Answer to Question #213806 in Mechanical Engineering for Shraddha

Question

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

Answer 1

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

$M = \frac{PV}{RT}$

$M_A = \frac{800*4}{0.287*375}=29.73 kg$

$M_B = \frac{450*4}{0.287*498}=12.59 kg$

$(M_Ac_vT_A+M_Bc_vT_B)= (M_Ac_v+M_Bc_v)T_F$

$(29.73*0.718*375+12.59*0.718*498)= (29.73*0.718+12.59*0.718)T_F$

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

$T_F=\frac{12506.53326}{30.38576} =411.5919^0C$

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

(b) The amount of entropy generation

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

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

$\Delta S = (29.73*0.1645)+(12.59*-0.2858)$

$\Delta S = 1.2924 kJ/kg.K$

(c) the irreversibilities in the process

$dU = TdS - dW = 1.2924*300- 300*0.2858 = 301.98 kJ$