• # question_answer A refrigerating system operating on reverse Brayton refrigeration cycle is used for maintaining 250 K. If the temperature at the exit of constant pressure cooling is 300 K and rise in the temperature of air in the refrigerator is so K, then the network of compression will be (assume air as the working substance with ${{c}_{p}}=\,\,1\,\,KJ\,per\,kg\,per{}^\circ C)$ A) 250 kJ/kg                      B) 200 kJ/kgC) 50 kJ/kg            D) 25 kJ/kg

Network of compression for the reversed Brayton refrigeration cycle$=({{h}_{2}}-{{h}_{1}})-({{h}_{3}}-{{h}_{4}})$ Now, $\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{3}}}{{{T}_{4}}}$ $\therefore \,\,\,\,\,\,\,\,\,\,{{T}_{2}}=250\times \frac{300}{200}=375\,K$             Net work $={{c}_{p}}[({{T}_{2}}-{{T}_{1}})-({{T}_{3}}-{{T}_{4}})]$                         $\,\,=0.1\,[(375-250)-(300-200)]$ $\,\,=1[125-100=1\times 25=25\,kJ/kg]$