5-58 Helium is compressed by a compressor

5-58 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input required is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats.

Properties The constant pressure specific heat of helium is c_p = 5.1926 kJ/kg·K (Table A-2a).

Analysis There is only one inlet and one exit, and thus . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Thus,

5-89 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the outlet temperature of the air are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.

Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg.C, respectively.

Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Then the rate of heat transfer from the exhaust gases becomes

The mass flow rate of air is

Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the air becomes

5-62E Air is expanded in an adiabatic turbine. The mass flow rate of the air and the power produced are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is well-insulated, and thus there is no heat transfer. 3 Air is an ideal gas with constant specific heats.

Properties The constant pressure specific heat of air at the average temperature of (800+250)/2=525°F is c_p = 0.2485 Btu/lbm·R (Table A-2Eb). The gas constant of air is R = 0.3704 psiaft³/lbmR (Table A-1E).

Analysis There is only one inlet and one exit, and thus . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

The specific volume of air at the exit and the mass flow rate are

Similarly at the inlet,

Substituting into the energy balance equation gives

5-68 Refrigerant-134a is throttled by a capillary tube. The quality of the refrigerant at the exit is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved.

Analysis There is only one inlet and one exit, and thus . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

since .

The inlet enthalpy of R-134a is, from the refrigerant tables (Table A-11),

The exit quality is