Question

An air-standard cycle is executed within a closed piston–cylinder system, and it consists of the following three processes: 1–2 V = Constant heat addition from 100 kPa and 30°C to 850 kPa 2–3 Isothermal expansion until V3 = 8.5V2 3–1 P = Constant heat rejection to the initial state Assume air has constant properties with cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

Required information An air-standard cycle is executed within a closed piston, cylinder system, and it consists of the following three processes: 1–2 V=Constant heat addition from 100 kPa and 30°C to 850 kPa 2-3 Isothermal expansion until V3 = 8.5V2 3-1 P=Constant heat rejection to the initial state Assume air has constant properties with Cy= 0.718 kJ/kg.K, Cp = 1.005 kJ/kg-K, R=0.287 kJ/kg.K, and k= 1.4. Determine the cycle thermal efficiency. The cycle thermal efficiency is

Answer 1

In this problem, first draw P-V diagram for the given cycle then use area under the curve for Calculation of thermal efficiency which is Calculated as below:-

Given data :- *T=constant 100 ka P, = 100 kPa Ti= 30°c= 303 K 1850k P₂ = 850 kla Vg = V2(8.5) for air, Co=0.718 kolkg_k cp = 1,005 kJ lig-k R = 0.287 KJ lig k y=1.4 solution :- Thermal efficiency of cycle (Vrh) :- hoa Area of cycle (1-2-3) Area under the care 2-3 "ith = (Area under the core 2-3) - (Area under the arre) (Area under the curre 2-3) th = - Area (1-3-4-5) Area (2-3-4-5-1) belere, Area (1-3-45) = P (V3_W,) (Vg = 8.682) Area (1-3-4-5) = P. (8.51-V2) (& V, = (2) = P, vz (8.5-1) = 7.5 P, V2 1-3

3- 4- therma u Wan & Area (2-3-4-5-1) = I go thermal proces work - Peva lim ( 13 ) = Pq Vz ln (8.582) = Pe V2 ln (8.5) - 2.14 Pave. - You (s - Home Eva) - (- your orders Von = (1 - 1:50) = (1-0.4123) Von = 0.5877 = 58-771.1 th = 7.591 V2 2.14 Pq V2 7.5 x 100 l 2:14x850)