Answer to Question #263453 in Mechanical Engineering for Rekteat

Question #263453

A steam engine isentropically expands 5 kg/sec of steam from 0.68 Mpa,

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

process (b) the final temperature, and (c) the work of a steady flow process.


1
Expert's answer
2021-11-10T00:49:04-0500

"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}"




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