A couter example Discuss the relationship of c,and c,of water Solution u=u(T,v) du= dT dv p δq_du+δw=du+ =0 for Cp dy C= dT dT dT dT 1:water:p=1.0 atm dv T emperature Pressure Density Enthalpy Cp (o (atm) (kg/m) (kJ/kg) (kJ/kg-K(kJ/kg-K) 1.0000 1.0000 999.90 4.2788 4.2148 4.2161 2 2.0000 1.0000 999.94 8.4933 4.2124 4.2130 3 3.0000 1.0000 999.97 12.705 4.2100 4.2102 +p 4 4.0000 1.0000 999.97 16.914 42075 4.2075 5 5.0000 1.0000 999.97 21.120 4.2049 4.2050 6 6.0000 1.0000 999.94 25.324 4.2022 4.2028 7.0000 1.0000 999.90 29.526 4.1994 4.2006 8 8.0000 1.0000 999.85 33.725 4.1965 4.1987 9.0000 1.0000 999.78 37.923 4.1936 4.1968 10.000 1.0000 999.70 42.119 4.1906 4.1952 Density of water reaches maximum at 4 C,so cp=c,at 4 C 上游充通大学 March 21,2018 6 SHANGHAI JLAO TONG UNIVERSITYMarch 21, 2018 6 Discuss the relationship of cp and cv of water d d p v T u u v c p T v T                          Density of water reaches maximum at 4 ℃, so cp= cv at 4 ℃ Solution d d v T u v c p v T                  2 1 d d v T u c p T                     A couter example  ,  d d d v T u u T v u u u T v T v                    δ d δ d d d d d d d d q u w u p v v p c T T T T T       =0 for cp