Question

please help with 5.34 , 5.35 , 5.36 and 5.45

4. Problem 5.34 5.34 For each of the cases below, determine if the heat engine satisfies the first law (energy equation) and if it violates the second law. a. Qu = 6kW, 0. = 4kW, W = 2 kW b. Qu = 6kW, = 0kW, W = 6kW c. Qu=6kW, 2. = 2 kW, W = 5kW d. Qu=6kW, L = 6kW, W = 0kW 5.35 For each of the cases in Problem 5.34, determine if a heat pump satisfies the first law (energy equation) and if it violates the second law. 5.36 Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 0.25 kg liq- uid water at 10°C. Assume that the refrigerator has B 3.5 and a motor-compressor of 750 W. How much time does it take if this is the only cooling load? 5. Problem 5.35 6. Problem 5.36 7. Problem 5.45 5.45 An ideal (Carnot) heat engine has an efficiency of 40%. If the high temperature is raised 15%, what is the new efficiency keeping the same low tempera- ture?

Answer 1

= 6kw i alt w = 6 kutoka 5.34 & QH > ai Energy Equation an = diton ( 1 Lan for good Lane w to or dito a.) ein = 6 kw * d (+0W = (U+27 Kw=6kw A dito & w to so it Obey felt & 2nd laule. & w=6kw b) dx = 6kw id=0 tw So (न obey 1 ht law day not obey lane *2=0) dy=6kw; ai li = 6kw but ang c ) 2 kw & W=5kw dztw = (2+1) Kw=7kw so it does not obey ft x and Law. d ) altw = 6 Ich 1 tt Law but day not or netolty so it obey grel Law ( qu=0 kW)

1:37 Similarly fe 535 2에 Law then 7 Both law and low an dito (fth Law) anto i 9,80; Qw to a), hy [obey both Law. b) robey - 1 tt law (disobey gnal Law 1 disoby + d) foluey Itt law [disobey - 50367. B=3.5 = °C lice) dw. W = 750 W Pay m. ute, m. 4 W2 = m. UztWz 1 By Energy a = m. Um mu,+W₂ al = m. 4₂ cmu + Po mi V-V) Q2 = m( uzte. 232-m14, the V) 02 = m 1 ha-hi) T, = 100 dm = m = 0.25 lcg sec dt comlo'wion) 2

% h, е Ар- - 91, 99 kf Thang ( fussing capacity 2 fe Т. с по ХА, - h; = - 3 33 42 к1 Kung 2 т, = o't So feat los q = m(har ha) 0» 25 ( -33-92 - 9 9 9) = –1-8ГkТ - о: 25 Noo -ф ß = 934-1 - 0 N 3. 6. 8 (0 x lo° ) ( М. ('. с 86-горию лю 26. ex |о Nom - 4 : P 50 36 Хес

4 given, n=40% = 0.4 (Tn). 0.15TH = 115TH = Tnto. new New Carnot cycle thermal efficiency is given by : Tu Tnews (Tahew 1.15.JT = 1-TL - قد we must find IĮ rete first TH T TH = l-4 = 1-0.4-0.6 Morew = |- Tu 1- (1.159TH 016 0.6 1.15 Per = 0.4781