Thermodynamics 1 Understanding and Using the Thermodynamic Property

Thermodynamics 1 Understanding and Using the Thermodynamic Property Tables 18 September 2020

Table Structure Each substance has its own table Tables are subdivided into sections Saturated Mixture Superheated Vapor Compressed Liquid Property values are given in each section

Saturated Mixture Table

Saturated Mixture Table

Saturated Mixture Table

Superheated Vapor Table

Compressed Liquid Table

Examples T = 120°C P = 198. 5 k. Pa

Examples T = 120°C P = 198. 5 k. Pa Answer: Saturated Mixture nf = 0. 001060 m 3 / kg nfg = 0. 89080 m 3 / kg ng = 0. 89186 m 3 / kg

Examples P = 100 k. Pa T = 99. 62°C

Examples P = 100 k. Pa T = 99. 62°C Answer: Saturated Mixture nf = 0. 001043 m 3 / kg nfg = 1. 69296 m 3 / kg ng = 1. 69400 m 3 / kg

Examples T = 120°C n = 0. 45 m 3 / kg

Examples P=const T T = 120° n = 0. 45 m 3/kg nf = 0. 001080 m 3/kg ng = 0. 50885 m 3/kg n

Examples T = 120°C n = 0. 45 m 3 / kg Answer: Because nf < ng Saturated Mixture P = Psat = 198. 5 k. Pa

Examples P = 200 k. Pa T = 150°C T ≠ Tsat

Examples P = 200 k. Pa T = 150°C

Examples P = 200 k. Pa T = 150°C Answer: Superheated Vapor n = 0. 95964 m 3/kg

Examples T = 150°C n = 0. 5 m 3/kg T ≠ Tsat n > ng

Examples P=const T T = 150° n = 0. 5 m 3/kg nf = 0. 001090 m 3/kg ng = 0. 39278 m 3/kg n

Examples T = 150°C n = 0. 5 m 3/kg 300 k. Pa < P < 400 k. Pa

Examples T = 150°C n = 0. 5 m 3/kg Must Interpolate

Examples – Interpolation P 400 P 300 n. 47084 0. 5 . 63388

Examples P = 500 k. Pa T = 80°C T ≠ Tsat

Examples P = 500 k. Pa T = 80°C

Examples P = 500 k. Pa T = 80°C Answer: Compressed Liquid n = 0. 001029 m 3/kg

Examples P = 10 k. Pa T = 100°C

Examples P = 10 k. Pa T = 100°C

Examples T = 100°C n = 0. 42 m 3 / kg Saturated Mixture