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# TEST BANK FOR An Introduction to Equilibrium Thermodynamics By Bernard Morrill (Auth.)

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**SOLUTION MANUAL FORAn Introduction toEquilibrium ThermodynamicsBernard MorrillProfessor of Mechanical EngineeringSwarthmore CollegePergamon Press Inc.New York • Toronto • Oxford • Sydney • Braunschweig11-1t = TOO Cst = 212°PsChapter 11-5t. = Ov l Ct± = 32°Ft°P = (32 + 1.8°C)1-2pV = n PTY = »P1-3k =JRN.k =k = 5.662x10"24 ft lbf/part. °Rk = 200!00 x 2 = 400 in-lbf33.33 ft-lbf1 = 0.0428 BTU1-6W = 4k x2ewherek1+ k 2ke " k1 k2 "k1k 2k -40x60e k1 +k2 40+60W = - i x 24 x (i)2 =W = -T2 = 0 .25 ft-:= 24 Ibf/in3.21 x 10~4 BTU= C1°C 1.8°Pw kx21-4¥ =k = 40W = - 45 in-lbf A r\-3.75 ft-lbfW = - 3.75/778 = - .00482 BTUW = 0.491 BTUW = p(v2 - V1) = 190.99(5-3)= 381.97 ft lbf190.99 lbf/ft2 P =21-8For a constant temperature processpV = CFrom initial conditionC = 200 x 144 x 3 = 86400 ft lbfThenW = C InV = 86400 x In 5/3 = 86400 x 0.5108W = 44135 ft lbf1-9 cont.c) z = J" y cos x dx + sin x dydz = y cos x dx + sin x dy1-10u = A ( RT + pB)du = A(Rdt + Bdp)1-11AU = AQ - AWAW = p (V9 - V. ) constant pressureprocessAW = 5.553 BTUAQ = - 8.5 BTUAU = - 8.5 + 5.553= - 2.947 BTUAu =Au = - 12.85 BTU/lbmcontinuedcos x = cos xdz is exact= 56.73 BTU1-92a) z = 3xy + 4xdz = (3y + 8x)dx + 3x dyIf exact then3 = 3dz is exactb) z = x2 + xy + yx 3dz = (2x + 3y + 3yx2)dx+ (3x + x5)dy3 + 3x2 = 3 + 3x2dz is exactm = 0.22<31-14 cont.= .375 x .171 x (425 - 70) + 40¥ = - 62.76 BTU1-121 HP = ^ Q ° ° = 42.42 BTU/mindV = dQ - d¥dU dQ_ d¥dt dt dt= - 30. + 42.4H = 12.4 BTU/min1-15cyAT = Aum cy _ J^4 _ .124 BTU/min°RM ~ 171 ~ *^2^ lbm/min1-13W = ^pdVFor constant energy process, T = CThenpV = Cv¥ = C J ^=Cln^-v1 1P, - - S^HP2* = * » ° P*„ P1V 1 28520 x 1 _ .0 = 778 ~ 778 36' 65W = 36.65 In 2.W = 40.26 BTU¥ = 31330 ft lbsQ = W = 40.26 BTU1-16u = (j(x1 fx2,x3)" - 1 2 , DX1 + ^ D J C 2 + % D X 3Extensive property conjugate to, v v^ax1 ' xiQ X 2 ' ^ X2•?S_ 1 „^ x 3 - *31-17Given pV = CRTThen IOT r"1 - "l (.)continued1-14RT 53.3 x 530 iD*-A V = A U -= JB O y 4kT - ,A Qcontinued4T pRwhere i/r(b)T2 = 801.71Au = c ? T = .171(801.7-530)Au = 46.46 BTU/lbm1-19continued1/2T1-20For adiabatic processr 2A U = J p dVp = CVV1-rAu ( 2 -r dV1-21Along 1-3 pathPi7 / "along 2-3 pathP2V2 = P3V3Divide 1st by 2nd V1-18T2 = T1= 530= 530 (.235)-286= 530 (4.25)*286bXJ = - 94.08 BTU1-19 cont.x 10(b) V_ = 3.715 ft3T(a) T2 = 602.2°R1-17 cont.starting with (a)RTV = —Pthen (a) becomescontinued**

## [Solved] TEST BANK FOR An Introduction to Equilibrium Thermodynamics By Bernard Morrill (Auth.)

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