Air at 480Kpa is expanded adiabatically to a pressure of 94Kpa. It is then heate
ID: 955777 • Letter: A
Question
Air at 480Kpa is expanded adiabatically to a pressure of 94Kpa. It is then heated at constant volume until it again attains its temperature, when the pressure is found to be 150KPa. it then undergoes an isothermal compression to return the air to its original state. Using this information sketch the p-v diagram and calculate the value of . If the initial temperature of air is 190oC determine the work done per kg during the adiabatic expansion and teh isotherman compression. Calculate the change in enthropy during the constant volume process.
Take R= 0.29KJ/KgK
Explanation / Answer
adiabatic isothermal
P1 = 480 kpa P1 = 94 k pa
P2 = 94 k pa P2 = 150 k pa
To find gamma
T1 = 190 degree centigrade = 273+190 = 363 K
R = 0.29kj/kg K
Let n =1
V1 = nRT/P1 = 1*0.29*1000*363/480*1000 =0.219
V2 = nRT/P2 = 1*0.29*1000*363/94*1000 = 1.1198
Now P2/P1 (V2/V1)^ =1 Therefore (94*1000/480*1000) ( 1.1198/.219) ^ =1
(5.113)^ = 1*480/94 = 5.106
Take logs
log(5.113)^ = log 5.106
=log(5.106 - log 5.113 = -5.949*10^-4
To find work done (isothermal)
w = -NRT lnVB/VA = -1*.29*1000*363ln(1.1198/0.219) = 17178.3128 J
Work done in adiabatic process = - f P1V1( (v2/V1)^1- -1)
= -(-2878 *94*1000* ((1.475/1.1198)^1-(-5.949*10^-4 -1)
= 2.70*10^8J
( f in above has been calculated here
Because = f+2/f
so -5.949*10^-4 = f+2/f
f(1+5.949*10^-4) = -2
f = -2878
Here V1 = 1.1198
V2 = nRT/P2 = 1*.29* 1000*363/150*1000 = 1.475 )
Change in entropy at constant volume = delta S = nRT in PA/PB = 1*0.29*1000* ln (94*10^3/150*10^3)
= -812.78 j/k
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