APSC252 Final Exam Formula Sheet Pressure, temperature and specific volume Absolute temperature: π[πΎ] = π[°πΆ] + 273.15 Pressure: π = πΉ/π΄ Absolute pressure: Pabs=Patm+ ρgh Gage pressure: Pgage=Pabs - Patm Specific volume: π 1 π£=π=π (π: volume) Two-phase mixture Quality: π₯ = ππ£ππππ’π ππππ₯π‘π’ππ Specific volume: π£ = π£π + π₯π£ππ = (1 − π₯)π£π + π₯π£π , π£ππ = π£π − π£π Specific internal energy: π’ = π’π + π₯π’ππ = (1 − π₯)π’π + π₯π’π , π’ππ = π’π − π’π Specific enthalpy: β = βπ + π₯βππ = (1 − π₯)βπ + π₯βπ , βππ = βπ − βπ Specific entropy: π = π π + π₯π ππ = (1 − π₯)π π + π₯π π , π ππ = π π − π π Compressed liquid (when tables are not available) π£ ≈ π£π@π π’ ≈ π’π@π β ≈ βπ@π π ≈ π π@π Ideal gas EOS: ππ£ = π π or ππ = ππ π (R: gas constant; π: volume; π£: specific volume) (T in Kelvin) Specific heats: ππ = ππ£ + π , π = ππ /ππ£ , ππ π ππ = π−1 , ππ£ = π−1 Specific enthalpy and specific internal energy: β = π’ + ππ£ = π’ + π π π’ = ππ£ π , Change in specific internal energy (assuming constant ππ£0 ) Change in specific enthalpy (assuming constant ππ0 ) β = ππ π π’2 − π’1 = ππ£0 (π2 − π1 ) β2 − β1 = ππ0 (π2 − π1 ) Change in specific entropy π 0 0 ) π 2 − π 1 = (π π2 − π π1 − π ln π2 • Accurate method • Approximate method (assuming constant ππ£0 and ππ0 ) 1 π 2 − π 1 = ππ0 ln Compressibility factor: π = π2 π1 ππ£ π π − π ln π2 π1 or π 2 − π 1 = ππ£0 ln ; reduced pressure and temperature: ππ = π2 π1 π ππ + π ln π£2 π£1 π , ππ = π π (T in Kelvin) (T in Kelvin) Solid and liquid Specific heats: π = ππ = ππ£ Change in specific internal energy or specific enthalpy: β2 − β1 ≈ π’2 − π’1 ≈ π(π2 − π1 ) π2 Change in specific entropy: π 2 − π 1 = πln (T in Kelvin) π1 1 Energy, work and heat Total energy: πΈ = π + πΎπΈ + ππΈ = ππ’ + ½ ππ 2 + πππ§ (V: velocity) Specific total energy: π = πΈ/π = π’ + ½ π 2 + ππ§ (V: velocity) Specific internal energy: π’ = π/π Specific heat transfer: π = π/π 2 Boundary work: 1W2= ∫1 πππ (area under the P- π diagram; π: volume) Specific work: π€ = π/π 2 1 Spring work: πspring = ∫1 πΉππ₯ = πΎ(π₯22 − π₯12 ) 2 Μ Shaft power: π = ππ (T: torque) Closed systems (control mass) 1st law: E2 − E1 = 1Q2 − 1W2 2nd law: π2 − π1 = ∑ ππ ππ or + ππππ U2 − U1 = 1Q2 − 1W2 (assuming οKE=0, οPE=0) (ππππ ≥ 0) (T in Kelvin) Volume flow rate and mass flow rate Volume flow rate: πΜ = πππ£π,π π΄ = πΜπ£ (π: volume, V: velocity, π£ : specific volume) Mass flow rate: πΜ = ππΜ = ππππ£π,π π΄ Steady-state, steady flow through a control volume (open system) Conservation of mass: ∑ πΜπ = ∑ πΜπ 1 1 1st law: πΜππ£ + ∑ πΜπ (βπ + ππ2 + ππ§π ) = πΜππ£ + ∑ πΜπ (βπ + ππ2 + ππ§π ) 2 2nd law: ∑ πΜπ π π − ∑ πΜπ π π = ∑ 2 πΜπ.π£. π Μ + ππππ Μ (ππππ ≥ 0) (T in Kelvin) Applications of the conservation of mass and the 1st law in different steady-state, steady flow devices (note: the following derived relations are valid only if the given assumptions can be applied.) • Throttling valve: πΜπ = πΜπ , βπ = βπ (assuming πΜππ£ = 0, πΜππ£ = 0, ΔPE = 0, ΔKE = 0) 1 1 • Nozzle and diffuser: πΜπ = πΜπ , βπ + ππ2 = βπ + ππ2 (assuming πΜππ£ = 0, πΜππ£ = 0, ΔPE = 0) 2 2 • Mixing chamber: ∑ πΜπ = ∑ πΜπ , πΜππ£ + ∑ πΜπ βπ = ∑ πΜπ βπ (assuming πΜππ£ = 0, ΔPE = 0, ΔKE = 0) • Heat exchanger: πΜπ = πΜπ (for each of the hot and cold streams, separately) πΜππ£ + ∑ πΜπ βπ = ∑ πΜπ βπ (assuming πΜππ£ = 0, ΔPE = 0, ΔKE = 0) • Turbine: πΜπ = πΜπ = πΜ , πΜπ βπππ‘ = πΜ(βπ − βπ ) (assuming πΜππ£ = 0, ΔPE = 0, ΔKE = 0) • Compressor: πΜπ = πΜπ = πΜ , πΜπ βπππ‘ = πΜ(βπ − βπ ) (assuming πΜππ£ = 0, ΔPE = 0, ΔKE = 0) 2 Processes • Isobaric process: P=const • Isochoric process: v=const • Isothermal process: T=const • Adiabatic process: heat transfer Q=0 (for closed systems) • Reversible process: ππππ = 0 (for closed systems) πΜππ£ = 0 (for control volumes) or Μ ππππ = 0 (for control volumes) or • Isentropic process: reversible adiabatic process • Polytropic process: πππ = ππππ π‘. (π: volume) Boundary work in a polytropic process o 1W2= π2 π2 −π1 π1 o 1W2= π1 π1 ππ π2 = π2 π2 ππ π2 = ππ π ππ π2 (π = 1) (π ≠ 1) 1−π π π 1 π 1 Heat engines • Any heat engine o Net work output: ππππ‘,ππ’π‘ = ππ» − ππΏ o Thermal efficiency: ππ‘β = ππππ‘,ππ’π‘ • Carnot heat engine: ππ‘β,πππ£ = 1 − ππ» ππΏ π = 1 − ππΏ π» (T in Kelvin) ππ» Refrigerators • Any refrigerator o Net work input: ππππ‘,ππ = ππ» − ππΏ o Coefficient of performance: πΆπππ = • Carnot refrigerator: πΆπππ ,πππ£ = ππΏ ππ» −ππΏ =π ππΏ ππππ‘,ππ 1 Coefficient of performance: πΆπππ»π = • Carnot heat pump: πΆπππ»π,πππ£ = ππ» ππ» −ππΏ =π π» /ππΏ −1 Heat pumps • Any heat pump o Net work input: ππππ‘,ππ = ππ» − ππΏ o (T in Kelvin) 1 ππ» ππππ‘,ππ 1 = 1−π /π πΏ π» ππΏ π» −ππΏ =π 1 π» /ππΏ −1 (T in Kelvin) =π ππ» π» −ππΏ 1 = 1−π πΏ /ππ» (T in Kelvin) ======================================================================================= 3
