This preview shows page 1. Sign up to view the full content.
Unformatted text preview: the facility pays
$0.075/kWh for electricity. SOLUTION We take the turbine as the system (Fig. 7–30). This is a control volume since mass crosses the system boundary during the process. We note that
·
·
·
there is only one inlet and one exit and thus m 1 m 2 m .
Assumptions 1 This is a steadyflow process since there is no change with time
at any point and thus mCV 0, ECV 0, and SCV 0. 2 The turbine is adiabatic and thus there is no heat transfer. 3 The process is reversible. 4 Kinetic
and potential energies are negligible.
Analysis The assumptions above are reasonable since a turbine is normally
wellinsulated and it must involve no irreversibilities for best performance and
thus maximum power production. Therefore, the process through the turbine
must be reversible adiabatic or isentropic. Then, s2 s1 and State 1: P1
T1 5 MPa
115 K State 2: P2
s2 1 MPa
s1 h1
s1
1 h2 232.2 kJ/kg
4.9945 kJ/kg · K
422.15 kg/s
222.8 kJ/kg Also, the mass flow rate of liquid methane is ·
m ·
r1V 1 (422.15 kg...
View
Full
Document
This document was uploaded on 11/28/2012.
 Spring '09

Click to edit the document details