Question

Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of

Answer 1

Answer:

option C

option A

Explanation:

Enthalpy gained by air= 1023-290

                                       = 733 kJ/kg

Rate of energy gain= mass flow rate × Enthalpy gained by air

                               = 0.1 × 733

                            = 73.3 kJ/s

rate of heat transfer between compressor and air= 77kW

Heat loss by air to surroundings= 77-73.3

                                                     =3.7kW

Enthalpy lost by steam in turbine= 1407.6-1236.4

                                                     = 171.2 Btu/lb

Rate of energy transfer to turbine= Enthalpy lost by steam× mass flow rate

                                                    = 171.2×5

                                                     = 856 Btu/s

Net rate of energy transfer to turbine=rate of  Energy transfer to turbine- rate of heat transfer to turbine

                          = 856-40

                         = 816 Btu/s