At these low temperatures specific heat does not

Info icon This preview shows pages 2–4. Sign up to view the full content.

View Full Document Right Arrow Icon
At these low temperatures, specific heat does not change dramatically w/ temperature, so we can use isentropic relationships shown in Chapter 5. approximately u 3 0 kJ kgair = T 3 1750K = Use P3 and v3 to look up remaining properties at state 3 in Figure 4-5 v 3 0.073 m 3 kgair = v 3 v 2 β = β v 3 v 2 = 2.13 = use cutoff ratio P 3 P 2 = Note h 3 513.575 kJ kgair = h 3 h s.2 h fu.knot + = h fu.knot 81.425 - kJ kgair = h fu.knot 51.9 - 1181 x b - ( 29 kJ kgair = Eq. 4.32 h s.2 595 kJ kgair = Need h.s.2 from Figure 4.3 at 820 K h 3 h s.2 h fu.knot + = Eq 4.28: 2 to 3 is constant pressure combustion!! Note, to determine Qin, need enthalpy values rather than internal energy values. Need Equations 4-28 and 4.32 for φ =0.4 and enthalpy
Image of page 2

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon