Exam (elaborations) TEST BANK FOR An Introduction to Equilibrium Thermodynamics By Bernard Morrill (Auth.) (Solution Manual)

Exam (elaborations) TEST BANK FOR An Introduction to Equilibrium Thermodynamics By Bernard Morrill (Auth.) (Solution Manual) SOLUTION MANUAL FOR An 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°0-5O/O ^in ~ ^out c ^in v ) Q. = Q .=100 U m ^out . ? T -2 T t, H lL 1000 „„, 1 s = 2 TW-TT = 1500 X 1 0 0 = 6 6 ' 7 n L ^ Qin " 1. - .6438 = 2 8 0 ' 7 7 1-28 ¥ net, = Qi.n - Qo ut. = 280.77 - 100 a) constant temperature process ¥n et. = 180.77 BTU C P = V v 1-26 (¥)T = c / " f = c m ^ v1 1 In general from equation 1.15.11 V2 ¥ork = J Q ^ - |Q2 ^mR^-T^lity- V2 (¥)T = m RT1 In — continued b) constant pressure process rv2 ( w ) P = p i Jvi d v = p i ( V V 7 1-28 cont. 1-29 cont. c) Adiabatic pV = C oO 1 Rl t process T2 dV C Ll-