm / ft3 1 gal = 3.78 L 1.2 kgm / m3 1 ft3 = 7.48 gal 1 m3 = 1000 L Specific Weight πΎ= πΉ% ππQ ππQ πQ = = = π π( π π( π( π Specific weight of water at sea level at standard conditions (Standard temperature of 39.2°F or 4°C and standard pressure of 1 atm) 62.4 lbf / ft3 1000 kgf / m3 9.8066 kN / m3 0.98066 dyne / cm3 Specific weight of water at sea level at standard conditions (Standard temperature of 70°F or 21.11°C and standard pressure of 1 atm) 0.075 lbm / ft3 1.2 kgm / m3 11.768 N / m3 Specific gravity of liquids π . π = π π] Specific gravity of gases π . π = π π$ Specific gravities of common liquids Water – 1.0 Mercury – 13.6 Oil – 0.8 Seawater – 1.03 Measuring pressure Standard atmospheric pressure at sea level Atmospheric Pressure (Barometer) 1 atm π$8# ππQ β$8# = π( Gage Pressure (Manometer) ππQ β% π% = π( 101.325 πΎππ$ 760 mmHg 33.9 ft π»O π 14.696 ππ π$ 1.0332 kgf / cm2 29.92 in Hg Vacuum ππQ β% π% = − π( 760 torrs 1.01325 bars Internal Energy Terms U = f (m, T) 1 BTU: amount of heat needed to raise the temperature of 1 lbm of water at 68°F by 1°F π’= π π Conversion 1 BTU = 1.055 kJ = 778 ft-lbf 1 kCal = 4.187 kJ = 428.1 kg-m 1 calorie: amount of heat needed to raise the temperature of 1g of water at 14.5°C by 1°C Thermodynamic Property Intensive Property: independent of the mass of the substance Extensive Property: dependent on the mass of the substance Law of Conservation of Mass E = mc2 Where c = speed of light in a vacuum = 2.9979 x 108 m/s Volume flow rate (πΜ) πΜ = Aπ Mass flow rate (πΜ) πΜ = TΜ g = ππ΄π Where: A = cross sectional area of the stream π = average speed Forms of Energy Potential Energy (PE) à f (m, z) ΔππΈ = πΜ V = πΜO π΄V πV πV = π΄O πO πO ππQ (π§O − π§V ) π( Kinetic Energy (KE) à f(m, π) Law of Conservation of Energy ΔπΎπΈ = V# O %& (πOO − πVO ) Internal Energy (U) à f(m, T) βπ = ππp (πO − πV ) Flow work/ energy (Wf) à f (P, V) βπ7 = (πO πO − πV πV ) Where: πp = specific heat at constant volume Enthalpy (H) H = U + Wf βπ» = ππs βπ Heat (Q) Q = ∫ πππ (for any ideal gas or vapor as working substance) Q = mcβT (for any ideal gas as working substance) Where: c = specific heat of gas (depending upon the process involved during the change of state) βT = change in absolute temperature πΜ is constant during a process: Q = πΜβt Where: βt = process time interval = t2 – t1 πΜ is not constant during a process: 82 π = w πΜππ‘ 8x Mechanical Work (W) W = πΉπ Mechanical Power (πΜ ) y zP = 8 = 8 = Fπ Non-flow work (Wn) Thermodynamic properties vs. energy transfer mechanisms Example of point function: change in volume O w ππ = πO − πV = βπ O Wn = ∫V πππ V Example of path function: work πΏπ = πV|O ≠ βπ Steady Flow Energy Equation Ein = Eout PE1 + KE1 + U1 + P1V1 + Q = PE2 + KE2 + U2 + P2V2 + Ws Notes: Heat(Q) • (+) if heat is added to the system by the surroundings • (–) if heat is rejected by the system to the surroundings Steady flow work (WS) • (+) if work is done by the system o Turbine • (–) if work is done to the system o Compressor o Pump Non-Flow Energy Equation Q = βπ + π/ Notes: Heat(Q) • (+) if heat is added to the system by the surroundings • (–) if heat is rejected by the system to the surrounding Non flow work (Wn) • (+) if work is done by the system • (–) if work is done to the system Power Conversion 1 hp equals 33000 ft-lbf /min 550 ft-lbf /sec 42.4 BTU/min 2544 BTU/hr 0.746 kW
