Question:

A thermal power plant is running with no reheat or regeneration. The specific enthalpy and specific entropy of steam at the turbine inlet are 3344 kJ/kg and 6.5 kJ/kg·K, respectively. The turbine isentropic efficiency is 0.9, and the mass flow rate of steam at the turbine inlet is 102 kg/s. The turbine power output is _________ MW (rounded off to 1 decimal place).


 

Show Hint

To calculate turbine power output, use the mass flow rate and the difference in enthalpy between the turbine inlet and exit. Account for the isentropic efficiency to determine the actual enthalpy at the turbine exit.
Updated On: Apr 15, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Given:
\[ h_1 = 3344 \, \text{kJ/kg}, \quad s_1 = 6.5 \, \text{kJ/kg$\cdot$K}, \quad \eta_{\text{turbine}} = 0.9, \quad \dot{m} = 102 \, \text{kg/s} \] From steam tables (assuming isentropic expansion), let: \[ h_{2s} = 2230 \, \text{kJ/kg} \] Using the isentropic efficiency formula: \[ \eta_{\text{turbine}} = \frac{h_1 - h_2}{h_1 - h_{2s}} \Rightarrow h_2 = h_1 - \eta_{\text{turbine}} \cdot (h_1 - h_{2s}) \] \[ h_2 = 3344 - 0.9 \cdot (3344 - 2230) = 3344 - 0.9 \cdot 1114 = 3344 - 1002.6 = 2341.4 \, \text{kJ/kg} \] Power output: \[ \dot{W}_{\text{turbine}} = \dot{m} \cdot (h_1 - h_2) = 102 \cdot (3344 - 2341.4) = 102 \cdot 1002.6 = 102265.2 \, \text{kW} \] \[ \dot{W}_{\text{turbine}} = \frac{102265.2}{1000} = \boxed{102.3 \, \text{MW}} \]
Was this answer helpful?
0
0

Top Questions on Thermodynamics

View More Questions

Questions Asked in GATE exam

View More Questions