reproacht3

2022-01-05

A $0.5-{m}^{3}$ 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.

Pademagk71

Beginner2022-01-06Added 34 answers

For this stationary closed system, the energy balance is expressed as

$E}_{\in}-{E}_{out}=\mathrm{\u25b3}{E}_{system$

therefore ${Q}_{\in}=\mathrm{\u25b3}U=m({u}_{2}-{u}_{1})$

use tables (Tables $A-11$ through $A-13$)

the properties of $R-134a$

${P}_{1}=160Kpa$

${x}_{1}=0.4$

so $\Rightarrow {v}_{f}=0.0007435,{v}_{g}\}=0.12355\frac{{m}_{3}}{kg}$

$v}_{f}=31.06,{u}_{fg}=190.31\frac{kj}{kg$

therefore $v}_{1}={v}_{f}+{x}_{1}{v}_{fg$

the equation's values are changed

${v}_{1}=0.0007435+0.4(0.12355-0.0007435)$

$\Rightarrow {v}_{1}=0.04987\frac{{m}^{3}}{kg}$

so, $u}_{1}={u}_{f}+{x}_{1}{u}_{fg$

we replace the values in the equation

${u}_{1}=31.06+0.4\left(190.31\right)$

$\Rightarrow {u}_{1}=107.19\frac{kj}{kg}$

therefore

${P}_{2}=700Kpa$

$({v}_{2}={v}_{1})\Rightarrow {u}_{2}=377.23\frac{kj}{kg}$

the mass of the refrigenant is determined

$m=\frac{{V}_{1}}{{v}_{1}}$

we replace the values in the equation

$m=\frac{0.5{m}^{3}}{0.04987\frac{{m}^{3}}{kg}}\Rightarrow m=10.03kg$

par (b)

${Q}_{\in}=m({u}_{2}-{u}_{2})$

we replace the values in the equation

$Q}_{\in}=\left(10.03kg\right)(377.23-107.19)\frac{kj}{kg$

${Q}_{\in}=2708kj$

Beverly Smith

Beginner2022-01-07Added 42 answers

First law:

For a rigid tank,

State 1:

State 2:

At

Since

Table A-13:

karton

Expert2022-01-11Added 613 answers

Step 1

Given,

Refrigrant is r-134a

From Refrigrantion tables

at P = 160 KPa, we get

Specific volume:

Internal Energy:

Step 2

NSK

Initial Specific Volume

Initial Specific Volume

Volume of Tank =0.5

Step 3

As the volume is constant, the refrigerant becomes supersaturated

At

We get

22+64

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