Question

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.
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​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

Answer 1

option C

option A

Step-by-step 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