Steam enters a turbine with a pressure of $2 \times 10^6 Pa$ and a temperature of $250^\circ C$. The steam exits the turbine as saturated vapor with a pressure of $15 \times 10^3 N/m^2$ and a velocity of $200 m/s$. The heat loss from the turbine casing to the surroundings is $160 kW$. The power output of the turbine is $3430 kW$ and the mass flow rate of the steam is $21960 kg/h$. We are asked to determine: (a) The dryness fraction (quality) of the steam at the turbine exit. (b) The cross-sectional area of the turbine exit.

1. Problem Description

Steam enters a turbine with a pressure of $2 \times 10^6 Pa$ and a temperature of $250^\circ C$. The steam exits the turbine as saturated vapor with a pressure of $15 \times 10^3 N/m^2$ and a velocity of $200 m/s$. The heat loss from the turbine casing to the surroundings is $160 kW$. The power output of the turbine is $3430 kW$ and the mass flow rate of the steam is $21960 kg/h$. We are asked to determine:

(a) The dryness fraction (quality) of the steam at the turbine exit.

(b) The cross-sectional area of the turbine exit.

2. Solution Steps

First, convert the mass flow rate from $kg/h$ to $kg/s$:

$\dot{m} = 21960 \, kg/h = \frac{21960}{3600} \, kg/s = 6.1 \, kg/s$

Next, apply the steady-flow energy equation (SFEE) to the turbine:

$\dot{m} \left( h_1 + \frac{V_1^2}{2} + gz_1 \right) = \dot{m} \left( h_2 + \frac{V_2^2}{2} + gz_2 \right) + \dot{W} + \dot{Q}$

Where:

$\dot{m}$ is the mass flow rate

$h_1$ is the specific enthalpy at the inlet

$V_1$ is the velocity at the inlet

$z_1$ is the height at the inlet

$h_2$ is the specific enthalpy at the outlet

$V_2$ is the velocity at the outlet

$z_2$ is the height at the outlet

$\dot{W}$ is the power output

$\dot{Q}$ is the heat transfer rate

Assuming the change in potential energy is negligible ($gz_1 \approx gz_2$), and the inlet velocity is very small ($V_1 \approx 0$), the equation simplifies to:

$\dot{m} h_1 = \dot{m} \left( h_2 + \frac{V_2^2}{2} \right) + \dot{W} + \dot{Q}$

We are given $\dot{W} = 3430 kW = 3430000 W$ and $\dot{Q} = -160 kW = -160000 W$ (heat loss). $V_2 = 200 m/s$. Therefore:

$h_1 = h_2 + \frac{V_2^2}{2} + \frac{\dot{W}}{\dot{m}} + \frac{\dot{Q}}{\dot{m}}$

$h_1 = h_2 + \frac{(200)^2}{2} + \frac{3430000}{6.1} + \frac{-160000}{6.1}$

$h_1 = h_2 + 20000 + 562295.08 - 26229.51$

$h_1 = h_2 + 556065.57 \, J/kg = h_2 + 556.066 \, kJ/kg$

From steam tables, at $P_1 = 2 \times 10^6 Pa = 2 MPa$ and $T_1 = 250^\circ C$, we can find $h_1$.

Looking up steam tables for saturated steam at 2 MPa, we find the saturation temperature is approximately 212.4 degrees Celsius. Since 250 degrees Celsius is above this, the steam at the inlet is superheated.

Looking up superheated steam tables at $P_1 = 2 MPa$ and $T_1 = 250^\circ C$, we find that $h_1 \approx 2903.5 \, kJ/kg$.

Therefore:

$2903.5 = h_2 + 556.066$

$h_2 = 2903.5 - 556.066 = 2347.434 \, kJ/kg$

At the exit, the pressure is $P_2 = 15 \times 10^3 Pa = 15 kPa$. From steam tables at 15 kPa, we find:

$h_f = 225.94 \, kJ/kg$

$h_g = 2598.3 \, kJ/kg$

$h_{fg} = 2372.4 \, kJ/kg$

(a) The dryness fraction (quality), $x_2$, is given by:

$h_2 = h_f + x_2 h_{fg}$

$2347.434 = 225.94 + x_2(2372.4)$

$x_2 = \frac{2347.434 - 225.94}{2372.4} = \frac{2121.494}{2372.4} \approx 0.894$

(b) To find the cross-sectional area at the exit, we use the equation:

$A_2 = \frac{\dot{m} v_2}{V_2}$, where $v_2$ is the specific volume at the exit.

$v_2 = v_f + x_2 v_{fg}$

At $P_2 = 15 kPa$:

$v_f = 0.001014 \, m^3/kg$

$v_g = 10.02 \, m^3/kg$

$v_{fg} = v_g - v_f = 10.02 - 0.001014 = 10.018986 \, m^3/kg$

Therefore:

$v_2 = 0.001014 + 0.894(10.018986) = 0.001014 + 8.955063 = 8.956077 \, m^3/kg$

$A_2 = \frac{(6.1 \, kg/s)(8.956077 \, m^3/kg)}{200 \, m/s} = \frac{54.6320697}{200} = 0.27316 \, m^2$

3. Final Answer

(a) The dryness fraction of the steam at the turbine exit is approximately 0.

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4. (b) The cross-sectional area of the turbine exit is approximately $0.273 \, m^2$.