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Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the temperature and pressure changes to 800 K and 8.74 * 10 N/m². Determine the change in following across the shockwave: a. Internal energy b. Enthalpy c. Entropy

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Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Step-by-step explanation:

adiabatic constant
\gamma =1.4

specific heat is given as
=(\gamma R)/(\gamma -1)

gas constant =287 J⋅kg−1⋅K−1


Cp = (1.4*287)/(1.4-1) = 1004.5 Jkg^(-1) k^(-1)

specific heat at constant volume


Cv = (R)/(\gamma -1) = (287)/(1.4-1) = 717.5 Jkg^(-1) k^(-1)

change in internal energy
= Cv(T_2 -T_1)


\Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^(-1)

change in enthalapy
\Delta H = Cp(T_2 -T_1)


\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^(-1)

change in entropy


\Delta S =Cp ln((T_2)/(T_1)) -R*ln((P_2)/(P_1))


\Delta S =1004.5 ln((800)/(295)) -287*ln((8.74*10^5)/(1.01*10^5))


\Delta S = 382.79 J kg^(-1) K^(-1)

User James Simm
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