Answer in Mechanics | Relativity for Martin Navarro #140245
Solution
a) for non flow process
This is isothermal process so
Change in internal energy
“Delta U=0”
“P_1V_1=nT_1\T_1=frac{P_1V_1}{n}”
Where n=0.3452Kg
After putting value of pressures and n then
“T_1=255.9K”
n = m / M
= ( 10 kg )/ ( 28.97 kg/kgmol )
= 0.3452 kgmol
Work done
W=nRT“ln{frac{P_1}{P_2}}”
W=0.3452“times” 8.314“times” 255.9“ln{frac{96KPa}{620KPa}}n\”
W=-1370KJ
Now heat
“Q=Delta U+Delta W\Q=-1370KJ”
Now entropy change
“Delta S=frac{Q}{T}=frac{-1370KJ}{255.9}”
“Delta S=-5.354KJ/kelvin”
b) now for steady flow
Work done
Ws=0.3452“times” 8.314“times” 255.9“ln{frac{96KPa}{620KPa}}n\”
“W_s=-1370KJ/min”
“Delta H=0”
“Delta KE=frac{10times(60^2-15^2) }{2}=14” KJ/min
“Delta PE=0”
Total heat
“Q=Delta H+Delta KE+Delta PE+” Ws
Q=-1353KJ/min
Now entropy change
“Delta S=frac{Q}{T}=frac{-1353KJ/min}{255.9Kelvin}”
“Delta S=-5.267” KJ/min-Kelvin
