At steady state, 500 kg/s of steam enters a turbine with specific enthalpy equal to 3500 kJ/kg and specific entropy equal to 6.5 kJ-kg^-1K^-1. It expands reversibly in the turbine to the condenser pressure. Heat loss occurs reversibly in the turbine at a temperature of 500 K. If the exit specific enthalpy and specific entropy are 2500 kJ/kg and 6.3 kJ-kg^-1K^-1, respectively, the work output from the turbine is _________ MW (in integer).

The correct answer is 450.

Step-by-step Explanation:

From the schematic diagram of the turbine, the following properties and data points are given:
- Inlet state (1): specific enthalpy $h_1=3500\text{kJ/kg}$, specific entropy $s_1=6.5\text{kJ/kg K}$, and steam mass flow rate $\dot{m}=500\text{kg/s}$.
- Exit state (2): specific enthalpy $h_2=2500\text{kJ/kg}$ and specific entropy $s_2=6.3\text{kJ/kg K}$.
- Heat transfer occurs reversibly at a boundary temperature of $T=500\text{K}$.

1. Calculation of Specific Heat Transfer ($q$):
Since the expansion and heat transfer processes in the turbine are reversible, the change in specific entropy of the steam is related to the specific heat transfer by the relation:
$s_2-s_1=\frac{q}{T}$
Rearranging the equation to solve for the specific heat transfer $q$:
$q=T(s_2-s_1)$
Substituting the given values:
$q=500\times (6.3-6.5)$
$q=500\times (-0.2)=-100\text{kJ/kg}$
(The negative sign indicates that heat is lost from the turbine, which is consistent with the $d{Q}_{\text{Loss}}$ label shown in the diagram).

2. Calculation of Specific Work Output ($w$):
Applying the steady flow energy equation (SFEE) to the turbine control volume (neglecting kinetic and potential energy changes):
$h_1+q=h_2+w$
Rearranging to solve for the specific work output $w$:
$w=(h_1-h_2)+q$
Substituting the values of enthalpy and the calculated specific heat transfer:
$w=(3500-2500)+(-100)$
$w=1000-100=900\text{kJ/kg}$

3. Calculation of Total Work Output Rate (Power Output, $P$):
The power output of the turbine is the product of the mass flow rate and the specific work output:
$P=\dot{m}\times w$
Substituting the values:
$P=500\text{kg/s}\times 900\text{kJ/kg}=450000\text{kW}$
Converting the power output from kilowatts (kW) to megawatts (MW):
$P=\frac{450000}{1000}=450\text{MW}$