 reproacht3

2022-01-05

$0.5-{m}^{3}$ rigid tank contains refrigerant-134a initially at 160 kPa and 40 percent quality. Heat is now transferred to the refrigerant until the pressure reaches 700 kPa.
Calculate (a)the amount of refrigerant in the tank by mass.
(b) the amount of heat transferred. Hence, show the process on a $P-v$ diagram with respect to saturation lines. For this stationary closed system, the energy balance is expressed as
${E}_{\in }-{E}_{out}=\mathrm{△}{E}_{system}$
therefore ${Q}_{\in }=\mathrm{△}U=m\left({u}_{2}-{u}_{1}\right)$
use tables (Tables $A-11$ through $A-13$)
the properties of $R-134a$
${P}_{1}=160Kpa$
${x}_{1}=0.4$
so $⇒{v}_{f}=0.0007435,{v}_{g}\right\}=0.12355\frac{{m}_{3}}{kg}$
${v}_{f}=31.06,{u}_{fg}=190.31\frac{kj}{kg}$
therefore ${v}_{1}={v}_{f}+{x}_{1}{v}_{fg}$
the equation's values are changed
${v}_{1}=0.0007435+0.4\left(0.12355-0.0007435\right)$
$⇒{v}_{1}=0.04987\frac{{m}^{3}}{kg}$
so, ${u}_{1}={u}_{f}+{x}_{1}{u}_{fg}$
we replace the values in the equation
${u}_{1}=31.06+0.4\left(190.31\right)$
$⇒{u}_{1}=107.19\frac{kj}{kg}$
therefore
${P}_{2}=700Kpa$
$\left({v}_{2}={v}_{1}\right)⇒{u}_{2}=377.23\frac{kj}{kg}$
the mass of the refrigenant is determined
$m=\frac{{V}_{1}}{{v}_{1}}$
we replace the values in the equation
$m=\frac{0.5{m}^{3}}{0.04987\frac{{m}^{3}}{kg}}⇒m=10.03kg$
par (b)
${Q}_{\in }=m\left({u}_{2}-{u}_{2}\right)$
we replace the values in the equation
${Q}_{\in }=\left(10.03kg\right)\left(377.23-107.19\right)\frac{kj}{kg}$
${Q}_{\in }=2708kj$ Beverly Smith

$V=0.5{m}^{3},R-134a,{P}_{1}=160kPa,{X}_{1}=0.4$
${P}_{2}=700kPa$, Find m, Q2, show P-U DIAG.
First law:
${Q}_{2}-{W}_{2}=m\left({U}_{2}-{U}_{1}\right)$
For a rigid tank, ${W}_{b}={S}_{1}^{2}PAV=0$
${Q}_{2}=m\left({U}_{2}-{U}_{1}\right)$
State 1: ${P}_{1}=160kPa,{X}_{1}=0.4$
${U}_{1}={U}_{f}+{X}_{1}\cdot {U}_{fg}=\left(31.09\right)+\left(0.4\right)\left(190.27\right)=107.2\frac{kj}{kg}$
${U}_{1}={U}_{f}+{X}_{1}\left({U}_{g}-{U}_{f}\right)$
$=\left(0.0007434\right)+\left(0.3\right)\left(0.12348-0.0007437\right)$
${V}_{1}=0.04984\frac{{m}^{3}}{kg}$
$m=\frac{V}{{V}_{1}}=\frac{\left(0.5{m}^{3}\right)}{\left(0.04984\frac{{m}^{3}}{kg}\right)}=10.03kg$
State 2: ${P}_{2}=700kPa,{V}_{2}={V}_{1}=0.04984\frac{{m}^{3}}{kg}$
At ${P}_{4at}=700kPa,{V}_{g}=0.029361\frac{{m}^{3}}{kg}$
Since ${V}_{2}>{V}_{g}$
Table A-13: $P=0.70mPa$
$\begin{array}{|cc|}\hline М\left(\frac{{m}^{3}}{kg}\right)& U\left(\frac{kJ}{kg}\right)\\ 0.047306& 347.41\\ 0.048497& 367.29\\ \hline\end{array}$
$U={U}_{1}+\left(\frac{{U}_{2}-{U}_{1}}{{V}_{2}-{V}_{1}}\right)\left(V-{V}_{1}\right)$ karton

Step 1
Given,
$\phantom{\rule{0ex}{0ex}}V=0.5{m}^{3}\phantom{\rule{0ex}{0ex}}{P}_{1}=160kPa\phantom{\rule{0ex}{0ex}}{x}_{1}=40\mathrm{%}=0.4\phantom{\rule{0ex}{0ex}}{P}_{2}=700kPa$
Refrigrant is r-134a
From Refrigrantion tables
at P = 160 KPa, we get
Specific volume:
${v}_{f}=0.0007437{m}^{3}/kg;{v}_{g}=0.12348{m}^{3}/kg$
Internal Energy:
${u}_{f}=31.09kJ/kg;{u}_{g}=190.27kJ/kg$
Step 2
NSK
Initial Specific Volume $\left(v\right)={v}_{f}+x.\left({v}_{g}-{v}_{f}\right)$
$v=0.0007437+0.4\left(0.12348-0.0007437\right)$
Initial Specific Volume $\left(v\right)=0.049409\frac{{m}^{3}}{kg}$
Volume of Tank =0.5

Step 3
$Q=m\left({u}_{2}-{u}_{1}\right)\phantom{\rule{0ex}{0ex}}{u}_{1}={u}_{f}+x.{u}_{fg}\phantom{\rule{0ex}{0ex}}{u}_{1}=31.09+0.4×190.27\phantom{\rule{0ex}{0ex}}{u}_{1}=107.198kJ/kg$
As the volume is constant, the refrigerant becomes supersaturated
At ${P}_{2}=700kPa$, and the specific volume is 0.04984 (approximately near to 0.0485)
We get $T={160}^{\circ }C$ hence ${u}_{2}=367.29kJ/kg$
$\phantom{\rule{0ex}{0ex}}Q=m\left({u}_{2}-{u}_{1}\right)\phantom{\rule{0ex}{0ex}}Q=10.031\left(367.29-107.198\right)\phantom{\rule{0ex}{0ex}}Q=2608.98kJ$

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