Question

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).

Answer 1

Step-by-step 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 } * \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)(kj)/(kg)`

`w_(c)=4 \cdot 013 \mathrm{kw}`

(b)

Refrigerator capacity

`Q_(i n)=\dot{m}\left(h_(1)-h_(4)\right)=\left((g \cdot s)/(60) k_(0) / s\right) *(234 \cdot 45-71 \cdot 33) (k J)/(k_(8))`

`Q_(i n)=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega`

`\ Q_(in) =6 \cdot 583 \text { tons }`

(c)

Cop:

`\beta=(\left(h_(1)-h_(4)\right))/(\left(h_(2)-h_(1)\right))=(Q_(i n))/(\omega_(c))=(23 \cdot 108)/(4 \cdot 013)`

`\beta=5 \cdot 758`