{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Thermodynamics filled in class notes_Part_14

# Thermodynamics filled in class notes_Part_14 - 35 2.3...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 35 2.3. RANKINE 5 Turbine Boiler 7 6 4 Feedwater Heater Condenser 2 Pump P2 3 Pump P1 1 Figure 2.7: Schematic for Rankine cycle with regeneration and open feedwater heating. First consider the low pressure pump. h2 = = = h1 + v1 (P2 − P1 ), kJ m3 191.8 + 0.00101 kg kg kJ 192.194 kg (2.157) ((400 kP a) − (10 kP a)) , (2.158) (2.159) The pump work is wP 1 = v1 (P1 − P2 ), m3 = 0.00101 kg kJ = −0.3939 . kg (2.160) ((10 kP a) − (400 kP a)) , (2.161) (2.162) Note, a sign convention consistent with work done by the ﬂuid is used here. At this point the text abandons this sign convention instead. CC BY-NC-ND. 18 November 2011, J. M. Powers. 36 CHAPTER 2. CYCLE ANALYSIS Now consider the turbine dmcv dt = m5 − m6 − m7 , ˙ ˙ ˙ (2.163) = m6 + m7 , ˙ ˙ m7 ˙ m6 ˙ + , m5 ˙ m5 ˙ (2.164) ˙ ˙ Qcv −Wcv + m5 h5 − m6 h6 − m7 h7 , ˙ ˙ ˙ (2.166) =0 m5 ˙ 1= dEcv dt = (2.165) =0 =0 ˙ Wcv = m5 h 5 − m6 h 6 − m7 h 7 . ˙ ˙ ˙ (2.167) On a per mass basis, we get, wt ˙ m6 ˙ m7 ˙ Wcv = h5 − h6 − h7 , m5 ˙ m5 ˙ m5 ˙ m6 ˙ m6 ˙ h7 , h6 − 1 − h5 − m5 ˙ m5 ˙ m6 ˙ m6 ˙ h5 − h6 + h6 − h7 , h6 − 1 − m5 ˙ m5 ˙ m6 ˙ m6 ˙ − 1− h7 , h5 − h6 + h6 1 − m5 ˙ m5 ˙ m6 ˙ h5 − h6 + 1 − (h6 − h7 ). m5 ˙ = = = = = (2.168) (2.169) (2.170) (2.171) (2.172) Now consider the feedwater heater. The ﬁrst law for this device gives dEcv dt = =0 =0 h3 = = = 604.73 kJ kg m6 ˙ m5 ˙ ˙ ˙ ˙ ˙ ˙ Qcv − Wcv +m2 h2 + m6 h6 − m3 h3 , = = (2.173) =0 m6 ˙ m2 ˙ h2 + h6 , m3 ˙ m3 ˙ m7 ˙ m6 ˙ h2 + h6 , m5 ˙ m5 ˙ m6 ˙ m6 ˙ h2 + h6 , 1− m5 ˙ m5 ˙ m6 ˙ kJ 1− 192.194 m5 ˙ kg 0.165451. CC BY-NC-ND. 18 November 2011, J. M. Powers. (2.174) (2.175) (2.176) + m6 ˙ m5 ˙ 2685.6 kJ kg , (2.177) (2.178) ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern