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

290 kJ/kg and exits at a higher pressure with a specific enthalpy of 1023 kJ/kg. The mass flow rate is 0.1 kg/s. Kinetic and potential energy effects are negligible and the air can be modeled as an ideal gas.

If the compressor power input is 77 kW, the rate of heat transfer between the air and its surroundings is

Option A: 3.7 kW from the surroundings to the air.

Option B: 150.3 kW from the air to the surroundings.

Option C: 3.7 kW from the air to the surroundings.

Option D: 150.3 kW from the surroundings to the air.

Steam enters a turbine operating at steady state with a specific enthalpy of 1407.6 Btu/lb and expands to the turbine exit where the specific enthalpy is 1236.4 Btu/lb . The mass flow rate is 5 lb/s. During this process, heat transfer to the surroundings occurs at a rate of 40 Btu/s


Neglecting kinetic and potential energy effects, the power developed by the turbine is

Option A: 816 Btu/s

Option B: 896 Btu/s

Option C: 656 Btu/s

Option D: 74.2 Btu/s

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

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explanation:

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explanation: