Question

a steam engine working on Rankine cycle operates between 1.96Mpa, 250degree celsius and 13.7Kpa. If engine consumes steam at the rate of 0.086kg per second, determine rankine cycle efficiency, neglecting pump work. Also, find Rankine cycle effeciency considering pump work.

Answer 1

For ,$P_1=P_2=1.96MPa;t_2=250^0\implies h_2=2904.9,S_2=6.559$

$From\space table\space A-1, For\space P_3=0.1\space bar$

$t_s=45.83^oC\space h_f=191.8\space h_fg=2392.9$

$S_f=0.6493\space S_fg=7.5018$

Since $From\space table\space A-1, For\space P_3=0.1\space bar$

$t_s=45.83^oC\space h_t=191.8\space h_fg=2392.9\space S_t=0.6493\space S_fg=7.5018$

$Since X_3=0.9,\space h_3=h_f4+x_3h_fg$

$=191.8+0.9(239.9)$

$=2345.4\space kj/kg=191.8+0.9(239.9)=2345.4 kj/kg$

$Also, since\space process\space 2-3_s\space is\space isentropic,\space S_2$

$i.e., 7.1251=S_fg4+X_3s\space S_fg3$

$=0.6493+X_3s(7.5018)=0.6493+X_3s$

$\therefore X_3s=0.863$

$\therefore h_3s=191.8+0.863(2392.9)=2257.43\space kj/kg$

$\eta_R=\frac{h_2-h_3}{h_2-h_{3s}}=\frac{2904.9-2345.4}{2904.9-2257.43}=0.86$

$W_p=v\int dp =0.0010102(19.6-0.137)*10^3=19.66kJ/kg$

$\eta_R=\frac{2904.9-2345.4-19.66}{2904.9-2257.43}=0.83$