Answer to Question #323218 in Mechanical Engineering for hoshikwon

Question

A steam engine isentropically expands 7.382 kg/sec of steam from 0.68 Mpa,

220°C, the exhaust is dry and saturated. Determine (a) the work of nonflow

process in kW and (b) the work of a steady flow process in kW. NOTE: - INDICATE THE HS and TS DIAGRAM - INDICATE THE INITIAL AND FINAL STATE OF THE SUBSTANCE

Answer 1

$state-1$

$p_1=2mpa$

$t_1=375^0c$

$h_1=3137.7+\frac{(3248.4-3137.7)}{(400-350)}\times(375-350)=3193.05\frac{kj}{kg}$

$s_1=6.9583+\frac{(7.1292-6.9583)}{(400-350)}\times(375-350)=7.0437\frac{kj}{kg,k}$

$state -2$

$T_2=95^0 C$

$S_2=S_1=7.0437\frac{kj}{kg,k}$

$Sf_2=1.2504$

$Sg_2=7.4151$

$hf_2=398.09$

$hfg_2=2269.6$

The dryness factor ,$x_2=\frac{s_2-sf_2}{sg_2-sf_2}=0.9397$

$h_2=hf_2+x_2hfg_2=2530.83\frac{kj}{kg}$

energy balance for steady flow process

$Ein-Eout=\triangle energy$

$h_1-W_{out}-h_2=0$

$W_{out}=h_1-h_2=3193.05-2530.83=662.22$

The non-flow process:

$Q=\triangle U+W$

$W=-\triangle U=-u_2-u_1$

$u_1=2860.5+\frac{(2945.9-2860.5)}{(400-350)}\times(375-350)=2903.2\frac{kJ}{kg}$

$u_2=398+(0.9397)(2102)=2373.25\frac{kJ}{kg}$

$w=2903.2-2373.25=529.95\frac{kJ}{kg}$