This preview shows pages 1–2. Sign up to view the full content.
Solution 2
The Carnot Cycle:
Data Given:
Pa = 3 MPa ; Pc = 100 KPa ; Tc = 20 C = 20 + 273 = 293 K ;
Compression ratio = 15 = Vc/Va
(a) Thermal efficiency =
Now,
(b) Work output = thermal efficiency * Qh
Since process
A to B is an isothermal process, hence
PV = constant
Therefore,
Now, process B to C is an adiabatic process, hence,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: And γ = 1.4 ( for air) Hence, 6.96 = 326.3 J/kgK Work done = 326.3 * 0.5 *0.5 = 81.75 KJ/kg (c) MEP = work done/ ( Vc – Va) Therefore, Vc = 15Va To calculate Va, we use the gas equation, Therefore, = 0.056 * 10^(3) Vc Va = 14Va MEP = Work done / 14Va = 156.5 kPa...
View
Full
Document
This note was uploaded on 10/19/2011 for the course CIVIL 441 taught by Professor Urvashi during the Spring '11 term at CUNY City Tech.
 Spring '11
 urvashi

Click to edit the document details