Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically to the exit, where it is saturated vapor at 0.8 bar, with a velocity of 150 m/s and a volumetric flow rate of 15 m3/s. Determine the power developed by the turbine, in kW.

 The power developed by the turbine is 5701 KW. The positive sign denote that the power is developed from the system.    

Given here,  

Inlet:

P1 = 40 bar ;

T1 = 500°C  

V1 = 200 m/s  

Exit:

At saturated vapour;

P2 = 0.8 bar and

V2 = 150 m/s  

Volumetric flow rate =

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state,

So from the table,  

The mass flow rate,

m = (AV) /v

Where, AV is the volumetric flow rate.

Thus, the mass flow rate at exit,  

m = 15/(2.087) = 7.17 kg/s

From energy equation,  

For adiabatic flow, potential energy can be taken to be zero,

So,  

From, table 2, at

P1 = 40 bar

T1 = 500°C

h1 - specific enthalpy = 3445.3 KJ/Kg

At P2 = 0.8 bar; specific enthalpy,  

h2 = 2665.8 KJ/Kg

So we now calculate power developed;  

The power developed by the turbine is 5701 KW. The positive sign denote that the power is developed from the system.

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W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

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