Gas Power Cycles
By: S K Mondal ∴ ∴
Chapter 13
p p2 = 3 T2 T3
or
η= 1−
p3 r r × T2 = eγ × T1 . rcγ - 1 = T1 e = T4 p2 rc rc Cp (T4 − T1 )
T3 =
Q2 = 1− Cv (T3 − T2 ) + RT3 In re Q1
r ⎛ ⎞ Cp ⎜ T1 . e − T1 ⎟ rc ⎝ ⎠ = 1− re r ⎛ ⎞ Cv ⎜ T1 . − T1 rcγ − 1 ⎟ + R .T1 e In re rc rc ⎝ ⎠ ⎛r ⎞ γ ⎜ e − 1⎟ r ⎝ c ⎠ = 1− r re ⎛ e γ −1 ⎞ ⎜ r − rc ⎟ + ( γ − 1) r In re c ⎝ c ⎠ γ(re − rc ) = 1− γ (re − rc ) + ( γ − 1) re l n re ∴
η= 1−
γ[re − rc ] (re − r ) + ( γ − 1) re l n re γ c
Given p1 = 1 bar = 100 kPa T1 = 40°C = 313 K
rc =8 and p3 = 100 bar = 10000 kPa p3 = 100 p1 1.4 (100 − 8) ∴η= 1− 1.4 (100 − 8 ) + (1.4 – 1 × In 100 128.8 = 1− 265.83 = 0.51548 = 51.548 % ∴ p3 = p1 . re ∴ re =
⇒ T3 = T1 ×
re 313 × 100 = = 3912.5 K 8 rc
T2 = T1 × rcγ − 1 = 719 K ∴ Heat addition, Q = Cv ( T3 –T2) + R T3 In re
= 0.718 (3912.5 – 719) + 0.287 × 3912.5 × ln 100 = 7464 kJ/kg ∴ Work, W = Q η = 3847.5 kJ/kg ∴ p m (V4 – V2) = W
∴ v 4 = 100 v 2
∴ p m (100 –1) v 2 = W ∴ pm (99) ×
v1 =W 8 Page 245 of 265
v2 =
v1 rc