To calculate the work produced by the turbine, we will apply the First Law of Thermodynamics for a steady-flow, adiabatic process. Since the problem states that we can neglect changes in kinetic and potential energy, and the process is adiabatic (no heat exchange), the first law simplifies to:

$W_{turbine} = h_1 - h_2$

Where:

• $W_{turbine}$ is the work output of the turbine per unit mass (in kJ/kg),
• $h_1$ is the specific enthalpy of steam at the inlet (in kJ/kg),
• $h_2$ is the specific enthalpy of steam at the outlet (in kJ/kg).

Step 1: Obtain the inlet conditions (state 1)

From the steam tables, for steam at 2 MPa (2000 kPa) and 300°C, we look up the specific enthalpy $h_1$.

• Pressure: $P_1 = 2 \, \text{MPa}$
• Temperature: $T_1 = 300^\circ C$

From the steam tables, for superheated steam at these conditions:

• $h_1 \approx 3021.4 \, \text{kJ/kg}$

Step 2: Obtain the outlet conditions (state 2)

At the outlet, the steam is at 30 kPa. Since no temperature is given, we assume the steam is saturated at the outlet. We can find the specific enthalpy corresponding to the saturated condition at 30 kPa.

From the steam tables:

• $P_2 = 30 \, \text{kPa}$
• Saturation temperature $T_{sat} = 69.12^\circ C$
• Enthalpy of saturated liquid $h_f = 340.5 \, \text{kJ/kg}$
• Enthalpy of saturated vapor $h_g = 2551.3 \, \text{kJ/kg}$

However, since we don’t know the final state (whether it's saturated or superheated), we need to use the isentropic expansion assumption. We calculate the entropy at the inlet, and then use that to find the final state.

Step 3: Entropy at inlet (state 1)

For superheated steam at 2 MPa and 300°C, the specific entropy is:

$s_1 = 6.743 \, \text{kJ/kg·K}$

Step 4: Entropy at outlet (state 2)

Since the process is adiabatic and reversible, $s_2 = s_1$. So at the outlet:

$s_2 = 6.743 \, \text{kJ/kg·K}$

From the steam tables for 30 kPa, the entropy values for saturated liquid and vapor are:

• $s_f = 1.092 \, \text{kJ/kg·K}$
• $s_g = 8.148 \, \text{kJ/kg·K}$

Now, we determine the quality $x_2$ of the steam at the outlet using the entropy relation:

$s_2 = s_f + x_2 (s_g - s_f)$

Solving for $x_2$:

$6.743 = 1.092 + x_2 (8.148 - 1.092)$

$x_2 \approx 0.886$

Step 5: Enthalpy at outlet (state 2)

Now that we have the quality $x_2$, we can calculate the enthalpy at the outlet:

$h_2 = h_f + x_2 (h_g - h_f)$

Substituting the values:

$h_2 = 340.5 + 0.886(2551.3 - 340.5)$

$h_2 \approx 2309.6 \, \text{kJ/kg}$

Step 6: Calculate the turbine work

Finally, the work produced by the turbine per kg of steam is:

$W_{turbine} = h_1 - h_2 = 3021.4 - 2309.6 = 711.8 \, \text{kJ/kg}$

Thus, the work output of the turbine is approximately 711.8 kJ/kg.

---

Would you like more details on any step?

Here are 5 related questions:

1. How is entropy used to determine the quality of steam during expansion?
2. What is the significance of assuming an adiabatic process in turbines?
3. How do steam tables help in solving thermodynamic problems?
4. What happens if kinetic energy is not neglected in this analysis?
5. How would changing the outlet pressure affect the turbine's work output?

Tip: Always ensure the correct phase (liquid, vapor, or a mixture) of the steam when using steam tables for accurate calculations.