Lista Ciclo Diesel e Ciclo Dual Máquinas Térmicas

Lista 2 \\quad T_{3} = 300 K, \\quad \\, \\text{sabendo então que} \\mu_{s} = 314,07 \\text{e} v_{3} = 623,72 \\quad Processos 1-2: \\quad v_{3} = \\frac{v_{h3}}{\\mu} = \\frac{6 \\sqrt{2}}{20} \\quad |v_{3}| = 31,06 \\quad \\text{logo}, T_{2} = 940 K \\text{e} u_{2} = 708,08 \\quad Processos 2-3: \\quad g_{transo} = u_{3} - u_{2} = \\quad 3000 = u_{3} - 708,08 \\quad (u_{3} = 2008,08) \\quad \\text{logo} \\quad T_{3} = 2290 K 2) \\mu = \\frac{v_{21}}{v_{23}} - \\left( \\frac{p_{21}}{p_{1}} \\right)^{\\frac{1}{k}} Se \\quad k = 1,4 : \\mu = \\frac{6 \\cdot 10^{-6}}{95 \\cdot 10^{-3}} \\left( \\frac{1}{\\sqrt{3}} \\right) \\mu = \\left( 6,3 \\cdot 15 \\right) \\mu = 19,29 \\approx 19,37 3) Achando \\quad T_{2}: \\frac{T_{2}}{T_{3}} = \\frac{p_{32}}{p_{12}} \\left( \\frac{K-1}{K} \\right) \\quad T_{2} = \\frac{6,06 \\cdot 290}{95 \\cdot 10^{-3}} \\left( \\frac{1}{\\sqrt{3}} \\right)^{\\frac{4-1}{3}} \\quad T_{2} = 290 \\cdot (6,73 \\cdot 15) \\; 0,285 \\quad T_{2} = 290 \\cdot 3.16 \\quad |T_{2}| = 94,51 K \\quad \\text{q_{transo} = h_{3} - h_{2} = g_{p}(T_{3}-T_{2})} \\quad g_{transo} = 2409,6 - 291,6 \\quad g_{transo} = 2188,44 Wliq = 1400.964 \u00b7 10^-4\n|Wliq| = 1.2\nPot = Wliq \u00b7 Necls \u00b7 RPM\n 60\n 1/2\nPot = 1.2 \u00b7 6 \u00b7 2000 \u00b7 1/2\nPot = 321 KW