# Tut1 - r l Santana-v7 ﬂ(PITIV/M’U/Hejc 1“...

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Unformatted text preview: \ \ r ' l Santana-v7.- ' ﬂ .. (PITIV/M’U./Hejc) / 1“. ProlW‘H-RS S‘an Vila P . ISA; M .V. v >V l, *' and 4:) from.“ closed 5 den. “nadir/oz: mmié/g Process ; w = _J Pdv 4: Myf-IW'J \ § ~ infra/V, P writ M t. 1.1 la (640‘ Mk. ——'—-"———__————-—‘___ [W'on ;l\¢ 5* '(‘MT bark) v.= o.oo¢vm’ Hamid from linur rmflle I): = 4.1— 5" Puma ‘5; V2. 305’?— ”‘7 - Thu P1 = M-F baw- dh’ MM fm95W-e (P;)."M5}‘é’, l( {hula Wregfed t (w: '= M. back t; max 541‘ kw” ad from;; I wk??? 2%: 7652? ‘ ' (\Lq’ 3 \$Wk; (1) Linear Reversible Expansion 1 4 2 To find the work done in this process, we can use either the analytical or the graphical method. oAnalytical Method \ P = mV + K (where m & K are the straight line equation constants) 4.2 = 0.004 m + K a (1) 1.4 = 0.02 m +.K 4 (2) Solving simultaneously gives m ='- 175 and K = 4.9 2 2 141-2 «J P.dV =-J (- 1'75v + 14.9) (W l 1 at 7 [V2]? + 459 mi) ==--87.5 [0.022 - 0.0042] + 4.9 [0.02 — 0.004]) = 49% ~oe336 + 0.0784) =-0.0448 bar m3 4. tofu/H: . ",3 w i 5&4480 N.m (Work done by the Fluid). CMOMI’N wiou. Graphical Method The work is the area under line 1 4 2, that is 12ab1 which is a trapezoid. Area of the trapezoid 12ab1 = 8 (2a + 1b) >< ab = \$2 (1.4 + 4,2) X (0.02 — 0.004) 0.0AA8 bar m3 = 4A80 N.m (Work done by the Fluid which is the same result obtained from the analytical a“ M solution. tv; . Midf- II (2) Reversible Cooling at Constant Pressure 2 a 3 You will notice that we require V3 to be able to obtain the work in the process 2 -> 3. In the process 3 —* l, the product of pressure and volume is always constant, so Plvl = Psva e—ﬁ" .o' =Wldj- M F’”“’55 V3 ll Plvl/Pa 3 vb I ll (4.2)(0.004)/(1.l+) 0.012 m3 3 ‘P.dV S N (1. 1| =-P [v]: =uP (V3 - V2) =-(1.£+)(0.012 - 0.02) = + 0.0112 bar In3 "II—- -= g]: 1120 N.m (Work done on the Fluid) tpocH-NQ WM (3) Reversible Compression with PV = Constant 3 4 1 1 hmﬂ W3_1 = J;an 3 . PI'V. : l -dV / 1 = Constant T =—Constant [103”,3 V]3 3 =-(4,2)(0.001+)(- 5.521 - [- 4.4231) = -|- 0.01845 bar m3 = \$1845 N.m (Work done on the Fluid) M So, Net work of the cycle =-£+480 + 1120 + 1816 =-1515 N.m «all: to". 57 1‘: wrkp‘y ﬂuid t it! SMMJI‘A \$‘_ Elk-1&1: A r3834 (Ian'st “r on” (no u! not. A: cw” «5.1L. nut-W t d‘. \$WIMJ;N I, m ' 1: i’dthM ﬁe wk m Jm‘y W: [NW-3. (Asian; 3‘! M #004” from: {a q rmp'lh ...
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## This note was uploaded on 07/26/2009 for the course ENG sem314 taught by Professor Dunno during the Spring '09 term at A.T. Still University.

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Tut1 - r l Santana-v7 ﬂ(PITIV/M’U/Hejc 1“...

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