15

A movable cover is located at the top of an opened metal box.  The metal box contains steam at 350^{\circ}C, and pressure of 2 MPa.  Currently the steam is about to be cooled.  Determine the compression work W_b for the steam and pressure and its final temperature where the mass of the movable box and metal box is 4kg.

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T = 350^\circ\text{C}
P = 2 \ \si\MPa
m = 4 \ \si\kg
\text{Final state is cooled to 64.28}\%

Given:

T_1=350^{\circ}C

P_1=2MPa

T_2=225^{\circ}C

P_2=2MPa

m=4kg

Find:

Determine the compression work W_b for the steam and pressure and its final temperature.

Solution:

Using table A-6

At First State

T_1=350^{\circ}C

P_1=2MPa

v_{f,1}=0.10381 m^3/kg

At Final State

T_2=225^{\circ}C

P_2=2MPa

v_{f,2}=0.10381 m^3/kg

a) the compression work

W_b=mP(v_1-V_2)

W_b=4(2000)(0.13860-0.10381)=278.32 kJ

b) the volume of the metal box at the final state is 64.28% from its initial volume, so the work will be as

W_b=mP(v_1-0.6428v_1)

W_b=4(2000)(0.13860-(0.6428*0.13860))=396.06336

P_2=2*0.6428=1.2856\cong 1300kPa

v_2=0.6428*0.13860=0.089092 m^3/kg

T_2=191.60^{\circ}C

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Thermodynamics Copyright © by Diana Bairaktarova (Adapted from Engineering Thermodynamics - A Graphical Approach by Israel Urieli and Licensed CC BY NC-SA 3.0) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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