목차
4.1. The group u+gz+V2/2 and h+gz+V2/2 occur in most thermodynamics problems To evaluate the relative magnitude terms, calculate gz and V2/2 in Btu/lbm and J/kg for the following z=10, 100, 1000, 10,000 ft; V= 10, 100, 1000, 10,000 ft/s. Show these result on a log-log pot
본문내용
PROBLEM
Several problems in this section deal with ideal gas. It may be shown that for an ideal gas the enthalpy and internal energy depend on temperature alone. If an ideal gas has an constant heat capacity (which may be assumed in all the ideal gas problems in this chapter), it is very convenient to choose an enthalpy datum that leads to h=C_p T and u=C_V T where T is the absolute temperature; these values may be used in ideal-gas problems in this chapter
4.1. The group u+gz+V2/2 and h+gz+V2/2 occur in most thermodynamics problems To evaluate the relative magnitude terms, calculate gz and V2/2 in Btu/lbm and J/kg for the following z=10, 100, 1000, 10,000 ft; V= 10, 100, 1000, 10,000 ft/s. Show these result on a log-log pot
1) potential energy
① Btu/lbm
1[Btu/lbm]=778.16[(lb_f∙ft)/lb_m ]×32.2[(lb_m∙ft)/(lb_f 〖∙s〗^2 )]
*1[Btu]=778.16[lb_f∙ft]
1[Btu/lbm]=25057[ft^2/s^2 ]
g_c=1=1/25057 [(Btu〖∙s〗^2)/(lbm〖∙ft〗^2 )]
Several problems in this section deal with ideal gas. It may be shown that for an ideal gas the enthalpy and internal energy depend on temperature alone. If an ideal gas has an constant heat capacity (which may be assumed in all the ideal gas problems in this chapter), it is very convenient to choose an enthalpy datum that leads to h=C_p T and u=C_V T where T is the absolute temperature; these values may be used in ideal-gas problems in this chapter
4.1. The group u+gz+V2/2 and h+gz+V2/2 occur in most thermodynamics problems To evaluate the relative magnitude terms, calculate gz and V2/2 in Btu/lbm and J/kg for the following z=10, 100, 1000, 10,000 ft; V= 10, 100, 1000, 10,000 ft/s. Show these result on a log-log pot
1) potential energy
① Btu/lbm
1[Btu/lbm]=778.16[(lb_f∙ft)/lb_m ]×32.2[(lb_m∙ft)/(lb_f 〖∙s〗^2 )]
*1[Btu]=778.16[lb_f∙ft]
1[Btu/lbm]=25057[ft^2/s^2 ]
g_c=1=1/25057 [(Btu〖∙s〗^2)/(lbm〖∙ft〗^2 )]
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