Exam (elaborations) TEST BANK FOR An Introduction to Equilibrium Thermodynamics By Bernard Morrill (Auth.) (Solution Manual) SOLUTION MANUAL FOR An
... [Show More] Introduction to Equilibrium Thermodynamics Bernard Morrill Professor of Mechanical Engineering Swarthmore College Pergamon Press Inc. New York • Toronto • Oxford • Sydney • Braunschweig 1 1-1 t = TOO C s t = 212°P s Chapter 1 1-5 t. = Ov l C t± = 32°F t°P = (32 + 1.8°C) 1-2 pV = n PT Y = » P 1-3 k = JR N. k = k = 5.662x10"24 ft lbf/part. °R k = 200 !00 x 2 = 400 in-lbf 33.33 ft-lbf 1 = 0.0428 BTU 1-6 W = 4k x2 e where k1+ k 2 ke " k1 k2 "k1k 2 k -40x60 e k1 +k2 40+60 W = - i x 24 x (i)2 = W = -T2 = 0 .25 ft-: = 24 Ibf/in 3.21 x 10~4 BTU = C 1°C 1.8°P w kx2 1-4 ¥ = k = 40 W = - 45 in-lbf A r -3.75 ft-lbf W = - 3.75/778 = - .00482 BTU W = 0.491 BTU W = p(v2 - V1) = 190.99(5-3) = 381.97 ft lbf 190.99 lbf/ft2 P = 2 1-8 For a constant temperature process pV = C From initial condition C = 200 x 144 x 3 = 86400 ft lbf Then W = C In V = 86400 x In 5/3 = 86400 x 0.5108 W = 44135 ft lbf 1-9 cont. c) z = J" y cos x dx + sin x dy dz = y cos x dx + sin x dy 1-10 u = A ( RT + pB) du = A(Rdt + Bdp) 1-11 AU = AQ - AW AW = p (V9 - V. ) constant pressure process AW = 5.553 BTU AQ = - 8.5 BTU AU = - 8.5 + 5.553 = - 2.947 BTU Au = Au = - 12.85 BTU/lbm continued cos x = cos x dz is exact = 56.73 BTU 1-9 2 a) z = 3xy + 4x dz = (3y + 8x)dx + 3x dy If exact then 3 = 3 dz is exact b) z = x2 + xy + yx 3 dz = (2x + 3y + 3yx2)dx + (3x + x5)dy 3 + 3x2 = 3 + 3x2 dz is exact m = 0.22< 3 1-14 cont. = .375 x .171 x (425 - 70) + 40 ¥ = - 62.76 BTU 1-12 1 HP = ^ Q ° ° = 42.42 BTU/min dV = dQ - d¥ dU dQ_ d¥ dt dt dt = - 30. + 42.4 H = 12.4 BTU/min 1-15 cyAT = Au m cy _ J^4 _ .124 BTU/min°R M ~ 171 ~ *^2^ lbm/min 1-13 W = ^pdV For constant energy process, T = C Then pV = C v ¥ = C J ^=Cln^- v1 1 P, - - S^HP2* = * » ° P* „ P1V 1 28520 x 1 _ . 0 = 778 ~ 778 36' 65 W = 36.65 In 2. W = 40.26 BTU ¥ = 31330 ft lbs Q = W = 40.26 BTU 1-16 u = (j(x1 fx2,x3) " - 1 2 , DX1 + ^ D J C 2 + % D X 3 Extensive property conjugate to , v v ^ax1 ' xi Q X 2 ' ^ X2 •?S_ 1 „ ^ x 3 - *3 1-17 Given pV = C RT Then I OT r"1 - "l (.) continued 1-14 RT 53.3 x 530 iD* -A V = A U - = JB O y 4kT - ,A Q continued 4 T p R where i/r (b) T2 = 801.71 Au = c ? T = .171(801.7-530) Au = 46.46 BTU/lbm 1-19 continued 1/2T 1-20 For adiabatic process r 2 A U = J p dV p = CV V1 -r Au ( 2 -r dV 1-21 Along 1-3 path Pi7 / " along 2-3 path P2 V 2 = P3V3 Divide 1st by 2nd V 1-18 T2 = T1 = 530 = 530 (.235)-286 = 530 (4.25)*286 bXJ = - 94.08 BTU 1-19 cont. x 10 (b) V_ = 3.715 ft3 T (a) T2 = 602.2°R 1-17 cont. starting with (a) RT V = — P then (a) becomes continued 5 1-21 cont. 2.5 V, = .0466 ft 3 = 2897 psia P 2 = 39.58 psia 1*U2 = l / 2 - 1 W 2 = C V * T = .171 (534.6 - 405.3) 1 AU2 = 22.11 BTU 1 &W2 =-22.11 BTU _ ^3V3 _ 39.58 x 144 x 10 3 ~ R " 53.3 T, = 1069.3°R 3 continued 1-22 cont. 2 AU3 = Cy(T3-T2) = .171(1069,3-534.6) 2 AU5 = 91.43 BTU 2*3 2W3 /3 J dv = P 2 ( V 3 - V 2 ) 2 Q 3 = 2 ^ U 3 + 2 A W 3 = 91.43 + 36.63 2Q3 3W1 128.06 BTU 0 3A u i =3*Q i = WV = .171 (405.3 - 1069.3) 5 ^ = _ 113.54 BTU 3AQ1 = - 113.54 Wnet = ^ W = ~2 2 - 1 1 + 36.63 + 0. = 14.52 BTU Qn e t = X Q = 0. + 128.06 - 113.54 = 14.52 BTU 1-23 W in W . = H-52 BTU net Qin = 2Q3 = 1 2 8 , 0 6 _ 14.52 " 128.06 continued x 100 15 1-22 P 2 = ^ 534.6°R = 36.63 BTU 6 1-23 cont. 1-26 cont. For larger engine 1-24 ^ c = ( i 3 ) l o o - ( i - f g ) 100 ^ c = 22.73$ For smaller engine V l/2 C O P ^ 510 o.u.r. - T - 6 6 0 _ 5 1 0 C.O.P. = 3.40 If ¥T = W0 then H L 2 T = 2 T - T 1-25 L wnet x2 520 I c Q.i n ' T1. ' 1460 1-27 c = .6438 ¥hen T„ = 1000°R and TT = 500°R ¥ . = K Q. net I c m h 1000-500 , / I L " 1000 ^1° [Show Less]