Answer in Mechanical Engineering for Shraddha #213806

6. Two vessels, A and B each of volume 4m3 are connected by a tube of negligible volume. Vessel 

A contains air at 0.8MPa, 1050C, while vessel B contains air at 0.45MPa, 2250C. When A is allowed 

to mix with B and assuming the mixing to be complete and adiabatic, determine (a) the final 

pressure and temperature of air after mixing, (b) the amount of entropy generation and (c) the 

irreversibilities in the process. Take, T0=300C

(a) The final pressure and temperature of the air after mixing

“M = frac{PV}{RT}”

“M_A = frac{800*4}{0.287*375}=29.73 kg”

“M_B = frac{450*4}{0.287*498}=12.59 kg”

“(M_Ac_vT_A+M_Bc_vT_B)= (M_Ac_v+M_Bc_v)T_F”

“(29.73*0.718*375+12.59*0.718*498)= (29.73*0.718+12.59*0.718)T_F”

“frac{left(29.73cdot :0.718+12.59cdot :0.718right)T_F}{30.38576}=frac{12506.53326}{30.38576}”

“T_F=frac{12506.53326}{30.38576} =411.5919^0C”

“P_FV_F =M_FRT_F implies P_F = frac{(29.73+12.59)*0.287*411.59}{8}=624.89 kPa”

(b) The amount of entropy generation

“Delta S=Delta S_A +Delta S_B = (m_A(C_p ln frac{T_F}{T_A} -R ln frac{P_F}{P_A} ))+(m_B(C_p ln frac{T_F}{T_B}-R ln frac{P_F}{P_B} ))”

“Delta S= (29.73(1.005 ln frac{411.59}{375} -0.287 ln frac{624.89*10^3}{800} ))+(12.59(1.005 ln frac{411.59}{498}-0.287ln frac{624.89*10^3}{450} ))”

“Delta S = (29.73*0.1645)+(12.59*-0.2858)”

“Delta S = 1.2924 kJ/kg.K”

(c) the irreversibilities in the process

“dU = TdS – dW = 1.2924*300- 300*0.2858 = 301.98 kJ”