This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Spring 2006 1 ME 201 Thermodynamics Homework 18 Solutions 1. Determine the work per mass output of an adiabatic turbine with isentropic efficiency 0.83 that has a steam input of 15 MPa and 650 C and an outlet pressure of 50 kPa. Solution: The actual work will be calculated from the isentropic efficiency by w act = s w ideal Hence we need to calculate the ideal work. We begin by setting up our table and working the 1 st law problem System Type: Control Volume (Turbine) Working Fluid: Steam (compressible Substance) Process: Isentropic (ideal) Adiabatic (actual) with s = 0.83 State 1 State 2s State 2a T 1 = 650 C T 2s = 81.32 C T 2a = 81.32 C P 1 = 15 MPa P 2s = 50 kPa P 2a = 50 kPa h 1 = 3712.1 kJ/kg h 2s = 2372.4 kJ/kg h 2a = 2600 kJ/kg s 1 = 6.8233 kJ/(kg K) s 2s = 6.8233 kJ/(kg K) s 2a = 7.4649 kJ/(kg K ) phase: sup.vap. phase: 2 phase with x 2s = 0.882 phase: 2 phase with x 2a = 0.98 Italicized values are from steam tables. Bold values are calculated. State1: Fixed State2s: UNKNOWN State2a: UNKNOWN & Q = 0 & W sh = UNKNOWN 1st Law: [ ] & & m h h = W 1 2 sh- ME 201 Thermodynamics Spring 1999 2 Approach: To fix the state2s we will use our process description which will gives us...
View Full Document
This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.
- Spring '06