
#1
Mar2808, 05:52 AM

P: 38

I have
P_1 = 1 x 10^5 Pa p_2 = 11.764x10^5 Pa V_1 0.826 m^3 V_2 0.142 m^3 and Expansion / Compression Index of 1.4 I need to calculate Work Done. First of all i identify 12 as an Isothermal Process, which gives equation. W = P_2 * V_2 * ln (V_2 / V_1) If i use that formula i get an answer. What i am unsure of is 1) Am i using the right formula based on what data i have, i only have data that i have above. 2) How and what do i use expansion and compression index for? Thanks for any help, needed urgent to solve a question. 



#2
Mar2808, 02:40 PM

HW Helper
P: 1,664

If you are given an index of 1.4, you are probably supposed to be looking at an adiabatic process for a diatomic gas. So you don't want to use the work formula for an isothermal process.
In a (reversible) adiabatic process, the relationship is [tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex], where gamma is the adiabatic index (in your case, [tex]\gamma = 1.4[/tex]). In such a process, Q = 0 ; you would still calculate work by integrating P dV between the two values for volume, but the result is now a little more complicated... 



#3
Aug3108, 11:14 AM

P: 6

well if u r talking abt the work done in a cycle then it is
Workdone/cycle = n/n1 x P1V1 x [ (P2/P1)power n1/n  1 ] but put pressure in KN/m2 this will give u the answer in KJ/minute, and if u divide the answer by 60 then it will b KiloJoules/Sec that is Watts, so that will b the Power in KW. 


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