Question

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.

Answer 1

Answer:

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.

Answer 2

 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 = $\bold {15 m^3/s}$

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

So from the table,  

$\bold {v_2 = 2.087\ m^3/kg}$

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,  

$\bold {Q-W = m(h1 - h2) + {(V_2^2 - \dfrac {V1^2} {2}} + g(Z2 - Z1)}$

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

So,  

$\bold {-W = m(h2 - h1) + (V2(^2) - \dfrac {V1^2} {2}}$

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;  

$\bold {W = - 7.17 (2665.8 - 3445.3) + \dfrac {150^2 - 200^2}{2000}} \\\\\bold {W =5701\ KW}$

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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