Question

Er<E EF E E> E) W>0) and Polytropic PathsSpcl Cases for Ideal Gas n 0 constant pressure n 1 constant temperature n k constant entropy, adiabatic (q 0) n constant volume and W<0 W 0 For air R 0.287 kJ/kg-K and k Cp/Cv 1.4 if pi 300 kPa, v, 0.861 m3/kg then T, 900 K For T Ta if pa /p 3 then va = m2/kg pv RT and (T, n/(n-1) p2 V1 For vi v if po /p1 3 then Ta = For pi pe if TT, = 2.2 then ve For T1 Ts if va /v 2 then pa For 1Q 0 uf pi/pe = 4 then Te = K m/kg pv C T p V2 kPa K WARNING: Do not use if n-0, n-1 orn For air, helium and carbon-dioxide approximated as a perfect gas (with T Tope) use data from the textbook Note at 300 K: Cvo. air= 0.71 75 kJ/kg-K, CVo,He 3.116 kJ/kg-K, . Cvo.co2 0.657 kJ/kg-K Find the change in specific internal energy and average Cv for a process from T: = 340 K to T2 = 540 K Air (T Tonpt 133 K) Helium (T>Tatpt 5.2 K) cO2 (T>Tontp 304 K) >>Find the change in specific internal energy and average Cv for a process from T, = 1340 K to T2 1540 K Air (T Totpt 133 K) Helium (T Ton5.2 K) CO2 (T Ton 304 K) (kJ/kg) and Cv,ave Au/AT (kJ/kg) and Cv,ave Au/AT (kJ/kg) and Cv, ave Au/AT kJ/kg-K kJ/kg-K kJ/kg-K u2-U1 u2-U U2 U (kJ/kg) and Cv,ave Au/AT = (kJ/kg) and Cv, ave Au/AT = (kJ/kg) and Cv,ave Au/AT kJ/kg-K kJ/kg-K _ kJ/kg-K U2-Ut U2-Ut U2-Ut

Answer 1

aix iven 3 oob- Fo 3 xP - 3 Pa 900 KPa then mR T O. 28z M3 900X0. 86 270o K

3) oob×とと=」 1980 K と=R。 0.287X 1980 30o m3 bor -T, - 900k X =ス×O-861 R To R TL 028テ×906 P 150 K Pa

adliatakio means P V Pv nR T 1-4 C. -o.H Te Pe 1-H Te TI fe -o.H Te (H) Te 60S.6S8 K