Question

Air is expanded reversibly and adiabatically in a turbine from 3.5 bar and 260 ℃ to 1 bar. The turbine

is insulated and the inlet velocity is negligible. The exit velocity is 150 m/s. Find the work output of the

turbine per unit mass of air flow. Take for air, Cp = 1.005 kJ/kg K, and = 1.4​

Answer 1

Answer:

Air is expanded reversibly and adiabatically in a turbine from 3.5 bar and 260 ℃ to 1 bar. The turbine

is insulated and the inlet velocity is negligible. The exit velocity is 150 m/s. Find the work output of the

turbine per unit mass of air flow. Take for air, Cp = 1.005 kJ/kg K, and = 1.4

Answer 2

Answer:

The work output of the turbine per unit mass of air flow is -749.62 J/kg. Note that the negative sign indicates that work is done on the turbine, which is consistent with the fact that the air is being expanded in the turbine.

Explanation:

We can solve this problem using the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the work done on the system plus the heat added to the system. For an adiabatic process, there is no heat added or removed from the system, so the change in internal energy is equal to the work done.

The work done by the turbine is given by:

W = h1 - h2

where h1 is the enthalpy of the air at the inlet conditions and h2 is the enthalpy of the air at the outlet conditions.

To find h1 and h2, we can use the following equations:

h1 = CpT1 + 0.5v1^2

h2 = CpT2 + 0.5v2^2

where Cp is the specific heat at constant pressure, T1 and T2 are the temperatures at the inlet and outlet, and v1 and v2 are the velocities at the inlet and outlet.

We can assume that the process is reversible, which means that the entropy remains constant. Therefore, we can use the following equation to relate the temperatures at the inlet and outlet:

T2/T1 = (P2/P1)^((γ-1)/γ)

where P1 and P2 are the pressures at the inlet and outlet, and γ is the ratio of specific heats.

Substituting the given values, we have:

T2/T1 = (1/3.5)^((1.4-1)/1.4) = 0.595

T2 = 0.595*T1

We also know that v1 = 0 and v2 = 150 m/s.

Using the equation for h1, we have:

h1 = CpT1 + 0.5v1^2 = Cp*T1

Using the equation for h2, we have:

h2 = CpT2 + 0.5v2^2 = Cp*(0.595T1) + 0.5(150)^2

Now we can substitute these values into the equation for the work done:

W = h1 - h2 = CpT1 - (Cp0.595*T1 + 11250)

Simplifying, we get:

W = CpT1(1 - 0.595) - 11250

W = 0.405CpT1 - 11250

Substituting the given values for Cp and T1, we get:

W = 0.4051.005(260) - 11250

W = -749.62 J/kg

Therefore, the work output of the turbine per unit mass of air flow is -749.62 J/kg. Note that the negative sign indicates that work is done on the turbine, which is consistent with the fact that the air is being expanded in the turbine.

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