Chapter 4: Homework Solution

Adapted by Diana Bairaktarova

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.

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

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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