This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MECH 213 September 19, 2007 First law o (Qin Qout) + (Win Wout) + zbar*min(h + ke + pe) zbar*mout(h + ke + pe) = [m(h + ke + pe)] o closed system have no mass change (Qin Qout) + (Win Wout) = m(h + ke + pe)] o stationary system has no velocity i.e. no ke or pe Qnet W = m*u +in in out +out Specific form q w = u differential form q w = du Q W = m*du o constant pressure in a stationary closed system (Only time enthalpy is used) Q W Wb = m*u Where Wb = m*p(v2 v1) Q W = m((u2 u1) + p(v2 v1)) = m[u2 + p2v2 u1 + p1v1] = m[h] Chapter 4 #39 o Insulated piston cylinder, 5 L of saturated water, 175 kPa, 8 amps for 45 minutes, evaporates, 400 kJ o Find voltage and draw pvdiagram Closed system Real pure substance Constant pressure process insulated Q = 0 W = [400kJ (volts)It] = m(h2 h1) h1 = hf h2 = hf +xhfg Chapter 4 #78 o Piston/cylinder, start at 100kPa, .5kg He, 25C, m 500kPa to raise o Find Q to lift piston Constant volume Ideal Gas pv=RT h = cpT u = cvT p1 = 100kPa T1 = 25 + 273 = 298K p2 = 500kPa v2 = v1 p1/T1 = p2/T2 T2 = T1(p2/p1) = 5T1 = 1490K Q W = mu Q = m*cv(T2 T1) Check UNITS ...
View
Full
Document
This note was uploaded on 04/07/2008 for the course MECH 213 taught by Professor Knisley during the Fall '08 term at Bucknell.
 Fall '08
 knisley

Click to edit the document details