Engineering: An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at 1.25 bar, and saturated liquid exits the condenser at 5 bar. The mass flow rate of refrigerant is 8.5 kg/min. Determine the magnitude of the compressor power input required, in kW (report as a positive number).

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An ideal vapor-compression refrigeration cycle - id5524548, dillonmichaeld8069

Ответ:

quantamagic

21.01.2022

Engineering

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

$\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}$

State 2:

$\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}$

State 3:

$\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}$

State 4:

Throttling process $h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}$

(a)

Magnitude of compressor power input

$\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}$