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

state1state-1

p1=2mpap_1=2mpa

t1=3750ct_1=375^0c

h1=3137.7+(3248.43137.7)(400350)×(375350)=3193.05kjkgh_1=3137.7+\frac{(3248.4-3137.7)}{(400-350)}\times(375-350)=3193.05\frac{kj}{kg}

s1=6.9583+(7.12926.9583)(400350)×(375350)=7.0437kjkg,ks_1=6.9583+\frac{(7.1292-6.9583)}{(400-350)}\times(375-350)=7.0437\frac{kj}{kg,k}

state2state -2

T2=950CT_2=95^0 C

S2=S1=7.0437kjkg,kS_2=S_1=7.0437\frac{kj}{kg,k}

Sf2=1.2504Sf_2=1.2504

Sg2=7.4151Sg_2=7.4151

hf2=398.09hf_2=398.09

hfg2=2269.6hfg_2=2269.6

The dryness factor ,x2=s2sf2sg2sf2=0.9397x_2=\frac{s_2-sf_2}{sg_2-sf_2}=0.9397

h2=hf2+x2hfg2=2530.83kjkgh_2=hf_2+x_2hfg_2=2530.83\frac{kj}{kg}

energy balance for steady flow process

EinEout=energyEin-Eout=\triangle energy

h1Wouth2=0h_1-W_{out}-h_2=0

Wout=h1h2=3193.052530.83=662.22W_{out}=h_1-h_2=3193.05-2530.83=662.22

The non-flow process:

Q=U+WQ=\triangle U+W

W=U=u2u1W=-\triangle U=-u_2-u_1

u1=2860.5+(2945.92860.5)(400350)×(375350)=2903.2kJkgu_1=2860.5+\frac{(2945.9-2860.5)}{(400-350)}\times(375-350)=2903.2\frac{kJ}{kg}

u2=398+(0.9397)(2102)=2373.25kJkgu_2=398+(0.9397)(2102)=2373.25\frac{kJ}{kg}

w=2903.22373.25=529.95kJkgw=2903.2-2373.25=529.95\frac{kJ}{kg}




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment