Answer to Question #213790 in Mechanical Engineering for SID

Question

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 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

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