University:
Viterbo University
Course:
PHYS 260 | University Physics I
Academic year:

2023
Views:

20

Pages:

11
Author:

TomAnd45

ITcl Sheet #8 the thermal efficreng of camat Engine:- PH) TW = 30% T: 300k Th? ec - ITd Th 1-ec= Thl 17h TC 300 S 1-ec -429k 1-03 Wag=250 J C= 25%, W b) e=W= 0.25= 250 Pr - 1000 J (6) Wang Pa-121 250 = 1000- Pc Q=750 J ⇒(2) 2 Part 6 kW, e=0.3 = 9000 J a) Pub) k? 1Pc) = 1-0.3=0-7 1QW= 9000=12857.14J 0.7 (a) Wang = 1911-191 Weng = 3857-14 J Part Weng tyde cycle ==3857.14=0.64 Sec cycle 6000 =>(b) 3) Camot Engine Pout-140 KW Lc = 30°C = 303 k Th=5Sic = 823 K Pont 140 13 Pin 303 |- 823 (T) Pin: 221 KW (KJIS) "Pin = 221+ 60x60x105 Pin = 7.9 +108 (J/hr) ITcle Pe = 7.94108303 823 P=2-900 (5/hr) 4Tc=240°C = 513 K Th=1890°c = 2163 K a) emax=? 6) e = 40%, Pout:? Pincl. 641 w Th = 1- 513 2163 everx = 0.76 enax: 76% (a) e=04 Wens - Pout = - Rul Pin Pout: 0.44 1.6x108 Pout = 6.4x105 W. -64(6) Th=250523 k Tc=98363K e: 22% enax Ph=? Weng: 11 KJ e: 0.22 emak e=0.22 [1-6] e: 0-221-3637 523. e= 0.067