Ideal Reheat Rankine Cycle Analysis for Steam Power Plant

Analyzing the thermal efficiency and mass flow rate of an ideal Rankine cycle with superheat and reheat using steam as the working fluid. The cycle involves stages of expansion, reheating, and condensing to generate a net power output of 100 MW. Detailed calculations for states of the cycle are provided, and the thermal efficiency is determined. The mass flow rate of steam is obtained based on the net power developed by the cycle.

• Rankine cycle
• Thermal efficiency
• Steam power plant
• Reheat
• Mass flow rate

danica Follow

Uploaded on Aug 29, 2024 | 1 Views

Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

Presentation Transcript

1. Reheat Rankine cycle Sheet -2 Q1- Steam is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam enters the first-stage turbine at 8.0 MPa, 480 C, and expands to 0.7 MPa. It is then reheated to 440 C before entering the second-stage turbine, where it expands to the condenser pressure of 0.008 MPa. The net power output is 100 MW. Determine (a) the thermal efficiency of the cycle, (b) the mass flow rate of steam, in kg/h, Solusion: To begin, we fix each of the principal states. Starting at the inlet to the first turbine stage, the pressure is 8.0 MPa and the temperature is 480 C, so the steam is a superheated vapor. h3= 3348.4 kJ/kg and s3= 6.6586 kJ/kg .K.

2. P3= 8.0 MPa and T3= 480C, so the steam is a superheated vapor. h3= 3348.4 kJ/kg s3= 6.6586 kJ/kg .K. State 4 p4 = 0.7 MPa s4 = s3 = 6.6586 kJ/kg .K

3. State 5 is superheated vapor p5 = 0.7 MPa T5 = 440 C, so from table, h5 =3353.3 kJ/kg s5 = 7.7571 kJ/kg .K

4. state 6, p6 = 0.008 MPa = 0.08 bar s6 = s5 =7.7571 kJ/kg .K h6 = 173.88 + (0.9382)2403.1 = 2428.5 kJ/kg State 1 is saturated liquid at 0.008 MPa = 0.08 bar , so h1= 173.88 kJ/kg. h2 = 181.94 kJ/kg. (a) The net power developed by the cycle is

5. (a) the thermal efficiency of the cycle (b) The mass flow rate of the steam can be obtained with the expression for net power given in part (a).