Respuesta :
Explanation:
Note: Refer the diagram below
Obtaining data from property tables
State 1:
[tex]\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}[/tex]
State 2:
[tex]\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}[/tex]
State 3:
[tex]\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}[/tex]
State 4:
Throttling process [tex]h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}[/tex]
(a)
Magnitude of compressor power input
[tex]\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}[/tex]
[tex]w_{c}=4 \cdot 013 \mathrm{kw}[/tex]
(b)
Refrigerator capacity
[tex]Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}[/tex]
[tex]Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega[/tex]
[tex]\ Q_{in} =6 \cdot 583 \text { tons }[/tex]
(c)
Cop:
[tex]\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}[/tex]
[tex]\beta=5 \cdot 758[/tex]
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